NCERT Solutions for Class 7 Maths Chapter 1 Integers

NCERT solutions for class 7 maths chapter 1 Integers- Class 6 CBSE NCERT maths introduced the chapter integers. The first chapter of class 7 starts with recollecting the concepts of integers studied in the previous class. The first exercise explained in the solutions of NCERT for class 7 maths chapter 1 integers is based on these recollected concepts. In NCERT class 7 maths chapter 1 integers, you will study more about the integers, their properties, and operations. There are a total of 4 exercises with 30 questions. All the questions including topic wise are explained in the CBSE NCERT solutions for class 7 maths chapter 1 integers. The NCERT solutions help students in their preparation of final examination. NCERT Class 7 maths chapter 1 integers is an important chapter for a math student. All the topics included in this chapter are extremely important for the students because these basic concepts of integers are also included in the higher classes. In this chapter, we will study important topics like properties of addition, subtraction, multiplication, and division of Integers, multiplication of a positive integer with a negative integer, multiplication of two or more than two negative integers and also learn closure under addition, subtraction and multiplication. The NCERT solutions for class 7 maths chapter 1 integers help students to understand how to apply the concepts in an application-level problem. Here you will get solutions to all four exercises of this chapter.

Exercise:1.1

Exercise:1.2

Exercise:1.3

Exercise:1.4

Topics of NCERT Grade 7 Maths Chapter 1 Integers-

1.1  Introduction

1.2  Recall

1.3   Properties of Addition and Subtraction of Integers

1.3.2  Closure under Subtraction

1.3.3  Commutative Property

1.3.4  Associative Property

1.4  Multiplication of Integers

1.4.1  Multiplication of a Positive and a Negative Integers

1.4.2  Multiplication of two Negative Integers

1.4.3  Product of three or more Negative Integers

1.5  Properties of Multiplication of Integers

1.5.1  Closure under Multiplication

1.5.2  Commutativity of Multiplication

1.5.3  Multiplication by Zero

1.5.4  Multiplicative Identity

1.5.5  Associativity for Multiplication

1.5.6  Distributive Property

1.5.7  Making Multiplication Easier

1.6 Division of Integers

1.7 Properties of Division of Integers

CBSE NCERT solutions for class 7 maths chapter 1 integers topic 1.2:

–3 and –2 are marked by E and F respectively. Which integers are marked by B, D, H, J, M and O?

First, we complete the number line.

Now, the integers marked by:

B = -6

D = -4

H = 0

J = 2

M = 5

O = 7

2. Arrange 7, –5, 4, 0 and – 4 in ascending order and then mark them on a number line to check your answer.

The given number are:  7, –5, 4, 0 and – 4

Arranging them in ascending order (increasing order)

$-5,-4, 0 , 4, 7$

On a number line, as we move towards right, the number increases.

3. We have done various patterns with numbers in our previous class. Can you find a pattern for each of the following? If yes, complete them:     (a)     7, 3, – 1, – 5, _____, _____, _____.     (b)     – 2, – 4, – 6, – 8, _____, _____, _____.     (c)     15, 10, 5, 0, _____, _____, _____.     (d)     – 11, – 8, – 5, – 2, _____, _____, _____. Make some more such patterns and ask your friends to complete them.

(a)     7, 3, – 1, – 5,  $\underline{-9} ,\underline{ -13},\underline{ -17},\underline{ -21}$

The pattern is : $7-4 =3; 3-4 =-1\ and\ so\ on$

(b)     – 2, – 4, – 6, – 8,  $\underline{-10},\underline{ -12},\underline{ -14},\underline{ -16}$

The pattern is : $-2-2 =-4; -4-2 =-6\ and\ so\ on$

(c)     15, 10, 5, 0, $\underline{5},\underline{ 10}, \underline{15}, \underline{20}$

The pattern is :  $15-10 =5; 10-5=5\ and\ so\ on$

(d)     – 11, – 8, – 5, – 2,  $\underline{1},\underline{ 4},\underline{ 7}, \underline{10}$

The pattern is :  $-11+3 =-8; -8+3=-5\ and\ so\ on$

Solutions of NCERT for class 7 maths chapter 1 integers topic 1.3.5

(a) a negative integer :  $-8 \ \&\ 3$
(b) zero : $-8 \ \&\ 8$
(c) an integer smaller than both the integers.      $-8 \ \&\ -3$
(d) an integer smaller than only one of the integers. $-8 \ \&\ 3$
(e) an integer greater than both the integers. $8 \ \&\ 3$

2.  Write a pair of integers whose difference gives (a) a negative integer.                                           (b) zero. (c) an integer smaller than both the integers.      (d) an integer greater than only one of the integers. (e) an integer greater than both the integers.

