Matt@VeritasPrep posted a reply to Which of the following fractions is the largest? in the Problem Solving forum
“For D to be the largest, it''d have to be bigger than B. To compare two fractions, we could test an inequality: 7/30 > 13/56 7*56 > 13*30 392 > 390 Success! This is true, so 7/30 IS > 13/56.”
December 8, 2017
Matt@VeritasPrep posted a reply to A certain taxi company charges $3.10 in the Problem Solving forum
“The trick is converting the problem to the units given. We want fifths of the mile, so 8 miles = 40 fifths of a mile. From there, our first fifth costs $3.10 and our remaining 39 fifths cost 40¢ each. That gives us a fare of $3.10 + 39*40¢, or $18.70.”
December 8, 2017
Matt@VeritasPrep posted a reply to A survey was conducted to find out how many people in... in the Problem Solving forum
“When we''ve got three groups A, B, and C with some overlap, we can say that Total = A + B + C + (people in no groups) - (people in exactly two groups) - 2*(people in exactly three groups) We''re told that the total = 144, A = (144 - 100), B = (144 - 89), and C = (144 - 91). (Remember that A, ...”
December 8, 2017
Matt@VeritasPrep posted a reply to Three runners A, B and C run a race with A finishing... in the Problem Solving forum
“One last approach that''s slow but almost foolproof is backsolving. I''ll use two answers to show what the right answer should look like and what a wrong answer could look like. Suppose I try answer A, and make the race an hour long. If the race is 40 meters long, then A runs 40 meters, B runs ...”
December 8, 2017
Matt@VeritasPrep posted a reply to Three runners A, B and C run a race with A finishing... in the Problem Solving forum
“An approach with less algebra would go something like this: When A finishes the race, we know A has run D meters B has run D - 20 meters C has run D - 34 meters At this point, B has 20 meters left to go and C has 34 meters left to go. When B finishes the race, we know C still has ...”
December 8, 2017
Matt@VeritasPrep posted a reply to Three runners A, B and C run a race with A finishing... in the Problem Solving forum
“Let''s start with a technical approach: Distance = Rate * Time We''ll call t A''s time to finish the race, and a, b, c the rates of the three runners. We know that D = a * t D - 20 = b * t D - 34 = c * t Now let''s look at what happens after A finishes. Let''s say that the time ...”
December 8, 2017
Matt@VeritasPrep posted a reply to A random 10-letter code is to be formed using the letters... in the Problem Solving forum
“A simple way: Start by treating the two I''s as one unit grafted together. If we then arrange the "nine" letters (A, B, C, D, E, F, G, H, and II), we''ve got 9! arrangements, so there are 9! ways to put the I''s together. But if we have no restrictions at all, we have 10!/2! ways ...”
December 8, 2017
Matt@VeritasPrep posted a reply to Solution Y is 40 percent sugar by volume, and solution X... in the Problem Solving forum
“Another approach would be finding the weighted average. If X is 20% sugar, Y is 40% sugar, and X + Y = 25% sugar, then .2x + .4y = .25(x + y) .2x + .4y = .25x + .25y .15y = .05x 3y = x We were told y = 150, so x = 3y = 3*150 = 450.”
December 8, 2017
Matt@VeritasPrep posted a reply to Solution Y is 40 percent sugar by volume, and solution X... in the Problem Solving forum
“If we''ve got 150 gallons of solution Y, then 40% of 150, or 60 gallons of it are sugar. From here, we''re adding x gallons of Solution X. 20% of that is sugar, so sugar / total = (60 + 0.2x) / (150 + x) is our ratio. We need that to equal 25%, or 1/4, so (60 + 0.2x) / (150 + x) ...”
December 8, 2017
Matt@VeritasPrep posted a reply to A basket contains 5 apples of which one is rotten... in the Problem Solving forum
“Another approach could use combinatorics: p(rotten) = (# of pairs with the rotten apple) / (total # of pairs) # of pairs with the rotten apple = 4, since the rotten apple could be paired with any of the other four apples total # of pairs = (5 choose 2) = 5!/(3! * 2!) = 10 So our ratio ...”
December 8, 2017
Matt@VeritasPrep posted a reply to A basket contains 5 apples of which one is rotten... in the Problem Solving forum
“Another approach: p(getting the rotten apple) + p(not getting the rotten apple) = 1, since one of those two things MUST happen. Subtracting from both sides, we have p(getting the rotten apple) = 1 - p(not getting the rotten apple) and p(not getting the rotten apple) = p(apple 1 ...”
December 8, 2017
Matt@VeritasPrep posted a reply to A basket contains 5 apples of which one is rotten... in the Problem Solving forum
“Here''s another way of thinking about it: Henry could select the rotten apple FIRST or SECOND. The probability that he selects it first = p(rotten apple) * p(fresh apple) = 1/5 * 4/4 = 1/5 The probability that he selects it second = p(fresh apple) * p(rotten apple) = 4/5 * 1/4 = 1/5 Adding ...”
December 8, 2017
Matt@VeritasPrep posted a reply to A gardener is going to plant 2 red rosebushes in the Problem Solving forum
“Another resurrected zombie thread - what is happening in the graveyard this week? :)”
December 8, 2017
Matt@VeritasPrep posted a reply to What is the smallest positive integer K such that the... in the Problem Solving forum
“Another approach (not quite as good, but you never know what''s helpful!): pull the integers roots out of 1575K. √1575K => √1575 * √K => √(25*9*7) * √K => √25 * √9 * √7 *√K => 5 * 3 * √7 * √K => 5 * 3 * √7K So if √7K = an integer, ...”
December 8, 2017
Matt@VeritasPrep posted a reply to What is the smallest positive integer K such that the... in the Problem Solving forum
“Here''s a trick: break 1575K into two identical square roots. For instance, say we have 36. 36 = 2 * 2 * 3 * 3 = (2 * 3) * (2 * 3), or two identical square roots. Now let''s try that with 1575K. 1575K => 25 * 63 * K => 5 * 5 * 3 * 3 * 7 * K => (5 * 3 * 7) * (5 * 3 * ...”
December 8, 2017
Matt@VeritasPrep posted a reply to When to STOP studying for the GMAT in the Helpful Resources forum
“Obligatory disclaimer of a follow up post: I''m not recommending that anyone buy Bitcoin! (In fact if it were me I''d be selling, but don''t take investment advice from me, for goodness sake!) It just seemed like a neat way to frame an idea I thought might be helpful.”
December 8, 2017
Matt@VeritasPrep posted a reply to When to STOP studying for the GMAT in the Helpful Resources forum
“Great post, Ceilidh! One thing I''d add: progress on the GMAT isn''t linear! It''s much harder to get from 650 to 700 than it is to get from 500 to 650. I tell students this a lot, but there''s a perfect illustration in the news this week - Bitcoin! Everyone''s heard about the unfathomable price ...”
December 8, 2017
Matt@VeritasPrep posted a reply to help in the GMAT Math forum
“No prob!”
December 7, 2017
Matt@VeritasPrep posted a reply to time speed distance question in the GMAT Math forum
“Fun follow up conceptual question: the harmonic mean of 20 and 30 just happens to be 24. Is this a shortcut to solve this problem, or just a coincidence? Explain your answer. :) *Jeopardy music*”
December 7, 2017
Matt@VeritasPrep posted a reply to Master List of Quant/GMAT Math Resources in the GMAT Math forum
“I see you''ve already done this, but it might be a good idea to copy the entire thread to the Helpful Resources subforum: it''ll have less of a chance of getting swept off the main page! (Though I suppose it could keep being resurrected :))”
December 7, 2017
Matt@VeritasPrep posted a reply to There are three blue marbles, three red marbles .... in the Problem Solving forum
“This isn''t really answerable because it isn''t clear if we''re replacing the marbles between draws. (It *sounds* like we aren''t, but it isn''t clear.) If we ARE replacing the marbles, then it''s just ((# of yellow marbles) / (# of marbles))³ If we AREN''T replacing the marbles, then it''s ...”
December 6, 2017
Matt@VeritasPrep posted a reply to If x >= 0 and x=root(8xy-16y^2), then, in terms of y, x=? in the Problem Solving forum
“A little easier: x² - 8xy + 16y² = 0 (x - 4y)² = 0 x = 4y”
December 6, 2017
Matt@VeritasPrep posted a reply to If x is an integer, how many possible values in the Problem Solving forum
“I''ve got a quick way! x² + 5|x| + 6 = 0 (|x| + 2) * (|x| + 3) = 0 (|x| + 2) = 0 or (|x| + 3) = 0 |x| = -2 or |x| = -3 But no absolute values are negative, so there are no real solutions to this equation.”
December 6, 2017
Matt@VeritasPrep posted a reply to Elana was working to code protocols for computer... in the Problem Solving forum
“Piggybacking on David''s answer, 154 * 18/4 => 154 * 4.5, so the answer has to be more than 150 * 4, or more than 600. From there it''s a cinch!”
December 6, 2017
Matt@VeritasPrep posted a reply to If x-y+z=-1, -x+y+z=1, and x+y-z=-1, then x+y+z=? in the Problem Solving forum
“We could also invoke the Lazy Testwriter Principle: the answers are probably friendly and/or small numbers, so try those. Let''s start with x-y+z=-1 This looks like -1 - 0 + 0, so let''s say x = -1, y = 0, and z = 0. If we plug those into our other two equations: -x + y + z => ...”
December 6, 2017
Matt@VeritasPrep posted a reply to If -6 and 3 are the solutions of the... in the Problem Solving forum
“Yet another way: If our solutions are -6 and 3, then we can say (-6)² -6 * b + c = 0 and 3² + 3b + c = 0 so: 36 - 6b + c = 0 and 9 + 3b + c = 0 Now just solve the two equations! Let''s multiply the bottom one by 2: 18 + 6b + 2c = 0 then add it to the top one: ...”
December 6, 2017
Matt@VeritasPrep posted a reply to If -6 and 3 are the solutions of the... in the Problem Solving forum
“Another way: Remember that the roots of (x + r) * (x + s) = 0 are x = -r and x = -s. We''re told that -r = -6 and -s = 3, so r = 6 and s = -3. Plugging those in, we have (x + 6) * (x + -3) = 0 and foiling x*x + 6x - 3x - 18 = 0 or x*x + 3x - 18 = 0 so b = 3, c = -18, and ...”
December 6, 2017
Matt@VeritasPrep posted a reply to If s is the product of the integers from 100 to 200 inclusiv in the Problem Solving forum
“s = 100 * 101 * ... * 199 * 200 t = 100 * 101 * ... * 199 * 200 * 201 t = s * 201 t/201 = s so: 1/t + 1/s is the same as 1/t + 1/(t/201) is the same as 1/t + 201/t is the same as 202/t”
December 6, 2017
Matt@VeritasPrep posted a reply to One side of a triangle has lenght 8 and a second side... in the Problem Solving forum
“Also, for those flashcarding this: Given two legs of a triangle, the area of the triangle has the range: 0 < area ≤ (given side * other given side) / 2”
December 6, 2017
Matt@VeritasPrep posted a reply to One side of a triangle has lenght 8 and a second side... in the Problem Solving forum
“Let''s avoid trig, since the GMAT doesn''t expect us to know it. Visualizing the triangle, we can make the area about as small as we''d like by simply extending the third side as far as it can go: https://s8.postimg.org/stqfnzyqp/Screen_Shot_2017-12-05_at_5.18.22_PM.png As our base ...”
December 6, 2017
Matt@VeritasPrep posted a reply to How many multiples of 7 are between 21 and 343, exclusive? in the Problem Solving forum
“21 = 7*3 343 = 7*49 So we want every multiple of 7 from 7*1 to 7*48, but we can''t have 7*1, 7*2, or 7*3. That means we have 48 - 3 = 45 multiples.”
December 5, 2017
Matt@VeritasPrep posted a reply to If a and b are nonnegative integers and 4a+3b=32... in the Problem Solving forum
“4a + 3b = 32 3b = 32 - 4a 3b = 4 * (8 - a) Since the right side is a multiple of 4, the left side must also be a multiple of 4. Since 3 can''t be a multiple of 4, b must be. Our integers have to be nonnegative, so 3b must be < 32, since a ≥ 0. That means b is a multiple of 4 ...”
December 5, 2017
Matt@VeritasPrep posted a reply to Probability Problem - Cumbersome Solution in the Problem Solving forum
“Another approach: Since p(R) = p(N), we can say: p(0 days of R) = p(5 days of R) p(1 day of R) = p(4 days of R) p(2 days of R) = p(3 days of R) and we know those six scenarios sum to 1. p(0 days of R) is easy = (1/2)⁵ p(1 day of R) is similar = (1/2)⁵ * 5 so our sum ...”
December 5, 2017
Matt@VeritasPrep posted a reply to If n = 2×3×5×7×11×13×17, then which of the following.. in the Problem Solving forum
“We know that n = 2 * (some huge odd number). From there: (i) For n to divide by 600, it must divide by 6 * 100, or 6 * 10 * 10, or 6 * 2 * 5 * 2 * 5. But we saw that n only has ONE factor of 2, and to divide by 600 it''d need THREE factors of 2, so (i) is NOT true. (ii) if n + 19 = 19 * ...”
December 5, 2017
Matt@VeritasPrep posted a reply to Fraction raised to a negative exponent in the GMAT Math forum
“We could prove that easily enough too! (1/x)⁻² = x² is the same as 1 = x⁻² * x² is the same as 1 = x⁰ which is true!”
December 5, 2017
Matt@VeritasPrep posted a reply to Fraction raised to a negative exponent in the GMAT Math forum
“You are correct! (1/x)⁻² = x²”
December 5, 2017
Matt@VeritasPrep posted a reply to If x>3,000, then x/(2x+1) is closest to... in the GMAT Math forum
“It is the night of the zombie threads tonight! Just pick a number. Say x = 5000. Then x/(2x + 1) = 5000/10001, which is really close to 5000/10000, or 1/2.”
December 5, 2017
Matt@VeritasPrep posted a reply to calculate large exponents in the GMAT Math forum
“To do it exactly we''d have to ask Shakuntala Devi! She''d be able to do it in a matter of seconds, but I don''t know how. She wrote quite a few books on mental math, though. I agree with Brent, we''d be best off estimating, but it''d still be tough. My rule of thumb is that (1 + small change) ...”
December 5, 2017
Matt@VeritasPrep posted a reply to Are last 50 probs OG 11 math 700-800 level? in the GMAT Math forum
“I''d add that GMAC doesn''t seem to like releasing their Devilish problems: the harder problems I saw on my exam were quite a bit worse (and stranger) than the problems in any of the OGs or the software.”
December 5, 2017
Matt@VeritasPrep posted a reply to Please help me with this question in the GMAT Math forum
“Let''s translate this to an equation. 90 = (1 + p/100) * 5 * .3 * 40 90 = (1 + p/100) * 60 1.5 = (1 + p/100) 0.5 = p/100 50 = p”
December 5, 2017
Matt@VeritasPrep posted a reply to Median Values Question in the Data Sufficiency forum
“OK, great! {6, 15, j, k} can have a median less than 10 only if 15 is the largest number in the set. (If it isn''t the largest, then our set would be {j, 6, 15, k} or {6, j, 15, k}, in which case the median is at least 10.5) From here, I''d try a few sets. We could have j = 2, k = 7, which ...”
November 10, 2017
Matt@VeritasPrep posted a reply to How can I solve the equation? in the Problem Solving forum
“Since x/y = 3, we know x = 3y. Replacing x with with 3y in our other fraction gives (x - y)/x => (3y - y)/3y => 2y/3y => 2/3”
November 10, 2017
Matt@VeritasPrep posted a reply to 2,600 has how many positive divisors? in the Problem Solving forum
“If it''s easier to see this visually, check out this great tutorial. In the future, remember that Google is your friend! In a pinch you can usually find walkthroughs of most basic formulas such as this without having to wait a few hours for an expert to reply.”
November 10, 2017
Matt@VeritasPrep posted a reply to A job opening was posted in September and again... in the Data Sufficiency forum
“To find the percentage change, all we need is the number of applicants in January. Once we have that, we can use the percent change formula: (New - Old)/Old * 100. S1: gives us the number of applicants in January, sufficient S2:: gives us the number of applicants in January, also sufficient!”
November 10, 2017
Matt@VeritasPrep posted a reply to If the function f is defined for all x by f(x) = . . . in the Data Sufficiency forum
“Let''s start by plugging 3 into the function: f(3) = 3²*a + 3*b - 43 f(3) = 9a + 3b - 43 So to solve, we need to know the value of 9a + 3b. We can write this as 3 * (3a + b), so 3a + b would do the trick too. S1:: not sufficient, doesn''t give 3a + b S2:: sufficient, DOES give the ...”
November 10, 2017
Matt@VeritasPrep posted a reply to What is 11% of x? in the Data Sufficiency forum
“To solve, we need the value of x. S1:: gives us the value of x, sufficient S2:: gives us the value of x, also sufficient If you''re having trouble with this one, I''d recommend doing more skillbuilding in algebra before trying too many CATs or assorted problems. Once your fundamentals are ...”
November 10, 2017
Matt@VeritasPrep posted a reply to Is k^2 + k – 2 > 0? in the Data Sufficiency forum
“k² + k - 2 > 0 becomes (k + 2) * (k - 1) > 0 That''ll be greater than 0 if either: (i) the smaller of the two terms, k - 1, is positive (ii) the larger of the two terms, k + 2, is negative So our question can be rephrased as "Is either k > 1 or -2 > k?" and from ...”
November 10, 2017
Matt@VeritasPrep posted a reply to What is the value of p^3 - q^3? in the Data Sufficiency forum
“Alternatively, if p - q = 0, we can say that p = q, and that p³ - q³ = p³ - p³ = 0.”
November 10, 2017
Matt@VeritasPrep posted a reply to What is the value of p^3 - q^3? in the Data Sufficiency forum
“Any time we have one cube minus another, we can factor like so: p³ - q³ => (p - q) * (p² + pq + q²) So if we''re told (p - q) = 0, p³ - q³ is 0 * (p² + pq + q²), or 0.”
November 10, 2017
Matt@VeritasPrep posted a reply to X is 45.6% larger than Y. What is the value of Y? in the Data Sufficiency forum
“From the stem we''ve got x = 1.46y. We need one other equation (technically one other independent linear equation) in x and y to solve. S1:: x = 1.3z and z = 1.12y This takes us in the wrong direction! We wanted another equation in x and y, but we got a new variable, z. That said, we now ...”
November 10, 2017
Matt@VeritasPrep posted a reply to 500 PS in the Problem Solving forum
“(90% of 60) + (5% of 40) => .9 * 60 + .05 * 40 => 54 + 2 => 56”
November 10, 2017
Matt@VeritasPrep posted a reply to (17*19*23*29)^31=n in the Problem Solving forum
“What a great question! We could test the principle ourselves with smaller digits. Say we have (2 * 3 * 4)⁵, or 24⁵. The four possible reductions are (1 * 3 * 4)⁵, or 12⁵ (2 * 2 * 4)⁵, or 16⁵ (2 * 3 * 3)⁵, or 18⁵ (2 * 3 * 4)⁴, or 24⁴ We want the LEAST possible ...”
November 10, 2017
Matt@VeritasPrep posted a reply to Which of the following are/is prime? in the Problem Solving forum
“Piggybacking on David''s answer, one neat way of finding a composite number is noticing that it consists of multiples of the same number glued together, so to speak. For example, say we have the number 28144977. We could illuminate that as 28144977, or four multiples of 7. That might not seem like ...”
November 10, 2017
Matt@VeritasPrep posted a reply to A machine at the golf ball in the Problem Solving forum
“I''d fall back on the trusty equation Work = Rate * Time. Here, we know that each machine follows the equation Work = Rate * Time 16 = Rate * 5 (since work = 16 golf balls and time = 5 minutes) Dividing both sides by 5, we get Rate = 16/5, so each machine works at a rate of 16/5 of a ...”
November 10, 2017
Matt@VeritasPrep posted a reply to If a, b, and c are integers and ab^2/c . . . in the Problem Solving forum
“If we''ve got ab² / c = even, then we''ve also got ab² = c * even That means that either a or b is even, since a * b * b = even. Once a or b is even, we know that even * any integer = even, so ab MUST be even.”
November 10, 2017
Matt@VeritasPrep posted a reply to Kim purchased n items from a catalog for $8 each. in the Problem Solving forum
“We''ve got ONE item that costs $8 + $3 and (n - 1) items that each cost $8 + $1, so our total is $11 + (n - 1)*$9, or $11 + $9n - $9, or $9n + $2”
November 10, 2017
Matt@VeritasPrep posted a reply to There are 5 competitors in a race (John, mary and... in the Problem Solving forum
“No uncertainty here: if I''m getting even odds, give me the faster runner for the max. B-)”
November 10, 2017
Matt@VeritasPrep posted a reply to Probability Disks in the Problem Solving forum
“p(zero blue disks) = p(first disk isn''t blue) * p(second disk isn''t blue) * p(third disk isn''t blue) * p(fourth disk isn''t blue) In any given bag, the ratio of blues to nonblues is 25 : 75, so the probability of getting a nonblue disk is 75/(75 + 25), or 3/4. With that in mind, p(zero blue) ...”
November 10, 2017
Matt@VeritasPrep posted a reply to Median Values Question in the Data Sufficiency forum
“It''s impossible to tell what the prompt is giving us. 15 * j * 6 * k is an expression, not a set, so we don''t know the set whose median we seek or how to do anything with 90jk. If the prompt is asking that''s a different story, but I''d need to know more before answering.”
