Chapter A.8, Problem 48AYU

To determine

To find: Required amount of pure cement to make desired mixture.

Answer to Problem 48AYU

  $5$ pounds.

Explanation of Solution

Given information:

Some amount of pure cement is added to $20$ pound of $25\%$ cement and $75\%$ sand mixture to produce a mixture that is $40\%$ cement.

Calculation:

As per the given information −

Some amount of pure cement is added to $20$ pound of $25\%$ cement and $75\%$ sand mixture to produce a mixture that is $40\%$ cement.

Therefore, there are $5$ pounds of pure cement and $15$ pounds of sand in original mixture.

Let us suppose $x$ pounds of pure cement is added to $20$ pounds of economy cement.

So, total amount of new mixture will be $(20+x)$ pounds out of which $40\%$ is pure cement.

Amount of sand remains same $15$ pounds which is $60\%$ of new mixture.

  $\begin{array}{l}60\%\text{ of }\left(20+x\right)=15\\ \Rightarrow \left(20+x\right)0.6=15\\ \Rightarrow 12+0.6x=15\\ \Rightarrow x=\frac{3}{0.6}\\ \Rightarrow x=5\end{array}$

Hence, required amount of pure cement to make $40\%$ cement mixture is $5$ pounds.