Chapter 1.9, Problem 3R

To determine

Find the required number.

Answer to Problem 3R

  $\frac{2}{33}$

Explanation of Solution

Given:

Three and two − third times a number equals two − ninths

Concept Used:

Convert the phrase ‘Three and two − third times a number equals two − ninths ’into equation.

Calculation:

Let the number be $\text{N}$

Three and two − third times a number: $3\frac{2}{3}\text{N}$

Three and two − third times a number equals two − ninths: $3\frac{2}{3}\text{N}=\frac{2}{9}$

Equation: $3\frac{2}{3}\text{N}=\frac{2}{9}$

Solve : Convert the mixed fraction in the left into improper fraction : $\frac{11}{3}\text{N}=\frac{2}{9}$

Use the cross multiplication method to simplify the equation. $\frac{a}{b}=\frac{c}{d}\Rightarrow a·d=b·c$

  $\begin{array}{l}3\frac{2}{3}\text{N}=\frac{2}{9}\\ \frac{11}{3}\text{N}=\frac{2}{9}\\ 9·11\text{N}=2·3\text{[Using cross Multiplication Method] }\\ \text{99N}\text{=}6\\ \frac{99\text{N}}{99}=\frac{6}{99}\text{[}\text{Divide}\text{both}\text{sides}\text{by}99\text{]}\\ \text{N}=\frac{2}{33}\end{array}$

The required number is $\frac{2}{33}$

Thus, the required number is $\frac{2}{33}$