Question

If each edge of a cuboid is increased by $50\%$, then the percentage increase in the surface area is

• A. $50$
• B. $125$
• C. $150$
• D. $300$

Answer 1

The correct option is B $125$

Old Surface Area $=2(lb+bh+lh)$
New dimensions $=(l+50\%l)\times (b+50\%b)\times (h+50\%h)$

$=\frac{3}{2}l\times \frac{3}{2}b\times \frac{3}{2}h$
New Surface Area $=2(\frac{3}{2}l\times \frac{3}{2}b+\frac{3}{2}h\times \frac{3}{2}b+\frac{3}{2}l\times \frac{3}{2}h)$
$=2\times \frac{9}{4}(lb+bh+lh)$
Change in surface area $=$ New S.A.$-$ Old S.A.
$=2\times \frac{5}{4}(lb+bh+lh)$
Percentage change in surface area $=\frac{2\times \frac{5}{4}(lb+bh+lh)}{2\times (lb+bh+lh)}\times 100$ $\%$
$=125$ $\%$