Q. 4: 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. 58
2. 50
3. 95
4. 85
Let the number be a and b.
So ab = 616
Given, (a^3 –b^3)/(a-b)^3 = 157/3
(a-b)*(a^2 + b^2 +ab)/(a-b)*(a^2 + b^2 -2ab) = 157/3
Let a^2 + b^2 =k
So (k+ab)/(k-2ab) = 157/3
3k + 3ab = 157k – 314ab
154k = 117ab = 317*616
k = 317*616/154 = 1268
As we know (a+b)^2 = (a^2 + b^2) + 2ab = k + 2ab = 1268 + 2*616
(a+b)^2 =2500 = 50^2
So a+b = 50
Inspiring Education… Assuring Success!!
Ⓒ 2020 – All Rights Are Reserved