# Ex.2.3 Q8 fractions-and-decimals Solutions-Ncert Maths Class 7

## Question

a) i) Provide the number in the box \( □\), such that \(\begin{align} \frac{2}{3}\times \square =\frac{10}{30}\end{align} \)

ii) The simplest form of the number obtained in \( □\) is ―.

b) i) Provide the number in the box \( □\), such that \(\begin{align} \frac{3}{5}\times \,\square =\frac{24}{75}\end{align} \)

ii) The simplest form of the number obtained in □ is ―.

## Text Solution

**Steps:**

(a)

**What is known?**

Equations.

**What is unknown?**

Value of the box.

**Reasoning:**

To make L.H.S. \(=\) R.H.S. we have to multiple numerator by \(5\) and denominator by \(10\).

i)\(\begin{align} \frac{2}{3}\times \square =\frac{10}{30}\end{align} \)

\[=\frac{2}{3}\times \frac{5}{10}=\frac{10}{30}\]

Therefore, the number in the box □, such that \(\begin{align} \frac{2}{3} \times \square = \frac{{10}}{{30}}{\text{ is }}\frac{5}{{10}}\end{align} \)

ii) The simplest form of the number obtained in \(\begin{align} \frac{5}{{10}}{\text{ is }}\frac{1}{2}\end{align} \) .

(b)

**What is known?**

Equations.

**What is unknown?**

Value of the box.

**Reasoning:**

To make L.H.S. \(=\) R.H.S. we have to multiple numerator by \(8\) and denominator by \(15\).

i) \(\begin{align} \frac{3}{5}\times \,\square =\frac{24}{75}\end{align} \)

\[\begin{align}=\frac{3}{5}\times \frac{8}{15}=\frac{24}{75} \\\end{align}\]

Therefore, the number in the box \( □\), such that \(\begin{align} \frac{3}{5} × □ = \frac{{24}}{{75}} \rm \,is\, \frac{8}{{15}}\end{align} \)

ii) As \(\begin{align} \frac{8}{{15}}\end{align} \) can’t be simplified further. Therefore, its simplest form is \(\begin{align} \frac{8}{{15}}\end{align} \) .