CAT 2019 Quant Question with Solution 28

Question:
If the rectangular faces of a brick have their diagonals in the ratio $3: 2 \sqrt{3}: \sqrt{15},$ then the ratio of the length of the shortest edge of the brick to that of its longest edge is

1. $\sqrt{3}: 2$
2. $2: \sqrt{5}$
3. $1: \sqrt{3}$
4. $\sqrt{2}: \sqrt{3}$

Let the edges of the brick be a, b, and c such that $a<b<c$

${{a}^{2}}+{{b}^{2}}={{3}^{2}}=9....(1)$

${{a}^{2}}+{{c}^{2}}={{\left( 2\sqrt{3} \right)}^{2}}=12...(2)$

${{b}^{2}}+{{c}^{2}}={{\left( \sqrt{15} \right)}^{2}}=15...(3)$

Adding all three equations. We get

$2\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)=9+12+15=36$

${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=18...(4)$

From (1) and (4), $c=3$

From (3) and (4), $a=\sqrt{3}$

Therefore, required ratio = $\frac{a}{c}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$

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