The Product of Two Positive Numbers is 616 GMAT Problem Solving

Question: The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157:3, then the sum of the two numbers is

1. 95
2. 85
3. 58
4. 50
5. 40

Correct Answer: D
Solution and Explanation:
Approach Solution 1:

a*b = 616
616 = 1*616
2*308
4*154
7*88
8*77
11*56
14*44
22*28
a^3−b^3/(a−b)^3= 157/3

It seems possible only with 22*28 because only these are so close that their cubes will be close to cube of their difference
Also, the difference of these numbers is a multiple of 3 which is necessary to meet the condition of denominator

11304/216= 157/3
i.e. a and b are 22 and 28
i.e. Sum= 22+28
= 50

Approach Solution 2:

(a^3-b^3)/(a-b)^3=157/3
(a-b)(a^2+ab+b^2)/(a-b)^3=157/3
(a^2+ab+b^2)/(a-b)^2=157/3
3(a^2+b^2)+3ab=157(a^2+b^2) -314ab
317ab =154(a^2+b^2)

Now ab =616
Therefore

a^2 +b^2 =1268

Therefore
a=28 b=22

Hence a+b=50

Approach Solution 3:
xy=616-----(1)
(x^3−y^3)/(x−y)^3= 157/3
(x−y)(x^2+xy+y^2)/(x−y)^3= 157/3
3(x^2+xy+y^2)= 157(x−y)^2
3x^2+3xy+3y^2=157x^2−314xy+154y^2
154x^2−317 xy+154y^2= 0
x^2+y^2= 317/154 xy
(x+y)^2= 2xy+317/154 xy
(x+y)^2= 625/154 xy
But xy= 616
Hence (x+y)^2= 25^2∗616/154
(x+y)^2= 25^2∗2^2
So
x+y=2∗25= 50

“The product of two positive numbers is 616. If the ratio of the”- is a topic of the GMAT Quantitative reasoning section of GMAT. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. GMAT Quant practice papers improve the mathematical knowledge of the candidates as it represents multiple sorts of quantitative problems.

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