November 10, 2017
Matt@VeritasPrep posted a reply to 3 integers: x, y, z if there is a multiple of 7? in the Data Sufficiency forum
“Let''s rephrase the prompt as S1:: x + y + z = 7 * some integer We could have 1 + 2 + 4, or we could have 7 + 14 + 21; NOT SUFFICIENT. S2:: xyz = 7 * some integer Since each side consists entirely of integers, each side must have the same prime factorization. Since 7 ...”
November 10, 2017
Matt@VeritasPrep posted a reply to Find absolute value:| xy | in the Data Sufficiency forum
“S1:: We could have x = 3, y = 1, or x = √10, y = 0. Not sufficient. S2:: Same idea! We could have x = √7, y = 1, or x = √6, y = 0. Not sufficient. S1 + S2:: Add the two together, yielding 2x² = 16, or x² = 8, or |x| = 2√2. Now subtract the first from the second, yielding 2y² = ...”
November 10, 2017
Matt@VeritasPrep posted a reply to Basketball Jersey DS in the Data Sufficiency forum
“S1: We could have the jersey numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, the average of which is about 50, or we could have 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, the average of which is 91. Conflicting ...”
November 10, 2017
Matt@VeritasPrep posted a reply to How many different equations can be found? in the Problem Solving forum
“Let me be helpful, though: this problem is basically a balls and urns problem, and you can read a lot about how to solve such problems here.”
November 10, 2017
Matt@VeritasPrep posted a reply to How many different equations can be found? in the Problem Solving forum
“Well, this isn''t a GMAT problem ... and even if it were, it isn''t well-formulated ... and even if it were, its solution involves a lot of tedious casework, so ... it''s a tough sell for us. :) I''m not sure where your problems are coming from, but I answered another one that wasn''t ...”
November 10, 2017
Matt@VeritasPrep posted a reply to OG 18, question-225 in the GMAT Math forum
“Absolutely the best GMAT advice you''re going to read today, people! If I were granted the power to impart one lasting lesson to every GMAT student, this would be the one. (Well, that and "don''t check your phone during the break!!" :))”
November 10, 2017
Matt@VeritasPrep posted a reply to How many integer values in the GMAT Math forum
“Whoops, should add a note explaining that last assertion. Since x * (x + p) = 24, we also have p = (24 - x)²/x. (It''s safe to divide by x since x = 0 is not a solution.) Any integer p that''s a solution to that quadratic is going to itself create a quadratic with two solutions. For instance, ...”
November 10, 2017
Matt@VeritasPrep posted a reply to How many integer values in the GMAT Math forum
“I''m going to assume that the problem also requires p to be an integer. x² + px - 24 = 0 x² + px = 24 x * (x + p) = 24 So x must be a positive or negative factor of 24. That gives us x = -24, -12, -8, -6, -4, -3, -2, -1, 1, 2, 3, 4, 6, 8, 12, and 24, or sixteen values, each of ...”
November 10, 2017
Matt@VeritasPrep posted a reply to probability in the GMAT Math forum
“One more approach: take it case by case! Case 1: We pull a multiple of 6 Case 2: We DON''T pull a multiple of 6, but we pull a multiple of 2 and a multiple of 3, and 2 * 3 = 6 From here, we can split things up into subcases: Case 1(i): We pull the multiple of 6 from the first deck ...”
November 10, 2017
Matt@VeritasPrep posted a reply to OG 18, question-364 in the GMAT Math forum
“Another tip: reformat the stem into something familiar so that your mind doesn''t keep wandering back to the unfamiliar symbol. If we grapple with the prompt for a bit, then rephrase this as the statements are a cinch, but if we don''t, it''s easy to keep staring at that ceiling function and ...”
November 10, 2017
Matt@VeritasPrep posted a reply to Formula for consecutive, even, odd integers in the GMAT Math forum
“I wouldn''t do too much with this thread, personally. The GMAT is all about thinking creatively, not formulaically, and the testwriters devise problems to test your ability to handle subtle deviations and curveballs from the normal, boring math drills. On the GMAT, if you don''t understand where a ...”
November 10, 2017
Matt@VeritasPrep posted a reply to Coin flip questions made easy in the GMAT Math forum
“Zombie bots run amok! What''s the probability that another one rises from the digital earth to haunt this thread? :D”
November 10, 2017
Matt@VeritasPrep posted a reply to Few questions after GMATprep trial exam in the GMAT Math forum
“Mystery solved - the prompts are here! I''ll chime in the relevant subthreads, created (of course!) by the great Brent.”
November 10, 2017
Matt@VeritasPrep posted a reply to GMATprep trial exam question 1 in the GMAT Math forum
“This one has always bugged me, as "the amount of her take-home pay that she didn''t save" seems too vague. (Are we talking monthly or yearly?) OfficialGMAT, any comment on this? Feels like it should be revised.”
November 10, 2017
Matt@VeritasPrep posted a reply to Official GMATprep trial exam question 2 in the GMAT Math forum
“Or cheat a bit and pick numbers that satisfy xy = 1, such as ... I dunno, x = 1, y = 1 :D It ain''t pretty, but it''s quick!”
November 10, 2017
Matt@VeritasPrep posted a reply to mixture problem. please hep with simple solution in the GMAT Math forum
“By the way, that is not a trivial GMAT problem! It takes more than the normal amount of of setup and legwork to sort out, and feels a bit more like an algebra homework problem than anything you''d encounter on this test.”
November 10, 2017
Matt@VeritasPrep posted a reply to mixture problem. please hep with simple solution in the GMAT Math forum
“Say that each liter in A costs $a and each liter in B costs $b. Then the price of the barrels is $140a and $60b, respectively. After we switch c liters from each barrel to the other, we''ve got $(140 - c)*a + cb in one barrel and $(60 - c)*b + c*a in the other. Since these create an equal price ...”
November 10, 2017
Matt@VeritasPrep posted a reply to What are the important topics one should focus on in quants? in the GMAT Math forum
“Arithmetic, algebra, and word problems. If you have a solid grasp of those, you''re in pretty good shape: most GMAT problems will test those (and only those) directly, and the other problems typically test them indirectly (you still need good arithmetic and algebra to do geometry, probability, etc.)”
November 10, 2017
Matt@VeritasPrep posted a reply to If 0 <X <1 and 1 <Y <2, will X> 1/2? 1) XY = in the GMAT Math forum
“... and here''s my solution! S1:: Since xy = 1, we know that x = 1/y. Suppose x ≤ 1/2. Then we''d have something like x = 1/2, 1/3, 1/4, 1/5, ..., whatever. But since x = 1/y, those give us y = 2, 3, 4, 5. ..., all of which are too big! So as x shrinks below 1/2, y grows beyond 2, ...”
November 10, 2017
Matt@VeritasPrep posted a reply to If 0 <X <1 and 1 <Y <2, will X> 1/2? 1) XY = in the GMAT Math forum
“I think we should reformat the question first, for those attempting it:”
November 10, 2017
Matt@VeritasPrep posted a reply to Official GMATprep trial exam question 3 in the GMAT Math forum
“Is it just me, or did the original post go up in smoke? I''m only seeing the replies to this (and a few other questions).”
November 10, 2017
Matt@VeritasPrep posted a reply to If 6xy=x^2y+9y, what is the value of xy? in the Data Sufficiency forum
“6xy = x²y + 9y y * 6x = y * (x² + 9) If y = 0, this statement will be true, so y = 0 is one solution. To find the other, we divide by y: 6x = x² + 9 0 = x² - 6x + 9 0 = (x - 3)² and x = 3 is our OTHER solution. With that, we know that either y = 0 or x = 3. S2 tells us ...”
October 13, 2017
Matt@VeritasPrep posted a reply to Is the integer p divisible by 4? in the Data Sufficiency forum
“If you''re multiplying five consecutive integers together, then at least two of them must be even. Since you''ve got at least two evens, you''ve got: 2*something * 2*something else => 4*something*something else => a multiple of 4”
October 13, 2017
Matt@VeritasPrep posted a reply to Is y = 6 ? in the Data Sufficiency forum
“S1: y² - 36 = 0 (y + 6) * (y - 6) = 0 y = 6 or y = -6 S2: (y - 1) * (y - 6) = 0 y = 1 or y = 6 S1 + S2: Only y = 6 fits both!”
October 13, 2017
Matt@VeritasPrep posted a reply to Is ab odd? (1) a is even (2) a is an integer in the Data Sufficiency forum
“If a is even, then a * b = even * b. From there, if b is an integer, we must have even * integer = even, but if b is something else (√2, π, ½, whatever) we could have any result we like, be it even, odd, not an integer, not a repeating decimal, etc.”
October 13, 2017
Matt@VeritasPrep posted a reply to The only articles of clothing. . . . . in the Data Sufficiency forum
“Let''s start with the ratio: we''ve got 9n shirts, 4n dresses, and 5n jackets. We also know n is an integer, since we can''t have fractional clothes! (Well, tell that to my closet, but hey.) Since we''re told in the prompt that 4n > 7, we know n ≥ 2. S1: 9n + 5n < 30 14n < ...”
October 13, 2017
Matt@VeritasPrep posted a reply to What is the greatest common factor of x and y ? in the Data Sufficiency forum
“With S1, we could have x = 8 and y = 32, or x = 256, y = 256, or whatever. They both divide by 4, so we know one common factor, but they may have an even GREATER common factor that we don''t know.”
October 13, 2017
Matt@VeritasPrep posted a reply to Integers in the Data Sufficiency forum
“S1: We could have 2 * 3, or 2 * 3 * 3, or 2 * 2 * 3, or ... S2: We could have 2⁵, or 2² * 3, or 3² * 2, or 3⁵ ... (If finding the number of unique factors of a number is a new idea, check out this great link.) S1+S2: We could have 2² * 3 or 3² * 2 ...”
October 13, 2017
Matt@VeritasPrep posted a reply to What is the remainder when 7^74 - 5^74 is divided by 24? in the Problem Solving forum
“I know I said above that I wouldn''t do this, but here''s another explanation I just couldn''t resist: 7⁷⁴ - 5⁷⁴ => (7² - 5²) * (7⁷² + 5² * 7⁷⁰ + 5⁴ * 7⁶⁸ + ... + 5⁷⁰ * 7² + 5⁷²) => 24 * (7⁷² + 5² * 7⁷⁰ + 5⁴ * 7⁶⁸ + ... + 5⁷⁰ * 7² + ...”
October 13, 2017
Matt@VeritasPrep posted a reply to What is the remainder when 7^74 - 5^74 is divided by 24? in the Problem Solving forum
“Just to follow up on my last post, that solution relies on multiplicative remainders, a topic that shouldn''t appear on the GMAT. To get a feel for how these work, consider something like this: 5 has remainder 1 when divided by 4. (For the rest of the post, I''ll use the term "mod 4" ...”
October 13, 2017
Matt@VeritasPrep posted a reply to Salesperson in the GMAT Math forum
“My pleasure! Thanks for thanking! :)”
October 13, 2017
Matt@VeritasPrep posted a reply to The sum of the squares of three numbers is 138, while the in the Problem Solving forum
“Those were the first squares that popped into my head - I was pretty surprised that there was another set of three squares that sums to 138! (which, apropos of nothing, has been a favorite number of mine ever since I heard that Misfits song about it 20 years ago ...)”
October 13, 2017
Matt@VeritasPrep posted a reply to Remainder in the Problem Solving forum
“Let''s call our number n. We''re told that n/5 = 8 + (remainder of n/34) To find that remainder, we can say that n = 34*something + r, or n = 34z + r, or r = n - 34z. Once we''ve got that, we can replace (remainder of n/34) with r, or with n - 34z: n/5 = 8 + n - 34z 34z = 8 + (4/5)n ...”
October 13, 2017
Matt@VeritasPrep posted a reply to Probability- hair treatment in the Problem Solving forum
“Not sure how this one got on the front page, but I do this stuff for my day job a lot. Here''s an R function I wrote that will answer all five parts: hairfun <- function(a,b,p,x){sum(sapply(a:b, function(p, x, k){p^k * (1-p)^(x-k) * choose(x, k)}, p=p, x=x))} where a = the minimum number ...”
October 13, 2017
Matt@VeritasPrep posted a reply to What is the remainder when 7^74 - 5^74 is divided by 24? in the Problem Solving forum
“This one requires remainder and/or factoring skills that the GMAT doesn''t expect of you, at least in 2017. The cleverest ways of solving this will use remainder properties that will spark more questions than they answer, so let me give a clumsier approach that''s more in keeping with the GMAT. ...”
October 13, 2017
Matt@VeritasPrep posted a reply to Toy Factory Machine in the Problem Solving forum
“And in case anybody wants the algebra, it''d look like this: We start with t machines that each produce (1/3) of a toy per minute. That nets us (t/3) toys per minute. After the replacement, we''ve got 0.6t machines that each produce (1/3) of a toy per minute and 0.4t machines that each ...”
October 13, 2017
Matt@VeritasPrep posted a reply to A company decreased the price in the Problem Solving forum
“We could make it easy on ourselves by imagining what would have if we had one of these products. Say I''ve got a $100 watch. The price drops by 50%, and now it''s a $50 watch. But I sell enough of them to still have $100, so I must have sold two of them. Aha! So selling TWICE as many at HALF ...”
October 13, 2017
Matt@VeritasPrep posted a reply to Which of the following points reflect to y=-x at (-3,2)? in the Problem Solving forum
“Final thought: this isn''t a properly formatted GMAT problem. "reflect to" isn''t really grammatical, and the prompt is too ambiguous to appear on the exam. I''m also not sure that the test has ever explicitly asked about a reflection across an axis or a line -- I can''t recall such a ...”
October 13, 2017
Matt@VeritasPrep posted a reply to Which of the following points reflect to y=-x at (-3,2)? in the Problem Solving forum
“Another way to think about this: y = -x means "take y, multiply by -1, and move it to x". If our first y is 2, we thus: take y: y = 2 multiply by -1: y = -2 move it to x: x = -2 And if our first x is -3, we do the same, since y = -x is the same as -y = x: take x: x = -3 ...”
October 13, 2017
Matt@VeritasPrep posted a reply to Which of the following points reflect to y=-x at (-3,2)? in the Problem Solving forum
“Since our equation is y = -x, just take the original coordinates and plug them into the formula. original x = -3 original y = 2 new y = -(original x) = -(-3) = 3 new x = -1 * (original y) = (-1) * 2 = -2 That gives us a reflected (x, y) of (-2, 3), and we''re set.”
October 13, 2017
Matt@VeritasPrep posted a reply to In the product 2^19*9^13*5^24 in the Problem Solving forum
“If it''s not clear why 10¹⁹ would determine the 20th place in our number above, take a look at few smaller examples. Suppose we have 10² * 5. That''ll give us 500, and the 5 becomes the THIRD digit from the right. Suppose we have 10⁴ * 17. That''ll give us 170,000, and the 7 (or the ...”
October 13, 2017
Matt@VeritasPrep posted a reply to In the product 2^19*9^13*5^24 in the Problem Solving forum
“It seems suspicious that we''d be asked for the 20th place, so let''s see if there''s some reason we''d get that easily. Noticing that we''ve got 2¹⁹, we might think of 10¹⁹, which would get us to the 20th place pretty nicely. (If it''s not clear why, see my post below.) And what do you ...”
October 13, 2017
Matt@VeritasPrep posted a reply to Find the sum of all the four digit numbers which are formed in the Problem Solving forum
“In the spirit of my other answers, here''s one more. Suppose I take 1256 and cycle it: 1256, 2561, 5612, 6125. That sum is 15554. The cycle is determined by the first arrangement of numbers (1256). By fixing one of those and holding the others constant, I can find the number of cycles. ...”
October 13, 2017
Matt@VeritasPrep posted a reply to Find the sum of all the four digit numbers which are formed in the Problem Solving forum
“Another quick and dirty approach that could game us to the right answer is estimation. We know that we''ve got 6 numbers are around 1000, 6 that are around 2000, 6 that are around 5000, and 6 that are around 6000. That means our sum should be more or less 6 * (1000 + 2000 + 5000 + 6000), or ...”
October 13, 2017
Matt@VeritasPrep posted a reply to Find the sum of all the four digit numbers which are formed in the Problem Solving forum
“Unnecessary but fun follow up: once we''ve got that sum for the units digit, by the way, we can use it for all the other columns like so: Units column = 84 Tens column = 84 Hundreds column = 84 Thousands column = 84 So the sum of those columns would be 84 + 840 + 8400 + 84000, or 93324. :)”
October 13, 2017
Matt@VeritasPrep posted a reply to Find the sum of all the four digit numbers which are formed in the Problem Solving forum
“The easiest way is to cheat with the units digit. If you''ve got 24 numbers that look like so: six of the form _ _ _ 1 six of the form _ _ _ 2 six of the form _ _ _ 5 six of the form _ _ _ 6 then when you add them up, your units digit is 6*1 + 6*2 + 6*5 + 6*6 => _ 4 (well, 84, but we ...”
October 13, 2017
Matt@VeritasPrep posted a reply to A set has exactly five consecutive positive in the Problem Solving forum
“Initial set: 1, 2, 3, 4, 5 Initial average: 3 New set: 1, 2, 3, 4 New average: 2.5 Decrease = (2.5 - 3)/3 = -.5/3 = -1/6 ≈ 16.67%”
October 12, 2017
Matt@VeritasPrep posted a reply to A collection of books went on sale. . . . in the Problem Solving forum
“You are right!”
October 12, 2017
Matt@VeritasPrep posted a reply to An air-conditioning unit costs $470 in the Problem Solving forum
“We could also use digit properties: 470 * .84 * 1.16 => 10 * (47 * .84 * 1.16) => Since 7 * 4 * 6 ends in 8, this gets us => 10 * (some decimal that ends in 8) Only A ends in 8, so we''re done!”
October 12, 2017
Matt@VeritasPrep posted a reply to For how many positive integer x is 130/x an integer? in the Problem Solving forum
“This is just a funky way of asking "How many distinct factors are there of 130?" To see a nice way of doing this, check out this link!”
October 12, 2017
Matt@VeritasPrep posted a reply to Fractions in the Problem Solving forum
“6/24 => (6*1)/(6*4) => 1/4 1/4 isn''t greater than itself, so E is out.”
October 12, 2017
Matt@VeritasPrep posted a reply to Probability in the Problem Solving forum
“Minimum sum = 3 Maximum sum = 11 Only squares in there: 4, 9 Possible sums: 1 + 3, 3 + 6, 4 + 5 So three possible pairs out of (6 choose 2) distinct pairs, or 3/(6!/4!2!), or 1/5”
October 12, 2017
Matt@VeritasPrep posted a reply to Automobile installment credit in the Problem Solving forum
“Looks good to me!”
October 12, 2017
Matt@VeritasPrep posted a reply to If 3 and 8 are the lengths of two sides of a triangular regi in the Problem Solving forum
“In a pinch, the rule to remember: |first side - second side| < third side < first side + second side”
October 12, 2017
Matt@VeritasPrep posted a reply to Barbara has 8 shirts and 9 pants. How many clothing combinat in the Problem Solving forum
“We could also compile the valid combinations as follows. 6 shirts can go with any pants: 6 * 9 2 shirts can go with exactly 6 pants: 2 * 6 6 * 9 + 2 * 6 => 66”
October 12, 2017
Matt@VeritasPrep posted a reply to The average price of an antique car in the Problem Solving forum
“Of course, we could also punish ourselves and find the exact answer. My usual approach is to chop the percentages into friendlier pieces, like 1% and 10%. For example: 11500 * 1.13 => 11500 * (1 + .1 + .01 + .01 + .01) => 11500 + 1150 + 115 + 115 + 115 => 12995 Now we''ve ...”
October 12, 2017
Matt@VeritasPrep posted a reply to The average price of an antique car in the Problem Solving forum
“Here''s another trick: 11500 * 1.13 * 1.2 => 13800 * 1.13 From here, notice that .03 * 800 = 24, so our answer MUST end in a 4. Only A fits, so we''re done!”
October 12, 2017
Matt@VeritasPrep posted a reply to The average price of an antique car in the Problem Solving forum
“11500 * 1.13 * 1.2 => 11500 * 1.13 * (6/5) => 13800 * 1.13 => 13800 * (1 + .13) => 13800 + 13800 * .13 => 13800 + 13800 * (≈1/8) => ≈ 13800 + 1725 => ≈ 15525 A''s the closest, so we''ll go with that!”
October 12, 2017
Matt@VeritasPrep posted a reply to Advanced overlapping sets in the Problem Solving forum
“Yup, you can do it visually! This will help you write the formulas yourself, and see what''s happening: https://s1.postimg.org/3p2qkuhb0r/Screen_Shot_2017-10-11_at_6.23.33_PM.png”
October 12, 2017
Matt@VeritasPrep posted a reply to How to Start with my GMAT QUANT Beginner in the GMAT Math forum
“It might be a good idea to do a quant diagnostic to see where you''re strong and where you''re weak. Knowing which topics to tackle first can save a lot of time!”
October 12, 2017
Matt@VeritasPrep posted a reply to Ambigious _ Question_Veritas Prep in the GMAT Math forum
“Yup, you''re definitely right about the names - we''ll edit the question to tidy that up.”
