Ex.1.4 Q1 Integers Solution - NCERT Maths Class 7

Go back to  'Ex.1.4'

Question

Evaluate each of the following: –

\({\rm{a)}}-30{\rm{ }} \div {\rm{ }}10\) \({\rm{b)\; }}50 \div {\rm{(-5)}}\)
\({\rm{c)\; }}\left( {-36} \right) \div \left( {-9} \right)\) \({\rm{d)\; }}\left( {{\rm{-49}}} \right){\rm{  \div  }}\left( {{\rm{-49}}} \right)\)
\({\rm{e)}}\,\,13 \div \left[ {\left( {-2} \right) + 1} \right]\) \({\rm{f)\; }}0 \div \left( {-12} \right)\)
\({\rm{g)\; }}\left( {-31} \right) \div \left[ {\left( {-30} \right) + \left( {-1} \right)} \right]\) \({\rm{h)\;}}\left[ {\left( {-36} \right) \div 12} \right] \div 3\)
\({\rm{i)\; }}\left[ {\left( {{\rm{-6}}} \right){\rm{  + 5}}} \right]{\rm{  \div  }}\left[ {\left( {{\rm{-2}}} \right){\rm{  + 1}}} \right]\)  

 

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

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

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

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

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

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

  
Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school