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Ratios problem

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by infiniti007 » Tue Oct 20, 2015 1:18 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 theCEO » Tue Oct 20, 2015 5:23 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
First number = 3x
Second number = 4x
Ratio of the 2 numbers= 3x:4x = 3:4

Add k to each number we get:
3x+k : 4x+k = 4:5

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

(3x+k) / (117) = 4/9 ---equation 1
(4x+k) / (117) = 5/9 ---equation 2

36x + 9k = 585 ---equation 1
27x + 9k = 468 ---equation 2
9x = 117 -> x = 13

Using equation 1:
36x + 9k = 585
36(13) + 9k = 585
9k = 585 - 468 = 117
k = 13

ans = b
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by Brent@GMATPrepNow » Tue Oct 20, 2015 7:36 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: B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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