(a) a negative integer :  $-8 \ \&\ 3 : -8 - 3 = -11$
(b) zero : $8 \ \&\ 8$
(c) an integer smaller than both the integers.      $-8 \ \&\ 3$
(d) an integer smaller than only one of the integers. $-8 \ \&\ -3$
(e) an integer greater than both the integers. $8 \ \&\ -3$

NCERT solutions for class 7 maths chapter 1 integers topic 1.4.1

1. Find:
(i) 6 × (–19)
(ii) 12 × (–32)
(iii) 7 × (–22)

Multiplying a positive integer and a negative integer, we multiply them as whole numbers and put a minus sign (–) before the product. We thus get a negative integer

(i) $6 \times (-19) = -(6\times19) = -114$

(ii) $12 \times (-32) = -(12\times32) = -384$

(iii) $7 \times (-22) = -(7\times22) = -154$

1.   Find:  (a) 15 × (–16)         (b) 21 × (–32)  (c) (– 42) × 12        (d) –55 × 15

We know, multiplication of a positive and negative integer is given by:

$a \times (- b) = (- a) \times b = - (a \times b)$

(a) $15 \times (-16) = -(15\times16) = -240$

(b) $21 \times (-32) = -(21\times32) = -672$

(c) $(- 42) \times 12 = -(42\times12) = -504$

(d) $%u201355 \times 15 = -(55\times15) = -825$$(-55)\times 15=-(55\times 15)=825$

2. Check if     (a)    25 × (–21) = (–25) × 21         (b)    (–23) × 20 = 23 × (–20)

Write five more such examples.

We know, when multiplying a positive integer and a negative integer, we multiply them as whole numbers and put a minus sign (–) before the product. We thus get a negative integer.

(a)    $25 \times (-21) = (-25) \times 21$

L.H.S = $25 \times (-21) = -(25\times21) = -525$

R.H.S = $(-25) \times 21 = -(25\times21) = -525$

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

(b)   $(-23) \times 20 = 23 \times (-20)$

L.H.S = $(-23) \times 20 = -(23\times20) = -460$

R.H.S = $23 \times (-20) = -(23\times20) = -460$

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

Five more examples:

$\\ 20 \times (-1) = (-20) \times 1 \\ 15 \times (-20) = (-15) \times 20 \\ 3 \times (-2) = (-3) \times 2 \\ (-10) \times 5= 10\times (-5) \\ (-50) \times 110= 50\times (-110)$

Given,

$\\ (-5) \times 4 = -20$

$\\ (-5) \times 3 = -15 = -20+5 \\ (-5) \times 2 = -10 = -20 +10 \\ (-5) \times 1 = -5 = -20+15 \\ (-5) \times 0 = 0 = -20 + 20 \\ (-5) \times (-1) = 5 = -20+25 \\ (-5) \times (-2) = 10 = -20+30$

Therefore,

$\\ (-5) \times (-6) = -20+50=30$

(ii) Starting from (– 6) × 3, find (– 6) × (–7)

Given,

$(- 6) \times 3 = -18$

$\\ (-6) \times 2 = -12 = -18 +6 \\ (-6) \times 1 = -6 = -18+12 \\ (-6) \times 0 = 0 = -18 + 18 \\ (-6) \times (-1) = 6 = -18+24 \\ (-6) \times (-2) = 12 = -18+30 \\ \\ (-6) \times (-3) = 18 = -18+36$

Therefore,

$\\ (-6) \times (-7) = -18+60=42$

Q. Find: (–31) × (–100), (–25) × (–72), (–83) × (–28)

We know,

Multiplication of two negative integers : $(- a) \times (- b) = a \times b$

$\\ (-31) \times (-100) = 31 \times100 = 3100 \\ \\ (-25) \times (-72) = 25\times 72 = 1800 \\ \\ (-83) \times (-28) = 83\times 28 = 2324$

Solutions of NCERT class 7 maths chapter 1 integers topic 1.4.3

We know,

If the number of negative integers in a product is even, then the product is a positive integer and if the number of negative integers in a product is odd, then the product is a negative integer.

Hence, the product $(-9) \times (-5) \times (- 6)\times(-3)$ is positive whereas the product $(-9) \times (-5) \times 6\times(-3)$ is negative

(a) 8 negative integers and 3 positive integers?

(b) 5 negative integers and 4 positive integers?

(c) (–1), twelve times?

(d) (–1), 2m times, m is a natural number?

We know,

If the number of negative integers in a product is even, then the product is a positive integer and if the number of negative integers in a product is odd, then the product is a negative integer

And, any number of positive integers will always give a positive integer.

So, the sign of the product will be decided by the number of negative integers.

(a) 8 negative integers and 3 positive integers: Even number of negative integers, hence product is positive.

(b) 5 negative integers and 4 positive integers: Odd number of negative integers, hence product is negative.