October 12, 2017
Matt@VeritasPrep posted a reply to What is the smallest integer that. . . . . in the Problem Solving forum
“They''re saying that the answer / (7 * some prime) = another prime And they want the answer to be the smallest number that solves the equation above. If we call the answer a, some prime p, and the other prime q, we''ve got: a / 7p = q a = 7pq To make this as small as possible, ...”
September 29, 2017
Matt@VeritasPrep posted a reply to One gallon of soft drink is made of 40% orange juice and 60% in the Problem Solving forum
“We''ve got 0.4 gallons of OJ and 0.6 gallons of water. When we add x gallons of OJ, we''ll have (x + 0.4) gallons of OJ and 0.6 gallons of water. The % of OJ at that point will be OJ / Total => (x + 0.4) / (x + 0.4 + 0.6) We want that to equal 60%: (x + 0.4) / (x + 1) = .6 ...”
September 29, 2017
Matt@VeritasPrep posted a reply to If n has 15 positive divisors. . . . in the Problem Solving forum
“To find the number of divisors of a number, find its prime factorization, count the number of each distinct prime, add one to each count, then multiply the counts together. For instance, the process for 60: 1) Find its prime factorization: 60 = 2 * 2 * 3 * 5 2) Count each distinct prime: ...”
September 29, 2017
Matt@VeritasPrep posted a reply to How many real roots does the equation . . . in the Problem Solving forum
“We could factor as follows: x²y + 16xy + 64y = 0 y * (x² + 16x + 64) = 0 y * (x + 8)² = 0 Since y < 0, x must = -8. Since the equation is presumably a quadratic in x, we''d have one real root x. (I think the idea is that y is some known negative and that x is the only unknown, ...”
September 29, 2017
Matt@VeritasPrep posted a reply to Journey of Two Trains in the Problem Solving forum
“Somehow I missed this one! Here goes nothing: Extend the tracks so that A has a 15 mile turnaround at each city and that B has a 10 mile turnaround at each city. Now force each train to run constantly, without stops, in Gomez Addams fashion. Call a full trip the time it takes a train to go ...”
September 27, 2017
Matt@VeritasPrep posted a reply to Roll Number of Jessica in the Problem Solving forum
“Here''s another approach that doesn''t involve guesstimation: The sum from 1 to n = n * (n + 1) / 2, so n * (n + 1)/2 - j = 458. Since n * (n + 1)/2 > 458, n ≥ 30. But n ≥ j, and if n > 30, then n * (n + 1)/2 - n > 458, meaning there is no solution j. So 30 ≥ n. Since n ...”
September 27, 2017
Matt@VeritasPrep posted a reply to Remainder from large integer in the Problem Solving forum
“Sure, but with respect: 1) This problem is harder and more technical than any official GMAT problem and not at all within the scope or spirit of the test. It''s "slightly difficult" relative to certain tests, such as our American AMC 12, but those tests require far more mathematical ...”
September 27, 2017
Matt@VeritasPrep posted a reply to Remainder from Sum containing Factorial in the Problem Solving forum
“Ah, but that isn''t a mistake! It''d be a mistake to say that 33 = -4 mod 29, but that isn''t what I said ;) Of course, I could see why it would seem that way from what I typed, but hey, I''m wriggling out of this one.”
September 27, 2017
Matt@VeritasPrep posted a reply to Remainder from Sum containing Factorial in the Problem Solving forum
“Every term other than 33!/29 will still have a 29 in its numerator, so every term other than 33!/29 is divisible by 29. That means that problem can be stated as (33*32*31*30*28*27*...*1) mod 29 => (33*32*31*30 * 28!) mod 29 => (-4 * -3 * -2 * -1 * 28!) mod 29 => (24 * ...”
September 27, 2017
Matt@VeritasPrep posted a reply to Remainder from large integer in the Problem Solving forum
“If you''re feeling too lazy, skeptical, or bored to check the above, here''s a computer verification: http://www.wolframalpha.com/input/?i=(sum+(451+-+n)*10%5E(3n+-+3)+from+n+%3D+1+to+n+%3D+150)+mod+91 (Apologies for the formatting - for some reason this wouldn''t cooperate with the url tag.)”
September 27, 2017
Matt@VeritasPrep posted a reply to Remainder from large integer in the Problem Solving forum
“For those wondering why that property holds, here''s an explanation. Each of our summands can be expressed in the form (451 - n) * 10³ⁿ⁻³, and we have all such summands from n=1 to n = 150. Notice that each of those summands is multiplied by 10 raised to some multiple of 3. Now notice ...”
September 27, 2017
Matt@VeritasPrep posted a reply to Remainder from large integer in the Problem Solving forum
“This isn''t anything close to a GMAT problem, so no students are likely to answer it. It doesn''t seem like any instructors are interested either, but I''m game. It took me a while to find an elegant solution, but here''s one that absolutely rules :) We want 301,302,303,...,449,450 mod 91. Let''s ...”
September 27, 2017
Matt@VeritasPrep posted a reply to mathematics in the Problem Solving forum
“Not really a GMAT question though :)”
September 27, 2017
Matt@VeritasPrep posted a reply to The sum of two numbers is 13. . . in the Problem Solving forum
“a + b = 13 ab = 30 a² + b² = (a + b)² - 2ab a² + b² = 13² - 2*30 a² + b² = 109”
September 27, 2017
Matt@VeritasPrep posted a reply to The “primeness” in the Problem Solving forum
“Find the prime factorization, then do largest prime - smallest prime. A: 10 = 2 * 5, largest - smallest = 5 - 2 = 3 B: 12 = 2 * 2 * 3, largest - smallest = 3 - 2 = 1 C: 14 = 2 * 7, largest - smallest = 7 - 2 = 5 etc. etc. You get the idea :)”
September 27, 2017
Matt@VeritasPrep posted a reply to What is the sum of the multiples of 7 from 84 to 140? in the Problem Solving forum
“We want 7*12 + 7*13 + ... * 7*20 To find this, we could add the first 20 multiples of 7, then subtract the ones we don''t want (the first 11 multiples of 7). That gives us: (7*1 + ... + 7*20) - (7*1 + ... + 7*11) From there, we could write this as: 7*(1 + 2 + ... + 20) - 7*(1 + 2 ...”
September 27, 2017
Matt@VeritasPrep posted a reply to If a, b are integers and |a-b|=8 in the Problem Solving forum
“You''re correct, -16 is the minimum. (Proving that is a bit more complicated, but with these answer choices we don''t really need the proof.) Either the OA is a typo or the stem itself is missing some qualifier.”
September 27, 2017
Matt@VeritasPrep posted a reply to If a and b are two consecutive in the Problem Solving forum
“We could also set a in terms of b at the outset: a = b + 2, since a > b and they''re consecutive evens. With that, replace a with b + 2 in the second inequality: 2b > 3 * (b + 2) 2b > 3b + 6 -6 > b”
September 27, 2017
Matt@VeritasPrep posted a reply to If a and b are two consecutive in the Problem Solving forum
“We want a common term in each inequality in order to chain them together, so multiply the first inequality by 3. With that, we''ve got 3a > 3b and 2b > 3a. That chains into 2b > 3a > 3b, or 2b > 3b. 2b > 3b isn''t possible for any positive b, so b is negative. But if b = ...”
September 27, 2017
Matt@VeritasPrep posted a reply to A team of copper miners in the Problem Solving forum
“If you find yourself getting lost on these, take them one chunk at a time rather than all at once. First chunk: The miners wanted to mine 1800 tons of ore in d days, or 1800/d per day. Second chunk: For the first d/3 days, the miners were only able to mine (1800/d) - 20 tons per day. ...”
September 27, 2017
Matt@VeritasPrep posted a reply to At a certain automobile dealership in the Problem Solving forum
“Footnote: this question is designed to waste time and test your speed. There are a couple like this (at least) on every exam, so if the clock is against you and you''re feeling overwhelmed, this is a good one on which to make a quick educated guess to get past.”
September 27, 2017
Matt@VeritasPrep posted a reply to At a certain automobile dealership in the Problem Solving forum
“Start with the Tajimas. We''ve got 205 of them. Call the hybrids h and the nonhybrids n. That gives us h + n = 205 n = 3h - 35 Solve that system: h = 60, n = 145 Since h = 60 and the ratio of h : nonhybrid Franks = 5 : 4, we know nonhybrid Franks = 48. We''ve got 75 total Franks, ...”
September 27, 2017
Matt@VeritasPrep posted a reply to If positive integers a and b are both odd, in the Problem Solving forum
“You asked about D specifically, so I''ll only address that. It actually comes down to the remainder of our odd integers when divided by 4. If they have different remainders (one has remainder 1, one has remainder 3), then the average of the two will be even. If they have the same remainder by 4 ...”
September 27, 2017
Matt@VeritasPrep posted a reply to A certain league has four divisions. in the Problem Solving forum
“An easy way to think of this is to think of the number of losses there would have to be to eliminate each team. Since only one team loses each game, the number of losses / 2 = the number of eliminations. We need to eliminate 8, 9, 10, and 11 teams during the divisional rounds, a process that ...”
September 27, 2017
Matt@VeritasPrep posted a reply to Two trains continuously travel between Washington DC in the Problem Solving forum
“Can do! https://s26.postimg.org/4o5h5vg8l/Screen_Shot_2017-09-20_at_5.52.35_PM.png”
September 21, 2017
Matt@VeritasPrep posted a reply to What is the smallest integer greater than 1 in the Problem Solving forum
“You could also find the LCM of 6, 8, and 10, then add 1 to it. :) This trick generalizes well to other such remainder problems too!”
September 21, 2017
Matt@VeritasPrep posted a reply to A number when divided by 105 leaves 99 as remainder. in the Problem Solving forum
“Let me also add a more conceptual approach. Let''s say that our number = 105*something + 99. When I divide that number by 21, I get (105*something + 99)/21 => 105*something/21 + 99/21 => 5*something + 99/21 Now we can see the remainder will come entirely from the second ...”
September 21, 2017
Matt@VeritasPrep posted a reply to A number when divided by 105 leaves 99 as remainder. in the Problem Solving forum
“We could cheat and say that a number = 105 + 99 a number = 204 From here, find the remainder when 204 is divided by 21, and you''re done!”
September 21, 2017
Matt@VeritasPrep posted a reply to A baseball player signs a new contract in the Problem Solving forum
“We could also say that 2% of his old salary = 500. From there .02s = 500 s = 25,000 His old salary is 25,000, and his new salary is 1.1s + 500, or 1.1*25,000 + 500, or 27,500 + 500, or 28,000”
September 21, 2017
Matt@VeritasPrep posted a reply to Trip Calculation in the GMAT Math forum
“II: Algebra We''ve got two equations: 1.5rt = 96, or rt = 64 and (r - 4) * (t + 16) = 1.5rt The second one gives us rt - 4t + 16r - 64 = 1.5rt 16r - 4t - 64 = 0.5rt We know rt = 64, so we can replace 0.5rt with 0.5*64, or 32: 16r - 4t - 64 = 32 16r - 4t = 96 ...”
September 21, 2017
Matt@VeritasPrep posted a reply to Trip Calculation in the GMAT Math forum
“Just in case anyone reading wants to actually solve, there are two approaches: I:: The Lazy Testwriter Principle The testwriter will probably make the answer an integer. (Here, in fact, the answers are all integers, so we should be OK.) With that in mind, once we have 1.5rt = 96 or ...”
September 21, 2017
Matt@VeritasPrep posted a reply to Trip Calculation in the GMAT Math forum
“You CAN use this equation: (R-4) * (T+16) = 1.5 R * T You''re just missing a crucial piece: the distance of 96. With that, we have (r - 4) * (t + 16) = 1.5rt and 1.5rt = 96 and we can solve.”
September 21, 2017
Matt@VeritasPrep posted a reply to Rubik's Cube in the GMAT Math forum
“When you want to remove all painted cube from the larger cube, you need to remove a painted face from both sides. So if I had eight cubes before, once I remove one from both sides (the furthest on the left and the furthest on the ride), I''ll be left with six cubes, or 8 - 2.”
September 21, 2017
Matt@VeritasPrep posted a reply to Salesperson in the GMAT Math forum
“We could also try to speed up the problem by thinking of the differences in the incomes. A has a $360 head start. After each salesperson sells $1000 of product, B has $80 and A has the same as (s)he started with, $360, since A gets no commission on the first $1000. From here, B needs to make ...”
September 21, 2017
Matt@VeritasPrep posted a reply to Salesperson in the GMAT Math forum
“Put the equation in words, then translate those words to math: A makes $360 + 6% of (Total - $1000) B makes 8% of Total We want these to be equal, so: $360 + 6% of (Total - $1000) = 8% of Total Now let''s assign a variable to the total (how about t?) and write the percentages as ...”
September 21, 2017
Matt@VeritasPrep posted a reply to Code in alphabetical order in the GMAT Math forum
“Let me add a step here, since calculating 26 choose 3 isn''t the most natural thing in the world: 26 choose 3 = 26! / (23! * 3!) = (26 * 25 * 24) / (3 * 2 * 1) = 26/2 * 24/3 * 25/1 = 13 * 8 * 25 = 13 * 200 = 2600”
September 21, 2017
Matt@VeritasPrep posted a reply to Can you solve this? in the Problem Solving forum
“17 + 3 = 20 games played 20 games played = (2/3) of all games, so 30 = all games 30 total games, and we want the team to finish with more than 22.5 wins, or with at least 23 wins. To reach that record, the team must win AT LEAST 6 more games. There are 10 games left, so the team can lose ...”
September 18, 2017
Matt@VeritasPrep posted a reply to Difficult Math Problems #28 - Consecutive integers in the Problem Solving forum
“It''s easiest to see with numbers first. If our first number is 1, then we start with 1, 2, 3, 4, 5 and the next six numbers are 6, 7, 8, 9, 10 Since each number is the second set is FIVE more than its counterpart in the first set (6 = 1 + 5, 7 = 2 + 5, 8 = 3 + 5, ...), we''re adding ...”
September 18, 2017
Matt@VeritasPrep posted a reply to Probability in the Problem Solving forum
“It seems like this is easy conceptually too. We want to pick both guys, so there''s only one way to do that: 2 choose 2. If we''re just picking two random guys, there are 6 choose 2, or 15 ways to do that, as shown in my previous post. From there: Our target = the one group Our total = 6 ...”
September 18, 2017
Matt@VeritasPrep posted a reply to Probability in the Problem Solving forum
“We could also do All Groups - Groups With Neither Guy - Groups With Only One of the Guys: All Groups = 6 choose 2 = 6!/(4!2!) = 15 Neither Guy = 4 choose 2 = 4!/(2!2!) = 6 Only One Guy = (4 choose 1) * (2 choose 1) = 8 If Only One Guy doesn''t make sense, I''m choosing one of the other four ...”
September 18, 2017
Matt@VeritasPrep posted a reply to Rubik's Cube in the GMAT Math forum
“Think of the cube inside the cube, none of whose faces will be red. To get to that cube inside the cube, we need to cut off a cube on either side of it. So if the larger cube is, say, 8 by 8 by 8, the cube inside will be 6 by 6 by 6. That means we''ll have (8*8*8) - (6*6*6) cubes with at least ...”
September 1, 2017
Matt@VeritasPrep posted a reply to Max purchased a guitar for a total of $624, in the Data Sufficiency forum
“Let''s rephrase the question. We know that Guitar + Tax = 624 If we say the tax is p% of the guitar, we''ve got Guitar * (1 + p%) = 624 or (1 + p%) = 624/Guitar p% = (624/Guitar) - 1 From here, we can write the question as: "Is 624/Guitar - 1 > .03"? S1 tells ...”
September 1, 2017
Matt@VeritasPrep posted a reply to OG2016 DS A box contains only in the Data Sufficiency forum
“The trick here: P(W) + P(B) + P(R) = 1 We want P(W) + P(B). Notice that by subtracting P(R) from both sides of the equation above, we have P(W) + P(B) = 1 - P(R) So knowing P(R) will give us P(W) + P(B)! Once we''re told in S2 that P(R) = 1/3, we''ve got P(W) + P(B) = 1 - 1/3 ...”
September 1, 2017
Matt@VeritasPrep posted a reply to DS in the Data Sufficiency forum
“Whoops, forgot S2! (x + 3)/3 = odd (x + 3) = 3 * odd (x + 3) = odd But this doesn''t help us: as seen above, we still need to know whether (x + 4) is divisible by 4 or not.”
September 1, 2017
Matt@VeritasPrep posted a reply to DS in the Data Sufficiency forum
“(x + 4) * (x² + 8x + 15) => (x + 4) * (x + 3) * (x + 5) This means we''re multiplying three consecutive numbers together: x + 3, x + 4, and x + 5. If (x + 3) is even, then (x + 5) is also even, and we''re multiplying two consecutive evens together, e.g. 2 * 4, 4 * 6, 6 * 8, etc. That ...”
September 1, 2017
Matt@VeritasPrep posted a reply to If x, y, and z are positive numbers, what is the value in the Data Sufficiency forum
“Just to elaborate on my last point, once we get (x + z)/2 = √(x² - z) we can say that (x + z)²/2² = x² - z (x² + 2xz + z²)/2² = x² - z x² + 2xz + z² = 4x² - z 3x² - 2xz - z² - z = 0 If we make this a quadratic in x, such that ax² + bx + c = 0 We can say ...”
September 1, 2017
Matt@VeritasPrep posted a reply to If x, y, and z are positive numbers, what is the value in the Data Sufficiency forum
“We want (x + z)/2. S1: x + z = 2y (x + z)/2 = y This is close! If we can figure out the value of y, we''ll have our answer. S2: x² - y² = z x² - z = y² √(x² - z) = y Ugh. This doesn''t give us y, although it does give us another equation. S1 + S2 We ...”
September 1, 2017
Matt@VeritasPrep posted a reply to Is the standard deviation of the numbers X, Y and Z in the Data Sufficiency forum
“As a tip for test takers, this seems to be a favorite property of the GMAC''s: SD of (a, b, c, ...) = SD of (a + x, b + x, c + x, ...) I''ve seen this on a few official questions, so keep it in mind on test day.”
September 1, 2017
Matt@VeritasPrep posted a reply to Tricky absolute question..... help needed in the Data Sufficiency forum
“It might be easier to do this conceptually. |x + y| > |x - y| can be stated as "x is further from -y than it is from y". S2 can be stated as "x is further from 0 than it is from y". If x is further from 0 than it is from y, x and y MUST be on the same side of zero, ...”
September 1, 2017
Matt@VeritasPrep posted a reply to Number Problem in the Problem Solving forum
“Couldn''t have said it better myself :)”
September 1, 2017
Matt@VeritasPrep posted a reply to Mind Exercise in the Problem Solving forum
“Get the slopes in the same form, y = mx + b: y = -2x + 2 and y = (1/2)x Two lines are perpendicular if the product of their slopes is -1. Since the slopes are -2 and 1/2, respectively, the lines must be perpendicular: -2 * (1/2) = -1.”
September 1, 2017
Matt@VeritasPrep posted a reply to A photographer will arrange 6 people of 6 different heights in the Problem Solving forum
“Let''s elaborate a little bit. We''ll call our six people A, B, C, D, E, and F with heights such that A > B > C > D > E > F. A is the tallest, so we''ll start with him. There''s nobody taller than A, so nobody can stand behind A, meaning A must go in the back row. Since there ...”
September 1, 2017
Matt@VeritasPrep posted a reply to Algebra in the Problem Solving forum
“The simplest way to address that question is to try numbers. |-3| = 3 = -(-3) |-10| = 10 = -(-10) etc. So if x is negative, |x| = -x”
September 1, 2017
Matt@VeritasPrep posted a reply to Algebra in the Problem Solving forum
“(S)he is asking why, for 0 > q, |q| = -q.”
September 1, 2017
Matt@VeritasPrep posted a reply to 1+2+2^2+2^3+2^4+2^5=? in the Problem Solving forum
“The way to solve this under test conditions, however, is to look for a pattern. 1 + 2 = 3 = 2*2 - 1 1 + 2 + 4 = 7 = 2*2*2 - 1 1 + 2 + 4 + 8 = 15 = 2*2*2*2 - 1 Aha! If I add the first n nonnegative powers of 2 on the left, I come out with 2ⁿ - 1 on the right. Under test conditions, ...”
September 1, 2017
Matt@VeritasPrep posted a reply to 1+2+2^2+2^3+2^4+2^5=? in the Problem Solving forum
“Here''s a little hocus pocus: 63 => 2⁶ - 1 => (2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵) * (2 - 1) => 2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵”
September 1, 2017
Matt@VeritasPrep posted a reply to Averages in the Problem Solving forum
“True, but ... 1) I didn''t say it was an official GMAT problem, only that it was in scope; 2) No poster would ever misabbreviate, mistype, or otherwise distort an answer choice on this forum :)”
September 1, 2017
Matt@VeritasPrep posted a reply to OG 13th edition #9 - Arithmetic in the Problem Solving forum
“√∛0.000064 => √∛64/1000000 => √∛(2⁶/10⁶) => √(2²/10²) => 2/10”
September 1, 2017
Matt@VeritasPrep posted a reply to Greatest prime factor... in the Problem Solving forum
“More legibly :) 4¹⁷ - 2²⁸ => (2²)¹⁷ - 2²⁸ => 2³⁴ - 2²⁸ => 2²⁸*2⁶ - 2²⁸*1 => 2²⁸ * (2⁶ - 1) => 2²⁸ * 63 => 2²⁸ * 3 * 3 * 7”
September 1, 2017
Matt@VeritasPrep posted a reply to Exponents - so simple I just don't get it in the Problem Solving forum
“Think of it this way: x + x + x + y + y is 3x''s and 2y''s If we''ve got two 2⁵s and three 3⁵s, then we''ve got 2*2⁵ + 3*3⁵ and from there we''re set!”
