Exercise 1.4 Integers - NCERT Solution Class 7

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Chapter 1 Ex.1.4 Question 1

Evaluate each of the following: –

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

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

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

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

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

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

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

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

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

 

Solution

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

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

Chapter 1 Ex.1.4 Question 2

Verify that \(a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)\) for each of the following values of \(a, b\) and \(c.\)

(a) \( a=12,b=-4,c=2\)

(b) \(a=\left( -10 \right),\,\,b=1,\,\,c=1\)

Solution

Video Solution

Steps:

Let,

\[a\div \left( b+c \right)\ne \left( a\div b \right)+\left( a\div c \right)\dots\left( 1 \right)\]

a) \(a = 12, b = –4, c = 2\)

Take L.H.S, \(a \div(b+c)\)

Putting the values of \(a, b\) and \(c,\) we get

\[\begin{align} a \div (b+c) &=12\div \left( -4+2 \right) \\ &=12\div \left( -2 \right) \\ &=12\times \frac{1}{\left( -2 \right)} \\ & =\frac{12}{\left( -2 \right)} \\ & =-6 \\ \end{align}\]

Now take R.H.S

\[\begin{align}&(a \div b)+(a \div c) \\ &\left[ 12\div (-4) \right]+\left( 12\div 2 \right) \\ \ \ \ &=\left[ 12\times \frac{(-1)}{4} \right]+\left( 12\times \frac{1}{2} \right) \\ &=\frac{\left( -12 \right)}{4}+\left( \frac{12}{2} \right) \\ & =-3+6 \\ & =3 \\ \end{align}\]

Putting the values of L.H.S and R.H.S in equation (\(1\)), we get

\[\begin{align} a(b+c)&\ne (a \div b)+(a \div c) \\  6&\ne 3 \\  \text{ L}\text{.H}\text{.S }&\ne \text{ R}\text{.H}\text{.S } \\\end{align}\]

Hence verified.

b) \(a =(-10),\;b=1,\;c=1\)

Take L.H.S, \(a (b + c)\)

Putting the values of \(a, b\) and \(c,\) we get

\[\begin{align}  &=(-10)(1+1) \\   &=(-10)(2) \\  &=(-10) \times \frac{1}{2} \\   &=-5 \\\end{align}\]

Take R.H.S,\((a \div b)+(a \div c)\)

Putting the values of \(a, b\) and \(c,\) we get

\[\begin{align}&= [(–10) ÷1] + [(–10) ÷1]\\ &= [(–10) × ] + [(–10) × ]\\ &= (–10) + (–10)\\ &= – 10 –10\\ &= –20\end{align}\]

Putting the values of L.H.S and R.H.S in equation (\(1\)), we get

\[\begin{align} a ÷ (b + c) &≠ (a ÷ b) + (a ÷ c)\\ –5& ≠ –20\end{align}\]

L.H.S \(≠\) R.H.S

 Hence verified

Chapter 1 Ex.1.4 Question 3

Fill in the blanks: –

a)  \(\begin{align}369 \div\underline{\qquad\qquad} = 369 \end{align}\)

b)  \(\begin{align}(–75) \div\underline{\qquad\qquad} = –1 \end{align}\)

c)  \(\begin{align}(–206) \div\underline{\qquad\qquad} = 1 \end{align}\)

d)  \(\begin{align}(–87)\div\underline{\qquad\qquad} = 87 \end{align}\)

e)  \(\begin{align}\underline{\qquad\qquad} \div1 = –87\end{align}\)

f)  \(\begin{align}\underline{\qquad\qquad} \div48 = –1\end{align}\)

g)  \(\begin{align}20\div \underline{\qquad\qquad} = –2\end{align}\)

h)  \(\begin{align}\underline{\qquad\qquad} \div (4) = –3\end{align}\)

 

Solution

Video Solution

Steps:

a)  \(\begin{align}369 \div\underline{\qquad1\qquad} = 369 \end{align}\)

b)  \(\begin{align}(–75) \div\underline{\qquad75\qquad} = –1 \end{align}\)

c)  \(\begin{align}(–206) \div\underline{\qquad(–206)\qquad} = 1 \end{align}\)

d)  \(\begin{align}(–87)\div\underline{\qquad(-1)\qquad} = 87 \end{align}\)

e)  \(\begin{align}\underline{\qquad(-87)\qquad} \div1 = –87\end{align}\)

f)  \(\begin{align}\underline{\qquad(-48)\qquad} \div48 = –1\end{align}\)

g)  \(\begin{align}20\div \underline{\qquad(-10)\qquad} = –2\end{align}\)

h)  \(\begin{align}\underline{\qquad(-12)\qquad} \div (4) = –3\end{align}\)

Chapter 1 Ex.1.4 Question 4

Write five pair of integers \((a, b)\) such that  \(a\div b = –3.\) One such pair is \((6, –2)\) because

\[6\div \left( -2 \right)=\left( 3 \right).\]

 

Solution

Video Solution

Steps:

i) \((–9) ÷ 3 = (–3)\)

ii) \(12 ÷ (–4) = (–3)\)

iii) \((–15) ÷ 5= (–3)\)

iv) \(18 ÷ (–6) = (–3)\)

v) \((–27) ÷ 9= (–3)\)

Chapter 1 Ex.1.4 Question 5

The temperature at \(12\) noon was \(10^\circ \rm{c}\)  above zero. If it decreases at the rate of \(2^\circ \rm{c}\) per hour until midnight, at what time would the temperature be \(8^\circ \rm{c}\) below zero? What would be the temperature at mid night?

