Two positive integers that have a ratio of 1 : 2 are increas

Question

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

KarishmaB · Accepted Answer

The question is based on the following concept: 
When you add the same number to the numerator and denominator of a fraction, the fraction increases if it is less than 1 (e.g. 1/2. If I add 3 to both numerator and denominator, it becomes 4/5) and decreases if it is more than 1 (e.g. 7/3. If I add 3 to both numerator and denominator, I get 10/6 = 5/3 )
So if the numbers are in the ratio 1:2, and you increase them in the ratio 1:1, i.e. you add the same number to both of them, the new ratio will be more than 1:2. e.g. add 3 to both the numbers. You get 4:5, a bigger ratio.
Only option (C) gives a ratio greater than 1:2.

KarishmaB · Answer

General piece of advice: Think more, solve less. Don't jump on making equations. Let that be the last resort at least during practice. (In the exam, do what comes to mind first.)
Analyze each statement step by step.e.g.

Two positive integers that have a ratio of 1 : 2 are increased in a ratio of 1 : 1.  
Here, I stop and say, "Ok. If the integers are 1 and 2 and are increased in ratio 1:1, let me say they increase by 3. So the number now are 4 and 5, ratio of 4/5 (since it is a ratio question)"

Which of the following could be the resulting integers? 
Of course there are infinite possibilities. A property that is common to all of them...

Look at the options - the ratios they represent. The first one is 2 and 4. A ratio of 1/2. This is not possible. If you add the same number to both, ratio will change. Go on to the next... You might be able to figure out that 15 and 28 have a better chance than all others. It might seem odd that this option gives a ratio greater than 1/2 while all others give less than 1/2. Then you can make equation for it to check if you wish.

Practice and make a note of all new concepts you come across.

oldstudent · Answer

Hi , can you please also help me to form an equation for this condition.
I formed:

5-y/13-y =1/2 -(Eq.1) (The resultant numerator/denominator is deducted by the same number y - ratio y:y=1:1 to get 1/2)
=> 10-2y = 13 - y
=>y = -3
substituting -3 in Eq (1)
8/16 = 1/2 - which suffices the condition.
similarly option C and option D also suffices the condition.
So I do not get where is the mistake in forming this equation. PLEASE help.

puneetpratik · Answer

I also followed a method similar to old student and was misguided. But I think this is a natural way of thinking, if one doesn't know the concept pointed by Karishma. Could you please tell us how to save our selves from such traps?

muralimba · Answer

Let me help u guys.

Lets say the #s, after increasing by P (adding P to both), are M and N

Then,

before increase, the number would be M-P and N-P and given that M-P / N-P = 1/2
Note that P is a positive # as it is siad that it is an INCREMENT.

cross mult. M-P / N-P = 1/2
==> 2M-2P=N-P ==> 2M-N=P and M and N are the #s after increasing by 1:1 that is P:P

hence all we have is 2M-N=P ==> double of the first # - the 2nd # is posive
cross verify with the answers for 2M-N
(A) 2 and 4 ==> 4-4 not positive
(B) 5 and 13 ==> 10-13 Not positive
(C) 15 and 28 ==> 30-28 YESSSSSSSS positive and hence the correct answer
(D) 33 and 68 ==> 66-68 Not positive
(E) 71 and 143 ==> 142-143 Not positive

Hope it helps. The trick here is to assume that the #s, after increasing by P (adding P to both), are M and N

KarishmaB · Answer

There is your problem. The initial ratio was  increased  so y cannot be negative.
Definitely, in option (D) too, y will come out to be negative.
Only option (C) will give you positive y because 15/28 is greater than 1/2.

jlgdr · Answer

Even if you didn't know how to solve one can easily eliminate A,B and E straight away just cause you are always going to have one even and one odd in the numerator and denominator. It is impossible to have both odd and even in both Cheers! J

kzivrev · Answer

Is this a proper GMAT wording, I found this problem to have bad wording, to me it was not clear that I have to add the same new number on the denominator and on the nominator. so not having that in mind it is taking you to a totoaly wrong direction

EMPOWERgmatRichC · Answer

Hi kzivrev,

I agree that this question is NOT worded particularly well (and not up to the typical 'standards' of the GMAT questions that you'll see on Test Day). The original post is from over 4.5 years ago and is likely from a source that is even older than that (and also unknown), so some deviation from the norm is to be expected. The basic concepts behind this question will show up on Test Day though, so while you can ignore this prompt, you still need to know the ideas behind it.

