NCERT Solutions for Class 7 Maths Chapter 1 Integers

1. Write down a pair of integers whose:
(a) sum is –7 (b) difference is –10 (c) sum is 0

Answer: A pair of integers whose:

(a) sum is –7 : $-8\mathrm{\&}1$

(b) difference is –10 : $-8\mathrm{\&}2:-8-2=-10$

(c) sum is 0 : $-9\mathrm{\&}9$

2. (a) Write a pair of negative integers whose difference gives 8.
(b) Write a negative integer and a positive integer whose sum is –5.
(c) Write a negative integer and a positive integer whose difference is –3.

Answer:

(a) A pair of negative integers whose difference gives 8 :

$-8\mathrm{\&}-16:$

$=-8-(-16)=-8+16=8$

(b) A negative integer and a positive integer whose sum is –5:

$-8\mathrm{\&}3$ :

$-8+3=-5$

(c) A negative integer and a positive integer whose difference is –3:

$-1\mathrm{\&}2$

$-1-(2)=-1-2=-3$

3. In a quiz, team A scored – 40, 10, 0 and team B scored 10, 0, – 40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?

Answer: Given,

Team A scored: $-40,10,0$

$\therefore$ Team A's total = $-40+10+0=-30$

Team B scored: $10,0,-40$

$\therefore$ Team B's total = $10+0+(-40)=10-40=-30$

Therefore, both the teams scored the same.

Yes, we can add integers in any order.

4. Fill in the blanks to make the following statements true:
(i) (–5) + (– 8) = (– 8) + (............)
(ii) –53 + ............ = –53
(iii) 17 + ............ = 0
(iv) [13 + (– 12)] + (............) = 13 + [(–12) + (–7)]
(v) (– 4) + [15 + (–3)] = [– 4 + 15] + ............

Answer:

(i) (–5) + (– 8) = (– 8) + ( $\underset{―}{-5}$ )

By the commutative property $a+b=b+a$

(ii) –53 + $\underset{―}{0}$ = –53

$0$ is the additive identity. The number added to $0$ gives the same number.

(iii) 17 + $\underset{―}{(-17)}$ = 0

By additive identity property.

(iv) [13 + (– 12)] + ( $\underset{―}{-7}$ ) = 13 + [(–12) + (–7)]

By associative property $(a+b)+c=a+(b+c)$

(v) (– 4) + [15 + (–3)] = [– 4 + 15] + $\underset{―}{(-3)}$

By associative property $(a+b)+c=a+(b+c)$

1. Find each of the following products:

(a) 3 × (–1)
(b) (–1) × 225
(c) (–21) × (–30)
(d) (–316) × (–1)
(e) (–15) × 0 × (–18)

Answer:

(a) $3\times (-1)=-3$

(b) $(-1)\times 225=-225$

(c) $(-21)\times (-30)=(21\times 30)=630$

The product of two negative integers is a positive integer

(d) $(-316)\times (-1)=(316\times 1)=316$

(e) $(-15)\times 0\times (-18)=[(-15)\times 0]\times (-18)$
$=0\times (-18)=0$

1. Find each of the following products:

(f) (–12) × (–11) × (10)
(g) 9 × (–3) × (– 6)
(h) (–18) × (–5) × (– 4)
(i) (–1) × (–2) × (–3) × 4
(j) (–3) × (–6) × (–2) × (–1)

Answer:

The product of an even number of negative integers is an even integer.

(f) $(-12)\times (-11)\times (10)=[-(12\times 11)]\times 10$
$=(-132)\times 10=-1320$

(g) $9\times (-3)\times (-6)=9\times [(-3)\times (-6)]$
$=9\times 18=162$

(h) $(-18)\times (-5)\times (-4)=(-18)\times [(-5)\times (-4)]$
$=(-18)\times 20=-360$

(i) $(-1)\times (-2)\times (-3)\times 4=[(-1)\times (-2)]\times [(-3)\times 4]$
$=2\times (-12)=-24$

(j) $(-3)\times (-6)\times (-2)\times (-1)=[(-3)\times (-6)]\times [(-2)\times (-1)]$
$=18\times 2=36$

2 (a). Verify the following:

18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]

Answer:

L.H.S = $18\times [7+(-3)]=18\times 4=72$

R.H.S = $[18\times 7]+[18\times (-3)]=126+(-54)=72$

L.H.S = R.H.S

Hence, verified.