(c) (–1), twelve times: Even number of negative integers, hence product is positive.

(d) (–1), 2m times, m is a natural number: Even number of negative integers, hence product is positive.

NCERT solutions for class 7 maths chapter 1 integers topic 1.5.6

L.H.S = $10 \times [(6 + (-2)] = 10\times4 =40$

R.H.S =  $10 \times 6 + 10 \times (-2) = 60 + (-20) = 40$

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

Hence, $10 \times [(6 + (-2)] = 10 \times [(6 + (-2)]$

(ii) Is (–15) × [(–7) + (–1)] = (–15) × (–7) + (–15) × (–1)?

L.H.S = $(-15) \times [(-7) + (-1)] = (-15)\times(-8) = 120$

R.H.S = $(-15) \times (-7) + (-15) \times (-1) = [-(15 \times 7)] + [-(15 \times 1) ]=(105) + 15 = 120$

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

Hence, $(-15) \times [(-7) + (-1)] = (-15) \times (-7) + (-15) \times (-1)$

(i) Is 10 × (6 – (–2)] = 10 × 6 – 10 × (–2)?

L.H.S = $10 \times [6 - (-2)] = 10\times[6+2] = 10\times8 = 80$

R.H.S = $10 \times 6- 10 \times (-2) = 60 - (-20) = 60+20 = 80$

$\therefore$ L.H.S = R.H.S

L.H.S = $(-15) \times [(-7) - (-1)] =(-15) \times [-7 +1]$

$= (-15) \times (-6) = -(15\times6) = -90$

R.H.S = $(-15) \times (-7) - (-15) \times (-1) = -(15\times7)-[-(15\times1)]$

$= -105+15 = -90$

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

Solutions for NCERT class 7 maths chapter 1 integers topic 1.5.7

We know,

Distributive law: for any integers a, b and c, $a \times (b + c) = a \times b + a \times c$

$\\ (- 49) \times 18 = (-50+1)\times18 = (-50)\times18 + 1\times18 \\ = -900 + 18 = -782 \\ \\ (-25) \times (-31) = (-25)\times((-30) + (-1)) \\ = (-25)\times(-30) + (-25)\times(-1) = 750+25 = 725 \\ \\ 70 \times (-19) + (-1) \times 70 = 70\times((-19)+ (-1)) \\ = 70\times(-20) = -1400$

Solutions for NCERT class 7 maths chapter 1 integers topic 1.6

Find:
(a) (–100) ÷ 5            (b) (–81) ÷ 9

(c) (–75) ÷ 5               (d) (–32) ÷ 2

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

(a) $(-100) \div 5 = -\frac{100}{5} = -20$

(b) $(-81) \div 9 = -(\frac{81}{9} )= -9$

(c) $(-75) \div 5 = -(\frac{75}{5} )= -15$

(d) $(-32) \div 2 = -(\frac{32}{2} )= -16$

Find:     (a) 125 ÷ (–25)        (b) 80 ÷ (–5)        (c) 64 ÷ (–16)

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.

(a) $125 \div (-25) = -\left (\frac{125}{25} \right ) = -5$

(b) $80\div (-5) = -\left (\frac{80}{5} \right ) = -16$

(c) $64\div (-16) = -\left (\frac{64}{16} \right ) = -4$

Find:    (a) (–36) ÷ (– 4)        (b) (–201) ÷ (–3)        (c) (–325) ÷ (–13)

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

(a)

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

(b)

$\\ (-201) \div (- 3) \\ = +\left (\frac{201}{3} \right ) \\ = 67$

(c)

$\\ (-325) \div (- 13) \\ = +\left (\frac{325}{13} \right ) \\ = 25$

Solutions of NCERT class 7 maths chapter 1 integers topic 1.7

(i) $1 \div a = 1$ is true only for $a =1$.

(ii) $a \div (-1) = - a$

Taking, $a =1; 1\div (-1) = -1$

$a =2; 2\div (-1) = -2$

$a =-1; (-1)\div (-1) = 1$

Hence, this is true for every integer.

NCERT solutions for class 7 maths chapter 1 integers exercise 1.1

(a) Observe this number line and write the temperature of the places marked on it.

(b) What is the temperature difference between the hottest and the coldest places among the above?

(c) What is the temperature difference between Lahulspiti and Srinagar?

(d) Can we say the temperature of Srinagar and Shimla taken together is less than the temperature at Shimla? Is it also less than the temperature at Srinagar?

