Search found 2630 matches
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.
- by Matt@VeritasPrep
Thu Dec 07, 2017 4:20 pm- Forum: Helpful Resources
- Topic: When to STOP studying for the GMAT
- Replies: 3
- Views: 3587
ok sure
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 spike of...
- by Matt@VeritasPrep
Thu Dec 07, 2017 4:16 pm- Forum: Helpful Resources
- Topic: When to STOP studying for the GMAT
- Replies: 3
- Views: 3587
- by Matt@VeritasPrep
Thu Dec 07, 2017 3:57 pm- Forum: GMAT Math
- Topic: help
- Replies: 12
- Views: 9630
subjected to this again
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*
- by Matt@VeritasPrep
Thu Dec 07, 2017 3:56 pm- Forum: GMAT Math
- Topic: time speed distance question
- Replies: 3
- Views: 3888
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 )
- by Matt@VeritasPrep
Thu Dec 07, 2017 3:53 pm- Forum: GMAT Math
- Topic: Master List of Quant/GMAT Math Resources
- Replies: 9
- Views: 5550
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 (# of yellow / ...
- by Matt@VeritasPrep
Tue Dec 05, 2017 6:36 pm- Forum: Problem Solving
- Topic: There are three blue marbles, three red marbles ....
- Replies: 3
- Views: 1069
- by Matt@VeritasPrep
Tue Dec 05, 2017 6:30 pm- Forum: Problem Solving
- Topic: If x >= 0 and x=root(8xy-16y^2), then, in terms of y, x=?
- Replies: 3
- Views: 1033
wv
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.
- by Matt@VeritasPrep
Tue Dec 05, 2017 6:27 pm- Forum: Problem Solving
- Topic: If x is an integer, how many possible values
- Replies: 3
- Views: 2339
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!
- by Matt@VeritasPrep
Tue Dec 05, 2017 6:19 pm- Forum: Problem Solving
- Topic: Elana was working to code protocols for computer...
- Replies: 4
- Views: 1158
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 => -(-1) + 0 + 0 => 1 and x...
- by Matt@VeritasPrep
Tue Dec 05, 2017 6:14 pm- Forum: Problem Solving
- Topic: If x-y+z=-1, -x+y+z=1, and x+y-z=-1, then x+y+z=?
- Replies: 4
- Views: 1133
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: 54 + 3c = 0 54 = -3c -18 = c Now plug c...
- by Matt@VeritasPrep
Tue Dec 05, 2017 6:02 pm- Forum: Problem Solving
- Topic: If -6 and 3 are the solutions of the...
- Replies: 4
- Views: 1177
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 b + c = -15.
- by Matt@VeritasPrep
Tue Dec 05, 2017 5:43 pm- Forum: Problem Solving
- Topic: If -6 and 3 are the solutions of the...
- Replies: 4
- Views: 1177
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
- by Matt@VeritasPrep
Tue Dec 05, 2017 5:32 pm- Forum: Problem Solving
- Topic: If s is the product of the integers from 100 to 200 inclusiv
- Replies: 3
- Views: 7684
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
- by Matt@VeritasPrep
Tue Dec 05, 2017 5:24 pm- Forum: Problem Solving
- Topic: One side of a triangle has lenght 8 and a second side...
- Replies: 3
- Views: 1078
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 approaches 8 + 5, ou...
- by Matt@VeritasPrep
Tue Dec 05, 2017 5:23 pm- Forum: Problem Solving
- Topic: One side of a triangle has lenght 8 and a second side...
- Replies: 3
- Views: 1078