September 1, 2017
Matt@VeritasPrep posted a reply to Mathematics in the Problem Solving forum
“Say you''ve got one pound of $5, three pounds of $15, and two pounds of $17.50. That gives you six pounds for $85, or $85/6 per pound for the lot, which works out to $14+(1/6) ... an unpayable price. :)”
August 31, 2017
Matt@VeritasPrep posted a reply to Geometry - Parallelogram Problem in the Problem Solving forum
“You can generally assume that most instructors'' replies are either: 1) right, or 2) wrong, but quickly corrected by another instructor looking to score on the first instructor From those two statements, it more or less follows that an instructor''s reply that hasn''t been corrected by ...”
August 31, 2017
Matt@VeritasPrep posted a reply to Points J (3, 1) and K (– 1, – 3) are two vertices of an in the Problem Solving forum
“Pretty much. (3, 1) and (-1, -3) are equidistant from (2, -2), and from there the solution follows. (You could also treat (2, -2) as the midpoint of the base, then work with isosceles properties.)”
August 31, 2017
Matt@VeritasPrep posted a reply to OG2016 PS If (4 - x)/(2 + x) = x in the Problem Solving forum
“By the way, in case anyone out there is scratching his/her head at this answer: REMEMBER THAT THE PROBLEM ISN''T ASKING US TO SOLVE FOR X :) The GMAC can be so mean ...”
August 31, 2017
Matt@VeritasPrep posted a reply to OG2016 PS If (4 - x)/(2 + x) = x in the Problem Solving forum
“(4 - x) / (2 + x) = x multiply both sides by (2 + x): 4 - x = x * (2 + x), or 4 - x = x² + 2x add x to both sides: 4 = x² + 3x subtract 4 from both sides: 0 = x² + 3x - 4 so x² + 3x - 4 = 0, and we''re done”
August 31, 2017
Matt@VeritasPrep posted a reply to OG2016 PS If x = -1, in the Problem Solving forum
“Replace every x in the equation with -1: ((-1)⁴ - (-1)³ + (-1)²)/(-1 -1) From there it''s pretty easy: (1 - (-1) + 1)/(-2) or -3/2”
August 31, 2017
Matt@VeritasPrep posted a reply to OG 196 (Properties of numbers/exponents) in the Problem Solving forum
“Maybe an easier way of thinking about this: To make the number as small as possible, we want to multiply the number with the largest absolute value (in this case, either -10 or 10) times itself pretty much every time, then end with the smallest negative number (should we need it) to force the ...”
August 31, 2017
Matt@VeritasPrep posted a reply to terminating decimal in the Problem Solving forum
“1/(2³5⁷) => 1/(2³5³5⁴) => 1/10³ * 1/5⁴ => 1/1000 * 1/25 * 1/25 => .001 * .04 * .04 The result will have as many decimal places as everything we''re multiplying together, or 3 + 2 + 2 decimal places, or 7. Since 1 * 4 * 4 = 16, and those numbers come at the ...”
August 31, 2017
Matt@VeritasPrep posted a reply to integer m in the Problem Solving forum
“Suppose we have no restrictions. We''d maximize everything: * Any number on the left side of the median = m * Any number on the right side of the median = the maximum = 2m We''ve got three on each side of the median, giving us: m, m, m, m, 2m, 2m, 2m Average = Sum/# = (2m + 2m + 2m + ...”
August 31, 2017
Matt@VeritasPrep posted a reply to OG Delete in the Problem Solving forum
“If only the OG were so easily vanquished ... :(”
August 31, 2017
Matt@VeritasPrep posted a reply to Easy one in the Problem Solving forum
“ftfy ^ ^”
August 31, 2017
Matt@VeritasPrep posted a reply to Pls explain in the Problem Solving forum
“|a - b| = the distance from a to b With that, we can say |x - 1| > 3 really means the distance from x to 1 is greater than 3 That means that x is greater than 4 (since 4 is exactly 3 units from 1) or that -2 > x (since -2 is exactly 3 units from 1). |x| > 3 is similar: ...”
August 31, 2017
Matt@VeritasPrep posted a reply to Distribution centers in the Problem Solving forum
“Whoops! Forgot to add that my general solution is for the version of the problem downthread (where the letters must be in alphabetical order). The version in which we''re picking colors with no restrictions is simpler, and Brent''s formula is dead on.”
August 31, 2017
Matt@VeritasPrep posted a reply to Distribution centers in the Problem Solving forum
“I''d also plug in the answers if you want a quick and dirty answer, but if we had more time (or the numbers were more unreasonable), we could derive a general equation. Say that we have n letters, subject to the restrictions on coding given in the stem. Since we have n letters, we can have n ...”
August 31, 2017
Matt@VeritasPrep posted a reply to OG2016 PS A positive number x in the Problem Solving forum
“a positive number x: x is multiplied by 2: 2x then this product is divided by 3: 2x/3 and the square root of this equals x: √(2x/3) = x From here, square both sides: 2x/3 = x² divide both sides by x: 2/3 = x”
August 31, 2017
Matt@VeritasPrep posted a reply to Time And Distance Problem in the Problem Solving forum
“Another idea that''s a little quicker (but harder to think up): Our guy travels somewhere 35 km at a rate between 4 and 5 km/r. His minimum time = distance/faster time = 35/5 = 7 His maximum time = distance/slower time = 35/4 = 8.75 He can''t travel 4 OR 5 km for the whole trip, since ...”
August 31, 2017
Matt@VeritasPrep posted a reply to Time And Distance Problem in the Problem Solving forum
“If you want to use a longer but more predictable approach: Say his first distance = r and his second distance = s. Say his time traveling the first distance = x and his time traveling the second distance = y. We know: r + s = 35 r = 4x s = 5y by substitution: 4x + 5y = 35 ...”
August 31, 2017
Matt@VeritasPrep posted a reply to If ab > cd and a, b, c and d are all greater than zero, w in the Problem Solving forum
“We could also use algebra to prove that C is impossible. d/a > b/c Since a and c are both positive, we can crossmultiply without changing the sign: cd > ab But this contradicts what we were told in the stem! We know ab > cd. Since C leads to an impossible inequality, it ...”
August 31, 2017
Matt@VeritasPrep posted a reply to If ab > cd and a, b, c and d are all greater than zero, w in the Problem Solving forum
“It''s probably easiest to try smart numbers here, with the goal of MAKING EACH ANSWER TRUE. If you''re unable to do this after a little effort, there''s a good chance that you''ve found the answer. For instance, trying A, we could say a = 10, b = 1, c = 3, d = 1/2. Then ab > cd, as given, but ...”
August 31, 2017
Matt@VeritasPrep posted a reply to OG 13 -120 in the Problem Solving forum
“Brent, resurrecting threads! A GMAT expert and a necromancer ... :D Another way to do this one: √(r/s) = s square both sides: r/s = s * s multiply both sides by s: r = s * s * s”
August 31, 2017
Matt@VeritasPrep posted a reply to Fence posts problem in the Problem Solving forum
“One of the keys here is that every corner post is counted TWICE, since it''s effectively a vertex of the rectangle. The other key is that a side with 6 posts really has 5 fencelengths of 15, since we only need 1 fence to connect 2 posts, 2 lines to connect 3 posts, etc.”
August 18, 2017
Matt@VeritasPrep posted a reply to Maths in the Problem Solving forum
“p(at least 6) = 1 - p(less than 6) less than 6 = p(exactly 2) + p(exactly 3) + p(exactly 4) + p(exactly 5) We can get exactly 2 in one way, exactly 3 in two ways, exactly 4 in three ways, and exactly 5 in four ways, so less than 6 = 1/36 + 2/36 + 3/36 + 4/36 = 10/36 = 5/18 and we want ...”
August 18, 2017
Matt@VeritasPrep posted a reply to OG If x is a multiple of 4 in the Problem Solving forum
“If x is a multiple of m and y is a multiple of n, then x*y must be a multiple of the LCM of m and n and ALL FACTORS of that LCM. Here the LCM of 4 and 6 is 12, so that''s our answer: all factors of 12.”
August 18, 2017
Matt@VeritasPrep posted a reply to For a positive integer n, when 12n is divided by 15 in the Problem Solving forum
“12n / 15 => 4n/5 => (4/5) * n Since n divides into 5 pieces, we can''t have 5 left over. (If this doesn''t make sense, imagine I have 10 cookies and 5 friends. Each friend gets 2 cookies and there are 0 left over.)”
August 18, 2017
Matt@VeritasPrep posted a reply to OG If n is the product in the Problem Solving forum
“Count the unique primes in 2 * 3 * 4 * 5 * 6 * 7 * 8”
August 18, 2017
Matt@VeritasPrep posted a reply to OG Square root of 16/81 in the Problem Solving forum
“The half power is the same as the square root, so for legibility''s sake let''s write: n = √16/81 n = 4/9 From here, √n = √4/9 = 2/3.”
August 18, 2017
Matt@VeritasPrep posted a reply to Minimum value of an expression in the Problem Solving forum
“Completing the square is a very clever idea - I remember how excited I was the first time I saw it, and I won''t apologize for that :D - but AFAIK it''s totally irrelevant to the GMAT, so you can safely ignore this question for the time being.”
August 18, 2017
Matt@VeritasPrep posted a reply to A certain compound X has a ratio of 2 oxygen for every in the Problem Solving forum
“Also, I would guess that the OA is C, using this logic: (2/5)x + (1/4)y = (3/10)*(x + y) which reduces to a ratio of 1:2. Easy mistake to make!”
August 18, 2017
Matt@VeritasPrep posted a reply to A certain compound X has a ratio of 2 oxygen for every in the Problem Solving forum
“Easier, I think: (2/7)x + (1/5)y = (3/13)*(x + y) multiply both sides by 7*5*13: 2*5*13x + 7*13*y = 3*7*5*(x + y) 130x + 91y = 105x + 105y 25x = 14y and we''re done!”
August 18, 2017
Matt@VeritasPrep posted a reply to OG Bouquets in the Problem Solving forum
“We could also say that we''ve got 100 flowers, forcing our bouquets to be some factor of 100. That leaves 5 and 10 as options, but 10 won''t work because we can''t divide 15 bulbs into 10 bouquets without cutting them into fractions.”
August 18, 2017
Matt@VeritasPrep posted a reply to OG 125% of in the Problem Solving forum
“1.25 * 5 = 1*5 + .25*5 = 5 + (1/4)*5 = 5 + 5/4 = 5 + 1.25 = 6.25 This seems more like a drill question in a skillbuilder than an actual GMAT question.”
August 18, 2017
Matt@VeritasPrep posted a reply to OG Quotient divided in the Problem Solving forum
“x / (2/3) = 9/2 (3/2) * x = 9/2 3x/2 = 9/2 3x = 9 x = 3 Very surprised this is an OG question - it feels at or below the minimum difficulty of the GMAT.”
August 18, 2017
Matt@VeritasPrep posted a reply to OG 1 < x < y < z in the Problem Solving forum
“We could also do algebra. Here''s how to compare D vs E, and from this you can extrapolate how to do the others: y * (x + z) vs z * (x + y) yx + yz vs xz + yz subtract yz from both sides: yx vs xz divide both sides by x: y vs z z is bigger, so the right side is bigger, so ...”
August 18, 2017
Matt@VeritasPrep posted a reply to OG Car X traveled north in the Problem Solving forum
“The car traveled for 3.75/5, or 3/4 of 2 hours, at a rate of 60mph, so the car traveled (3/4) * 120, or 90 miles. (That 2 hours part is easy to overlook: I guess having 45 as an answer was too evil even for the GMAT!)”
August 18, 2017
Matt@VeritasPrep posted a reply to Rate problem from official guide in the Problem Solving forum
“... because you''ve outstickled the stickliest of us on the forum, time and again! :) I hope I''m still the most indignant about ² over ^2, though. I can''t see how in 2017 any expert imagines him/herself halfway credible if he/she won''t bother or can''t learn how to type that.”
August 18, 2017
Matt@VeritasPrep posted a reply to OG Cheryl purchaded 5 identical hollow pine doors in the Problem Solving forum
“Say the price of each oak door = x and the price of each pine door = y. We''re given x = 2y, so Cheryl should''ve paid 5*y + 6*2y, or 17y. She got 25% each oak door, so she actually paid 5y + 4.5*2y, or 14y. y = $40, making the total order 14*$40, or $560.”
August 18, 2017
Matt@VeritasPrep posted a reply to OG In the xy-plane, the Origin is the midpoint in the Problem Solving forum
“If we''ve got two points (a, b) and (c, d), then their midpoint = ((a + b)/2, (c+d)/2). Since the origin has the coordinates (0, 0), we need (a + b) = 0 and (c + d) = 0. This is possible with (r + -r) and (s + -s), so our missing coordinate is (-r, -s).”
August 18, 2017
Matt@VeritasPrep posted a reply to OG Sum of fractions in the Problem Solving forum
“1/9 = (1/3) * (1/3) 1/12 = (1/4) * (1/3) 1/15 = (1/5) * (1/3) 1/18 = (1/6) * (1/3) So everything on the right side is (1/3) of everything on the left side, meaning the left side is 3 times bigger than the right, and r = 3.”
August 18, 2017
Matt@VeritasPrep posted a reply to OG Which of the following is greatest in the Problem Solving forum
“Also, a follow up question for all those browsing the forum: if 13 > x > 0, for what positive value of x is x * √(13 - x) the greatest? Your answers will be scored by how elementary they are, the simpler (no calculus!) the better. :D”
August 18, 2017
Matt@VeritasPrep posted a reply to OG Which of the following is greatest in the Problem Solving forum
“Since all of these are > 1, we know that if x > y, x² > x² and vice versa. With that in mind, we can square each answer and take the biggest one, which is B.”
August 18, 2017
Matt@VeritasPrep posted a reply to OG The estimated number of homeschooled students in the Problem Solving forum
“% greater = 100 * ((new value - old value)/(old value) - 1) so 100 * ((181 - 79) / 79 - 1) => 100 * ((102 / 79) - 1) => Since 102/79 is pretty close to 4/3, we can say that this is about 100 * (4/3 - 1) or 33% D is the closest to this, so we''re good to go.”
August 18, 2017
Matt@VeritasPrep posted a reply to OG Number of packages shipped in the Problem Solving forum
“I''m having a hard time reconciling these two statements :) I don''t mean to goof on Jay, though: I think he''s giving excellent advice. Asking a lot of questions is usually a red flag that a student isn''t doing enough to work through the questions on his/her own and really learn the material, ...”
August 18, 2017
Matt@VeritasPrep posted a reply to OG The sum of the weekly salaries in the Problem Solving forum
“Say our salaries are a, b, c, d, e. We know that (a + b + c + d + e) = 3250. If everybody''s salary goes up by 10%, then we just multiply each side by 1.1: 1.1 * (a + b + c + d + e) = 1.1 * 3250 Since the average = 3250/5, we can see that average went up by 10% too, and we''re set.”
August 18, 2017
Matt@VeritasPrep posted a reply to OG Last week Chris earned in the Problem Solving forum
“Second option: plug in! Since answer C is in the middle of the answers, we can start there. If it works we''re done, and if doesn''t we''ll know whether to go up (to D) or down (to B). C = $15, so Chris would make $15*40 + $22*8. That''s not enough money, so we need to go UP. We then try ...”
August 18, 2017
Matt@VeritasPrep posted a reply to OG Last week Chris earned in the Problem Solving forum
“First option: algebra! Say Chris worked h hours in all. Total $ = $40 * x + $22 * (h - 40) We know h = 48 and total $ = $816, so 816 = 40x + 22*(48 - 40) 816 = 40x + 176 640 = 40x 16 = x”
August 18, 2017
Matt@VeritasPrep posted a reply to fractional exponents ... in the GMAT Math forum
“Very likely, they are totally in bounds. I wouldn''t lose too much sleep over them, though: just remember how exponents interact (addition and subtraction) and you should be OK.”
August 18, 2017
Matt@VeritasPrep posted a reply to can we consider SQRT(n) be negative? in the GMAT Math forum
“AFAIK this is a convention dating back to the ancient Greeks, who didn''t use/believe in/consider negative numbers, but did make great use of square roots. √16 = 4, always and forever, because we used the √ symbol. On the other hand, x * x = 16 has two roots, x = 4 and x = -4.”
August 18, 2017
Matt@VeritasPrep posted a reply to help in the GMAT Math forum
“Neat follow up question for anyone who wants to delve further into similar ideas: https://www.beatthegmat.com/triangle-inside-a-rectangle-t16268.html”
August 18, 2017
Matt@VeritasPrep posted a reply to Integer Properties. in the GMAT Math forum
“Brent, you are no pun at all :p”
August 18, 2017
Matt@VeritasPrep posted a reply to Is |X| < 1 ? in the GMAT Math forum
“My pleasure!”
August 18, 2017
Matt@VeritasPrep posted a reply to Negative Values in the GMAT Math forum
“We could take a conceptual approach too. Remembering that |x - y| = the distance from x to y on the number line, we could say A:: |a + b| - |a - b| < 0 |a + b| < |a - b| the distance from a to -b < the distance from a to b This is definitely possible. B:: |a + b| < ...”
August 18, 2017
Matt@VeritasPrep posted a reply to Negative Values in the GMAT Math forum
“Let me add a (semi-)algebraic approach too. Suppose that each answer is negative. For instance, for answer B that would give us |a + b| - |a| < 0, or |a + b| < |a| This is obviously possible, if a = 3 and b = -1, so a negative result could work here. Continuing this way ...”
August 18, 2017
Matt@VeritasPrep posted a reply to Negative Values in the GMAT Math forum
“If you try numbers, remember that you don''t have to try each answer every time! Once you get a negative result for an answer, that answer is eliminated, and you only have to doubletest the answers that showed positive.”
August 18, 2017
Matt@VeritasPrep posted a reply to Co-Ordinate Geometry in the GMAT Math forum
“Sure! It can''t be on the x-axis, since the lines will only intersect at one point, and we''ve already got an intersection between the two: x = 2 and y = 0 (the x-axis). Two straight lines can only intersect at one point - this is one of the cornerstones of Euclidean geometry. (Well, they can also ...”
August 18, 2017
Matt@VeritasPrep posted a reply to How to prepare for Quant in the GMAT Math forum
“Hi Tapas! You still out there? I''m ready and standing by with an answer :)”
August 18, 2017
Matt@VeritasPrep posted a reply to From 45/45/46 in Quant to 41/41 in the GMAT Math forum
“I''ll second that: don''t think of the GMAT as a test of rules. For almost everyone, the GMAT is the first exposure to real mathematical thinking, which isn''t about memorizing and repeating a bunch of formulas you were given in class, but actually solving problems in novel and creative ways without ...”
August 18, 2017
Matt@VeritasPrep posted a reply to Changes in preparation in the last 4 weeks in the GMAT Math forum
“Also, if you find that your hit rate isn''t improving through sheer volume of practice questions, stop and linger a bit on some of the problems you''ve already tried. My rule of thumb: if I can''t explain this question very, very clearly to a reasonably bright person who doesn''t know math, then I ...”
August 18, 2017
Matt@VeritasPrep posted a reply to Help: best resources for Quant prep in the GMAT Math forum
“I''ve also noticed over the years that students who start out above average (or better) in verbal and average (or worse) in math tend to improve much more than students who start strong in math and weak in verbal, so take heart! You can definitely improve, it''s just a matter of acquainting yourself ...”
August 18, 2017
Matt@VeritasPrep posted a reply to Jonathan buys c chairs and t tables for his newly set up res in the GMAT Math forum
“One nitty footnote to my last post, though (I didn''t want to be a wet blanket, but I must): we''re assuming that (a - c) doesn''t equal 0, or we can''t divide to set up that ratio.”
August 18, 2017
Matt@VeritasPrep posted a reply to Jonathan buys c chairs and t tables for his newly set up res in the GMAT Math forum
“This is a neat point. If we have ax + by = cx + dy, we can say that: ax - cx = dy - by x * (a - c) = y * (d - b) x/y = (d - b)/(a - c) Exactly as Balla told us! :)”
August 18, 2017
Matt@VeritasPrep posted a reply to OG Square root of Sci Notation Q in the Problem Solving forum
“The answers differ by powers of 10, so we can afford some clumsy approximation: √4.8 * √10⁹ ≈ 2 * √10⁸ * √10 ≈ 2 * 10⁴ * 3 ≈ 6 * 10000 Only B is at all close.”
August 7, 2017
Matt@VeritasPrep posted a reply to Tough Geometry Question in the Problem Solving forum
“Two important takeaways from my approach: 1) Drawing a well-chosen line can make seemingly impossible geometry questions very accessible, so when in doubt, ask yourself if you can draw something; 2) Always remind yourself that a GMAT problem can''t rely on anything beyond the basics you were ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Tough Geometry Question in the Problem Solving forum
“Not that I''m against the Ailles rectangle, but I don''t think we should assume students would know that or dream it up. How about this, a very GMAT appropriate way that only requires drawing a few lines and knowing the ratios of 45-45-90 and 30-60-90s. No algebra is used here, and the steps are ...”
