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## NCERT Solutions for Class 7 Maths Chapter 1 Integers

NCERT Solutions for Class 7 Maths Chapter 1 Integers are available here. When students feel stressed about searching for the most comprehensive and detailed NCERT Solutions for Class 7 Maths , we at BYJUβS have prepared step-by-step solutions with detailed explanations. We advise students who want to score good marks in Maths, to go through these solutions and strengthen their knowledge.

## Download Exclusively Curated Chapter Notes for Class 7 Maths Chapter β 1 Integers

Download most important questions for class 7 maths chapter β 1 integers.

Chapter 1 β Integers contains 4 exercises, and the NCERT Solutions available on this page provide solutions to the questions present in the exercises. Now, let us have a look at some of the concepts discussed in this chapter.

- Introduction of Integers
- Properties of Addition and Subtraction of Integers
- Multiplication of Integers
- Multiplication of a Positive and Negative Integer
- Multiplication of Two Negative Integer
- Properties of Multiplication of Integers
- Division of Integers
- Properties of Division of Integers

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## Access Exercises of NCERT Solutions for Class 7 Maths Chapter 1 Integers

Exercise 1.1 Solutions

Exercise 1.2 Solutions

Exercise 1.3 Solutions

Exercise 1.4 Solutions

## Access answers to Maths NCERT Solutions For Class 7 Chapter 1 β Integers

Exercise 1.1 Page: 4

1. Following number line shows the temperature in degree celsius (c o ) at different places on a particular day.

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

By observing the number line, we can find the temperature of the cities as follows:

The temperature in Lahulspiti is -8 o C

The temperature in Srinagar is -2 o C

The temperature in Shimla is 5 o C

The temperature in Ooty is 14 o C

The temperature in Bengaluru is 22 o C

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

From the number line, we observe that

The temperature at the hottest place, i.e., Bengaluru, is 22 o C

The temperature at the coldest place, i.e., Lahulspiti, is -8 o C

Temperature difference between hottest and coldest place is = 22 o C β (-8 o C)

= 22 o C + 8 o C

Hence, the temperature difference between the hottest and the coldest place is 30 o C.

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

From the given number line,

β΄ The temperature difference between Lahulspiti and Srinagar is = -2 o C β (8 o C)

= β 2 O C + 8 o C

(d) Can we say the temperature of Srinagar and Shimla, taken together, is less than the

temperature in Shimla? Is it also less than the temperature in Srinagar?

The temperature in Srinagar =-2 o C

The temperature in Shimla = 5 o C

The temperature of Srinagar and Shimla, taken together, is = β 2 o C + 5 o C

β΄ 5 o C > 3 o C

So, the temperature of Srinagar and Shimla, taken together, is less than the temperature at Shimla.

3 o > -2 o

No, the temperature of Srinagar and Shimla, taken together, is not less than the temperature of Srinagar.

2. In a quiz, positive marks are given for correct answers and negative marks are given

for incorrect answers. If Jackβs scores in five successive rounds were 25, β 5, β 10,

15 and 10, what was his total at the end?

From the question,

Jackβs score in five successive rounds are 25, -5, -10, 15 and 10

The total score of Jack at the end will be = 25 + (-5) + (-10) + 15 + 10

= 25 β 5 β 10 + 15 + 10

β΄ Jackβs total score at the end is 35.

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

The temperature on Monday in Srinagar = -5 o C

The temperature on Tuesday in Srinagar dropped by 2 o C = Temperature on Monday β 2 o C

= -5 o C β 2 o C

The temperature on Wednesday in Srinagar rose by 4 o C = Temperature on Tuesday + 4 o C

= -7 o C + 4 o C

Thus, the temperature on Tuesday and Wednesday was -7 o C and -3 o C, respectively.

4. A plane is flying at the height of 5000 m above sea level. At a particular point, it is exactly above a submarine floating 1200 m below sea level. What is the vertical distance between them?

The plane is flying at a height = 5000 m

Depth of submarine = -1200 m

The vertical distance between plane and submarine = 5000 m β (- 1200) m

= 5000 m + 1200 m

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

Withdrawal of the amount from the account is represented by a negative integer.

Then, the deposit of the amount to the account is represented by a positive integer.

Total amount deposited in bank account by the Mohan = βΉ 2000

The total amount withdrawn from the bank account by the Mohan = β βΉ 1642

Balance in Mohanβs account after the withdrawal = amount deposited + amount withdrawn

= βΉ 2000 + (-βΉ 1642)

= βΉ 2000 β βΉ 1642

Hence, the balance in Mohanβs account after the withdrawal is βΉ 358.

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

From the question, it is given that

A positive integer represents the distance towards the east.