2 (b). Verify the following:

[(–21) × [(– 4) + (– 6)] = [(–21) × (– 4)] + [(–21) × (– 6)]

Answer:

L.H.S =

$(-21)\times [(-4)+(-6)]=(-21)\times (-10)=210$

R.H.S

$[(-21)\times (-4)]+[(-21)\times (-6)]$
$=(21\times 4)+(21\times 6)$
$=84+126=210$

Therefore, L.H.S = R.H.S

Hence verified.

3. (i) For any integer a, what is (–1) × a equal to?
(ii) Determine the integer whose product with (–1) is
(a) –22 (b) 37 (c) 0

Answer:

(i) For any integer a, $(-1)\times a=-a$ , i.e, the additive inverse of the given integer.

(ii) The integer whose product with $(-1)$ gives the following are:
$$\begin{gathered} (a)-22:22 \\ (b)37:-37 \\ (c)0:0 \end{gathered}$$

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

Answer:

Given,

$-1\times 5=-5$
$-1\times 4=-4=-5+1$
$-1\times 3=-3=-4+1$
$-1\times 2=2=-3+1$
$-1\times 1=-1=-2+1$
$-1\times 0=0=-1+1$

Therefore,

$-1\times (-1)=0+1=1$

1. Evaluate each of the following:
(a) (–30) ÷ 10
(b) 50 ÷ (–5)
(c) (–36) ÷ (–9)
(d) (– 49) ÷ (49)
(e) 13 ÷ [(–2) + 1]
(f ) 0 ÷ (–12)
(g) (–31) ÷ [(–30) + (–1)]
(h) [(–36) ÷ 12] ÷ 3
(i) [(– 6) + 5)] ÷ [(–2) + 1]

Answer: Points to keep in mind:

When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put a minus sign (–) before the quotient.

When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put a minus sign (–) before the quotient.

When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put a positive sign (+).

(a) $(-30)÷10=-\left(\frac{30}{10}\right)=-3$

(b) $50÷(-5)=-\left(\frac{50}{5}\right)=-10$

(c) $(-36)÷(-9)=+\left(\frac{36}{9}\right)=4$

(d) $(-49)÷49=-\left(\frac{49}{49}\right)=-1$

(e) $13÷[(-2)+1]=13÷(-1)=-\left(\frac{13}{1}\right)=-13$

(f ) $0÷(-12)=-\left(\frac{0}{12}\right)=0$

(g) $(-31)÷[(-30)+(-1)]=(-31)÷(-31)=+\left(\frac{31}{31}\right)=1$

(h) $[(-36)÷12]÷3=[-\left(\frac{36}{12}\right)]÷3$
$=(-3)÷3=-\left(\frac{3}{3}\right)=-1$

(i) $$\begin{gathered} [(-6)+5]÷[(-2)+1] \\ =(-1)÷(-1) \end{gathered}$$
$=+\left(\frac{1}{1}\right)=1$

2. Verify that $a÷(b+c)\ne (a÷b)+(a÷c)$ for each of the following values of a, b and c.
(a) a = 12, b = – 4, c = 2 (b) a = (–10), b = 1, c = 1

Answer:

$a÷(b+c)\ne (a÷b)+(a÷c)$

(a) a = 12, b = – 4, c = 2

L.H.S = $a÷(b+c)$

$$\begin{gathered} =12÷[(-4)+2] \\ =12÷(-2) \\ =-\left(\frac{12}{2}\right) \\ =-6 \end{gathered}$$

R.H.S = $(a÷b)+(a÷c)$

$=[12÷(-4)]+(12÷2)$
$=[-\left(\frac{12}{4}\right)]+\left(\frac{12}{2}\right)$
$$\begin{gathered} =(-3)+6 \\ =3 \end{gathered}$$

Therefore. $L.H.S\ne R.H.S$

Hence verified.

(b) a = (–10), b = 1, c = 1

L.H.S = $a÷(b+c)$

$$\begin{gathered} =(-10)÷[1+1] \\ =(-10)÷2 \end{gathered}$$
$$\begin{gathered} =-\left(\frac{10}{2}\right) \\ =-5 \end{gathered}$$

R.H.S = $(a÷b)+(a÷c)$

$$\begin{gathered} =[(-10)÷1]+((-10)÷1) \\ =[-\left(\frac{10}{1}\right)]+[-\left(\frac{10}{1}\right)] \end{gathered}$$
$$\begin{gathered} =(-10)+(-10) \\ =-20 \end{gathered}$$

Therefore. $L.H.S\ne R.H.S$

Hence verified.

