Question

I thought of a number, divided it by $1\frac{1}{2}$, then added $2\frac{1}{2}$ to the result, then multiplied the sum by $1\frac{2}{3}$ . Then I subtracted $1\frac{2}{3}$ and got the result $2\frac{5}{6}$. What was my number?

Answer 1

Step-1: Forming an expression:

A number when divided it by $1\frac{1}{2}$, then added $2\frac{1}{2}$ to the result, then multiplied the sum by $1\frac{2}{3}$ . Then subtracted $1\frac{2}{3}$ and got the result $2\frac{5}{6}$.

Let the assumed number be $x$.

$\Rightarrow \left\{\left[\left(x÷1\frac{1}{2}\right)+2\frac{1}{2}\right]\times 1\frac{2}{3}\right\}-1\frac{2}{3}=2\frac{5}{6}$

Step-2: Find the value of $x$:

Consider, $\left\{\left[\left(x÷1\frac{1}{2}\right)+2\frac{1}{2}\right]\times 1\frac{2}{3}\right\}-1\frac{2}{3}=2\frac{5}{6}$.

Let us convert the mixed fraction to a proper fraction.

$\Rightarrow \left\{\left[\left(x÷\frac{3}{2}\right)+\frac{5}{2}\right]\times \frac{5}{3}\right\}-\frac{5}{3}=\frac{17}{6}$

Now add $\frac{5}{3}$ on both sides of the above equation.

$$\begin{gathered} \Rightarrow \left\{\left[\left(x÷\frac{3}{2}\right)+\frac{5}{2}\right]\times \frac{5}{3}\right\}-\cancel{\frac{5}{3}}+\cancel{\frac{5}{3}}=\frac{17}{6}+\frac{5}{3} \\ \Rightarrow \left[\left(x÷\frac{3}{2}\right)+\frac{5}{2}\right]\times \frac{5}{3}=\frac{17+10}{6} \\ \Rightarrow \left[\left(x÷\frac{3}{2}\right)+\frac{5}{2}\right]\times \frac{5}{3}=\frac{27}{6} \end{gathered}$$

Now divide $\frac{5}{3}$ on both sides of the above equation.

$$\begin{gathered} \left[\left(x÷\frac{3}{2}\right)+\frac{5}{2}\right]\times \frac{5}{3}÷\frac{5}{3}=\frac{27}{6}÷\frac{5}{3} \\ \Rightarrow \left[\left(x÷\frac{3}{2}\right)+\frac{5}{2}\right]\times \frac{\cancel{5}}{\cancel{3}}\times \frac{\cancel{3}}{\cancel{5}}=\frac{27}{\cancel{6}}\times \frac{\cancel{3}}{5} \\ \Rightarrow \left(x÷\frac{3}{2}\right)+\frac{5}{2}=\frac{27}{2\times 5} \\ \Rightarrow \left(x÷\frac{3}{2}\right)+\frac{5}{2}=\frac{27}{10} \end{gathered}$$

Let us subtract $\frac{5}{2}$ on both sides of the above equation.

$$\begin{gathered} \left(x÷\frac{3}{2}\right)+\cancel{\frac{5}{2}}-\cancel{\frac{5}{2}}=\frac{27}{10}-\frac{5}{2} \\ \Rightarrow x÷\frac{3}{2}=\frac{27-25}{10} \\ \Rightarrow x÷\frac{3}{2}=\frac{2}{10} \\ \Rightarrow x÷\frac{3}{2}=\frac{1}{5} \end{gathered}$$

Now multiply $\frac{3}{2}$ on both sides of the above equation.

$\begin{array}{rcl}x÷\frac{3}{2}\times \frac{3}{2}& =& \frac{1}{5}\times \frac{3}{2}\\ x\times \frac{\cancel{2}}{\cancel{3}}\times \frac{\cancel{3}}{\cancel{2}}& =& \frac{3}{10}\\ x& =& \frac{3}{10}\end{array}$

Therefore, the required value of $x$ as $\frac{3}{10}$.