Ex.1.4 Q1 Integers Solution - NCERT Maths Class 7

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Question

Evaluate each of the following: –

$${\rm{a)}}-30{\rm{ }} \div {\rm{ }}10$$

$${\rm{b)\; }}50 \div {\rm{(-5)}}$$

$${\rm{c)\; }}\left( {-36} \right) \div \left( {-9} \right)$$

$${\rm{d)\; }}\left( {{\rm{-49}}} \right){\rm{ \div }}\left( {{\rm{-49}}} \right)$$

$${\rm{e)}}\,\,13 \div \left[ {\left( {-2} \right) + 1} \right]$$

$${\rm{f)\; }}0 \div \left( {-12} \right)$$

$${\rm{g)\; }}\left( {-31} \right) \div \left[ {\left( {-30} \right) + \left( {-1} \right)} \right]$$

$${\rm{h)\;}}\left[ {\left( {-36} \right) \div 12} \right] \div 3$$

$${\rm{i)\; }}\left[ {\left( {{\rm{-6}}} \right){\rm{ + 5}}} \right]{\rm{ \div }}\left[ {\left( {{\rm{-2}}} \right){\rm{ + 1}}} \right]$$

Video Solution
Integers
Ex 1.4 | Question 1

Text Solution

Steps:

a) $$-30{\rm{ }} \div {\rm{ }}10$$

\begin{align}\left( { - 30} \right) \div 10 &= \left( { - 30} \right) \times \frac{1}{{10}}\\& = \frac{{ - 30}}{{10}}\\& = - 3\end{align}

b) $$50 \div {\rm{(-5)}}$$

\begin{align}50 \div \left( { - 5} \right) &= 50 \times \frac{1}{{\left( { - 5} \right)}}\\ &= \frac{{50}}{{ - 5}}\\& = - 10\end{align}

c) $$\left( {-36} \right) \div \left( {-9} \right)$$

\begin{align}\left( { - 36} \right) \div \left( { - 9} \right) &= \left( { - 36} \right) \times \frac{1}{{\left( { - 9} \right)}}\\ &= \frac{{36}}{9}\\ &= 4\end{align}

d) $$\left( {{\rm{-49}}} \right){\rm{ \div }}\left( {{\rm{-49}}} \right)$$

\begin{align}{\rm{(}} - {\rm{49)}}\,{\rm{ \div }}\,{\rm{49 }}&=\,{\rm{(}} - {\rm{49)}}\,{\rm{ \times }}\,\frac{{\rm{1}}}{{{\rm{(49)}}}}{\rm{ }}\\&=\frac{{ - {\rm{49}}}}{{{\rm{49}}}}\\&= - {\rm{1 }}\end{align}

e) $$13 \div \left[ {\left( {-2} \right) + 1} \right]$$

\begin{align}13 \div {\rm{[(}} - {\rm{2)}}\,{\rm{ + }}\,{\rm{1]}} &= 13 \div \left[ { - 2 + 1} \right]\\ &= 13 \div \left[ { - 1} \right]\\ &= 13 \times \frac{1}{{\left( { - 1} \right)}}\\ &= - 13\end{align}

f) $$0 \div \left( {-12} \right)$$

\begin{align}0 \div ( - 12)&= 0 \times \frac{1}{{\left( { - 12} \right)}}\\ &= \frac{0}{{ - 12}}\\ &= 0\\ &\quad \quad \left[ {0{\rm{ }} \div {\rm{ Number}} = {\rm{ }}0} \right]\end{align}

g) $$\left( {-31} \right) \div \left[ {\left( {-30} \right) + \left( {-1} \right)} \right]$$

\begin{align}\left( {-31} \right)&\div \left[ {\left( {-30} \right) + \left( {-1} \right)} \right]\\&=\left( {-31} \right) \div \left[ {-30-1} \right]\\ &= \left( {-31} \right) \div \left( {-31} \right)\\ &= \left( { - 31} \right) \times \;\frac{1}{{\left( { - 31} \right)}}\\ &= \frac{{ - 31}}{{ - 31}}\;\\&= 1\end{align}

h) $$\left[ {\left( {-36} \right) \div 12} \right] \div 3$$

\begin{align}{\rm{[(}} - {\rm{36)}}\,{\rm{ \div 12]}}& \div 3 \\&= \left[ {( - 36) \times \frac{1}{{12}}} \right] \div 3\\ &= \left[ {\left( {\frac{{ - 36}}{{12}}} \right)} \right] \times \frac{1}{3}\\&= \left( { - 3} \right) \times \frac{1}{3}\\&= - 1\\\end{align}

i) $$\left[ {\left( {{\rm{-6}}} \right){\rm{ + 5}}} \right]{\rm{ \div }}\left[ {\left( {{\rm{-2}}} \right){\rm{ + 1}}} \right]$$

\begin{align}\left[ \left( -6 \right)\text{ +5} \right]\!\!\div\!\!\text{ }\left[ \left( -2 \right)\text{ +1} \right]&=-1\div -1\\&=\frac{-1}{-1}=1\end{align}

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