# Exercise 1.3 Integers - NCERT Solution Class 7

## Chapter 1 Ex.1.3 Question 1

Find each of the following products:

a) $$3 \times (–1)$$

b) $$(–1) \times 225$$

c) $$(–21) \times(–30)$$

d) $$(–316) \times (–1)$$

e) $$(15) \times 0 \times (–18)$$

f) $$(–12) \times (–11) \times 10$$

g) $$9 \times(–3) \times (–6)$$

h) $$(–18) \times (–5) \times (–4)$$

i) $$(–1) \times (–2) \times(–3)\times 4$$

j) $$(–3) \times (–6) \times (–2) \times (–1)$$

### Solution

Steps:

a) $$3 × (–1) = \,– 3$$

b) $$–1 × 225 =\, –225$$

c) $$–21 × (–30) = 630$$

d) $$–316 × (–1) = 316$$

e) $$–15 × 0 × (–18) = 0$$

f) $$–12 × (–11) × (–10) = 1320$$

g) $$9 × (–3) × (–6) = 162$$

h) $$–18 × (–5) × (–4) = \,–360$$

i) $$–1 × (–2) × (–3) × 4 =\, –24$$

j) $$–3 × (–6) × (–2) × (–1) = 36$$

## Chapter 1 Ex.1.3 Question 2

Verify the following:

(a)

$$18 \times [ 7 +(-3) ] =[ 18 \times 7]+[18 \times (-3)]$$

(b)

\begin{align}&\left( -21 \right) \!\times\! \left[ {\left( -4 \right)\!+\! \left( -6 \right)} \right] \\&= \!\left[ {\left( -21 \right) \!\times\! \left( -4 \right)} \right] \!+\! \left[ {\left( -21 \right) \!\times\! \left( -6 \right)} \right]\end{align}

### Solution

Steps:

a)
$$18 \!\times\![ 7 + (-3)]= [ 18\times 7]\!+\! [ 18 \times(-3)]$$

L.H.S

\begin{align} &= 18 \times [ {{\rm{ 7 + }}\left( {{\rm{-3}}} \right)}]\left(\text{Open brackets} \right)\\&= {\rm{ }}18 \times {\rm{ }}4 \\&= 72\end{align}

R.H.S

\begin{align} &=[ 18\!\times\!7] \!+\![18 \!\times\!(-3)](\text{Open brackets})\\&= [18 \times7 ] + [ 18 \times (-3)]\\&= 126 + \left( {-54} \right) \\&= 72\end{align}

\begin{align}18 \!\times\![7 +( -3)] &\!=\! [18 \times 7] \!+\! [18 \!\times \!(-3)]\\72 &\!=\! 72\\\rm{L.H.S} &\!=\! \rm{R.H.S}\text{(Hence verified)}\end{align}

b)
\begin{align}&{\rm{}}(-21) \times [ (-4) + ( -6)] \\&= [(-21) \times(-4)] + [(-21) \times(-6)]\end{align}

L.H.S
\begin{align} &= (-21)\!\times\![(-4) \!+\! ( -6)](\text{Open brackets})\\&=[-21 \times(- 4 - 6)]\\&= - 21 \times - 10 \\&= 210\end{align}

R.H.S
\begin{align}&=[(-21) \times(-4)] +[(-21)\times(-6)]\\&= {\rm{ }}\left[ {-21 \times -4} \right]{\rm{ }} + {\rm{ }}\left[ {-21{\rm{ }} \times {\rm{ }}-6} \right]\\&= {\rm{ }}\left[ {84} \right]{\rm{ }} + {\rm{ }}\left[ {126} \right]\\& = 210\end{align}

\begin{align}\left( {-21} \right) \times \left[ {\left( {-4} \right) + \left( {-6} \right)} \right] &= \,\left[ {\left( {-21} \right) \times \left( {-4} \right)} \right] + \left[ {\left( {-21} \right){\rm{ }} \times {\rm{ }}\left( {-6} \right)} \right]\\210\, &= \,210\\ {\rm{L.H.S}}\,{\rm{ }}&= {\rm{R.H.S}}\text{(Hence verified.)}\end{align}

## Chapter 1 Ex.1.3 Question 3

i) For any integer $$a,$$ what is $$\left( {-1} \right) \times a$$ equal to?

ii) Determine the integer whose product with $$(–1)$$ is

(a) $$-22$$

(b) $$37$$

(c) $$0$$

### Solution

Steps:

(i) $$\left( {{\rm{-1}}} \right){{ a}}\,{{ = }}\,{{-a}}$$ where $$a$$ is any integer

(ii) a) $${{? \times }}\left( {{\rm{-1}}} \right){\rm{ = -22}}$$

Let $$x$$ be the required integer

\begin{align}x \times \left( {-1} \right) &= -22\qquad \dots\left( 1 \right)\\-x &= -22\\x &= 22\end{align}