(a) The temperature of the places in degree Celsius (°C) marked on it   are :

 Places Temperature Lahaul Spiti $-8^{\circ} C$ Srinagar $-2^{\circ} C$ Shimla $5^{\circ} C$ Ooty $14^{\circ} C$ Bangalore $22^{\circ} C$

(b) The hottest temperature is $22^{\circ} C$ and coldest is $-8^{\circ} C$

The temperature difference between them = $[22-(-8)]^{\circ} C = (22+8)^{\circ} C = 30^{\circ} C$

(c) The temperatures at Lahulspiti is $-8^{\circ} C$ and Srinagar is $-2^{\circ} C$

The temperature difference between Lahulspiti and Srinagar is  = $[(-2)-(-8)]^{\circ} C = (-2+8)^{\circ} C = 6^{\circ} C$

(d) The temperatures at Srinagar is $-2^{\circ} C$ and Shimla is $5^{\circ} C$

The temperature of Srinagar and Shimla took together = $[(-2)+5]^{\circ} C = 3^{\circ} C$

which is less than the temperature of Shimla.

But, it is not less than the temperature at Srinagar.

Given, Jack’s scores in five successive rounds are $25, - 5, - 10, 15\ and\ 10$

Therefore, his total score = $\\ 25 +(- 5) +(- 10)+ 15+10$

$\\ = 25 - 5 - 10+ 15+10 \\ = 35$

His total score at the end was  $35$

3. At Srinagar temperature was – 5°C on Monday and then it dropped by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day?

The temperature of Srinagar on Monday = $-5^{\circ} C$

According to question,

The temperature of Tuesday = $[(-5)-2]^{\circ} C$

$\therefore$ The temperature of Srinagar on Tuesday = $[(-5)-2]^{\circ} C = -8^{\circ} C$

Again, according to question

The temperature of Wednesday= $[(-8)+4]^{\circ} C$

$\therefore$ The temperature of Srinagar on Wednesday = $[(-8)+4]^{\circ} C = -4^{\circ} C$

The plane is flying at the height of 5000 m above the sea level

The distance of the plane from sea level = $5000\ m$

Also, the submarine floating 1200 m below the sea level

The distance of the submarine from sea level = $-1200\ m$

Distance between them = $[5000-(-1200)]\ m$

$6200\ m$

5.  Mohan deposits Rs 2,000 in his bank account and withdraws Rs 1,642 from it, the next day. If withdrawal of amount from the account is represented by a negative integer, then how will you represent the amount deposited? Find the balance in Mohan’s account after the withdrawal.

Given,

The amount deposited in bank account = $Rs.\ 2000$

Amount withdrawn from bank account = $Rs.\ 1642$

Balance in account = Amount deposited in bank account + Amount withdrawn from the bank account

$\\ = 2000 + (-1642) \\ = 2000 - 1642 \\ = 358$

Therefore,balance in Mohan’s account after the withdrawal is $Rs.\ 358$

6. Rita goes 20 km towards east from a point A to the point B. From B, she moves 30 km towards west along the same road. If the distance towards east is represented by a positive integer then, how will you represent the distance travelled towards west? By which integer will you represent her final position from A?

We will represent the distance travelled towards west with negative integer.

Distance travelled towards east = $20\ km$

Distance travelled towards west = $30\ km$

Distance travelled from A = $[20+(-30)]\ km$

$-10\ km$

As the distance is negative, so final position of Rita from A is towards the west direction.

7(i). In a magic square each row, column and diagonal have the same sum. Check if the following is a magic square.

 5 -1 -4 -5 -2 7 0 3 -3

Taking Rows-

$5-1-4 = 0$

$-5-2+7 = 0$

$0+3-3 = 0$

Taking Columns-

$5-5+0 = 0$

$-1-1+3 = 0$

$-4+7-3 = 0$

Taking Diagonals-

$-4-2+0 = -6$

$5-2-3 = 0$

As the sum of one of the diagonals is not equal to 0, it is not a magic square.

 1 -10 0 -4 -3 -2 -6 4 -7

Taking Rows-

$1-10+0 = -9$

$-4-3-2 = -9$

$-6+4-7= -9$

Taking Columns-

$1-4-6=-9$

$-10-3+4 = -9$

$0-2-7= -9$

Taking Diagonals-

$0-3-6= -9$

$1-3-7=-9$

As the sum of all the rows, columns and diagonals are equal, this is a magic square.

8. Verify $a - (-b) = a + b$ for the following values of a and b.

(i)     $a = 21, b = 18$

(ii)     $a = 118, b = 125$

(iii)     $a = 75, b = 84$

(iv)    $a = 28, b = 11$

(i)     $a = 21, b = 18$

L.H.S = $a-(-b) = 21-(-18)$

$= 21+18 = 39$

R.H.S = $a+b= 21+18 = 39$

$\therefore$ L.H.S = R.H.S

Hence verified.

(ii)     $a = 118, b = 125$

L.H.S = $a-(-b) = 118-(-125)$

$= 118+125 = 243$

R.H.S = $a+b= 118+125 = 243$

$\therefore$ L.H.S = R.H.S

Hence verified.