 

Solution

Video Solution

Steps:

The temperature at \(12\) noon \(=10^\circ \rm{c}\) (given)

The temperature decreases \( 2^\circ \rm{c} = 1\) hour (given)

The temperature decreases \(1^\circ \rm{c} =\) \(1 \over 2\)hour

The temperature decreases \(18^\circ \rm{c}=\frac{1}{2} \times 18\)

(From \(10^\circ \rm{c}\) to \( 8^\circ \rm{c}\) below zero \(=9 \) hours)

Total time

\[\begin{align}&= 12\, \rm{noon}+9 \,\rm{hours}\\ &= 21 \,\rm{hours}\\&= 9\, \rm{pm}\end{align}\]

Thus, at \(9\) pm temperature would be \(8^\circ \rm{c}\) below zero.

ii) The temperature at \(12\) noon \(=10^\circ \rm{c}\)

The temperature decreases \(=2^\circ \rm{c}\) every hour

The temperature decreases in \(12\) hours \(=\;–2^\circ \rm{c} ×12 =24 ^\circ \rm{c}\) 

At midnight, the temperature will be \(=10^\circ \rm{c}+(–24)^\circ \rm{c} =–14^\circ \rm{c}\)

Therefore, the temperature at mid night will be \(14^\circ \rm{c}\) below \(0.\)

Chapter 1 Ex.1.4 Question 6

In a class test \((+3)\) marks are given for every correct answer and \((–2)\) marks are given for every incorrect answer and no mark for not attempting any question.

i) Radhika scored \(20\) marks. If she has got \(12\) correct answers, how many questions has she attempted incorrectly?

ii) Mohini scores \(–5\) marks in this test, though she has got \(7\) correct answers. How many questions has she attempted incorrectly?

 

Solution

Video Solution

Steps:

Given: –

Marks given for every correct answer \(=3\) marks

Marks given for every incorrect answer\( =\;–2\)

Marks for not attempting any question  \(=0\)

i)Total score of Radhika \(= 20\) marks

 No. of correct answers given by Radhika \(=12 \) answers

Marks obtained for correct answers

\[\begin{align}&=\text{ 3  }\!\!\times\!\!\text{  12 } \\  &= \text{36 marks}  \end{align}\]

 Marks obtained for incorrect answers \(=\) Total score \(-\) Marks obtained for \(12\) correct answers 

\[\begin{align}& =2036 \\ & =16\,\text{marks} \end{align}\]

Marks obtained for every incorrect answer \( =\;–2 \) marks

Thus, number of incorrect answers

\[\begin{align}&= -16 \div (-2) \\ &=8 \end{align}\]

Therefore, she attempted \(8\) questions wrongly.

ii) Total score of Mohini \(=\;–5\)

No. of correct answers given by Mohini \(=7\)

Marks obtained for correct answers

\[\begin{align}&=\text{7  }\!\!\times\!\!\text{  3 } \\  & = \text{21}\,\text{marks}  \end{align}\]

 Marks obtained for incorrect answers \(=\) Total score \(-\)  Marks obtained for correct answers

\[\begin{align} & =-5- 21 \\ & =-26\end{align}\]

Marks obtained for every incorrect answer \(=\; –2\) marks

Thus, number of incorrect answers \(= (–26) ÷ (–2) = 13\)

Therefore, she attempted \(13\) questions wrongly.

Chapter 1 Ex.1.4 Question 7

An elevator descends into a mine shaft at the rate of \(6 \text{ m/min}\). If the descent starts from \(10\;\rm{m}\) above the ground level, how long will it take to reach \(-350\;\rm{m}\)?

 

Solution

Video Solution

Steps:

  Total distance covered by an elevator 

\[\begin{align}& =10-(-350)\text{m} \\  & =10+\left( 350 \right) \\  & =360\,\text{m}  \end{align}\]

So, time taken to cover a distance of \(6 \rm{m}=1\) min

Therefore, time taken to cover \(360 \text{ m}\)

\[\begin{align}&=\text{ 360  }\!\!\div\!\!\text{  6 } \\  & =\text{ 60}\,\text{minutes}  \end{align}\]

(\(1\) hour \(= 60\) minutes)

Thus, the elevator will reach \(–350\;\rm{m}\) from \(10\;\rm{m}\) in \(1\) hour.

  
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