3. Fill in the blanks:
(a) 369 ÷ _____ = 369
(b) (–75) ÷ _____ = –1
(c) (–206) ÷ _____ = 1
(d) – 87 ÷ _____ = 87
(e) _____ ÷ 1 = – 87
(f) _____ ÷ 48 = –1
(g) 20 ÷ _____ = –2
(h) _____ ÷ (4) = –3

Answer:

(a) Given, $369÷\underset{―}{}=369$

A number divided by 1 gives the number itself.

$369÷\underset{―}{1}=369$

(b) $(-75)÷\underset{―}{}=-1$

The product is negative, therefore there must be an odd number of negative integers.

$(-75)÷\underset{―}{75}=-1$

(c) $(-206)÷\underset{―}{}=1$

A number divided by itself gives 1

$(-206)÷\underset{―}{(-206)}=1$

(d) $(-87)÷\underset{―}{}=87$

The product is positive, therefore there must be an even number of negative integers.

$(-87)÷\underset{―}{(-1)}=87$

(e) $\underset{―}{}÷1=-87$

A number divided by 1 gives the number itself.

$\underset{―}{(-87)}÷1=-87$

(f) $\underset{―}{}÷48=-1$

The product is negative, therefore there must be an odd number of negative integers.

$\underset{―}{(-48)}÷48=-1$

(g) $20÷\underset{―}{}=-2$

The product is negative, therefore there must be an odd number of negative integers.

$20÷\underset{―}{(-10)}=-2$

(h) $\underset{―}{}÷(4)=-3$

The product is negative, therefore there must be an odd number of negative integers.

$\underset{―}{\left(\frac{-4}{3}\right)}÷(4)=-3$

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

Answer:

Five pairs of integers $(a,b)$ such that $a÷b=-3$ are:

$(-6,2);(9,-3);(-9,3);(3,-1);(-3,1)$

5. The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at midnight?

Answer: Given,

The temperature at 12 noon = ${10}^{\circ }C$

Final temperature = $-8^{\circ }C$

$\therefore$ The decrease in temperature = $10-(-8)={18}^{\circ }C$

Time taken for the temperature to decrease by $2^{\circ }C$ = $1hour$

$\therefore$ Time taken for the temperature to decrease by ${18}^{\circ }C$ =

$=\frac{1}{2}\times 18hour$

$=9hours$

Now,

Time until midnight = $12hours$

$\therefore$ The decrease in temperature in $12hours$ = $2\times 12={24}^{\circ }C$

Therefore, the temperature at midnight = $(10-24{)}^{\circ }C=-{14}^{\circ }C$

Therefore, the temperature at midnight will be ${14}^{\circ }C$ below zero.

6. In a class test (+ 3) marks are given for every correct answer and (–2) marks are given for every incorrect answer and no marks 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?

Answer: Given,

Marks for every correct answer = $+3$

Marks for every wrong answer = $-2$

(i) According to the question,

Marks obtained by Radhika = $20$

Number of correct answers = $12$

$\therefore$ Marks obtained for correct answers = $12\times 3=36$

$\therefore$ Marks obtained for incorrect answers = Total marks – Marks obtained for correct answers

$$\begin{gathered} =20-36 \\ =-16 \end{gathered}$$

$\therefore$ Number of incorrect answers = $(-16)÷(-2)$

$$\begin{gathered} =\frac{16}{2} \\ =8 \end{gathered}$$

Therefore, Radhika attempted 8 questions incorrectly.

(ii)

According to the question,

Marks obtained by Mohini = $-5$

Number of correct answers = $7$

$\therefore$ Marks obtained for correct answers = $7\times 3=21$

$\therefore$ Marks obtained for incorrect answers = Total marks – Marks obtained for correct answers

$$\begin{gathered} =(-5)-21 \\ =-26 \end{gathered}$$

$\therefore$ Number of incorrect answers = $(-26)÷(-2)$

$$\begin{gathered} =\frac{26}{2} \\ =13 \end{gathered}$$

Therefore, Mohini attempted 13 questions incorrectly.

7. An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m?

Answer: Given,

Initial height = $10m$

Final depth = $-350m$

$\therefore$ Total distance the elevator has to descend = Final position - Initial position

= $|(-350)-10|=360m$

Also,

Time the elevator takes to descend $6m$ = $1min$

$\therefore$ Time the elevator takes to descend $360m$ =

$\frac{1}{60}\times 360min=60min$

$=1hour$