Then, the distance travelled towards the west will be represented by a negative integer.

Rita travels a distance in the east direction = 20 km

Rita travels a distance in the west direction = β 30 km

β΄ Distance travelled from A = 20 + (- 30)

Hence, we will represent the distance travelled by Rita from point A by a negative integer, i.e., β 10 km

7. In a magic square, each row, column and diagonal have the same sum. Check which of the following is a magic square .

First, we consider the square (i)

By adding the numbers in each row, we get

= 5 + (- 1) + (- 4) = 5 β 1 β 4 = 5 β 5 = 0

= -5 + (-2) + 7 = β 5 β 2 + 7 = -7 + 7 = 0

= 0 + 3 + (-3) = 3 β 3 = 0

By adding the numbers in each column, we get

= 5 + (- 5) + 0 = 5 β 5 = 0

= (-1) + (-2) + 3 = -1 β 2 + 3 = -3 + 3 = 0

= -4 + 7 + (-3) = -4 + 7 β 3 = -7 + 7 = 0

By adding the numbers in diagonals, we get

= 5 + (-2) + (-3) = 5 β 2 β 3 = 5 β 5 = 0

= -4 + (-2) + 0 = β 4 β 2 = -6

Because the sum of one diagonal is not equal to zero.

So, (i) is not a magic square

Now, we consider the square (ii)

= 1 + (-10) + 0 = 1 β 10 + 0 = -9

= (-4) + (-3) + (-2) = -4 β 3 β 2 = -9

= (-6) + 4 + (-7) = -6 + 4 β 7 = -13 + 4 = -9

= 1 + (-4) + (-6) = 1 β 4 β 6 = 1 β 10 = -9

= (-10) + (-3) + 4 = -10 β 3 + 4 = -13 + 4

= 0 + (-2) + (-7) = 0 β 2 β 7 = -9

= 1 + (-3) + (-7) = 1 β 3 β 7 = 1 β 10 = -9

= 0 + (-3) + (-6) = 0 β 3 β 6 = -9

(ii) square is a magic square because the sum of each row, each column and the diagonal is equal to -9.

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

(i) a = 21, b = 18

a = 21 and b = 18

To verify a β (- b) = a + b

Let us take Left Hand Side (LHS) = a β (- b)

= 21 β (- 18)

Now, Right Hand Side (RHS) = a + b

By comparing LHS and RHS

Hence, the value of a and b is verified.

(ii) a = 118, b = 125

a = 118 and b = 125

= 118 β (- 125)

= 118 + 125

By comparing LHS and RHS,

(iii) a = 75, b = 84

a = 75 and b = 84

= 75 β (- 84)

(iv) a = 28, b = 11

a = 28 and b = 11

= 28 β (- 11)

9. Use the sign of >, < or = in the box to make the statements true.

(a) (-8) + (-4) [ ] (-8) β (-4)

Let us take Left Hand Side (LHS) = (-8) + (-4)

Now, Right Hand Side (RHS) = (-8) β (-4)

LHS < RHS

-12 < -4

β΄ (-8) + (-4) [<] (-8) β (-4)

(b) (-3) + 7 β (19) [ ] 15 β 8 + (-9)

Let us take Left Hand Side (LHS) = (-3) + 7 β 19

= -3 + 7 β 19

Now, Right Hand Side (RHS) = 15 β 8 + (-9)

= 15 β 8 β 9

-15 < -2

β΄ (-3) + 7 β (19) [<] 15 β 8 + (-9)

(c) 23 β 41 + 11 [ ] 23 β 41 β 11

Let us take Left Hand Side (LHS) = 23 β 41 + 11

Now, Right Hand Side (RHS) = 23 β 41 β 11

LHS > RHS

β 7 > -29

β΄ 23 β 41 + 11 [>] 23 β 41 β 11

(d) 39 + (-24) β (15) [ ] 36 + (-52) β (- 36)

Let us take Left Hand Side (LHS) = 39 + (-24) β 15

= 39 β 24 β 15

Now, Right Hand Side (RHS) = 36 + (-52) β (- 36)

= 36 β 52 + 36

β΄ 39 + (-24) β (15) [<] 36 + (-52) β (- 36)

(e) β 231 + 79 + 51 [ ] -399 + 159 + 81

Let us take Left Hand Side (LHS) = β 231 + 79 + 51

= β 231 + 130

Now, Right Hand Side (RHS) = β 399 + 159 + 81

= β 399 + 240

-101 > -159

β΄ β 231 + 79 + 51 [>] -399 + 159 + 81

10. 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.

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

Let us consider steps moved down are represented by positive integers, and then steps moved up are represented by negative integers.