August 7, 2017
Matt@VeritasPrep posted a reply to M is a positive integer less than 100 in the Problem Solving forum
“Fear not, help is on the way! Superscripts Anonymous is always in session here. (If you forget the codes, just bookmark the link and copypaste the relevant superscript for each problem.) The only ^s we should see on this board are in goofy ASCII smiley faces perpetrated by yours truly! ...”
August 7, 2017
Matt@VeritasPrep posted a reply to M is a positive integer less than 100 in the Problem Solving forum
“m³ = n² So we''re looking for numbers that are both cubes and squares. If a number is a cube, then it has some multiple of 3 of each of its prime factors. If a number is a square, then it has some multiple of 2 of each of its prime factors. If a number is both a cube AND a square, ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Percent problem........OG2018 in the Problem Solving forum
“We know (A and B) is a subset of A, and we want a ratio, so we can say (A and B) / A = our ratio 3 / 45 = our ratio 1/15 = our ratio From here, we want 1/15 = x/100. Multiplying both sides by 100 gives us x = 100/15 = 20/3 = 6 + (2/3) ≈ 6.67%”
August 7, 2017
Matt@VeritasPrep posted a reply to OG The average score on a test taken by 10 students in the Problem Solving forum
“Sum = Average * # of Things in the Set For the whole group, Sum = 10x. For five of the group, the sum = 5*8, so the sum for the other five = 10x - 40. The average for those five is the sum divided by five, or (10x - 40)/5, or 2x - 8.”
August 7, 2017
Matt@VeritasPrep posted a reply to Rate problem from official guide in the Problem Solving forum
“Another approach: Bob is going to run for 50 minutes at (60/8) miles per hour, so he''ll be able to run (5/6) * (60/8) => 25/4 miles. Adding in the distance Bob has ALREADY run (3.25 miles), he''ll run a total of 38/4 miles in all. Half of that, or 19/4, will be south. He''s already ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Rate problem from official guide in the Problem Solving forum
“Let me be useful and add an explanation too! Bob will run x more miles south, then turn around, run those x miles again north, then run the final 3.25 miles north. That means that Total distance to run = x + x + 3.25 Total time = 50 minutes = 5/6 of an hour We know Distance = Rate * Time, ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Rate problem from official guide in the Problem Solving forum
“Somewhere on the interwebs, Ceilidh is smiling! :D”
August 7, 2017
Matt@VeritasPrep posted a reply to OG On Saturday morning, Malachi will begin camping in the Problem Solving forum
“We need (Saturday no rain) * (Sunday no rain) * (Monday YES rain) => .8 * .8 * .2 => .128”
August 7, 2017
Matt@VeritasPrep posted a reply to Problem solving in the Problem Solving forum
“This one really is the riddle of the sphinx”
August 7, 2017
Matt@VeritasPrep posted a reply to GMATPREP QP1 - A machine has two flat circular plates of the in the Problem Solving forum
“Another trick, since the images are to scale, is to look at the points that seem closest (but not equal). Looking at the plates, it''s clear that the top hole is equal. The holes immediately to the right of the top hole on each plate are remarkably close: visually the closest. On the first ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Standard Deviation in the Problem Solving forum
“This is such a great lesson to learn from this problem. On the GMAT, of all places, is the devil IS in the details: subtle misreadings are deadly, and many of the questions are written to ENCOURAGE such misreadings!”
August 7, 2017
Matt@VeritasPrep posted a reply to I Suck at Probability need desperate help in the Problem Solving forum
“Yeah, this one is all about strategic guessing - it almost punishes you for doing it "properly".”
August 7, 2017
Matt@VeritasPrep posted a reply to Melting cube! in the Problem Solving forum
“I''m not sure about the stem here: it''s not clear whether the cube is made by gluing the cubes together in some way (like a tower of cubes) or by breaking in down into 216 unit cubes, then using those to form a giant cube. I guess the intention is the second ... but even then ''melting'' gives such ...”
August 7, 2017
Matt@VeritasPrep posted a reply to An operation Ф is defined by the equation x Ф y in the Problem Solving forum
“A much lamer approach: x²/4 - xy + y² => 186²/2² - 186*91 + 91² => 93² - 186*91 + 91² => 93² - 2*93*91 + 91² => (93 - 91)² => 2²”
August 7, 2017
Matt@VeritasPrep posted a reply to An operation Ф is defined by the equation x Ф y in the Problem Solving forum
“x Ф y = x²/4 - xy + y² 4 * (x Ф y) = x² - 4xy + 4y² 4 * (x Ф y) = (x - 2y)² x Ф y = (x - 2y)² / 4 From there, plug and chug, and we''re set.”
August 7, 2017
Matt@VeritasPrep posted a reply to Mixture Problem............OG Problem in the Problem Solving forum
“A grocer has 400 pounds of coffee in stock, 20 percent of which is decaffeinated. If the grocer buys another 100 pounds of coffee of which 60 percent is decaffeinated, what percent, by weight, of the grocer’s stock of coffee is decaffeinated? One last approach! If we want to find 20% of ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Mixture Problem............OG Problem in the Problem Solving forum
“Another approach: We know 80% of the coffee is 20% decaf, and that 20% of the coffee is 60% decaf. Since we have four times as much of the 20% as we do of the 60%, the average must be (4/5) of the way from 60 to 20. 60 - (4/5) * (60 - 20) = 28, so that''s our answer: 28%.”
August 7, 2017
Matt@VeritasPrep posted a reply to Mixture Problem............OG Problem in the Problem Solving forum
“One approach: We want ((20% of 400) + (60% of 100)) / 500, or (.2 * 400 + .6 * 100) / 500. Multiplying by 10/10 gives us (2 * 400 + 6 * 100) / 5000, or 1400/5000, or 14/50, or 28/100, or 28%.”
August 7, 2017
Matt@VeritasPrep posted a reply to OG Quantitative Review in the GMAT Math forum
“It also isn''t a bad idea to practice categorizing them yourself to see whether you''re recognizing problems well and thinking of which topics to draw on when attacking a question.”
August 7, 2017
Matt@VeritasPrep posted a reply to Word Problem involving Stats in the GMAT Math forum
“Just reading this problem reminded me of the bad old days when I had to grade stacks of papers like these, and, well ... ‹’’›(Ͼ˳Ͽ)‹’’›”
August 7, 2017
Matt@VeritasPrep posted a reply to Weighted Average problem in the GMAT Math forum
“We could do it algebraically too, though I like the conceptual approach better. Let''s call m the male average and f the female average. We know that the total number of tickets can be represented in two ways: (Average) * (# of People) OR (Female Average) * (# of Women) + (Male ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Statistics in the GMAT Math forum
“If the mean is 21 and the sd is 6, then mean = 21 one sd below = 21 - 6 = 15 two sds below = 21 - 2*6 = 9 so the answer should be between 9 and 15.”
August 7, 2017
Matt@VeritasPrep posted a reply to Apply now for the 2012 Beat The GMAT Scholarship! 1 wk left! in the GMAT Math forum
“It seems to have expired in 2012. (Mods, it might be a good idea to unsticky this thread?)”
August 7, 2017
Matt@VeritasPrep posted a reply to basic math in the GMAT Math forum
“Certainly! I don''t know if I deserve being ''dear'', but hey, I''m flattered.”
August 7, 2017
Matt@VeritasPrep posted a reply to How to prepare for Quant in the GMAT Math forum
“How close are you to your goals? What have you scored so far, and what score would you need to be done with the exam?”
August 7, 2017
Matt@VeritasPrep posted a reply to Geometry in the GMAT Math forum
“Equations are the best way here, but we could always go for another set of them from the ones given above. k = the slope of the line, and since k = 0 isn''t an option in the answers, know the line ISN''T horizontal. We also know that k is defined, so the line can''t be perfectly vertical either. ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Line Segment in the GMAT Math forum
“It isn''t worded properly, but yeah, that''s the idea: each of the three diamonds is a rhombus with side lengths of 4.”
August 7, 2017
Matt@VeritasPrep posted a reply to Jonathan buys c chairs and t tables for his newly set up res in the GMAT Math forum
“By the way, if you like S1, the fancy name for these equations - algebraic problems that only consider integer solutions - is Diophantine equations. This is a rich and ancient topic in math, definitely worth Googling and exploring should it strike your fancy.”
August 7, 2017
Matt@VeritasPrep posted a reply to Jonathan buys c chairs and t tables for his newly set up res in the GMAT Math forum
“One last, very crude approach to S1: 5c + 6t = 50 c + 1.2t = 10 We know that c is an integer, so 1.2t must also be an integer. 1.2t will only be an integer if t is a multiple of 5, since .2 * 5 = 1. But if t > 5, then 1.2t > 10, forcing c to be negative, which is impossible. So ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Jonathan buys c chairs and t tables for his newly set up res in the GMAT Math forum
“A light bulb just flickered and here''s another quick and dirty approach to S1: Start with 5c + 6t = 50. That gives us: 6t = 50 - 5c 6t = 5 * (10 - c) Since 6t = 5 * something = some multiple of 5, we know that either 6 or t is a multiple of 5. 6 obviously isn''t, so t must be. ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Jonathan buys c chairs and t tables for his newly set up res in the GMAT Math forum
“Here''s another approach that I think gets through S1 more quickly. (It isn''t very elegant, but it''s great for timed test conditions.) Let''s start by sorting out the prompt. We know that he spent $40c on chairs and $68t on tables, and we''re asked for 40c / (40c + 68t). OK! S1: He ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Co-Ordinate Geometry in the GMAT Math forum
“Another trick: notice where the line intercepts the x-axis, then work from there. We know (2, 0) is on the line and on the x-axis. We know that for all x-values < 2, the line is below the x-axis, and that for all x-values > 2, the line is above the x-axis. "Below the x-axis" ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Integer Properties. in the GMAT Math forum
“One last way to get there (and to get a feel for what''s happening) is trying numbers. A simple case if a = √2 and b = √2. That gives us [a] = 1 and [b] = 1 and [a] + [b] = 2. OK, too big, but let''s try monkeying around with the numbers. For simplicity''s sake, we''ll make a > b. If ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Integer Properties. in the GMAT Math forum
“Another way we could get there: ab = 2 a = 2/b With that, we have and . That''s a contradiction, so [b] = 1 and [2/b] = 0 is impossible. The second case ([b] = 0 and [2/b] = 1) works much the same way.”
August 7, 2017
Matt@VeritasPrep posted a reply to Integer Properties. in the GMAT Math forum
“Let''s assume that both a and b are positive. (If they''re both negative, then obviously their floors can''t sum to 1.) If a > 0, b > 0, and ab = 2, then AT LEAST ONE of a and b must be ≥ √2. (If they''re both < √2, then their product is less than √2 * √2.) If one of them is ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Integer Properties. in the GMAT Math forum
“But they didn''t read him his copyrights, so he''s off the hook :D”
August 7, 2017
Matt@VeritasPrep posted a reply to Is |X| < 1 ? in the GMAT Math forum
“Another way, if the first is too technical: S1: x > x / |x| |x| * x > x Now take two cases. If x > 0, then we divide both sides to get |x| > 1. If x < 0, then we divide both sides to get |x| < 1. So |x| will be < 1 if and only if x < 0. S2 I''d ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Is |X| < 1 ? in the GMAT Math forum
“Here''s one way: S1: Since |x| must be > 0, we can safely multiply both sides by it: x < x * |x| Now subtract x from both sides: 0 < x * |x| - x Now factor out x: 0 < x * (|x| - 1) This gives us two possibilities: (i) x AND |x| - 1 are both positive ...”
August 7, 2017
Matt@VeritasPrep posted a reply to Train A leaves New York in the Data Sufficiency forum
“I hate to resurrect this thread, but I was asked about this problem today and realized there wasn''t an expert answer. How I''d do it: Let b = Train B''s rate. We''re going to solve this problem entirely in terms of this variable. When the two trains meet, Train B has traveled for (1/6) ...”
July 26, 2017
Matt@VeritasPrep posted a reply to Complex rate problem in the Data Sufficiency forum
“I hate to resurrect this thread, but I was asked about it today and realized there wasn''t an expert answer. How I''d do it: Let b = Train B''s rate. We''re going to solve this problem entirely in terms of this variable. When the two trains meet, Train B has traveled for (1/6) of an hour ...”
July 26, 2017
Matt@VeritasPrep posted a reply to Day 6- FDP problem in the GMAT Math forum
“You could also remember that p/100 = p% So if you''ve got 1000/10, that''s equivalent to 10000/100, or 10000%. It''s not the best worded problem, though, since %s are usually taken from some base (e.g. "1000/10 is what percent of x?"). Simply saying "what percent is ...”
July 24, 2017
Matt@VeritasPrep posted a reply to A taxi company charges $1.75 for the first in the Problem Solving forum
“1.75 + .15q = 4.90 subtract 1.75: .15q = 3.15 multiply by 100: 15q = 315 divide by 15: q = 21 So we could travel an extra 21 quarter miles, or 5.25 miles. Adding the first 1/4 mile to the trip (the one for which we paid $1.75), we have a total of 5.50 miles.”
July 24, 2017
Matt@VeritasPrep posted a reply to The average of seven numbers in the Problem Solving forum
“Let''s say the "first four numbers" (whatever they mean by that) = a, b, c, d and that the last four numbers = d, e, f, g. We know that (a + b + c + d)/4 = 19 and (d + e + f + g)/4 = 24. From there, a + b + c + d = 76 and d + e + f + g = 96. We''re also told (indirectly, but I think ...”
July 24, 2017
Matt@VeritasPrep posted a reply to Percentage question in the Problem Solving forum
“Am I the only one who can''t see the poll in the header? I see it in the title, but the thread shows as a normal thread, with no poll option or results.”
July 24, 2017
Matt@VeritasPrep posted a reply to If the square root of p^2 is an integer greater than 1 in the Problem Solving forum
“Just a friendly (weekly) PSA to all instructors that ² is much nicer to read than ^2 :)”
July 24, 2017
Matt@VeritasPrep posted a reply to Jeff drives three times farther in the Problem Solving forum
“We could also say that Amy drives (1/3) of Jeff''s D in (5/6) of the time, so Amy would drive Jeff''s D in 3 * (5/6), or 15/6, or 5/2 of the time. Since Rate and Time are reciprocal, if Amy''s T = (5/2) of Jeff''s, then Amy''s R = (2/5) of Jeff''s. (2/5) of 40 is 16, and we''re done.”
July 24, 2017
Matt@VeritasPrep posted a reply to Jeff drives three times farther in the Problem Solving forum
“D = RT For Jeff, D = 40 * (3/5 of an hour) = 24 miles For Amy, D = R * (1/2 of an hour). We know Amy''s D = 8, since it''s 1/3 of Jeff''s D. That leaves 8 = R * 1/2, or R = 16.”
July 24, 2017
Matt@VeritasPrep posted a reply to Which of the following equations has in the Problem Solving forum
“Another way to get there: A and B are two sides of the same coin (2x = y and 2y = x), so they''ll both have the same number of solutions. D has an infinite number of solutions: {1,2} and {2, 3} and {3,4} and ... E is trickier, but xy = 1 has x = 1, y = 1 and x = -1, y = -1 as solutions, ...”
July 24, 2017
Matt@VeritasPrep posted a reply to Which of the following equations has in the Problem Solving forum
“I''d go right to C: that √5 can only be turned into an integer by being multiplied by something else with a √5 in it (which wouldn''t be an integer), or by 0. So x must be 0. From there, y must also be 0, and the only integer solution to C is x = 0, y = 0.”
July 24, 2017
Matt@VeritasPrep posted a reply to OG Negative Exponents Q in the Problem Solving forum
“Let''s rephrase the question first. We''re asked to solve for x in the following: (2⁻¹⁴ + 2⁻¹⁵ + 2⁻¹⁶ + 2⁻¹⁷)/5 = x * 2⁻¹⁷ Let''s multiply both sides by 5: (2⁻¹⁴ + 2⁻¹⁵ + 2⁻¹⁶ + 2⁻¹⁷) = 5x * 2⁻¹⁷ Then divide both sides by 2⁻¹⁷: ...”
July 24, 2017
Matt@VeritasPrep posted a reply to here are 5 locks and 5 keys and each of the 5 keys matches in the Problem Solving forum
“I actually took the first interpretation: they do mean that each key is a skeleton key that works in any of the five locks. From there, I took the minimum as the fewest (cleverly chosen) trials we''d need to run to prove this to a skeptical observer and the maximum as the worst (= longest) scenario ...”
July 24, 2017
Matt@VeritasPrep posted a reply to An army’s recruitment process included n rounds of selecti in the Problem Solving forum
“It might also help to see this with fractions: (2/5)ᵃ * (1/2)ᵇ * (7/10)ᶜ = 1400/100000 = 7/500 From here, notice that c = 1, since we have exactly one 7 in the numerator. That gives us (2/5)ᵃ * (1/2)ᵇ * (7/10) = 7/500 or (2/5)ᵃ * (1/2)ᵇ = 70/3500 = 1/50 From here, ...”
July 24, 2017
Matt@VeritasPrep posted a reply to An army’s recruitment process included n rounds of selecti in the Problem Solving forum
“Let''s start with everything in terms of selection percentage. First a rounds: 40% selected Next b rounds: 50% selected Last c rounds: 70% selected We know that in the end 1.4% were selected, so .4ᵃ * .5ᵇ * .7ᶜ = .014 From here, we can cheat a bit. .4 * .5 * .7 = .14, so we''re ...”
July 24, 2017
Matt@VeritasPrep posted a reply to Alice, Bobby, Cindy, Daren and Eddy participate in in the Problem Solving forum
“Another way would be finding the total number of arrangements, then finding the fraction that meet our condition (A > B > C). If we had no restrictions at all, we''d have 5! or 120 arrangements. But we need A > B > C, or ABC in that order. If we have three people, we have 3!, or ...”
July 24, 2017
Matt@VeritasPrep posted a reply to Alice, Bobby, Cindy, Daren and Eddy participate in in the Problem Solving forum
“I''ve got a few ways. Let''s start with one in which we place Bobby. Since Bobby finishes behind someone and ahead of someone else, he can only finish 2rd, 3rd, or 4th. Case 1: Bobby finishes second This means Alice is first, Bobby is second, and we have 3! arrangements for everyone else. ...”
July 24, 2017
Matt@VeritasPrep posted a reply to OG Marbles in Bags Q in the Problem Solving forum
“We know that (.108*37 + (2/3)x + .5*32)/ (37 + 32 + x) = 1/3 Assuming that .108 is rounded and treating .108*37 as 4, this gives us (4 + (2/3)x + 16)/(69 + x) = 1/3 From here, crossmultiply and solve: 3 * (20 + (2/3)x) = 69 + x 60 + 2x = 69 + x x = 9”
July 24, 2017
Matt@VeritasPrep posted a reply to OG Root Q in the Problem Solving forum
“One last approach, and probably the worst of the five, is to use the x² term to help out. Working with the answers, we can rewrite all of them in terms of x²: A) x² = 1 - 2x B) x² = 2x - 1 C) x² = -2x - 1 D) x² = 1 + 2x E) x² = x + 1 From here, compute x²: (1 + √2)² => ...”
July 23, 2017
Matt@VeritasPrep posted a reply to OG Root Q in the Problem Solving forum
“A pretty crude option would be guesstimation. √2 is about 1.4, so our root is about 2.4. Plugging 2.4 into the answers should give me one quadratic that actually works and four that don''t. A:: 2.4² + 2*2.4 - 1 = 0, no, left side way too positive B:: 2.4² - 2*2.4 + 1 = 0, no, the first ...”
July 23, 2017
Matt@VeritasPrep posted a reply to OG Root Q in the Problem Solving forum
“We could also the factor that we have in a different way. If the roots of my quadratic are r and s, I know that (x - r) * (x - s) = 0 I know that one of the roots, let''s say r, is 1 + √2, so (x - (1 + √2)) * (x - s) = 0 Looking at my answer choices, I notice that none of them ...”
July 23, 2017
Matt@VeritasPrep posted a reply to OG Root Q in the Problem Solving forum
“Another approach: using the factor we have. We know from the quadratic formula that the solutions to a quadratic of the form ax² + bx + c = 0 are x = (-b ± √(b² - 4ac))/2a I know that a = 1 (since there''s no coefficient on x² in any of the answers) and that one value of x = 1 + ...”
July 23, 2017
Matt@VeritasPrep posted a reply to OG Root Q in the Problem Solving forum
“I''ve got a few ways to solve this, but I''ll start by invoking my favorite, the Lazy Testwriter Principle. If 1 + √2 is a root, then the laziest other root we could possibly dream up is 1 - √2. The sum of the roots is (1 + √2) + (1 - √2), or 2. The product of the roots is (1 + √2) * ...”
July 23, 2017
Matt@VeritasPrep posted a reply to OG Negative Numbers Q in the Problem Solving forum
“Here''s a hack for you. The testwriters know that many people will accidentally choose the LESSER number, so they''ll put both the lesser and the greater in the answer choices. From there, look for two numbers that fit. -20 is four less than 2 * -8, so those must be the two numbers, and hey hey, ...”
July 23, 2017
Matt@VeritasPrep posted a reply to OG Joe and Sally in the Problem Solving forum
“Joe and Sally? I feel like I''ve walked into a time warp, come on OG!”
July 23, 2017
Matt@VeritasPrep posted a reply to Og Inequality Q in the Problem Solving forum
“I''d simplify the inequalities then graft them together. The first one: x + 6 > 10 => x > 4 The second one: 5 ≥ x - 3 => 8 ≥ x Since x is the middle term, we use it to connect them: 8 ≥ x > 4.”
