Question 79

# 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

Solution

Assume the numbers are a and b, then ab=616

We have, $$\ \ \frac{\ a^3-b^3}{\left(a-b\right)^3}$$ = $$\ \frac{\ 157}{3}$$

=> $$\ 3\left(a^3-b^3\right)\ =\ 157\left(a^3-b^3+3ab\left(b-a\right)\right)$$

=> $$154\left(a^3-b^3\right)+3*157*ab\left(b-a\right)$$ = 0

=> $$154\left(a^3-b^3\right)+3*616*157\left(b-a\right)$$ = 0        (ab=616)

=>$$a^3-b^3+\left(3\times\ 4\times\ 157\left(b-a\right)\right)$$    (154*4=616)

=> $$\left(a-b\right)\left(a^2+b^2+ab\right)\ =\ 3\times\ 4\times\ 157\left(a-b\right)$$

=> $$a^2+b^2+ab\ =\ 3\times\ 4\times\ 157$$

Adding ab=616 on both sides, we get

$$a^2+b^2+ab\ +ab=\ 3\times\ 4\times\ 157+616$$

=> $$\left(a+b\right)^2=\ 3\times\ 4\times\ 157+616$$ = 2500

=> a+b=50

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