Practicing Success
If both length and breadth of a cuboid is increased by 50 percent, then by how much percent its height should be reduced so that its volume remains same? |
62.34% 55.55% 37.25% 48.75% |
55.55% |
Let the dimensions of the cuboid be l, b, h 50% = $\frac{1}{2}$ The changed dimensions $\frac{3}{2}$l, $\frac{3}{2}$b, h' Since volume remains the same, hence Volume ⇒ lbh = $\frac{3}{2}$l.$\frac{3}{2}$b.h' h' = $\frac{4}{9}$h Height decreased by = 9R - 4R = 5R Percentage decreased in height = $\frac{5}{9}$ x 100 = 55.55%
Alternate Solution: Using Ratio method:
Now volume: $\frac{2×2×H}{3×3×h}$ = $\frac{1}{1}$ $\frac{H}{h}$ = $\frac{9}{4}$ Height decreased by = 9R - 4R = 5R Percentage decreased in height = $\frac{5}{9}$ x 100 = 55.55% |