July 23, 2017
Matt@VeritasPrep posted a reply to Approach is faulty: Experts please comment in the Problem Solving forum
“Another idea would be treating Michael and Anthony as one person, then assigning the committees from there. If Michael and Anthony are one person, then we have five people to assign to the committees. Our Michael/Anthony siamese twin can go to either of the two groups (2!), then we simply choose ...”
July 23, 2017
Matt@VeritasPrep posted a reply to mixture problem - rabbit & proteins in the Problem Solving forum
“One last idea, that tends to work in a pinch, is trying the answers. We may as well start with 150: if it''s too high, we can move to A&B, and if it''s too low, we can move to C&D. If we plug in 150, our rabbit got 10% of 150 + 15% of 150, or 37.5 grams of protein. Slightly too high! ...”
July 23, 2017
Matt@VeritasPrep posted a reply to mixture problem - rabbit & proteins in the Problem Solving forum
“We could also think of it conceptually. If our rabbit is getting 10% from X and 15% from Y, his protein total is .1x + .15y, so .1x + .15y = 38. We also know that our rabbit ate 300 grams of food, so x + y = 300. If I multiply the first equation by 10, I have x + 1.5y = 380. From there, ...”
July 23, 2017
Matt@VeritasPrep posted a reply to mixture problem - rabbit & proteins in the Problem Solving forum
“Another approach: 38/300 = 12 + (2/3)% If we had exactly half of each food, then we''d expect an average of 12.5%. We''re SLIGHTLY higher than that, so must have SLIGHTLY more Food Y and SLIGHTLY less Food X. Slightly less than half of 300 grams should be slightly less than 150, and of ...”
July 23, 2017
Matt@VeritasPrep posted a reply to beat this probability Qs in the Problem Solving forum
“As a fun aside, this problem is a nice primer on the perils of insider trading. Let''s say that I''m a bookmaker and decide to take bets on this marble drawing. Prior to Mary''s draw, John is 62.5% to draw blue, so the fair odds on that are 3 to 5: if you want to bet that John will draw blue, I''ll ...”
July 23, 2017
Matt@VeritasPrep posted a reply to beat this probability Qs in the Problem Solving forum
“Another way to demonstrate this: Suppose that there are x marbles that I want in a jar that contains y marbles in all. My friend gets to pick and keep a marble first, then it''s my turn. If my friend gets one of the nice marbles, then my chance of getting one on my pick is (x - 1)/(y - 1). ...”
July 23, 2017
Matt@VeritasPrep posted a reply to basic math in the GMAT Math forum
“Here''s the simplest way, I think: call this prime p. Since our number = 2 * 3 * p, it has the following factors: 1, 2, 3, p, 2*3, 2*p, 3*p, 2*3*p Since p is a two digit number, everything here with p in it will be (at least) two digits. So the factors p, 2p, 3p, and 6p are all greater than ...”
July 23, 2017
Matt@VeritasPrep posted a reply to Good tips for someone who hasnt studied math for 10 years in the GMAT Math forum
“Hey pcm182! Just checking in: did you have any luck with either recommendation?”
July 23, 2017
Matt@VeritasPrep posted a reply to How to prepare for Quant in the GMAT Math forum
“Just a little piece of advice: 2 hours with full energy and attention is better than 6 hours tired after work! (I don''t know your situation, but that is just my experience.)”
July 23, 2017
Matt@VeritasPrep posted a reply to How to prepare for Quant in the GMAT Math forum
“If you just need to work on your arithmetic and algebra, I would start with some proper math books. They''re not any different from the GMAT material, but they tend to be cheaper. There are also *so many* of them that there are bound to be a few that are much much better than any test prep material, ...”
July 23, 2017
Matt@VeritasPrep posted a reply to Weighted Average problem in the GMAT Math forum
“Another way to think of this: since there are twice as many women as men, the average should be twice as close to the female average as it is to the male average. With that in mind, if the difference between the female average and the overall average is 4 (or 70 - 66), the difference between the ...”
July 23, 2017
Matt@VeritasPrep posted a reply to Horrible at Arithmetic, is there hope? in the GMAT Math forum
“Another quick point: if you''ve got a good memory and an eye for details (and as a CPA, I''d assume that you do), you''ll find that a lot of arithmetic is just memorization. One of the great calculators in the world, Shakuntala Devi, was actually faster than most contemporary computers but she ...”
July 23, 2017
Matt@VeritasPrep posted a reply to Horrible at Arithmetic, is there hope? in the GMAT Math forum
“You can totally improve! Day 5 is a little early to declare the patient DOA. :) I''ve found that *most* of my students these days are better at the conceptual side of math than the computational side, and the conceptual side is much more important for this test. I think you''ll find that the ...”
July 23, 2017
Matt@VeritasPrep posted a reply to Is there an easier way to solve Q 15 diagnostic test OG'16? in the Problem Solving forum
“Another quick and dirty approximation: 2 * 19 ≈ 40 3 * 17 ≈ 50 5 * 13 ≈ 60 7 * 11 ≈ 80 From there, we''ve got 40 * 50 * 60 * 80 => 4 * 5 * 6 * 8 * 10 * 10 * 10 * 10 4 * 5 = 20 and 6 * 8 ≈ 50, so we''ve got 20 * 50 * 10⁴ 2 * 10 * 5 * 10 * 10⁴ or 10⁷. ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Probabilty Help in the Problem Solving forum
“Half the city are female, so that''s 50%. That includes all the female victims, leaving us with only the males to add. Of the male 50%, 14% are victims, so 14% * 50%, or 7%. Summing those, we get 57%.”
July 5, 2017
Matt@VeritasPrep posted a reply to Probabilty Help in the Problem Solving forum
“This is unanswerable because we don''t know how the person is chosen. Assuming (s)he is chosen at random, it''s about 60%, since we''re given 60% of the sample knows a victim of domestic or sexual abuse. We''d want to know whether knowing yourself counts (some of the abuse survivors may not know any ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Probabilty Help in the Problem Solving forum
“Much like the first problem, this is 40 choose 12, or 40! / (12! * 28!). Again, you should be able to leave the answer in that form.”
July 5, 2017
Matt@VeritasPrep posted a reply to Probabilty Help in the Problem Solving forum
“Assuming you have no restrictions on the jurors, this is 300 choose 40, or 300! / (40! * 260!). I doubt your prof would want you to actually work out that number. :)”
July 5, 2017
Matt@VeritasPrep posted a reply to Word Problem in the Problem Solving forum
“Think of the frame around a painting. The frame is the walkway and the painting is the garden.”
July 5, 2017
Matt@VeritasPrep posted a reply to Trailiiiiiing zer0s in the Problem Solving forum
“Practice practice practice! Technology is all computers anyway, and the stronger a conceptual problem solver you are the harder you''ll be to automate and replace! :)”
July 5, 2017
Matt@VeritasPrep posted a reply to Simple fraction in the Problem Solving forum
“1/(.75 - 1) => 1/(-.25) => -(1/.25) => -1/(1/4) => -4 I wouldn''t worry about speed here since this is about as quick a GMAT problem as I can recall - is it from the OG?”
July 5, 2017
Matt@VeritasPrep posted a reply to 12n/m = ? in the Problem Solving forum
“Alternatively, just pick valid numbers for n and m, then plug them into 12n/m. Say n = 3 and m = 2. From there, 12n/m = 12*3/2, or 18.”
July 5, 2017
Matt@VeritasPrep posted a reply to 12n/m = ? in the Problem Solving forum
“(1/4)n = (3/8)m n = (3/2)m Substituting (3/2)m for n, we get 12n/m => 12 * (3/2)m / m => 12 * (3/2) => 18.”
July 5, 2017
Matt@VeritasPrep posted a reply to R = ? in the Problem Solving forum
“p - (1 - p²)/p = r/p multiply both sides by p: p² - (1 - p²) = r do the subtraction on the left side: 2p² - 1 = r”
July 5, 2017
Matt@VeritasPrep posted a reply to Must be a multiple of what number? in the Problem Solving forum
“Alternatively, start with 3x + 4y = 200 3x = 200 - 4y 3x = 4 * (50 - y) From this, we know x is a multiple of 4. We''re told in the prompt that it''s a multiple of 5. Since it''s a multiple of 4 and of 5, it must also be a multiple of 4*5, or 20. That tells us x is also a multiple of ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Must be a multiple of what number? in the Problem Solving forum
“We can write "y is a multiple of 5" as y = 5m, where m is some integer whose value we don''t need. From there: 3x + 4y => 3x + 4*5m => 3x + 20m We''re told this equals 200, so 3x + 20m = 200 3x = 200 - 20m 3x = 20 * (10 - m) So x must be a multiple ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Standard deviation table in the Problem Solving forum
“Or, might we say, a copywrong? (I''ll show myself out ...)”
July 5, 2017
Matt@VeritasPrep posted a reply to Closest to 1/2? in the Problem Solving forum
“As an aside, if you remember your calculus from high school, the pattern here points to a limit: (n + 1) / (2n + 1) approaches 1/2 as n approaches ∞. (In other words, the longer we follow this pattern, the closer we get to 1/2, though we can never *quite* get there.)”
July 5, 2017
Matt@VeritasPrep posted a reply to Closest to 1/2? in the Problem Solving forum
“You don''t! :) Look for the pattern. If the numerator is half the denominator, rounded up, the fraction gets smaller the larger the numerator gets: 2/3 = .66666666... 3/5 = .6 4/7 = .571428... 5/9 = .55555555... etc. You don''t even need to calculate 4/7 (or any that come after) - ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Algebra question in the Problem Solving forum
“As noted in the other answers, the key takeaway here is that these equations play nicely together to give you exactly what you seek: (x + y)/3. GMAT questions are often contrived like this, so look for the Easter Egg! If you think formulaically the equations can take forever to solve, but if you ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Exchange Question in the Problem Solving forum
“That said, here''d be my way. Rose got $500 * (4/5) => €400. Rose spent (3/4) * €400 => €300. Rose was left with €100, for which she recieved $(1.2 * 100), or $120. This seems like an easy problem by the GMAT''s standards, so if it''s giving you a lot of trouble brush ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Exchange Question in the Problem Solving forum
“Define "fast". At some point it''ll come down to how quickly you can do the calculations, and any time spent thinking of a way to avoid them is time that you might have been able to spend solving the problem. Don''t grow sick longing for the ideal - it''s nice to have a clever way of ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Boat fuel use rate in the Problem Solving forum
“We want miles/gallons. We''re told that miles = 32 and gallons = 24, so miles/gallons = 32/24 = 4/3.”
July 5, 2017
Matt@VeritasPrep posted a reply to Coins and total value in the Problem Solving forum
“One last approach (that isn''t as good). Start with an easy number: 16 dimes. That gives you $1.60. From there, trade in one dime for one quarter until you get to $2.35. Every time you exchange a dime for a quarter, you''ll net 15¢. To get to $2.35, you need to make ($2.35 - $1.60) / 15¢ ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Coins and total value in the Problem Solving forum
“Another approach too! Let''s start by assuming we only have quarters. If we have 9 of them, we''ve got $2.25, so 9 quarters + 1 dime = $2.35. From there, we can trade quarters in for 2.5 dimes to try to get to 16 coins. If we trade 2 quarters for 5 dimes, we now have 7 quarters + 6 dimes ... ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Coins and total value in the Problem Solving forum
“Let''s try it without algebra, as requested. (This is also faster than most of the methods above, I think.) Start with the average. $2.35 / 16 ≈ 15¢, which is 5¢ away from a dime and 10¢ away from a quarter. By that logic, we should have about twice as many dimes as quarters, since the ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Waxing floors and getting paid in the Problem Solving forum
“Another idea: by the Lazy Testwriter Principle, which I invoke endlessly on this forum (guess who''s been a testwriter!), the amount that Makoto was paid should be divisible by 15, since a Lazy Testwriter would make her hourly wage an integer. Only D is so divisible, making it very likely the ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Waxing floors and getting paid in the Problem Solving forum
“Have to remark how surprised I was that this (splendidly) titled thread actually contained GMAT content. :D Anyway, my approach: Makoto worked 15 of the 65 hours worked in all, so she''d be paid 15/65, or 3/13, of the total. 3/13 * $780 is $180, and we''re done!”
July 5, 2017
Matt@VeritasPrep posted a reply to Papgust's GMAT MATH FLASHCARDS directory in the GMAT Math forum
“Once we''re armed with those, we can use the powers of an integer to determine the units digit. For instance, suppose I ask for the units digit of 3²³. We know that the units digits cycle in a block of 4: 3, 9, 7, 1, 3, 9, 7, 1, ... With that in mind, we only need to know the remainder when 23 ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Papgust's GMAT MATH FLASHCARDS directory in the GMAT Math forum
“powers of 8: 8, 64, 512, 4096, 32768, 262144, 2097152, 16777216, ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Papgust's GMAT MATH FLASHCARDS directory in the GMAT Math forum
“powers of 7: 7, 49, 343, 2401, 16807, 117649, 823543, 5764801, ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Papgust's GMAT MATH FLASHCARDS directory in the GMAT Math forum
“powers of 3: 3, 9, 27, 81, 243, 729, 2187, 6561, ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Papgust's GMAT MATH FLASHCARDS directory in the GMAT Math forum
“powers of 2: 2, 4, 8, 16, 32, 64, 128, 256, ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Papgust's GMAT MATH FLASHCARDS directory in the GMAT Math forum
“Piggybacking on my last post, a more memorable formula (in my opinion) is (aᵖ + aᵖ⁻¹ + ... + a¹ + 1) * (bᵒ + bᵒ⁻¹ + ... + b¹ + 1) * (cʳ + cʳ⁻¹ + ... + c¹ + 1) I prefer this one - it shows a little more clearly where the formula comes from.”
July 5, 2017
Matt@VeritasPrep posted a reply to Papgust's GMAT MATH FLASHCARDS directory in the GMAT Math forum
“But the fun part is proving it! :) How would we know that such a formula works?”
July 5, 2017
Matt@VeritasPrep posted a reply to Papgust's GMAT MATH FLASHCARDS directory in the GMAT Math forum
“What sort of writer? There are so many freelancers out there, and plenty of sites on which they advertise.”
July 5, 2017
Matt@VeritasPrep posted a reply to Counting in the Helpful Resources forum
“A quaint one too - who gets invited to anything any more? It seems like people just digitally announce events (or maybe I''ve been living in California too long ...)”
July 5, 2017
Matt@VeritasPrep posted a reply to GMAT time strategies in the Helpful Resources forum
“Knowing when to punt a question is important too - don''t assume you''ll solve everything, or that time spent on a problem is necessarily time well spent!”
July 5, 2017
Matt@VeritasPrep posted a reply to My Struggle w/ the GMAT but still admitted to top school in the GMAT Math forum
“Excellent work! Might want to shift this post into the more relevant subforums, though :)”
July 5, 2017
Matt@VeritasPrep posted a reply to categories for level of difficulty? in the GMAT Math forum
“Most of all, don''t sweat the difficulties too much - focus more on learning. If the problem seems hard, it has something to teach you, and if it seems easy it either IS easy and is a waste of time or merely SEEMS easy and is a great chance to practice the suspicion of the obvious that you''ll need ...”
July 5, 2017
Matt@VeritasPrep posted a reply to Problem on mile markers in the GMAT Math forum
“"runs east to west" is the key: we''d expect the mile markers to begin in the east (with mile marker 0 or mile marker 1), then rise as we head westward (to mile markers 2, 3, 4, ...)”
July 5, 2017
Matt@VeritasPrep posted a reply to How to pass CFA Exam scoring 70% all 10 topics in the GMAT Math forum
“Not exactly the right forum for this!”
July 5, 2017
Matt@VeritasPrep posted a reply to help in the GMAT Math forum
“Nice approach! For anyone needing a visual, this is the same approach the great Brent uses above.”
July 5, 2017
Matt@VeritasPrep posted a reply to Probability in the Data Sufficiency forum
“It might also be because probability is hard to teach and whatever resource you started with didn''t give you a solid grounding in the basics. If you have time, check out Freund''s Introduction to Probability. Unlike most books titled ''An Introduction to Probability'' (because of my day job, I have ...”
June 23, 2017
Matt@VeritasPrep posted a reply to PQP is a three-digit number having digits P and Q; in the Data Sufficiency forum
“PQP * P = RQS5 Since the right side ends in 5, the left side must also end in 5. This means P is 5. From there, 5Q5 * 5 = RQS5 (500 + 10Q + 5) * 5 = 1000R + 100Q + 10S + 5 2500 + 20 = 1000R + 50Q + 10S 2520 = 1000R + 50Q + 10S The maximum values for Q and S are each 9, so the ...”
June 23, 2017
Matt@VeritasPrep posted a reply to PQ and QP represent two-digit numbers in the Data Sufficiency forum
“PQ = 10P + Q QP = 10Q + P so A slightly more thorough breakdown of S1:: PQ + QP = 10P + Q + 10Q + P => 11P + 11Q => 11 * (P + Q) RSR = 100R + 10S + R = 101R + 10S Since 11*(P+Q) is at most 11*(9+9), we know that it''s less than 200. From that 101R + 10S must also be ...”
June 23, 2017
Matt@VeritasPrep posted a reply to DS + nth power in the Data Sufficiency forum
“I love the enthusiasm, but no need to resurrect the thread :)”
June 23, 2017
Matt@VeritasPrep posted a reply to What is the remainder in the Data Sufficiency forum
“I often like to do it algebraically. "n divided by m has remainder r" can be written as n = m*x + r, where x is some integer whose value we don''t necessarily care about. For this problem, that gives us S1:: n = 6x + 5 S2:: 3n = 6y + 3, or n = 2y + 1 Taking the statements ...”
June 23, 2017
Matt@VeritasPrep posted a reply to If p is a positive integer, is p a prime number? (1) p and p in the Data Sufficiency forum
“Other GMAT-relevant properties of 0: * It''s even * It''s neither positive nor negative * It should ALWAYS be considered when a problem uses the phrase "nonpositive" or "nonnegative" * 0! = 1 * 0 is a multiple of every integer * 0 is not a factor of any integer * If x ...”
June 23, 2017
Matt@VeritasPrep posted a reply to If p is a positive integer, is p a prime number? (1) p and p in the Data Sufficiency forum
“Every integer is a factor of 0 except 0 itself. Say m and n are integers. If m / n = an integer, then n is a factor of m. 0 / anything other than 0 = 0, an integer. It follows that anything other than 0 is a factor of 0.”
June 23, 2017
Matt@VeritasPrep posted a reply to If a, b and c are single digit numbers in the Data Sufficiency forum
“If an integer is divisible by 9, then the sum of its digits is divisible by 9, and vice versa. S1:: 5 + a + b = 9 * something Since 0 < a + b ≤ 18, we can only have 5 + a + b = 9 or 5 + a + b = 18. That gives a + b = 4 or a + b = 13. Not sufficient. S2:: Same logic, b + c = ...”
June 23, 2017
Matt@VeritasPrep posted a reply to If n is a prime number.........Manhattan question in the Data Sufficiency forum
“Piggybacking a bit, the second statement can be written this way: xⁿ ⁻ ¹ > (xⁿ ⁻ ¹)² We know that n is an odd prime, or n = 2m + 1, where m is some positive integer whose value we don''t care about. Substituting, we have x²ᵐ⁺¹ ⁻ ¹ > (x²ᵐ⁺¹ ⁻ ¹)² or ...”
June 23, 2017
Matt@VeritasPrep posted a reply to Right Circular Cone in the Problem Solving forum
“I almost forgot the most salient point (no pun intended): cones are the white whale of the GMAT. I''ve heard them mentioned and seen them in problems from a few test prep providers, but I''ve never seen an official problem that required you to know anything specific to cones. (I can''t even remember ...”
June 23, 2017
Matt@VeritasPrep posted a reply to Man cycling and buses in the Problem Solving forum
“A follow up to test your understanding! :) Suppose I said, "Ha! I''ve got a much easier way! Why not just use a weighted average? There are three buses coming every 4 minutes from one direction, and one bus coming every 12 minutes from the other! (3*4 + 1*12) / 4 buses = 6, so that''s ...”
June 23, 2017
Matt@VeritasPrep posted a reply to Right Circular Cone in the Problem Solving forum
“It''s definitely allowed, at least in practice - it seems to happen almost every day.”
June 23, 2017
Matt@VeritasPrep posted a reply to Right Circular Cone in the Problem Solving forum
“In the 2D world, a polygon inscribed in a circle has each vertex on that circle''s circumference, while a circle inscribed in a polygon is tangent to each side of the polygon. (See some images here.) From there, you can probably guess how to generalize this to 3D, 4D, etc.”
June 23, 2017
Matt@VeritasPrep posted a reply to Tough overlapping sets problem in the Problem Solving forum
“Let''s call the answer x. We know that x = 60% of all backup+standard. We know that backup+standard = 18% of (120 + y), where y is the number of two door cars with backup cameras. So x = 60% of 18% of (120 + y), or 10.8% of (120 + y), or 12.96 + .108y. y is at most 165, so our answer ...”
June 23, 2017
Matt@VeritasPrep posted a reply to For a non-negative integer n, in the Problem Solving forum
“Look for the pattern too. 2⁰, 2², 2⁴, ... all have remainder 1 2¹, 2³, 2⁵, ... all have remainder 2 So we need to have an even exponent. III follows from there.”