Putting the value of $$x$$ in equation ($$1$$), we get

\begin{align}x \times \left( {-1} \right) = -22\\22 \times \left( {-1} \right) = -22\end{align}

b) $$? × (–1) = 37$$

Let $$y$$ be the required integer

\begin{align}y \times \left( {-1} \right)&= 37\dots(1)\\-y &= {\rm{ }}37\\y &= -{\rm{ }}37 \end{align}

Putting the value of $$y$$ in equation (1), we get

$–37 × (–1) = 37$

c) $$? × (–1) =0$$

Let $$z$$ be the required integer

\begin{align}z \times \left( {-1} \right) = 0 & \left( 1 \right)\\-z = 0\\z = 0\end{align}

Putting the value of z in equation ($$1$$), we get

\begin{align}0 × (–1) = 0\\ ? × (–1) =0 \\0 × (–1) =0 \end{align}

(Multiplying any number by $$0$$, we get the product $$0$$)

Hence (c) $$0$$ is required integer.

## Chapter 1 Ex.1.3 Question 4

Starting from $$(–1) × 5,$$ write various products showing some pattern to show $$(–1) × (–1) =1.$$

### Solution

Steps:

\begin{align}\left( {-1} \right) \times {\rm{ }}5 &= -5\\\left( {-1} \right) \times 4 &= -4\\\left( {-1} \right) \times 3 &= -3\\\left( {-1} \right) \times 2 &= -2\\\left( {-1} \right) \times 1 &= -1\\\left( {-1} \right) \times 0 &= 0\\\left( {-1} \right) \times -&1 = {\rm{ }}1\end{align}

Thus, it is clear that when we multiply a negative integer by a positive integer the result is a negative integer whereas if we multiply a negative integer by a negative integer the result is always a positive integer.

## Chapter 1 Ex.1.3 Question 5

Find the product, using suitable properties: –

a) $$26 × (–48) + (–48) × (–36)$$

b) $$8× 53 × (–125)$$

c) $$15× (–25) × ( –4) × (–10)$$

d) $$(–41) × 102$$

e) $$625 ×–35 × (–625) × 65$$

f) $$7 × (50–2)$$

g) $$(–17) × (–29)$$

h) $$(–57) × (–19) + 57$$

### Solution

Steps:

(a) $${{26 \times }}\left( {{\rm{-48}}} \right)\,{\rm{ + }}\left( {{\rm{-48}}} \right){\rm{ \times }}\left( {{\rm{-36}}} \right)$$

Using distributive property, we get

\begin{align}{{{(a}}\,{ \times b)}}\,\,{{ + }}\,\,{{(b}}\,{{ \times }}\,{{c)}}\,&= \,{{b \times }}\,{{(a + c)}}\\ \,\,&=\,{{(}} - {{48)}}\,\,{{ \times }}\,{{[26}}\,{{ + }}\,{{(}} - {{36)]}}\\ \,\, &= \,{{(}} - {{48)}}\,\,{{ \times }}\,\,{{[26}} - {{36]}}\\\,\,&=\,{{(}} - {{48)}}\,\,{{ \times }}\,{{ [(}} - {{10)]}}\\&=\,{{480}}\end{align}

(b) $${{8 \times 53 \times }}\left( {{\rm{-125}}} \right)$$

Using associative property, we get

\begin{align}({a \times b}) \times c &= a \times ( {b \times c})\\&= 8 \times 53 \times \left( {-125} \right)\\ &= 8 \times -6625\\ &= -53000\end{align}

(c) $${{}}\,\,\,15 \times \left( {-25} \right) \times \left( {-4} \right) \times \left( {-10} \right)$$

Using associative property, we get

\begin{align} &= 15 \times \left[ {\left( {-25} \right) \times \left( {-4} \right) \times \left( {-10} \right)} \right]\\ &= 15 \times \left[ {100 \times \left( {-10} \right)} \right]\\ &= 15 \times \left[ {1000} \right]\\ &= -15000\end{align}

(d) $${\rm{}}\left( {-41} \right){\rm{ }} \times {\rm{ }}102$$

Using distributive law, we get

\begin{align}&=\,{{(}} - {{41)}}\,{{ \times }}\,{{(100}}\,{{ + }}\,{{2)}}\dots\,{{[a}}\,{{ \times }}\,{{(b}}\,{{ + }}\,{{c)}}]=\,{{(a}}\,{{ \times }}\,{{b}}\,{{ + }}\,{{a}}\,{{ \times }}\,{{c)]}}\\&={{{(}} - {{41)}}\,{{ \times }}\,{{100}}\,{{ + }}\,{{(}}\, - {{41) \times }}\,{{2}}}\\&= - {{4100}}\, - {{82}}\\&=\, - {{4182}}\end{align}

(e) $${{625}}\,{{ \times }}\left( {{{-35}}} \right){{ + }}\left( {{{-}}\,{{625}}} \right){{ \times }}\,{{65}}$$