(iii)     $a = 75, b = 84$

L.H.S = $a-(-b) = 75-(-84)$

$= 75+84 = 159$

R.H.S = $a+b= 75+84 = 159$

$\therefore$ L.H.S = R.H.S

Hence verified.

(iv)    $a = 28, b = 11$

L.H.S = $a-(-b) = 28-(-11)$

$= 28+11 = 39$

R.H.S = $a+b= 28+11 = 39$

$\therefore$ L.H.S = R.H.S

Hence verified.

(Note: On a number line, the number on the right is greater than a number towards its left.)

(a) (– 8) + (– 4)           (–8) – (– 4)

L.H.S = $(- 8) + (- 4) = -8-4 = -12$

R.H.S = $(- 8) - (- 4) = -8+4 = -4$

Clearly, $R.H.S > L.H.S$
(b) (– 3) + 7 – (19)            15 – 8 + (– 9)

L.H.S = $(- 3) + 7 - (19) = 4-19 = -15$

R.H.S = $15 - 8 + (- 9) = 7-9= -2$

Clearly, $R.H.S > L.H.S$

(c) 23 – 41 + 11         23 – 41 – 11

L.H.S = $23 - 41 + 11 = -18+11= -7$

R.H.S = $23 - 41 - 11 = -18-11= -29$

Clearly, $L.H.S > R.H.S$

(d) 39 + (– 24) – (15)        36 + (– 52) – (– 36)

L.H.S = $39 + (- 24) - (15) = 39-24-15 = 15-15 = 0$

R.H.S = $36 + (- 52) - (- 36) = 36-52+36 = -16+36 = 20$

Clearly, $R.H.S > L.H.S$

(e) – 231 + 79 + 51       –399 + 159 + 81

L.H.S = $- 231 + 79 + 51 = -152+51 =-101$

R.H.S = $-399 + 159 + 81=-240+81 = -159$

Clearly, $L.H.S > R.H.S$

10.(i)  A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the ninth step.

He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level?

Let the number of steps moved down be represented by positive integers and the number of steps moved up by negative integers.

Given,

Monkey is at the step = 1

Now,

The step after:

1st jump  = $1 + 3 = 4$

2nd jump  = $4 + (-2) = 2$

3rd jump  = $2 + 3 = 5$

4th jump  = $5 + (-2) = 3$

5th jump  = $3 + 3 = 6$

6th jump = $6 + (-2) = 4$

7th jump  = $4 + 3 = 7$

8th jump  = $7 + (-2) = 5$

9th jump  = $5 + 3 = 8$

10th jump = $8 + (-2) = 6$

11th jump = $6 + 3 = 9$

Therefore, after 11 jumps, the monkey reaches the water level which is at the ninth step.

After drinking water, he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. In how many jumps will he reach back the top step?

Let the number of steps moved down be represented by positive integers and the number of steps moved up by negative integers.

Given,

Monkey is at the step = 9

Now,

The step after:

1st jump  = $9 + (-4) = 5$

2nd jump  = $5 + 2 = 7$

3rd jump  = $7 + (-4) = 3$

4th jump  = $3 + 2 = 5$

5th jump  = $5 + (-4) = 1$

Therefore, after 5 jumps, the monkey will reach back at the top.

If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers, represent his moves in part (i) and (ii) by completing the following; (a) – 3 + 2 – ... = – 8 (b) 4 – 2 + ... = 8. In (a) the sum (– 8) represents going down by eight steps. So, what will the sum 8 in (b) represent?

If we represent the down steps by negative and up steps by positive then:

Moves in (i):

$- 3 + 2 - 3 + 2 - 3 + 2 - 3 + 2 - 3 + 2 - 3 = -8$

Also,

Moves in part (ii):

$4 - 2 + 4 - 2 + 4 = 8$

Therefore, the sum +8 represents going up by eight steps.

CBSE NCERT solutions for class 7 maths chapter 1 integers exercise  1.2

A pair of integers whose:

(a) sum is –7  : $-8\ \&\ 1$

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

(c) sum is 0 : $-9\ \&\ 9$

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.

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

$-8 \ \& -16 :$

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

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

$-8\ \&\ 3$:

$-8 + 3 = -5$

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

$-1\ \&\ 2$

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

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?

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 team scored same.

Yes, we can add integers in any order.

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

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

(ii)     –53 + $\underline{0}$ = –53

$0$ is the additive identity. The number added to $0$ gives the same number.
(iii)     17 + $\underline{(-17)}$ = 0

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

By associative property $(a+b)+c=a+(b+c)$
(v)     (– 4) + [15 + (–3)] = [– 4 + 15] + $\underline{(-3)}$

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

NCERT solutions for class 7 maths chapter 1 integers exercise 1.3

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

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

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

(c) $(-21) \times (-30) = (21\times30) = 630$

Product of two negative integers is a positive integer

(d) $(-316) \times (-1) = (316\times1) = 316$

(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)

Product of even number of negative integers is an even integer.

