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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.

by Matt@VeritasPrep

Thu Dec 07, 2017 5:56 pm
Forum: Problem Solving
Topic: Which of the following fractions is the largest?
Replies: 3
Views: 3015

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.

by Matt@VeritasPrep

Thu Dec 07, 2017 5:54 pm
Forum: Problem Solving
Topic: A certain taxi company charges $3.10
Replies: 6
Views: 3861

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, B, C are ...

by Matt@VeritasPrep

Thu Dec 07, 2017 5:28 pm
Forum: Problem Solving
Topic: A survey was conducted to find out how many people in...
Replies: 4
Views: 1374

yeah

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 20, and...

by Matt@VeritasPrep

Thu Dec 07, 2017 5:18 pm
Forum: Problem Solving
Topic: Three runners A, B and C run a race with A finishing...
Replies: 4
Views: 1722

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 21 meters left to ...

by Matt@VeritasPrep

Thu Dec 07, 2017 5:08 pm
Forum: Problem Solving
Topic: Three runners A, B and C run a race with A finishing...
Replies: 4
Views: 1722

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 between A's finishing the ...

by Matt@VeritasPrep

Thu Dec 07, 2017 4:56 pm
Forum: Problem Solving
Topic: Three runners A, B and C run a race with A finishing...
Replies: 4
Views: 1722

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 of arranging 10 letter...


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.


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) = 1/4 Then just cros...


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 is 4/10, or 2/5.

by Matt@VeritasPrep

Thu Dec 07, 2017 4:31 pm
Forum: Problem Solving
Topic: A basket contains 5 apples of which one is rotten...
Replies: 6
Views: 2353

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 fresh) * p(...

by Matt@VeritasPrep

Thu Dec 07, 2017 4:29 pm
Forum: Problem Solving
Topic: A basket contains 5 apples of which one is rotten...
Replies: 6
Views: 2353

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 those up, ...

by Matt@VeritasPrep

Thu Dec 07, 2017 4:27 pm
Forum: Problem Solving
Topic: A basket contains 5 apples of which one is rotten...
Replies: 6
Views: 2353

Another resurrected zombie thread - what is happening in the graveyard this week? :)

by Matt@VeritasPrep

Thu Dec 07, 2017 4:26 pm
Forum: Problem Solving
Topic: A gardener is going to plant 2 red rosebushes
Replies: 8
Views: 3071

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, we're set. If K = 7, then √7K = √49 = 7,...

by Matt@VeritasPrep

Thu Dec 07, 2017 4:25 pm
Forum: Problem Solving
Topic: What is the smallest positive integer K such that the...
Replies: 4
Views: 6085

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 * K) So if K = 7, we'll have two id...

by Matt@VeritasPrep

Thu Dec 07, 2017 4:23 pm
Forum: Problem Solving
Topic: What is the smallest positive integer K such that the...
Replies: 4
Views: 6085