CAT 2018 Set-1 | Question: 79

Question

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 1

Let the three numbers be $x,y$  and $z.$

Let, ashok took $z = 73 $ instead of $ z = 37 $

So, the final product will be

$ xy(73) – xy(37) = 720 $

$ \Rightarrow xy(73 -37) = 720 $

$ \Rightarrow xy(36) = 720 $

$ \Rightarrow \boxed {xy = 20}  \quad \longrightarrow (1)$

We know that, $\text{AM}$ and $\text{GM}$ are arithmetic and geometric mean respectively between two numbers $x$ and $y$, then

$ \boxed {\text{AM} \geqslant \text{GM}}$

$ \Rightarrow \frac{(x+y)}{2} \geqslant \sqrt{xy}$

$ \Rightarrow x+y \geqslant 2 \sqrt{xy} $

$ \Rightarrow (x+y)^{2} \geqslant 4xy$

$ \Rightarrow x^{2} + y^{2} + 2xy \geqslant 4xy $

$ \Rightarrow x^{2} + y^{2}  \geqslant 2xy $

$ \Rightarrow x^{2} + y^{2}  \geqslant 2(20) $

$ \Rightarrow \boxed {x^{2} + y^{2}  \geqslant 40} $

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$\textbf{Short Method:}$

The minimum possible sum of the squares of the two numbers when $ x = y. $

From equation $ (1), x = y = \sqrt {20}$

so, $ x^{2} + y^{2} = (\sqrt {20})^{2} + ( \sqrt {20})^{2} $

$ \Rightarrow x^{2} + y^{2} = 20 + 20 = 40 $

$\therefore$ The minimum possible value of the sum of squares of the other two numbers is $40.$

Correct Answer $:40$