GMAT assassins aren't born, they're made,
Rich

jayanthjanardhan · Answer

hey, sorry, not sure what i am missing, but how 15/28 greater than 1/2?

Bunuel · Answer

(1/2 = 14/28) < 15/28

jayanthjanardhan · Answer

ok never mind!...just one of those days!..thanks bunuel!

EMPOWERgmatRichC · Answer

Hi jayanthjanardhan,

Bunuel used a common denominator to prove that 15/28 is greater than 1/2; that particular 'math skill' will be quite useful on Test Day, so you should make sure that you're comfortable doing that type of math by hand.

There is another way to prove that 15/28 is greater than 1/2 though...

15/30 = 1/2

Since the denominator (28) of 15/28 is SMALLER than 30, the fraction must be BIGGER than 1/2.

You're going to come across some questions on the GMAT that are more about estimation and pattern-matching than strict calculation skills - understanding math 'patterns' and using them to your advantage can save you time, make questions easier AND raise your Quant score.

GMAT assassins aren't born, they're made,
Rich

chetan2u · Answer

Hi,
Just another way to look at it..
Ratio is 1:2... let them be x and 2x..
Add y to each.. x+y and 2x+y..
Difference in the two resulting integers is 2x+y-(x+y) = x..
Let's check the numbers..

(A) 2 and 4 ... 4-2=2.. but here x+y is 2, so x is not 2
(B) 5 and 13 ...13-5=8...again x should be8 but x+y is only 5
(C) 15 and 28 .....28-15=13.. x is 13 and x+ y =15.. possible.. 13+2 and 13*2+2.. 
(D) 33 and 68 ...68-33=35 again x is greater than x+y...
(E) 71 and 143 ..... 143-71=72...again x is greater than x+y

Only C possible

Manonamission · Answer

Fantastic.. between is this just an observation with numbers or there is an underlying concept behind it? How can I think in such a way?

danigonzalez01 · Answer

Solved by brute force:

The expresion should be such that:

(x + y)/(2x + y)

So, I started subtracting, in each answer, 1 on the numerator and one on the denominator to see if I could reach the answer.

When I got to 15/28, subtracting two yielded 13/26 so I knew that was the answer:

(13 + 2)/(2*13 +2) = 15/28

Is there a way to solve this using generic expresions or formulas?

GMATGuruNY · Answer

Increase the original ratio -- 1:2 -- in increments of 1:
 2:3
3:4
4:5
5:6
6:7 
In each of the blue cases above, the second value is less than twice the first value.
Only in C is the second value (28) less than twice the first value (15).

C

Alternate approach:
Let the original integers be x and 2x and the increase in each integer be d.
Resulting ratio  = \frac{x+d}{2x+d}

We can PLUG IN THE ANSWERS, which represent the value of the resulting ratio.

A:
 \frac{x+d}{2x+d} = \frac{2}{4}

4x+4d=4x+2d

2d=0

d=0 
Not viable, since d must be positive.
Eliminate A.

B:
 \frac{x+d}{2x+d} = \frac{5}{13}

13x+13d=10x+5d

8d=-3x

x=-\frac{8}{3}d 
Not viable, since x and d must both be positive.
Eliminate B.

C:
 \frac{x+d}{2x+d} = \frac{15}{28}

28x+28d=30x+15d

13d=2x

x=\frac{13}{2}d 
This works:
If  d=2 , then  x=\frac{13}{2}*2=13 
Implication:
Original integers -->  x=13  and  2x=26 
New integers -->  x+d=13+2=15  and  2x+d=26+2=28 
Resulting ratio -->  \frac{15}{28}

C

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Two positive integers that have a ratio of 1 : 2 are increas

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