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by infiniti007 » Tue Oct 20, 2015 12:53 pm
The ratio of two positive numbers is 3 to 4. If k is added to each number the new ratio will be 4 to 5, and the sum of the numbers will be 117. What is the value of k?

A.) 1
B.) 13
C.) 14
D.) 18
E.) 21
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by Brent@GMATPrepNow » Tue Oct 20, 2015 1:06 pm
infiniti007 wrote:The ratio of two positive numbers is 3 to 4. If k is added to each number the new ratio will be 4 to 5, and the sum of the numbers will be 117. What is the value of k?

A.) 1
B.) 13
C.) 14
D.) 18
E.) 21
The ratio of two positive numbers is 3 to 4.
Let 3x = the smaller number
Let 4x = the larger number

Aside: Notice that this ensures that their ratio will equal 3/4, since 3x/4x = 3/4

If k is added to each number the new ratio will be 4 to 5 . . .
When k is added to each value, the NEW values are 3x + k and 4x + k
So, (3x + k)/(4x + k) = 4/5
Cross multiply to get: 16x + 4k = 15x + 5k
Rearrange terms to get: x = k


. . . and the sum of the numbers will be 117
The two NEW values are 3x + k and 4x + k
So, we can write: (3x + k) + (4x + k) = 117
Simplify to get 7x + 2k = 117

Since x = k, we can take the above equation and replace x with k to get...
7k + 2k = 117
Simplify: 9k = 117
Solve: k = 13

Answer: A

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by GMATGuruNY » Tue Oct 20, 2015 1:19 pm
The problem above is the same as the following:
The ratio of two positive numbers is 4 to 5, and the sum of the two numbers is 117. If k is SUBTRACTED FROM each number, the new ratio will be 3 to 4. What is the value of k?

1
13
14
18
21
Let x = the smaller number and y = the larger number.
To determine the values of x and y, use a RATIO BOX:
Image

Since x:y = 4:5, the total for the ratio is 9:
Image

Since the actual total of x and y is 117, we get:
Image

Since (actual total)/(ratio total) = 117/9 = 13, the multiplier for the box is 13, as follows:
Image
Thus, x = 52 and y = 65.

From here, we can PLUG IN THE ANSWERS for the value of k.
When the correct value is subtracted from x and y, the resulting ratio will be 3 to 4.

Answer choice A: k=1
(52-1)/(65-1) = 51/64.
Eliminate A.

Answer choice B: k=13
(52-13)/(65-13) = 39/52 = 3/4.
Success!

The correct answer is B.
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by Pazoki » Fri Jun 17, 2016 7:52 am
infiniti007 wrote:The ratio of two positive numbers is 3 to 4. If k is added to each number the new ratio will be 4 to 5, and the sum of the numbers will be 117. What is the value of k?

A.) 1
B.) 13
C.) 14
D.) 18
E.) 21
hi all
is my way correct?
K/k add to 3/4 to get 5/4 and sum of the 4/5's numerator and denominator must be 117.then K can be any no. but resulted ratio parts must be sum up to 117. so we just add ratio parts to get 4+5=9 then 117/9=13 answer "B" is correct.
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by OptimusPrep » Sun Jun 19, 2016 2:56 am
Pazoki wrote:
infiniti007 wrote:The ratio of two positive numbers is 3 to 4. If k is added to each number the new ratio will be 4 to 5, and the sum of the numbers will be 117. What is the value of k?

A.) 1
B.) 13
C.) 14
D.) 18
E.) 21
hi all
is my way correct?
K/k add to 3/4 to get 5/4 and sum of the 4/5's numerator and denominator must be 117.then K can be any no. but resulted ratio parts must be sum up to 117. so we just add ratio parts to get 4+5=9 then 117/9=13 answer "B" is correct.
thanks
Hi Pazoki,

If you add k/k to 3/4, it means that you are adding 1 to 3/4

Whereas the question says to add k to each of numerator and denominator.
Assuming the numbers to be 3x and 4x.

We are given that (3x + k) / (4x + k) = 4/5
Does this help?
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by Matt@VeritasPrep » Thu Jun 23, 2016 4:16 pm
Pazoki wrote:
infiniti007 wrote:The ratio of two positive numbers is 3 to 4. If k is added to each number the new ratio will be 4 to 5, and the sum of the numbers will be 117. What is the value of k?

A.) 1
B.) 13
C.) 14
D.) 18
E.) 21
hi all
is my way correct?
K/k add to 3/4 to get 5/4 and sum of the 4/5's numerator and denominator must be 117.then K can be any no. but resulted ratio parts must be sum up to 117. so we just add ratio parts to get 4+5=9 then 117/9=13 answer "B" is correct.
thanks
This is a clever idea, but you're assuming here that number means integer, and that each piece of the ratio must be a factor of 117. If we make that assumption, we don't even need to bother with the other steps, we can just say 4n + 5n = 117, and n = 13. If we DON'T make that assumption, then we have to actually solve.
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by Brent@GMATPrepNow » Sat Jan 13, 2018 8:44 am
infiniti007 wrote:The ratio of two positive numbers is 3 to 4. If k is added to each number the new ratio will be 4 to 5, and the sum of the numbers will be 117. What is the value of k?

A.) 1
B.) 13
C.) 14
D.) 18
E.) 21
The ratio of two positive numbers is 3 to 4.
Let 3x = the smaller number
Let 4x = the larger number

Aside: Notice that this ensures that their ratio will equal 3/4, since 3x/4x = 3/4

If k is added to each number the new ratio will be 4 to 5 . . .
When k is added to each value, the NEW values are 3x + k and 4x + k
So, (3x + k)/(4x + k) = 4/5
Cross multiply to get: 16x + 4k = 15x + 5k
Rearrange terms to get: x = k


. . . and the sum of the numbers will be 117
The two NEW values are 3x + k and 4x + k
So, we can write: (3x + k) + (4x + k) = 117
Simplify to get 7x + 2k = 117

Since x = k , we can take the above equation and replace x with k to get...
7k + 2k = 117
Simplify: 9k = 117
Solve: k = 13

Answer: B

Cheers,
Brent
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by Scott@TargetTestPrep » Sun Aug 11, 2019 6:51 pm
infiniti007 wrote:The ratio of two positive numbers is 3 to 4. If k is added to each number the new ratio will be 4 to 5, and the sum of the numbers will be 117. What is the value of k?

A.) 1
B.) 13
C.) 14
D.) 18
E.) 21
Since the ratio of the numbers is 3 to 4, the two numbers can be expressed as 3x and 4x for some positive value x.

Since the ratio after k is added to each number is 4 to 5, we can create the following equation:

(3x + k)/(4x + k) = 4/5

15x + 5k = 16x + 4k

k = x

Since the sum of the two numbers after adding k to each of them is 117, we have:

(3x + k) + (4x + k) = 117

(3x + x) + (4x + x) = 117

9x = 117

x = 13

Thus, k = x = 13.

Answer: B

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