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Positive Integers

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Positive Integers

by manik11 » Tue Feb 23, 2016 5:35 am
Two positive integers that have a ratio of 1 : 2 are increased in a ratio of 1 : 1. Which of the following could be the resulting integers?
A) 2 and 4
B) 5 and 13
C) 15 and 28
D) 33 and 68
E) 71 and 143

OA : C

Hi Experts! Is there a strategy to follow on this type of questions or should I just brute force my way out of this. It took me more than 3 minutes to get the correct answer though?

Thanks!
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by DavidG@VeritasPrep » Tue Feb 23, 2016 5:52 am
manik11 wrote:Two positive integers that have a ratio of 1 : 2 are increased in a ratio of 1 : 1. Which of the following could be the resulting integers?
A) 2 and 4
B) 5 and 13
C) 15 and 28
D) 33 and 68
E) 71 and 143

OA : C

Hi Experts! Is there a strategy to follow on this type of questions or should I just brute force my way out of this. It took me more than 3 minutes to get the correct answer though?

Thanks!
Call the initial numbers x and 2x. If they're increased by a 1 to 1 ratio, we'll add y to both. Now the terms are x + y and 2x + y. Now back-solve and see what works.

Try B: x + y = 5;
2x + y = 13

Subtract the second from the first. -x = -8. x = 8. So y = -3. We were told that the numbers were increased by the same amount. (We'll assume the question implies that the terms are increased by a positive quantity.) So this isn't correct.

Try C: x + y = 15;
2x + y = 28

Subtract: -x = -13; x = 13. y = 2. So the original numbers were 13 and 26. When each was increased by 2, you get 15 and 28. C is the answer.
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by GMATGuruNY » Tue Feb 23, 2016 5:57 am
manik11 wrote:Two positive integers that have a ratio of 1 : 2 are increased in a ratio of 1 : 1. Which of the following could be the resulting integers?
A) 2 and 4
B) 5 and 13
C) 15 and 28
D) 33 and 68
E) 71 and 143
An alternate approach is to PLUG IN THE ANSWERS.
When the two values in the correct answer choice are decreased by the SAME AMOUNT, a ratio of 1:2 will be yielded.

Strategy:
Subtract 1 from the two values.
Subtract 2 from the two values.
Subtract 3 from the two values.
And so on.
Look for a resulting ratio of 1:2.

A: 2 and 4
1 and 3.
Doesn't work.

B: 5 and 13
4 and 12
3 and 11
2 and 10
1 and 9.
Doesn't work.

C: 15 and 28
14 and 27
13 and 26.
Success!
13:26 = 1:2.

The correct answer is C.
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by [email protected] » Tue Feb 23, 2016 9:57 am
Hi manik11,

When reviewing questions that took 'too long' to answer, it's important to define WHAT you were doing during that time. You state that it took more than 3 minutes to solve this problem, but what work were you actually doing during that time? Was it real work or were you just staring at your pad trying to come up with something?

'Brute force' can be a remarkably useful approach on a number of Quant questions on Test Day, but it is still based on your ability to quickly get to work and put numbers on the pad. Consider Mitch's example - you can't tell (just by looking at them) which answer will be correct, but you can quickly go about 'subtracting 1' from each option (over-and-over) until you either have the correct answer OR you run out of options and prove that an answer is incorrect.

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by manik11 » Tue Feb 23, 2016 10:44 pm
[email protected] wrote:Hi manik11,

When reviewing questions that took 'too long' to answer, it's important to define WHAT you were doing during that time. You state that it took more than 3 minutes to solve this problem, but what work were you actually doing during that time? Was it real work or were you just staring at your pad trying to come up with something?

Rich
Hi Rich,
I started off exactly the way Mitch did in his solution. But by the time I had finished ruling out option B I was already thinking if my approach was correct (long-winded). Though it shouldn't have taken long to subtract 1 from the the numbers and check the ratio, my apprehension about my approach made me think if I was on the right track.
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by MartyMurray » Wed Feb 24, 2016 6:02 am
manik11 wrote:Hi Rich,
I started off exactly the way Mitch did in his solution. But by the time I had finished ruling out option B I was already thinking if my approach was correct (long-winded). Though it shouldn't have taken long to subtract 1 from the the numbers and check the ratio, my apprehension about my approach made me think if I was on the right track.
While finding a super efficient way to do something is obviously optimal, much of the time, if you have an approach and the approach is going to get you to the answer, the thing to do at that point is to GO WITH IT.
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by Matt@VeritasPrep » Tue Mar 01, 2016 11:47 pm
We basically have an initial ratio of x : 2x and a new ratio of (x + y) : (2x + y).

Since (2x + y) - (x + y) = x, we can find the differences between the answer choices to see if they yield solutions for x and y.

A:: 4 - 2 = 2, so x = 2. But then y = 0, which isn't possible. (0 / 0 ≠ 1 / 1).

B:: 13 - 5 = 8, so x = 8. But then y = -3, which means we aren't increasing.

C:: 28 - 15 = 13, so x = 13. This gives y = 2; SUCCESS!

We don't have to try any more values at this point, but if we wanted to be extra double sure, we could:

D:: 68 - 33 = 35, so x = 35 and y = -2, yuck.

E:: 143 - 71 = 72, so x = 72 and y = -1, again, yuck.
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