# 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

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

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

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

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

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

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

\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

\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 $$=7$$

\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

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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