(f) $\\ (-12) \times (-11) \times (10) =[ -(12\times11)]\times10 = (-132)\times10 = -1320$

(g) $9 \times (-3) \times (- 6) =9 \times [(-3) \times (- 6)] = 9 \times18 = 162$

(h) $(-18) \times (-5) \times (-4) = (-18) \times [(-5) \times (-4)] = (-18) \times20 = -360$

(i)

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

(j)

$\\ (-3) \times (-6) \times (-2) \times (-1) = [(-3) \times (-6) ]\times [(-2) \times (-1)] \\ = 18 \times2 = 36$

2 (a). Verify the following:

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

L.H.S = $18 \times [7 + (-3)] = 18\times4 = 72$

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)]

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.

(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:
$\\ (a)\ -22 : 22 \\ (b)\ 37 : -37 \\ (c)\ 0 : 0$

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

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,

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

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

Using suitable properties of multiplication:

(a)

$\\ 26 \times (- 48) + (- 48) \times (-36) \\ = (- 48) \times 26 + (- 48) \times (-36)\ \ [a\times b = b\times a] \\ = (- 48) \times [26 + (-36)] \ \ [a\times b + a\times c = a\times(b+c)] \\ = (-48)\times(-10) = 480$

(b)

$\\ 8 \times 53 \times (-125) \\ = [8 \times 53 ]\times (-125) \ \[a\times b = b\times a] \\ = 53 \times[ 8 \times (-125)] \ \ a\times(b\times c) =( a\times b)\times c \\ = 53\times(-1000) = -53000$

(c)

$\\ 15 \times (-25) \times (- 4) \times (-10) \\ = 15 \times [(-25) \times (- 4)] \times (-10) \\ = 15 \times 100\times (-10) \\ = -15000$

(d)

$\\ (- 41) \times 102 \\ =(- 41) \times(100+2) \\ =(- 41) \times100+(- 41) \times2\ \ [a\times(b+c)=(a\times b) + (a\times c)] \\ = (-4100) + (-82) \\ = -4100-82 = -4182$

(e)

$\\625 \times (-35) + (- 625) \times 65 \\ = (-625) \times 35 + (- 625) \times 65 \ \[a\times b = b\times a] \\ = (-625) \times (35 +65) \ \ [(a\times b) + (a\times c) = a\times(b+c)] \\ = (-625) \times (100) = -62500$

(f)

$\\ 7 \times (50 - 2) \\ = (7 \times 50) - (7 \times2) \ \ [a\times(b-c)=(a\times b) - (a\times c)] \\ = 350 -14 = 336$
(g)

$\\ (-17) \times (-29) \\ =(-17) \times (-30 + 1) \\ = (-17) \times (-30) +(-17) \times 1 \ \ [a\times(b+c)=(a\times b) + (a\times c)] \\ = 510 -17 = 493$

(h)

$\\ (-57) \times (-19) + 57 \\ = (-57) \times (-19) + (-57) \times(-1) \\ = (-57) \times [(-19) +(-1)] \ \ [(a\times b) + (a\times c) = a\times(b+c)] \\ = (-57) \times (-20) = 1140$

6. A certain freezing process requires that room temperature be lowered from 40°C at the rate of 5°C every hour. What will be the room temperature 10 hours after the process begins?

Given,
Initial temperature = $40^{\circ}C$

Also,

Change in temperature per hour = $-5^{\circ}C$

Therefore,Change in temperature after 10 hours = $(-5)\times10 = -50^{\circ}C$

Hence,

Final temperature = $40^{\circ}C + (-50^{\circ}C) = -10^{\circ}C$

7 (i). 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 questions not attempted.

(i) Mohan gets four correct and six incorrect answers. What is his score?

Given,

Total questions = $10$

Marks for correct answer = $+5$

Marks for incorrect answers = $-2$

Marks for unattempted question = $0$

According to question,

Mohan gets four correct and six incorrect answers

Total marks = $4(+5)+6(-2) = 20 -12 = 8$

Therefore, Mohan scored 8 marks.

7(ii). 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 questions not attempted.

(ii) Reshma gets five correct answers and five incorrect answers, what is her score?

Given,

Total questions = $10$

Marks for correct answer = $+5$

Marks for incorrect answers = $-2$

Marks for unattempted question = $0$

According to question,

Total marks = $5(+5)+5(-2) = 25 -10 = 15$

Therefore, Reshma scored 15 marks.

7 (iii). 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 questions not attempted.

(iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score?