Using distributive property, we get

\begin{align} &=\,{625}\,\times[{ (}-{35)}\,{+}\,{(}-{65) }]\,{ }\dots[{ a\times b}\,{+}\,{a\times c}\,{=}\,{a(b+c) }] \\ &=\,{625}\times[\,-{35}\,-{65 }] \\ &=\,{625}\times[-{100 }] \\ &={-62500} \\\end{align}

(f) $$7\times \left( 50-2 \right)$$

Using distributive property, we get

\begin{align}&=7\times { }\,{(50}-{2)}\,{ }\dots[{ a\times}\,{ }{ }\,{(b}-{c)}\,{=}\,{a }\times{ b}-{a }\times{ c }]{ } \\ &=7\times { 50}-{7 }\times { 2} \\ &=350-{14} \\ &=336\end{align}

(g) $$\left( \text{-17} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{-29} \right)$$

Using distributive property, we get

\begin{align} &= \,( - 17)\, \times \,\,[( - 30)\, + 1]\dots[a\, \times \,(b\, + \,c)={a\times b}+{a \times c}]\\ &= \,( - 17)\, \times \,( - 30)\, + \,( - 17)\, \times 1\\ &= \,510\, + ( - 17)\\ &= \,510 - 17\\ &=\,493\end{align}

(h)$${{}}\left( {-57} \right) \times \left( {-19} \right) + 57$$

Using distributive property, we get

\begin{align}&={{{ }} \,\left( { - {{57}}} \right){{ \times (}} - {{19) + 57\,\, 1[a \times b}}\,{{ + }}\,{{a \times c}}\,\,={{ }}\,{{a \times (b + c)]}}}\\&={{{ }}\,{{57 \times 19 + 57 \times 1}}}\\&={{{ }}\,{{57 \times (19 + 1)}}}\\&={{{ }}\,{{57 \times 20}}}\\&={{{ 1140}}}\end{align}

## Chapter 1 Ex.1.3 Question 6

A certain freezing process requires that room temperature be lowered from $$40^\circ \rm C$$  at the rate of $$5^\circ{\rm C}$$  every hour. What will be the temperature $$10$$ hours after the process begins?

### Solution

Steps:

Present temperature of room $$=$$ $$40^\circ \rm C$$

Decrease in temperature after every hour $$= 5^\circ{\rm C}$$

Temperature of room after $$10$$ hours

\begin{align}&=40^{\circ} \mathrm{c}+10 \times(-5)^{\circ} \mathrm{c}\\&=40^{\circ} \mathrm{c}-50^{\circ} \mathrm{c}\\&=-10^{\circ} \mathrm{c}\end{align}

Thus, the room temperature after $$10$$ hours is $$–10^\circ{\rm C},$$ after the process begins.

## Chapter 1 Ex.1.3 Question 7

In a class test containing $$10$$ questions, $$5$$ marks are awarded for every correct answer and $$(–2)$$ marks are awarded for every incorrect answer and $$0$$ for question not attempted.

i) Mohan gets $$4$$ correct and $$6$$ incorrect answers. What is his score?

ii) Reshma gets $$5$$ correct and $$5$$ incorrect answers. What is her score?

iii) Heena gets $$2$$ correct and $$5$$ incorrect answers out of $$7$$ questions she attempts. What is her score?

### Solution

Steps:

Given: –

Total number of questions $$=$$ $$10$$

Marks awarded for every correct answer $$=$$ $$5$$

Marks awarded for every incorrect answer $$=$$ $$(–2)$$

Marks for not attempted question $$=$$ $$0$$

i) Marks obtained by Mohan by $$4$$ correct answers $$= 4 × 5 = 20$$ marks

Marks for 6 incorrect answers$$= 6× (–2) = –12$$

Total score of Mohan = Marks for correct answers + marks for incorrect answers

\begin{align}&= 20 + \left( {-12} \right)\\&= 20-12\\&= 8\,{\text{marks}}\end{align}

Thus, Mohan gets $$8$$ marks in the class test.

ii) Marks obtained by Reshma for $$5$$ correct answers $$= 5×5 = 25$$ maks

Marks for 5 incorrect answers $$= 5× (–2) = \,–10$$marks

$$\therefore$$ Total score of Reshma $$=$$ Marks for correct answers $$+$$ marks for incorrect answers

\begin{align} &= 25 + \left( {-10} \right)\\ &= 25-10\\ &= 15\,{\rm{marks}}\end{align}

(iii) Marks obtained by Heena for $$2$$ correct answer $$=2\times 5 = 10$$ marks

Marks for $$5$$ incorrect answer $$=5\times (-2)=-10$$ marks

Marks for not attempted question $$=3\times0=0$$ marks

$$\therefore$$ Total score of Heena $$=$$ Marks for correct answers $$+$$ marks for incorrect answers$$+$$Marks for not attempted questions.

\begin{align}&= 25 + \left( {-10} \right) + 0\\&= 25-10 + 0\\ &= 15\,{\rm{marks}}\end{align}

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