June 23, 2017
Matt@VeritasPrep posted a reply to Interest and principle in the Problem Solving forum
“Good question to ask! This concept will be important in your MBA as well - you need to know the terms, how often the interest is compounded, etc.”
June 23, 2017
Matt@VeritasPrep posted a reply to Another Tough overlapping problem in the Problem Solving forum
“A (relatively) quick approach would be to backsolve. We know that at most 128 fourth graders are taking Chinese, so the answer must be < 128. From there, 10 and 128 seem the most likely answers (min and max of the remaining options). I''d try one, see if it fits the parameters, and if it ...”
June 23, 2017
Matt@VeritasPrep posted a reply to How may times vs how many times more in the Problem Solving forum
“For the most part, "how many times" = "how many times more". Logically I see your point, but idioms/usage conquer all, and in English in 2017 the equation above holds. (In percent problems, though, the distinction is important: "percent of" ≠ "percent ...”
June 23, 2017
Matt@VeritasPrep posted a reply to Cloudy and Sunny in the Problem Solving forum
“R can be expressed in terms of T and D, so depending on what facts you''re given in the stem there might be some way to use T and D in a similar fashion.”
June 23, 2017
Matt@VeritasPrep posted a reply to Need some help..! in the Problem Solving forum
“5M - 4N = 20 5M = 20 + 4N 5M = 4 * (5 + N) Now we can use a neat trick. Since 5M = some multiple of 4, we know that 5M is a multiple of 4. Since 5 contains no prime factors of 4, we can say that M must contain them, so M = some multiple of 4. C isn''t a multiple of 4, so that CAN''T be the ...”
June 23, 2017
Matt@VeritasPrep posted a reply to If x ≥  0.9, in the Problem Solving forum
“Since x is a range, 1/√x should also have a range. That means the answer is likely to be either the smallest or the largest of the five options. With that in mind, try A: 1/√x = 1.02 1 = 1.02√x 1/1.02 = √x That number on the right I''d guesstimate at .98ish, so squaring it ...”
June 23, 2017
Matt@VeritasPrep posted a reply to OG - On Saturday morning, Malachi will begin a in the Problem Solving forum
“Oh goodness, zillions. Can you give me a little more direction? I have so many, not sure where to begin.”
June 23, 2017
Matt@VeritasPrep posted a reply to Probability Problem in the Problem Solving forum
“No prob :)”
June 23, 2017
Matt@VeritasPrep posted a reply to What is the lowest positive integer in the Problem Solving forum
“Are stats and geometry that common these days? It seems to me that arithmetic and algebra, in some guise, account for the lion''s share of the problems, with stats and geometry being ≤ 5 problems each (stats maybe even lower).”
June 23, 2017
Matt@VeritasPrep posted a reply to If p is the sum of the reciprocals in the Problem Solving forum
“Would add that I don''t think this would be on the GMAT (or at least, not in such a close form) since there''s a retired question that asks almost exactly the same thing.”
June 23, 2017
Matt@VeritasPrep posted a reply to If w, x, y and z are integers in the Problem Solving forum
“I think this is a similar, but shorter and easier way: 924 = 2² * 3 * 7 * 11 Since there are exactly four factors greater than 1, one and only one of w, x, y, and z must contain two prime factors. (We''re assigning five prime factors to four variables, with no empty variables (e.g. = 1) ...”
June 23, 2017
Matt@VeritasPrep posted a reply to The function f is defined by in the Problem Solving forum
“The function f is defined by f „(x) =… ƒ -1/x for all non-zero numbers x. If f „(a) …= ƒ-1/2 and f „(ab)… = ƒ1/6, then b ƒ= f(a) = -1/2 and, by definition, f(a) = -1/a so a = 2 f(ab) = f(2b) = 1/6 and, by definition f(2b) = -1/2b so b = -3”
June 23, 2017
Matt@VeritasPrep posted a reply to A marketing class of a college in the Problem Solving forum
“It''s easier visually, I think: https://s13.postimg.org/vhnpl8k3n/Screen_Shot_2017-06-22_at_5.58.59_PM.png We know that x + z = 10 and y + z = 6. To minimize x + y + z, make z (the common element) as large as possible. So if z = 6, y = 0, and x = 4, we''ve got the minimum. Filling that ...”
June 23, 2017
Matt@VeritasPrep posted a reply to Statistics in the Problem Solving forum
“One way of thinking about it: The SD is NEVER negative, no matter what the set, and the SD = 0 if and only if all the terms in the set are the same. Since there are different terms in this set, 0 is impossible, and all the negative answers are always impossible.”
June 23, 2017
Matt@VeritasPrep posted a reply to The function f is defined by in the Problem Solving forum
“p = √q - 20 p + 20 = √q (p + 20)² = q”
June 23, 2017
Matt@VeritasPrep posted a reply to area in the Problem Solving forum
“I love that, nice hack!”
June 23, 2017
Matt@VeritasPrep posted a reply to Is anyone using GMAT PrepNow for preparation. in the GMAT Math forum
“You know, I don''t think I''ve ever told you this, but because of your name and Canadianness, I connected you with the only other Canadian Brent I''ve heard of and now have this visual of you running your GMAT empire from a gas station in Saskatchewan, patiently indulging all the kooks, nits, and ...”
June 23, 2017
Matt@VeritasPrep posted a reply to أخبار الموضة in the GMAT Math forum
“I knew automation was going to claim my job, but I didn''t realize it would be today :B”
June 23, 2017
Matt@VeritasPrep posted a reply to Struggling with this GMAT Prep Quant Question - any help? in the GMAT Math forum
“I like Brent''s method above, but if you find yourself in a pinch on test day without a conceptual solution, you could always look for a pattern here. Suppose I take smaller versions of the function given, like h(2), h(4), and h(6). h(2) = 2 h(2) + 1 = 3 h(4) = 2 * 4 h(4) + 1= 2 * 4 + 1 ...”
June 22, 2017
Matt@VeritasPrep posted a reply to Is anyone using GMAT PrepNow for preparation. in the GMAT Math forum
“I haven''t used the course, but I''ll vouch for Brent as my favorite person on the forum!”
June 22, 2017
Matt@VeritasPrep posted a reply to question about combinations and permusation in the GMAT Math forum
“You know, I pondered that riff myself, but I couldn''t find any languages/cultures in which Mark and Dave were female names. (You''d think Davé would be a female name somewhere, but I couldn''t find it.)”
June 22, 2017
Matt@VeritasPrep posted a reply to GMAT Arithmetic practice in the GMAT Math forum
“Nice idea!”
June 22, 2017
Matt@VeritasPrep posted a reply to أخبار الموضة in the GMAT Math forum
“"Join us and enjoy the day of shopping, surprises and gifts. Marie Claire Arabia celebrates the first Arab shoes, an exclusive event organized by Leaville Chos Store." Surprisingly legible bot translation!”
June 22, 2017
Matt@VeritasPrep posted a reply to how many zeros? in the GMAT Math forum
“Cosign! I think it''s 3/(16 * 9 * 25), but I''m not sure.”
June 22, 2017
Matt@VeritasPrep posted a reply to Is n/12 an integer? in the Data Sufficiency forum
“S1: n²/144 = integer (n/12)² = integer n/12 = √integer n = 12*√integer This is pretty close to what we want, but we don''t know if √integer is itself an integer! For instance, we could have n² = 288, in which case n²/144 = 2, but n = 12√2. Not sufficient! S2: n/6 = ...”
June 9, 2017
Matt@VeritasPrep posted a reply to Can n /192 be an integer? in the Data Sufficiency forum
“S1: Since 192 is divisible by 16, any multiple of 192 must be a multiple of 16. n isn''t a multiple of 16, so it can''t be a multiple of 192: SUFFICIENT S2: Same idea. 192 is divisible by 48, so any multiple of 192 is a multiple of 48. n isn''t a multiple of 48, so it can''t be a multiple ...”
June 9, 2017
Matt@VeritasPrep posted a reply to What is the value of x? in the Data Sufficiency forum
“Other way round. We''ve got 3x ≥ |y| and |y| ≥ 0 So we combine the two inequalities 3x ≥ |y| ≥ 0 then drop the middle term 3x ≥ 0 and divide x ≥ 0”
June 9, 2017
Matt@VeritasPrep posted a reply to What is the value of x? in the Data Sufficiency forum
“S1: 3x ≥ |y| Since |y| ≥ 0, we know that 3x ≥ 0, and that x ≥ 0. That''s helpful, but not sufficient. S2: |5x - 1| = x + 7 This has two solutions, either 5x - 1 = x + 7 (which gives x = 2) or -(5x - 1) = x + 7 (which gives x = -1) Since x could be 2 or -1, ...”
June 9, 2017
Matt@VeritasPrep posted a reply to Is x > y? in the Data Sufficiency forum
“S1: |x| > |y| In words, this says "x is further from 0 than y is". This is true if x = 3 and y = 2, but it''s also true if x = 3 and y = -2. Not sufficient! S2: x + y > 0 x > -y OK, so x is greater than -y ... but we want to know if it''s greater than y ...”
June 9, 2017
Matt@VeritasPrep posted a reply to equilateral triangle pS in the Problem Solving forum
“This thread is a great reminder that there are easy hotkeys that make explanations easier to read. For square roots, use CTRL+V: √ For squares, use ALT+0178: ²”
June 9, 2017
Matt@VeritasPrep posted a reply to [b][m][fraction](a/b)/c [/fraction][/m][/b] In the expressio in the Problem Solving forum
“(a/b)/c => a/(b * c) Since we''re restricted to positive integers, we want a to be as small as possible and b*c to be as large as possible. This is easily done by making a the smallest option (3) and b and c the other options (4 and 7), leaving us with 3 / (4*7), or 3/28.”
June 9, 2017
Matt@VeritasPrep posted a reply to Leila in the Problem Solving forum
“P(at least 3) = P(exactly 3) + P(exactly 4) P(exactly 4) is easy => 1/5 * 1/5 * 1/5 * 1/5 P(exactly 3) is trickier, since we have four different arrangements to consider. (The miss could come on any of the four throws.) That means we''ve got four orders: Hit, Hit, Hit, Miss Hit, Hit, ...”
June 9, 2017
Matt@VeritasPrep posted a reply to Ramon wants to cut a rectangular board into identical square in the Problem Solving forum
“I''d start by looking for a square that''s a factor of the area of the rectangle: 18 * 30 = 2 * 3 * 3 * 2 * 3 * 5 So 2 * 2 * 3 * 3, or 36, is a factor of 540. That suggests that we might be able to form squares of area 36. Looking at our rectangle dimensions (18 and 30), we notice that ...”
June 9, 2017
Matt@VeritasPrep posted a reply to A vendor sells only Product A, for $6, and Product B, for $2 in the Problem Solving forum
“This problem appears quite closely modeled on a famous official problem about newspaper revenue that can be found here: http://www.beatthegmat.com/newspapers-a-b-t30515.html If anyone is keen to practice the topic, that problem would be a good follow-up.”
June 9, 2017
Matt@VeritasPrep posted a reply to Best GMAT math flashcards? in the GMAT Math forum
“I would be wary of relying too much on flashcards. I''ve found that when I try to learn from flashcards, I *do* memorize a bunch of useful formulas, but I *don''t* hone the ability to recognize when and how to use them! It''s important to learn from problems as much as you can, and to remember ...”
June 6, 2017
Matt@VeritasPrep posted a reply to question about combinations and permusation in the GMAT Math forum
“Hey, if they''re chairman and co-chairman of, say, the Flat Earth Society, I''d think we want the most problematic leadership possible. ó‿ó”
June 6, 2017
Matt@VeritasPrep posted a reply to Profit and loss in the Problem Solving forum
“Costs: $40 per item + $3000 Profits: $60 per item If our guy sells x items and makes $1000, we know that Profits - Costs = $1000 60x - (40x + 3000) = 1000 20x - 3000 = 1000 20x = 4000 x = 200”
June 6, 2017
Matt@VeritasPrep posted a reply to M is the sum of the reciprocals of the consecutive integers in the Problem Solving forum
“Also, when in doubt, try one (or more) of the following: 1) Try a smaller version of the same pattern to get a feel for what''s going on; 2) Approximate; 3) Backsolve from the answers It''s amazing how many GMAT problems respond to one of those three approaches.”
June 6, 2017
Matt@VeritasPrep posted a reply to Profit and loss in the Problem Solving forum
“He paid 2 * $16, or $32. He got back 1.5 * $12 + 0.5 * $4, or $20. He ended up with 20/32, or 5/8 of what he started with, meaning that he lost 3/8, or 37.5%.”
June 6, 2017
Matt@VeritasPrep posted a reply to Probability Question- from a bag containing.... in the Problem Solving forum
“Also, the y ≥ 0 isn''t terribly important, at least for GMAT purposes. I needed to specify that so I could safely cross-multiply the inequality. (If y ≥ 0, then 12 + y > 0, and I know that when I multiply by (12 + y), I''m multiplying by a positive number.)”
June 6, 2017
Matt@VeritasPrep posted a reply to Probability Question- from a bag containing.... in the Problem Solving forum
“We know y ≥ 0 because we can''t have a negative number of yellow marbles. (y = the # of yellows) When I manipulated both sides, I did the following: 12 / (12 + y) < 2/5 cross multiply: 12 * 5 < 2 * (12 + y) simplify: 60 < 24 + 2y subtract: 36 < 2y ...”
June 6, 2017
Matt@VeritasPrep posted a reply to OG - On Saturday morning, Malachi will begin a in the Problem Solving forum
“Absolutely! The art of the best official questions is testing precisely this. (The idea: if you can see these sorts of shortcuts on a math test, you''ll also see them in the business world, quantitative finance, real estate, wherever.)”
June 6, 2017
Matt@VeritasPrep posted a reply to Problem solving : profit and loss in the Problem Solving forum
“If that last fraction is tough to manipulate without a calculator, here''s the trick: 30 / (28 + 4/7) => 30 / ((28*7 + 4)/7) => 210/(28*7 + 4) => 105/(14*7 + 2) => 105/100”
June 6, 2017
Matt@VeritasPrep posted a reply to Problem solving : profit and loss in the Problem Solving forum
“Every kg he sells is 26/(26 + 30) parts $20 and 30/(26 + 30) parts $36. The cost of each kg to the trader is thus 26/56 * $20 + 30/56 * $36, or $28 + (4/7). When he sells the mix for $30 per kg, he gets $30 / ($28 + 4/7) more than he paid, or 1.05 times what he paid, or 5% more.”
June 6, 2017
Matt@VeritasPrep posted a reply to Problem solving : profit and loss in the Problem Solving forum
“If he''s paying $6.40 per liter and making a 37.5% profit, he''s selling each liter of milk for $8.80. If he''s selling a liter of milk and water for $8.00, then the mixture he''s selling is $8/$8.80 milk, or 10/11 milk, so the ratio of milk to water is 10 : 1.”
June 6, 2017
Matt@VeritasPrep posted a reply to If two of the four expression x+y, x+5y... in the Problem Solving forum
“On the modern (2017) GMAT, maybe 65th-70th percentile?”
June 6, 2017
Matt@VeritasPrep posted a reply to A breakfast that consists of 1 ounce of corn puffs in the Problem Solving forum
“I''d put it in words first before assigning variables. This makes it easier to follow your work in case you end up confusing yourself. 1 oz of corn + 8 ounces of X = 257 1 oz of corn + 8 ounces of Y = 185 If we subtract the second equation from the first, we get 8 ounces of X - 8 ounces ...”
June 6, 2017
Matt@VeritasPrep posted a reply to GMAT Prep If n= 1/3 in the Problem Solving forum
“In general, we could say that x + x² + x³ => x * (1 + x + x²) = > x * (1 + x * (1 + x)) which makes funkier numbers a little easier to approximate.”
June 6, 2017
Matt@VeritasPrep posted a reply to Counting in the Problem Solving forum
“Not 100% sure of this, but I think this Q (or one very much like it) is in the one of the Exam Packs from mba.com.”
June 6, 2017
Matt@VeritasPrep posted a reply to Combinations in the Problem Solving forum
“You could also think about arranging only Frankie and Joey before arranging everyone else. There are only two options (FJ and JF), so half of the arrangements will have FJ and the other half will have JF.”
June 6, 2017
Matt@VeritasPrep posted a reply to Combinations in the Problem Solving forum
“That''s a great question! Trial and error is typically how I do it, but one sign that slots are a good idea is a set of conditions on where certain people/things can be in the arrangement. (That isn''t foolproof, but it often works.)”
June 6, 2017
Matt@VeritasPrep posted a reply to Probablity/P&C problem Exam Pack1 in the Problem Solving forum
“Footnote: it''s definitely a good idea to practice as many probability questions as you can both ways: arranging and/or analyzing. Most problems that can be solved either way have one approach that is quite a bit simpler and easier than the other.”
June 6, 2017
Matt@VeritasPrep posted a reply to Probablity/P&C problem Exam Pack1 in the Problem Solving forum
“Typically this means either finding all the arrangements, then finding the number of arrangements that do what you want OR thinking step by step about what needs to go where. If we try it the first way (arrangements), we have RRWW in some order. There are 4!/2!2! = 6 ways to arrange those, of ...”
June 6, 2017
Matt@VeritasPrep posted a reply to Probability Problem in the Problem Solving forum
“Suppose you''ve arranged your six people: ABCDEF. When you go to place G, you have the following empty spaces: _ A _ B _ C _ D _ E _ F _ There are seven _''s, so we''ve got 7 ways to place that person. If we chose the second one, for instance, our order would become AGBCDEF. Once you''ve ...”
June 6, 2017
June 6, 2017
Matt@VeritasPrep posted a reply to There are 10 intermediate stations including two junctions A in the Problem Solving forum
“Agreed, you''ll want to clarify whether one can stop at a junction, and whether the junctions are considered stations or not.”
June 6, 2017
Matt@VeritasPrep posted a reply to Is someone able to solve this OG13 problem within 2 minutes? in the Problem Solving forum
“It really depends how engaged you feel with the problem: is your mind zeroing in on the answer, or are you drifting off into anxiety, daydreams, or confusion? I find that if I''m engaged, the answer often comes to me five seconds AFTER I''ve finally told myself "just give up" - it''s ...”
June 6, 2017
Matt@VeritasPrep posted a reply to Probability - 2 Pairs of Socks in the Problem Solving forum
“That''s if we want to get a match immediately, e.g. RR. If we just want a match at some point, we could have RWWR, RWRW, RRWW, etc. As long as one of the three remaining socks matches the first R, we''re good.”
June 6, 2017
Matt@VeritasPrep posted a reply to Probability question in the Problem Solving forum
“I like it!”
June 6, 2017
Matt@VeritasPrep posted a reply to fraction properties - If p and q are positive integers, and in the Problem Solving forum
“If we wanted a friendly introduction to get a feel for this, we could look at some fractions. Suppose I start with 1/2. Adding to the num and denom, I get 2/3, 4/5, 5/6, 6/7, 7/8, ... Hmm! So they keep getting bigger. What if I pick a more random fraction to start with? 1/5, 3/7, 5/9, ...”
June 6, 2017
Matt@VeritasPrep posted a reply to fraction properties - If p and q are positive integers, and in the Problem Solving forum
“We also need to do the other case, n > m > 0. This time, the inequality looks different: (m + 1)/(n + 1) - 1 > m/n - 1 (It''s backwards this time because each side of the inequality will be negative.) From there: (m + 1)/(n + 1) > m/n n * (m + 1) > m * (n + 1) ...”
June 6, 2017
Matt@VeritasPrep posted a reply to fraction properties - If p and q are positive integers, and in the Problem Solving forum
“Here''s a neat proof: Let''s say m > n > 0. If we can reduce m/n - 1 > (m + 1)/(n + 1) - 1 to m > n, we can say that adding 1 to the numerator and the denominator pushes the fraction closer to 0. (From there, we can say that adding 1 again will push it even closer, etc.) m/n ...”
June 6, 2017
Matt@VeritasPrep posted a reply to fraction properties - If p and q are positive integers, and in the Problem Solving forum
“Lord knows I need it! ⊙﹏⊙”
June 6, 2017
Matt@VeritasPrep posted a reply to Divisor rules - If 15 and 4 are factors of positive integer in the Problem Solving forum
“If we want to illustrate Brent''s point algebraically: x = d * n, where n is some integer we don''t care about y = d * m + r, where m is some integer we don''t care about, and r is the nonzero remainder when y is divided by d x + y => d*n + d*m + r => d*(n + m) + r So ...”
June 6, 2017
Matt@VeritasPrep posted a reply to Divisor rules - If 15 and 4 are factors of positive integer in the Problem Solving forum
“Just to illustrate: Suppose we take 3 and 5. If we think of 5 in terms of 3, we could say 5 = 3 + 2. So when we add them together, we get 3 + 5 => 3 + (3 + 2) and the resulting number WON''T be divisible by 3, because of that remainder of 2 that''s hanging around. Another ...”
June 6, 2017
Matt@VeritasPrep posted a reply to OG 2015 -> OG 2017 Questions in the Problem Solving forum
“There are a number of resources out there that give the changes from OG 13 to the 2016 edition, but not (AFAIK) from OG 13 to the 2017 edition. An easy patch for this would be to go to Google Books, load up the new OG, then search the text from the question you''re working on. The search result ...”
