# CAT 2018 Question Paper | Quants Slot 2

###### CAT Previous Year Paper | CAT Number Theory Questions | Question 10

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Question 10: While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is:
[TITA]

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### Explanatory Answer

##### Method of solving this CAT Question from Coordinate Geometry: A clear understanding of graphs and equations will help to solve.

Let the three real numbers be a, b, c. c is taken as 73 instead of 37
So, 73ab = 37ab + 720
36 ab = 720
ab = $$frac{720}{36}$ = 20 We know, AM ≥ GM $\frac{$a^2 + b^2$}{2}) ≥ ab
(a2 + b2) ≥ 2ab
(a2 + b2) ≥ 40
(Or)
ab = 20
(a,b) = (√20, √20)
(a2 + b2) = 20 + 20 = 40

The question is " While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is: The minimum possible value is 40 "

##### Hence, the answer is 40.

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