Given,

Total questions = $10$

Marks for correct answer = $+5$

Marks for incorrect answers = $-2$

Marks for unattempted question = $0$

According to question,

Heena gets two correct and five incorrect answers.

Therefore, number of unattempted questions = $10 -7 =3$

Total marks = $2(+5)+5(-2)+3(0) = 10 -10+0 = 0$

Therefore, Heena scored 0 marks.

8 (a). A cement company earns a profit of Rs. 8 per bag of white cement sold and a loss of Rs. 5 per bag of grey cement sold.

(a) The company sells 3,000 bags of white cement and 5,000 bags of grey cement in a month. What is its profit or loss?

Given,

Profit earned by selling 1 bag of white cement = $Rs.\ 8$

Profit earned by selling 3000 bags of white cement= $Rs.\ (3000\times8) = Rs.\ 24000$

Loss on 1 bag of grey cement = $-Rs.\ 5$

Loss of 5000 bags of grey cement = $-Rs.\( 5000\times5) = -Rs.\ 25000$

Total earnings = $Rs.\ 24000 -Rs.\ 25000 = -Rs.\ 1000$

This means that the company incurred a loss of $Rs.\ 1000$

8 (b). A cement company earns a profit of Rs. 8 per bag of white cement sold and a loss of Rs. 5 per bag of grey cement sold.

(b) What is the number of white cement bags it must sell to have neither profit nor loss, if the number of grey bags sold is 6,400 bags.

Given,

Profit earned by selling 1 bag of white cement = $Rs.\ 8$

Loss on 1 bag of grey cement = $-Rs.\ 5$

$\therefore$ Loss on 6400 bags of grey cement = $-Rs.\( 6400\times5) = -Rs.\ 32000$

To be neither in profit nor loss, the profit from white cement must be $Rs.\ 32000$

$\therefore$ Number of white cement bags to be sold = $Rs.\ 32000\div Rs.\ 8$

$\\ = \frac{32000}{8} \\ = 4000$

Therefore the company has to sell 4000 white cement bags.

9. Replace the blank with an integer to make it a true statement.  (a) (–3) × _____ = 27            (b) 5 × _____ = –35  (c) _____ × (– 8) = –56         (d) _____ × (–12) = 132

(a)Given,  $(-3) \times \_\_\_\_\ = 27$

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

$(-3) \times \underline{(-9)}= 27$

(b) Given, $5 \times \_\_\_\_\_ \ = -35$
The product is negative, therefore there should be odd number of negative integers.

$5 \times \underline{(-7)} = -35$

(c) $\_\_\_\_ \ \times (- 8) = -56$

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

$\underline{7}\times (- 8) = -56$

(d) $\_\_\_\ \times (-12) = 132$

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

$\underline{(-11)} \times (-12) = 132$

Solutions for NCERT class 7 maths chapter 1 integers exercise 1.4

(h) [(–36) ÷ 12] ÷ 3    (i) [(– 6) + 5)] ÷ [(–2) + 1]

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

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

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

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

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

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

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

(h)

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

(i)

$\\ \ [(- 6) + 5] \div [(-2) + 1] = \\ = (-1) \div (-1)= +\left (\frac{1}{1} \right ) = 1$

2. Verify that $a \div (b + c) \neq (a\div b)+ (a\div c)$ 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

$a \div (b + c) \neq (a\div b)+ (a\div c)$

(a) a = 12, b = – 4, c = 2

L.H.S = $\\ a \div (b + c)$

$\\ = 12 \div [(-4)+2] \\ = 12 \div (-2) \\ = - \left (\frac{12}{2} \right ) \\ = -6$

R.H.S = $(a\div b)+ (a\div c)$

$\\ = [12\div (-4)]+ (12\div 2) \\ = \left [-\left (\frac{12}{4} \right ) \right ] + \left (\frac{12}{2} \right ) \\ = (-3)+6 \\ = 3$

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

Hence verified.

(b) a = (–10), b = 1, c = 1

L.H.S = $\\ a \div (b + c)$

$\\ = (-10) \div [1+1] \\ = (-10) \div 2 \\ = - \left (\frac{10}{2} \right ) \\ = -5$

R.H.S = $(a\div b)+ (a\div c)$

$\\ = [(-10)\div 1]+ ((-10)\div 1) \\ = \left [-\left (\frac{10}{1} \right ) \right ] + \left [-\left (\frac{10}{1} \right ) \right ] \\ = (-10)+(-10) \\ = -20$

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

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

(a) Given, $369 \div \underline{\ \ \ \ \ \ } = 369$

A number divided by 1 gives the number itself.

