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CAT 2018 Slot 1 — QA Question 28

Number TheoryEasy

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 & solution

Correct answer: 40

Answer: 40

Solution

Easy

The error in one factor changed the product by 720720; this gives the product of the other two numbers. With their product fixed, minimise the sum of their squares using x2+y2≥2xyx^2+y^2\ge 2xy (AM–GM).

1

Translate the error into the product of the other two numbers. Let the other two reals be xx and yy.

xy⋅73−xy⋅37=720(increase in product)⇒ xy (73−37)=720⇒ 36 xy=720⇒ xy=20\begin{aligned} &xy\cdot 73 - xy\cdot 37 = 720 \quad\text{(increase in product)}\\ &\Rightarrow\ xy\,(73-37)=720\\ &\Rightarrow\ 36\,xy = 720\\ &\Rightarrow\ xy = 20 \end{aligned}
2

Minimise x2+y2x^2+y^2. Since the square of a real number is non-negative, (x−y)2≥0(x-y)^2\ge 0.

(x−y)2≥0⇒ x2+y2≥2xy⇒ x2+y2≥2(20)=40(from step 1)\begin{aligned} &(x-y)^2 \ge 0\\ &\Rightarrow\ x^2+y^2 \ge 2xy\\ &\Rightarrow\ x^2+y^2 \ge 2(20) = 40 \quad\text{(from step 1)} \end{aligned}

Equality (the minimum) holds when x=y=20x=y=\sqrt{20}.

min⁡(x2+y2)=40\min\left(x^2+y^2\right) = 40

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CAT 2018 Slot 1 QA Q28: While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the prod — Solution | TheCATExam