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Minimizing Sum of Squares of Factors

CAT 2019 Slot 1 · Quantitative Ability · Easy · Number Theory

This is an easy Quantitative Ability question from the CAT 2019 Slot 1 paper. It tests Number Theory. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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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:

Answer

40

Explanation

Let the three real numbers be a,b,ca, b, c. Given (73−37)bc=720⇒36bc=720⇒bc=20(73 - 37)bc = 720 \Rightarrow 36bc = 720 \Rightarrow bc = 20. We want to minimize b2+c2b^2 + c^2. Since (b−c)2≥0⇒b2+c2≥2bc=2(20)=40(b - c)^2 \ge 0 \Rightarrow b^2 + c^2 \ge 2bc = 2(20) = 40. Minimum value = 40.

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