$369 \div \underline{1} = 369$

(b) $(-75) \div \underline{\ \ \ \ \ } = -1$

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

$(-75) \div \underline{75} = -1$

(c) $(-206) \div \underline{\ \ \ \ }= 1$

A number divided by itself gives 1

$(-206) \div \underline{(-206)}= 1$

(d) $(-87) \div \underline{\ \ \ \ }= 87$

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

$(-87) \div \underline{(-1) }= 87$

(e)$\underline{\ \ \ \ } \div 1 = - 87$

A number divided by 1 gives the number itself.

$\underline{(-87)} \div 1 = - 87$

(f) $\underline{\ \ \ \ \ } \div 48 = -1$

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

$\underline{(-48)} \div 48 = -1$

(g) $20\div \underline{\ \ \ \ \ } = -2$

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

$20\div \underline{(-10)} = -2$

(h) $\underline{\ \ \ \ \ } \div (4) = -3$

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

$\underline{\left (\frac{-4}{3} \right ) } \div (4) = -3$

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

Five pairs of integers $(a, b)$ such that $a \div b = -3$ 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 mid-night?

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$$1\ hour$

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

$=\frac{1}{2}\times18\ hour$

$= 9\ hours$

Now,

Time until midnight = $12\ hours$

$\therefore$ The decrease in temperature in $12\ hours$ = $2\times12 = 24^{\circ}C$

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

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

Given,

Marks for every correct answer = $+3$

Marks for every wrong answer = $- 2$

(i) According to question,

Marks obtained by Radhika = $20$

Number of correct answers = $12$

$\therefore$ Marks obtained for correct answers = $12\times3 = 36$

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

$\\ = 20 - 36 \\= - 16$

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

$\\ = \frac{16}{2} \\ = 8$

Therefore, Radhika attempted 8 questions incorrectly.

(ii)

According to question,

Marks obtained by Mohini = $-5$

Number of correct answers = $7$

$\therefore$ Marks obtained for correct answers = $7\times3 = 21$

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

$\\ = (-5) - 21 \\= - 26$

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

$\\ = \frac{26}{2} \\ = 13$

Therefore, Mohini attempted 13 questions incorrectly.

Given,

Initial height = $10\ m$

Final depth = $-350\ m$

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

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

Also,

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

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

$\frac{1}{60}\times360\ min = 60\ min$

$= 1\ hour$

NCERT Solutions for Class 7 Maths - Chapter-wise

 Chapter No. Chapter Name Chapter 1 NCERT Solutions for class 7 maths chapter 1 Integers Chapter 2 CBSE NCERT solutions for class 7 maths chapter 2 Fractions and Decimals Chapter 3 NCERT solutions for class 7 maths chapter 3 Data Handling Chapter 4 Solutions of NCERT for class 7 maths chapter 4 Simple Equations Chapter 5 CBSE NCERT solutions for class 7 maths chapter 5 Lines and Angles Chapter 6 NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties Chapter 7 Solutions of NCERT for class 7 maths chapter 7 Congruence of Triangles Chapter 8 NCERT solutions for class 7 maths chapter 8 comparing quantities Chapter 9 CBSE NCERT solutions for class 7 maths chapter 9 Rational Numbers Chapter 10 NCERT solutions for class 7 maths chapter 10 Practical Geometry Chapter 11 Solutions of NCERT for class 7 maths chapter 11 Perimeter and Area Chapter 12 CBSE NCERT solutions for class 7 maths chapter 12 Algebraic Expressions Chapter 13 NCERT solutions for class 7 maths chapter 13 Exponents and Powers Chapter 14 Solutions of NCERT for class 7 maths chapter 14 Symmetry

Important points of Class 7 Maths Chapter 1 Integers-

• Integers are closed for both addition and subtraction. That is, if a and b are any integers then a + b and a – b are also integers.
• The addition is commutative for the integers, such that $\dpi{100} a + b = b + a$ for all integers 'a' and 'b'.
• The addition is associative for the integers, such that, $\dpi{100} (a + b) + c = a + (b + c)$ for all the integers 'a', 'b' and 'c'.
•  Integer 0 is the identity under addition. That is, $\dpi{100} a + 0 = 0 + a = a$  for every integer 'a'.
• Integers are closed under multiplication. That is,$\dpi{100} a \times b$ is an integer for any two integers 'a' and 'b'.
• Multiplication is commutative for the integers. That is, $\dpi{100} a \times b= b \times a$ for any integers 'a' and 'b.'
• The integer 1 is the identity under multiplication, such that, $\dpi{100} 1 \times a= a \times 1 = a$ for any integer 'a'.
• Multiplication is associative for the integers, such that, $\dpi{100} (a \times b) \times c = a \times (b \times c)$ for any three integers a, b and c.

If a student is familiar with the CBSE NCERT solutions for class 7 maths chapter 1 integers, then good marks in the exam is not a difficult task. Also, the NCERT  solutions for class 7 maths chapter 1 integers help in solving homework problems