V₁, V₂, V₃ and V₄ are the volumes of four cubes of side lengths x cm, 2 x cm, 3 x cm and 4 x cm respectively… · ABC26GN4851

Subject: General Aptitude · Chapter: Volume and Surface Area

$V_{1}, V_{2}, V_{3}$ and $V_{4}$ are the volumes of four cubes of side lengths $x \mathrm{cm}, 2 x \mathrm{cm}, 3 x \mathrm{cm}$ and $4 x \mathrm{cm}$ respectively. Some statements regarding these volumes are given below (1) $V_{1}+V_{2}+2 V_{3}<V_{4}$ (2) $V_{1}+4 V_{2}+V_{3}<V_{4}$ (3) $2\left(V_{1}+V_{3}\right)+V_{2}=V_{4}$ Which of these statements are correct?

(a)1 and 2

(b)2 and 3

(c)1 and 3

(d)1, 2 and 3

Answer

Answer (as printed): D

Explanation

Clearly, we have: $V_{1}=x^{3}, V_{2}=(2 x)^{3}=8 x^{3}, V_{3}=(3 x)^{3}=27 x^{3}$, $V_{4}=(4 x)^{3}=64 x^{3}$. $V_{1}+V_{2}+2 V_{3}=x^{3}+8 x^{3}+2 \times 27 x^{3}=63 x^{3}<V_{4}$. $V_{1}+4 V_{2}+V_{3}=x^{3}+4 \times 8 x^{3}+27 x^{3}=60 x^{3}<V_{4}$. $2\left(V_{1}+V_{3}\right)+V_{2}=2\left(x^{3}+27 x^{3}\right)+8 x^{3}=64 x^{3}=V_{4}$.

Explanation as extracted from the printed page; notation may be imperfect.

Study loop for Volume and Surface Area

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