Exercise 1.3 Integers - NCERT Solution Class 7

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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

Video 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) \(–3 × (–6) × (–2) × (–1) = 36\)

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

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

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

 

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

Video 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.

Question 4

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

Solution

Video 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.

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

Video 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

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

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

Video 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.

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

Video 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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