Q40. - Geometry

Question

If the length and width of a rectangular solid are each decreased by 20%, by what percent must the height be increased for the volume to remain unchanged? Give your answer to the nearest whole percent.

Step-by-Step Solution

Answer

The height must be increased by$56.25\%$  for the volume to remain unchanged.

Step 1. Given information.

The length and width of a rectangular solid area each decreases by 20%.

Step 2. Write the concept.

To find the increased height percent, equal the volumes before decreasing the length-width and after decreasing the length width.

Step 3. Determine the value.

Let the height increased x%.

$\begin{array}{c}{v}_a=v_b\\ b_a{h}_a=b_b{h}_b\end{array}$

 

Substitute all the values,

$\begin{array}{c}(l-\frac{20l}{100})(w-\frac{20w}{100})(h+\frac{xh}{100})=lwh\text{ [substitute]}\\ \left(\frac{80l}{100}\right)\left(\frac{80w}{100}\right)\left(\frac{(100+x)h}{100}\right)=lwh\text{ [simplify]}\\ \left(\frac{80}{100}\right)\left(\frac{80}{100}\right)\left(\frac{100+x}{100}\right)=1\end{array}$

 

Then,

$\begin{array}{c}\frac{100+x}{100}=\frac{10000}{6400}\text{ [cross multiplication]}\\ \frac{100+x}{100}=1.5625\text{ [divide]}\\ 100+x=156.25\text{ [multiply]}\\ x=56.25\%\text{ [subtract]}\end{array}$

 

Thus, the height must be increased by $56.25\%$ for the volume to remain unchanged.