Question

On each face of a cuboid, the sum of its perimeter and its area is written. Among the six numbers so written, there are three distinct numbers and they are $16,24$ and $31$. The volume of the cuboid lies between.

• A. $7$ and $14$
• B. $14$ and $21$
• C. $21$ and $28$
• D. $28$ and $35$

Answer 1

Answer: (D)

$28$ and $35$

Let the sides of the cubiod be $a,b,c$

$2(a+b)+ab=\mathrm{16.....}\left(i\right)$

$2(b+c)+bc=24$

$\Rightarrow c=\frac{24-2b}{2+b}.....\left(ii\right)$

$2(c+a)+ca=31$

$\Rightarrow c=\frac{31-2a}{2+a}.....\left(iii\right)$

From $\left(ii\right)$ and $\left(iii\right)$

$\frac{24-2b}{2+b}=\frac{31-2a}{2+a}$

$\Rightarrow 4a=2+5b.....\left(iv\right)$

using $\left(iv\right)$ in $\left(i\right)$

$2(\frac{2+5b}{4}+b)+\left(\frac{2+5b}{4}\right)\times b=16\Rightarrow b^2+4b-12=0\Rightarrow b=2,-6$

Negative value is not possible as side can't be negative

$\therefore b=2$

Using $b$ in $\left(i\right)$ and $\left(ii\right)$ we get $a=3$ and $c=5$ respectively

Volume of cubiod $=abc=3\times 2\times 5=30$

Hence, option $D$ is correct.