The ratio of two positive numbers is 3 to 4. If k is added

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

I think I approached this in a slightly different manner than Bunuel:

9x = 117
7x = 117 - 2k

Combine the equations:

2x = 2k

k = x = 117 / 9 = 13

Took about the same time as Bunuel but I found it easier.

Thanks

We have \(\frac{3x}{4x} = \frac{3}{4}\)................ (1)

Also, \(\frac{3x+k}{4x+k} = \frac{4}{5}\) .............. (2)

Looking at (1) & (2), we know that denominator of (1) = numerator of (2)

So we can directly equate:

4x = 3x + k

x = k = 13 = Answer

Some general question to the number properties in this case.
I thought when we have a ratio of two numbers, like 3 to 4, this means --> 3:4 = 3/7 and 4/7
Why can we here express this in the fraction 3x/4x and not 3x/(4x+3x) and 4x/(3x+4x) ???

Hi LaxAvenger,

You could refer to both parts of the ratio as you've described, but it would create more work for you overall. In these types of ratio question, when taking a 'math' approach, it's almost always easiest to do the work in 'part to part' format and not "part-to-whole TO "part-to-whole."

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

Bunuel,

I couldnt understand why you took a/b = 3x / 4x
I started with a/b = 3/4 and completely lost the plot.
Is this wrong approach?

Hi sagar2911,

When we're told that the ratio of A:B is 3 to 4, that means that A = some multiple of 3 and B = the equivalent multiple of 4.

For example:
A = 3 and B = 4
A = 6 and B = 8
A = 9 and B = 12
Etc.

Mathematically, since we don't know WHICH multiple we're dealing with, we have to assign a variable to it. Thus...

A/B = 3X/4X

This allows us to construct more complex equations that are based on that original ratio.

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

Hi All,

Many Test Takers would approach this question algebraically, but you can also approach this question by working back from the final piece of information and TESTing THE ANSWERS.

We're told that two numbers will end up in the ratio of 4:5 and their sum will be 117

4:5 is the same as saying 4X:5X

4X + 5X = 117
9X = 117
X = 13

Thus, the two "end" numbers are 4(13) and 5(13): 52 and 65

The question asks for the value of K - the value that was added to the ORIGINAL 2 numbers to get us to 52 and 65.

We know from the original ratio of 3:4 that the first number is a multiple of 3 and the second number is the equivalent multiple of 4. So we're looking for whichever answer, when subtracted from 52 and 65, gives us a number that's a multiple of 3 and another that's the SAME multiple of 4....

Answer A:
52-1 = 51 = 3(17)
65-1 = 64 = 4(16)
NOT the same multiple.

Answer B:
52-13 = 39 = 3(13)
65-13 = 52 = 4(13)
This IS the SAME multiple, so this must be the answer.

Final Answer:

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

9x=117
x=13
7x=91
91+2k=117
k=13
B

Solution:

We can let the original two numbers be 3x and 4x and create the equations:

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

and

3x + k + 4k + k = 117

Simplifying the first equation, we have:

16x + 4k = 15x + 5k

x = k

Substituting k for x in the second equation, we have:

3k + k + 4k + k = 117

9k = 117

k = 13

Answer: B

A student ask me whether we're obligated to use the "multiplier strategy" (e.g., ratio = 3x/4x) for this question.
The answer is "No," and what follows is a solution that uses three variables.

Let A = the smaller of the two numbers
Let B = the bigger number

The ratio of two positive numbers is 3 to 4.
We can write: A/B = 3/4
Cross multiply: 4A = 3B
Rewrite as: 4A - 3B = 0

If k is added to each number the new ratio will be 4 to 5...
We can write: (A + k)/(B + k) = 4/5
Cross multiply: 5(A + k) = 4(B + k)
Expand: 5A + 5k = 4B + 4k
Rewrite as: 5A - 4B = -k

...the sum of the numbers will be 117
We can write: (A + k) + (B + k) = 117
Simplify: A + B + 2k = 117
Rewrite as: A + B = 117 - 2k

We now have the following system:
4A - 3B = 0
5A - 4B = -k
A + B = 117 - 2k

Solve, solve, solve....
A = 39, B = 52 and k = 13

Answer: B

x / y = 3 / 4

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

x + y + 2k = 117

Solve for x in the first:

x / y = 3 / 4

x = 3y / 4

Plug that into our second equation and solve for y

[(3y / 4) + k] / (y + k) = 4 / 5

5 * (3y / 4 + k) = 4(y + k)

15y + 20k = 16y + 16k

4k = y

if y = 4k and x = 3y / 4, then x = 3(4k) / 4 = 3k

Now plug that into our final equation

x + y + 2k = 117

3k + 4k + 2k = 117

9k = 117

k = 13

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