Question

If length, breadth and height of a cuboid is increased by $x\%,y\%$ and $z\%$ respectively then its volume is increased by _____

• A. $[x+y+z+\frac{xy+xz+yz}{100}+\frac{xyz}{(100{)}^2}]$ $\%$
• B. $[x+y+z+\frac{xy+xz+yz}{100}]$ $\%$
• C. $[x+y+z+\frac{xyz}{(100{)}^2}]$ $\%$
• D. None of these

Answer 1

Answer: (A)

$[x+y+z+\frac{xy+xz+yz}{100}+\frac{xyz}{(100{)}^2}]$ $\%$

Let the original length be $l$.
The new length will be $l(1+\frac{x}{100})$
Similarly the new breadth and height will be $b(1+\frac{y}{100})$ and $h(1+\frac{z}{100})$
The original volume $=lbh$
The new volume $=l(1+\frac{x}{100})\times b(1+\frac{y}{100})\times h(1+\frac{z}{100})$
The new volume $=lbh\frac{(100+x)(100+y)(100+z)}{\left({100}^3\right)}$
$\%$ Change in volume $=\frac{\text{New volume}-\text{Old volume}}{\text{Old volume}}\times 100$
New volume $-$ Old volume $=lbh\frac{(100+x)(100+y)(100+z)}{\left({100}^3\right)}-lbh$
New volume $-$ Old volume $=lbh(\frac{(100+x)(100+y)(100+z)}{\left({100}^3\right)}-1)$
New volume $-$ Old volume $=lbh\left(\frac{{100}^3+{100}^2(x+y+z)+100(xy+yz+zx)+xyz-{100}^3}{{100}^3}\right)$
New volume $-$ Old volume $=lbh\left(\frac{{100}^2(x+y+z)+100(xy+yz+zx)+xyz}{{100}^3}\right)$
$\%$ change in volume $=\frac{lbh\left(\frac{{100}^2(x+y+z)+100(xy+yz+zx)+xyz}{{100}^3}\right)}{lbh}\times 100$
$\%$ change in volume $=\left(\frac{{100}^2(x+y+z)+100(xy+yz+zx)+xyz}{{100}^2}\right)$
$\%$ change in volume $=(x+y+z)+\frac{xy+yz+zx}{100}+\frac{xyz}{{100}^2}$
Hence, option $A$ is correct.