May 26, 2017
Matt@VeritasPrep posted a reply to fraction properties - If p and q are positive integers, and in the Problem Solving forum
“Would add that the algebra here is friendly enough -- thank you, Brent, for making these positive integers, greatly simplifying the inequality multiplication :) -- that I think it''s quicker to do that than to play around with numbers.”
May 26, 2017
Matt@VeritasPrep posted a reply to fraction properties - If p and q are positive integers, and in the Problem Solving forum
“Case III: (p - 1)/(q + 1) > (p - 1)/q q * (p - 1) > (p - 1) * (q + 1) pq - q > pq - q + p - 1 1 > p Oof! This won''t work for any positive integer p, since it forces p to be less than 1.”
May 26, 2017
Matt@VeritasPrep posted a reply to fraction properties - If p and q are positive integers, and in the Problem Solving forum
“Case II: (p + 1)/q > (p - 1)/q (p + 1) > (p - 1) Well ... duh. So this case will hold for any positive integers p and q, whether or not q > p.”
May 26, 2017
Matt@VeritasPrep posted a reply to fraction properties - If p and q are positive integers, and in the Problem Solving forum
“We could also take each fraction and try to show that it reduces to p < q, or to anything else that would work with positive integers p and q. Let me do this in three separate posts, since each case will be hard to read. Case I: The goal: transform (p + 1)/(q + 1) > p/q into q > p. ...”
May 26, 2017
Matt@VeritasPrep posted a reply to Divisor rules - If 15 and 4 are factors of positive integer in the Problem Solving forum
“We could also be a little more rigorous and try to disprove (D) as a possibility altogether. If 60n - m = 70, then we know m = 60n - 70 = 10 * (6n - 7). Since m = 10 * something, m must be divisible by 10. But we''re told that 5 is not a factor of m, so this is impossible, and we''re set.”
May 26, 2017
Matt@VeritasPrep posted a reply to Divisor rules - If 15 and 4 are factors of positive integer in the Problem Solving forum
“K = 60*n, where n is some integer whose value we don''t care about. From here, let''s try each answer: (A) 60n - m = 34, how about n = 1 and m = 26? (B) 60n - m = 44, how about n = 2 and m = 76? (C) 60n - m = 54, how about n = 1 and m = 6? (D) uh oh! (E) 60n - m = 83, how about n = 2 and ...”
May 26, 2017
Matt@VeritasPrep posted a reply to How many points with Integer x and Y co-ordinates lie within in the Problem Solving forum
“This is really more of a Calc I HW problem than a GMAT question, in my opinion.”
May 26, 2017
Matt@VeritasPrep posted a reply to A group of 630 children is arranged in rows for a photograph in the Problem Solving forum
“We could go a little further to find all possible values of r. Once we have the equation n r + 1.5 r² - 1.5 r - 630 = 0, we can isolate r in terms of n: r = (1/6) * (±√(4n² - 12n + 15129) - 2n + 3) Any positive integer n that makes that beastly equation in ( )s come out to a positive ...”
May 26, 2017
Matt@VeritasPrep posted a reply to A group of 630 children is arranged in rows for a photograph in the Problem Solving forum
“If we wanted to do this algebraically, we could! Suppose r = the number of rows and n = the number of students in the least populated row. We know that n + (n + 3) + (n + 6) + ... + (n + 3*(r - 1)) = 630. Simplifying this, we get n*r + 3 + 6 + ... + 3*(r - 1), or n*r + 3*(1 + 2 + ... + r - ...”
May 26, 2017
Matt@VeritasPrep posted a reply to A group of 630 children is arranged in rows for a photograph in the Problem Solving forum
“A neat arithmetic way of working this out: Since our rows are evenly spaced, we know that the mean and the median number in each row must be the same. If we take the total and divide by the number of rows, that number will be both the mean and the median. For A, this gives 630/3 = 210, ...”
May 26, 2017
Matt@VeritasPrep posted a reply to A group of 630 children is arranged in rows for a photograph in the Problem Solving forum
“Since nobody else answered this ... Suppose r is our smallest row. The numbers of children in each row are thus r, r + 3, r + 6, ..., r + 3*(n - 1), where n is the number of rows. If we try answer B, we find that r + r + 3 + r + 6 + r + 9 = 630, which gives us a solution. If we try ...”
May 26, 2017
Matt@VeritasPrep posted a reply to Pleas help--Probability problem in the Problem Solving forum
“We do want to consider the earlier events. In your problem, we''re doing that by adding one woman to Group B.”
May 26, 2017
Matt@VeritasPrep posted a reply to Probability question in the Problem Solving forum
“If it''s still not making sense, check out a few vids on YouTube that explain this concept, which has a great name (the Pigeonhole Principle). It is absolutely astounding how many applications this has in math, in so many unexpected places. Here are a few (not written by me):”
May 26, 2017
Matt@VeritasPrep posted a reply to Probability question in the Problem Solving forum
“Here''s a visual that might help. Think of the boxes that the pigeons are in as the months. Notice how once we fill the boxes, we still have two pigeons left over. If we want to put those pigeons in a box, that box MUST now contain at least two pigeons, so there MUST be at least one box with at ...”
May 26, 2017
Matt@VeritasPrep posted a reply to Probability question in the Problem Solving forum
“Whenever we see "at least" in a probability question, we want to think about solving the easy way: finding the probability that we DON''T HAVE at least 3 (or whatever), then subtracting that from 1. (We do this because this approach usually simplifies the calculations.) So if we say, ...”
May 26, 2017
Matt@VeritasPrep posted a reply to Probability GMAT Prep in the Problem Solving forum
“You''re right, yup! We need to consider two scenarios (exactly one odd and exactly three odds), and one of those scenarios is order-dependent: exactly one odd has to be arranged OEE, EOE, and EEO.”
May 26, 2017
Matt@VeritasPrep posted a reply to tricky probability - A drawer contains 8 pairs of socks. For in the Problem Solving forum
“Totally forgot to add: in Brent''s question, n = 8 and x = 6. (Duh!) Too funny that that both problems involved socks ... what were the odds? (Conditional probability says > 1/2, with all the sock talk on the boards lately.)”
May 26, 2017
Matt@VeritasPrep posted a reply to tricky probability - A drawer contains 8 pairs of socks. For in the Problem Solving forum
“This is a special case of a formula I posted yesterday for a related question. I hate quoting myself, but at least here it''s relevant (and not some glib one-liner, the most loathsome self-quote :)):”
May 26, 2017
Matt@VeritasPrep posted a reply to tricky probability - A drawer contains 8 pairs of socks. For in the Problem Solving forum
“Here''s a boring solution: P(at least one) = 1 - P(none) = 1 - all socks from different pairs = 1 - (1/1)*(14/15)*(12/14)*(10/13)*(8/12)*(6/11) = 1 - 32/143 = 111/143”
May 26, 2017
Matt@VeritasPrep posted a reply to Probability-Socks in the Problem Solving forum
“We could also approach this question algebraically. Let''s say that Tony has 2n socks, n unique and distinguishable pairs. For the first sock, Tony can pull anything. For the second sock, he must get something that doesn''t match the first one, so (2n - 2)/(2n - 1), since he''s got (2n - ...”
May 25, 2017
Matt@VeritasPrep posted a reply to Probability-Socks in the Problem Solving forum
“The key is given in the prompt: the pairs are unique. Since the pairs are unique and Tony owns them, he''d presuambly be able to recognize a match when we saw two identical socks in front of him. It''s definitely good to mind details on the GMAT, but don''t drive yourself too crazy with the ...”
May 25, 2017
Matt@VeritasPrep posted a reply to If y = 2z, and y and z are both positive integers in the Problem Solving forum
“One other trick that just occurred to me: If you''re trying to backsolve from the answers, note that C implies E. (If x is divisible by 2z, then x must be divisible by z.) With that in mind, you don''t even need to bother with E!”
May 25, 2017
Matt@VeritasPrep posted a reply to If y = 2z, and y and z are both positive integers in the Problem Solving forum
“It seems to me that the question really tests the ability to either quickly pick smart numbers that rule out four of the answers and/or the ability to quickly whip up an equation that will show the impossibility of one of the answers. If I take the first approach, some easy sets are: {1} vs ...”
May 25, 2017
Matt@VeritasPrep posted a reply to Probability-Coins in the Problem Solving forum
“You''re on the right track, but you''re ignoring one key condition: Harvey might''ve stopped earlier. With that in mind, we think of probability as Target Outcome / All Possible Outcomes Our target = HHT, and all possible outcomes = T, HT, and HHT. From here, we have 1/8 / (1/2 + 1/4 + ...”
May 25, 2017
Matt@VeritasPrep posted a reply to A researcher has determined that she requires a minimum of n in the Problem Solving forum
“We could also always pick our own numbers: this is a great, mostly foolproof way to get the answer on the test when you either can''t set up the algebra or fear that you''ve written the wrong equation. Suppose n = 50 and p = 20. Then we''re going to need 125 surveys. Plugging n = 50 and p = ...”
May 25, 2017
Matt@VeritasPrep posted a reply to A researcher has determined that she requires a minimum of n in the Problem Solving forum
“If we send out x surveys, we know that (1 - p/100), or (100 - p)/100 people will respond. We want that number to be at least 2n, so x * (100 - p)/100 = 2n x = 2n * 100 / (100 - p) and our answer is A.”
May 25, 2017
Matt@VeritasPrep posted a reply to Digits at ten's place in the Problem Solving forum
“As a footnote, let me add the pattern with powers of 21. Notice that 21 * 21 = (20 + 1) * 21 = 20*21 + 21 = 420 + 21 = 441. See how we''ve got 20 at the end of the first number? Now move to the next power: 441 * 21 => (440 + 1) * 21 => 440*21 + 21 => 9240 + 21 => 9261. Now ...”
May 25, 2017
Matt@VeritasPrep posted a reply to Digits at ten's place in the Problem Solving forum
“If the problem is actually 7²⁸¹ * 3²⁶⁴ - which would be pretty sloppy posting, but fair enough, formatting errors happen - then I''d do it somewhat differently. 7²⁸¹ * 3²⁶⁴ => 7¹⁷ * 7²⁶⁴ * 3²⁶⁴ => 7¹⁷ * 21²⁶⁴ From here, we can work with a ...”
May 25, 2017
Matt@VeritasPrep posted a reply to Digits at ten's place in the Problem Solving forum
“If you''re wondering how I can simplify like that, consider a related problem. Say I have one big number ending in 81 and another big number ending in 64. I can write the first as 100x + 81, since everything greater than 81 is in the hundreds place or further left. (For instance, 7281 = ...”
May 25, 2017
Matt@VeritasPrep posted a reply to Digits at ten's place in the Problem Solving forum
“A much easier solution, by the way: 7281 * 3264 will have the same tens digit as 81 * 64. 81 * 64 = (80 + 1) * (60 + 4) = 4800 + 60 + 320 + 4 The tens digit comes from 60 + 320 = 380, so our tens digit is 8.”
May 25, 2017
Matt@VeritasPrep posted a reply to Digits at ten's place in the Problem Solving forum
“[quote="paiboi782"] 7²⁰ᴷ => 49¹⁰ᴷ => 2401⁵ᴷ 2401 to any (positive integer) power is going to end in 01, so it will always have remainder 1 when divided by 100. This is not at all relevant to the GMAT in 2017 (or any prior year).”
May 25, 2017
Matt@VeritasPrep posted a reply to Arabian Horses - Good One! in the Problem Solving forum
“Nah, definitely fair game. The GMAT would use smaller numbers, though, something like 4 horses for 2 carts or 6 horses for 3.”
May 25, 2017
Matt@VeritasPrep posted a reply to Arabian Horses - Good One! in the Problem Solving forum
“It''s a great example of how difficulty is determined as much by obscurity as by the challenges of the problem itself. If you''ve done a little combinatorics, the problem is pretty easy, but enough of the GMAT test taking population HASN''T done combinatorics that the problem is assigned a high ...”
May 25, 2017
Matt@VeritasPrep posted a reply to Arabian Horses - Good One! in the Problem Solving forum
“If the groups ARE distinct, the problem is easier: just (9 choose 3) * (6 choose 3), then you''re set.”
May 25, 2017
Matt@VeritasPrep posted a reply to Arabian Horses - Good One! in the Problem Solving forum
“That''s a lot more complicated! We''d need to know whether the groups are distinct, but assuming that they''re not: First, choose 3 people for the first group: (9 choose 3) Next, choose 3 people for the second group: (6 choose 3) The three people remaining form the third group, so we''re ...”
May 25, 2017
Matt@VeritasPrep posted a reply to Molly worked at an amusement park over the summer.Every two in the Problem Solving forum
“So with my last point in mind, this is not exactly a geometric series. For it be geometric, the ratio between each pair of consecutive terms would be constant (2/1 = 4/2 = 8/4 = ...), and here that''s not the case: 161/160 ≠ 322/161. (I get the idea, but it doesn''t apply perfectly here.) All ...”
May 25, 2017
Matt@VeritasPrep posted a reply to Molly worked at an amusement park over the summer.Every two in the Problem Solving forum
“For those who aren''t familiar with the lingo here, geometric series are generated by multiplying by a constant, while arithmetic series are generated by adding a constant. A geometric series would be something like: 1, 2, 4, 8, 16, ... (multiplying by two every time) An arithmetic series ...”
May 25, 2017
Matt@VeritasPrep posted a reply to Molly worked at an amusement park over the summer.Every two in the Problem Solving forum
“You can use progressions, but you don''t need to - GMAT problems are designed not to benefit from higher math. (Not that an arithmetic series is particularly difficult; it''s just higher than the math on the GMAT.)”
May 25, 2017
Matt@VeritasPrep posted a reply to Which of the following is the greatest? (A) 1/√2+1/√4+1/ in the Problem Solving forum
“We could also just rough it out conceptually: (A) is the sum of four fractions between 1/1 and 1/3. (B) is the sum of four fractions from 1/4 to 1/64. Smaller than A. (C) is the sum of four fractions from 1/4 to 1/256. Smaller than A. (D) is 1/2 + 1/12. 1/√2 is bigger than 1/2 and ...”
May 12, 2017
Matt@VeritasPrep posted a reply to If 8 teams participate in Indian Premier League and every te in the Problem Solving forum
“We''ve got eight teams and we want to choose all possible groups of two teams. This is a combination, 8 choose 2, so we turn to our combinations formula: x choose y = x! / (y! * (x - y)!) 8 choose 2 = 8! / 2!6! = 8*7/2 = 28 Each match will be played twice, so we double this: 28 * 2, ...”
May 12, 2017
Matt@VeritasPrep posted a reply to help !! in the Problem Solving forum
“We could also just try to solve for each answer. (A): 1 / (1 - |x|) = 2 1 = 2 * (1 - |x|) 1 = 2 - 2|x| 2|x| = 1 |x| = 1/2 so x = 1/2 or -1/2 (B) 1 / (1 - |x|) = 1.5 1 = 1.5 - 1.5|x| 1.5|x| = .5 |x| = 1/3 (C) 1 / (1 - |x|) = 1 1 = 1 - |x| |x| = 0 ...”
May 12, 2017
Matt@VeritasPrep posted a reply to Solution in the Problem Solving forum
“x² - 7x = 144 x² - 7x - 144 = 0 (x + 9) * (x - 16) = 0 So x = -9 or x = 16. If y = -9 and n = 1, then x = -9: cross out (A). If y = 16 and n = 1, then x = 16: cross out (B). If y = 4 and n = 2, then x = 16: cross out (C). By process of elimination, (D) is the right ...”
May 12, 2017
Matt@VeritasPrep posted a reply to A fair coin will be tossed twice and a fair die with sides n in the Problem Solving forum
“P(at least one) = 1 - P(none) So P(≥1H) = 1 - P(0H) = 1 - 1/2*1/2 = 3/4 and P(≥1 1+) = 1 - P(0 1+) = 1 - 1/6 * 1/6 = 35/36 Then multiply these together: 3/4 * 35/36, for 35/48.”
May 12, 2017
Matt@VeritasPrep posted a reply to A certain club has 10 members-including Harry. in the Problem Solving forum
“Because to be selected secretary, he must NOT be selected treasurer. So just as you did with the first step, you need to have (8/9)*(1/8) for the last piece. Basic idea: President = 1/10 Treasurer = Not President * Yes Treasurer = (9/10)*(1/9) Secretary = Not President * Not ...”
May 12, 2017
Matt@VeritasPrep posted a reply to tricky VIAC (Variables in the answer choices) question in the Problem Solving forum
“We could also cheat with a pattern: n² - (n + 1)² => n² - (n² + 2n + 1) => -(2n + 1) => -(n + n + 1) for any value of n. Since we''ve got 1² - 2² + 3² - 4² ..., we''ve really got -(1 + 2) -(3 + 4) .... -(99 + 100), or -1 -2 -3 -4 .... - 99 - 100, or -(1 + 2 + ...”
May 12, 2017
Matt@VeritasPrep posted a reply to tricky VIAC (Variables in the answer choices) question in the Problem Solving forum
“I''m thinking ... 1² - 2² + 3² - 4² ± ... + 99² - 100² => (1² + 3² + 5² + ... + 99²) - (2² + 4² + ... + 100²) => (1² + 3² + 5² + ... + 99²) - ((1+1)² + (3+1)² + ... + (99+1)²) => (1² + 3² + 5² + ... + 99²) - (1² + 3² + ... + 99² + 2*1 + 2*3 + ... ...”
May 12, 2017
Matt@VeritasPrep posted a reply to help !! in the Problem Solving forum
“Since we''ve got x + 2y = a x - 2y = b we know that a = b + 4y, or a - b = 4y Likewise, adding the two equations together, we know that a + b = (x + 2y) + (x - 2y) = 2x So we can now replace (a + b) with 2x and (a - b) with 4y. Working with the answers, this gives us: (A) 2 * ...”
May 12, 2017
Matt@VeritasPrep posted a reply to Median @statistics in the GMAT Math forum
“I think you may be mistaking {1, 7, 9, 15, 18} for the set, but those are just five numbers, each of which COULD be added to the set.”
May 12, 2017
Matt@VeritasPrep posted a reply to question about combinations and permusation in the GMAT Math forum
“The order definitely matters! If Mark is the chairman and Dave is the co-chairman, we''ve got a different leadership team than we''d have if Dave is the chairman and Mark is the co-chairman: in the first case, Mark has final say, but in the second, Dave does. So many of the problems in the world ...”
May 12, 2017
Matt@VeritasPrep posted a reply to median / range problem in the GMAT Math forum
“Typically yes, but remember that here the greatest number can be written as a function of the smallest number: Greatest = Smallest + 25 So you want the Smallest to be as BIG as possible to make the Greatest also as big as possible.”
May 12, 2017
Matt@VeritasPrep posted a reply to Please Explain OG15 PS#218 and DS#72 in the GMAT Math forum
“Also be sure to post the text of the problem itself, that saves a lot of time and helps other students who don''t have that edition of the OG.”
May 12, 2017
Matt@VeritasPrep posted a reply to Two methods to calculate total average speed in the GMAT Math forum
“That does seem a bit much, though: I wouldn''t recommend memorizing formulas for unusual situations. It''s better to get a sense of where averages come from, then (re)generate a useful approach like this when the situation demands it.”
May 12, 2017
Matt@VeritasPrep posted a reply to Understanding Absolute Values in the GMAT Math forum
“Some of the crucial properties: |x - y| is the distance between x and y on the number line |x + y| is the distance between x and -y on the number line |x| ≥ 0 for all x |x| = 0 if and only if x = 0 |x| = √x² |x| = y has two solutions: x = y and -x = y”
May 12, 2017
Matt@VeritasPrep posted a reply to Understanding Absolute Values in the GMAT Math forum
“Here''s a good walkthrough.”
May 12, 2017
Matt@VeritasPrep posted a reply to DS- French/Japanese in the GMAT Math forum
“We could also skip the matrix and just do a little algebra. Say we have f = only French j = only Japanese b = both n = neither We know that 4% of French = Both, so .04*(f + b) = b, or .04f = .96b. From there: S1: b = 16 So b = 16, f = 384. Doesn''t give us j, so NOT ...”
May 12, 2017
Matt@VeritasPrep posted a reply to Hi in the GMAT Math forum
“No problem! :D I think everyone who answers questions on the forum loves such genuine thanks!”
May 12, 2017
Matt@VeritasPrep posted a reply to Important Tricks/shortcuts on GMAT Quant in the GMAT Math forum
“More than anything else, it''s important to try problems at a level that''s challenging but accessible FOR YOU, and to stick with those until you get good enough to try the harder ones. I see a lot of people waste time on needlessly hard problems, or kid themselves with very easy ones, but the ...”
May 12, 2017
Matt@VeritasPrep posted a reply to Exponents and Roots are Killing Me (and my score)!!!!!!!!! in the GMAT Math forum
“Let me add a mathematical approach too: If (x+1) * (|x| - 1) > 0, then either both (x+1) and (|x| - 1) are positive or both (x+1) and (|x| - 1) are negative. Let''s examine the first case: both are positive. Since (x+1) > 0, we know x > -1. Since |x| - 1 > 0, we know |x| > 1. ...”
May 12, 2017
Matt@VeritasPrep posted a reply to Exponents and Roots are Killing Me (and my score)!!!!!!!!! in the GMAT Math forum
“The good news is that exponents and roots are taught in every algebra I (and II) course out there, so there is a lot of practice material. I''d try Forgotten Algebra: it has great explanations and plenty of drills to get you up to speed. There are also loads of great web resources: Khan Academy, ...”
May 12, 2017