Initially, the monkey is sitting on the topmost step, i.e., the first step

In 1 st jump, the monkey will be at step = 1 + 3 = 4 steps

In 2 nd jump, the monkey will be at step = 4 + (-2) = 4 β 2 = 2 steps

In 3 rd jump, the monkey will be at step = 2 + 3 = 5 steps

In 4 th jump, the monkey will be at step = 5 + (-2) = 5 β 2 = 3 steps

In 5 th jump, the monkey will be at step = 3 + 3 = 6 steps

In 6 th jump, the monkey will be at step = 6 + (-2) = 6 β 2 = 4 steps

In 7 th jump, the monkey will be at step = 4 + 3 = 7 steps

In 8 th jump, the monkey will be at step = 7 + (-2) = 7 β 2 = 5 steps

In 9 th jump, the monkey will be at step = 5 + 3 = 8 steps

In 10 th jump, the monkey will be at step = 8 + (-2) = 8 β 2 = 6 steps

In 11 th jump, the monkey will be at step = 6 + 3 = 9 steps

β΄ The monkey took 11 jumps (i.e., the 9 th step) to reach the water level.

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

Initially, the monkey is sitting on the ninth step, i.e., at the water level,

In 1 st jump, the monkey will be at step = 9 + (-4) = 9 β 4 = 5 steps

In 2 nd jump, the monkey will be at step = 5 + 2 = 7 steps

In 3 rd jump, the monkey will be at step = 7 + (-4) = 7 β 4 = 3 steps

In 4 th jump, the monkey will be at step = 3 + 2 = 5 steps

In 5 th jump, the monkey will be at step = 5 + (-4) = 5 β 4 = 1 step

β΄ The monkey took 5 jumps to reach back to the top step, i.e., the first step.

(iii) 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 parts (i) and (ii) by completing the following: (a) β 3 + 2 β β¦ = β 8 (b) 4 β 2 + β¦ = 8. In (a), the sum (β 8) represents going down by eight steps. So, what will the sum 8 in (b) represent?

The number of steps moved down is represented by negative integers, and the number of steps moved up by positive integers.

Monkey moves in part (i)

= β 3 + 2 β β¦β¦β¦.. = β 8

Then, LHS = β 3 + 2 β 3 + 2 β 3 + 2 β 3 + 2 β 3 + 2 β 3

= β 18 + 10

β΄ Moves in part (i) represent the monkey going down 8 steps because itβs a negative integer.

Monkey moves in part (ii)

= 4 β 2 + β¦β¦β¦.. = 8

Then, LHS = 4 β 2 + 4 β 2 + 4

β΄ Moves in part (ii) represent the monkey going up 8 steps because itβs a positive integer.

Exercise 1.2 Page: 9

1. Write down a pair of integers whose:

(a) sum is -7

= β 4 + (-3)

= β 4 β 3 β¦ [β΅ (+ Γ β = -)]

(b) the difference is β 10

= -25 β (-15)

= β 25 + 15 β¦ [β΅ (- Γ β = +)]

(c) sum is 0

2. (a) Write a pair of negative integers whose difference gives 8

= (-5) β (- 13)

= -5 + 13 β¦ [β΅ (- Γ β = +)]

(b) Write a negative integer and a positive integer whose sum is β 5.

(c) Write a negative integer and a positive integer whose difference is β 3.

Solution :-

= β 2 β (1)

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

The score of team A = -40, 10, 0

Total score obtained by team A = β 40 + 10 + 0

The score of team B = 10, 0, -40

Total score obtained by team B = 10 + 0 + (-40)

= 10 + 0 β 40

Thus, the score of both the A team and B team is the same.

Yes, we can say that we can add integers in any order.

4. Fill in the blanks to make the following statements true.

(i) (β5) + (β 8) = (β 8) + (β¦β¦β¦β¦)

Let us assume the missing integer be x,

= (β5) + (β 8) = (β 8) + (x)

= β 5 β 8 = β 8 + x

= β 13 = β 8 + x

By sending β 8 from RHS to LHS, it becomes 8

= β 13 + 8 = x

Now, substitute the x value in the blank place.

(β5) + (β 8) = (β 8) + (- 5) β¦ [This equation is in the form of the Commutative Law of Addition]

(ii) β53 + β¦β¦β¦β¦ = β53

= β53 + x = β53

By sending β 53 from LHS to RHS, it becomes 53

= x = -53 + 53

= β53 + 0 = β53 β¦ [This equation is in the form of the Closure Property of Addition]

(iii) 17 + β¦β¦β¦β¦ = 0

= 17 + x = 0

By sending 17 from LHS to RHS, it becomes -17

= x = 0 β 17

= 17 + (-17) = 0 β¦ [This equation is in the form of Closure Property of Addition]

= 17 β 17 = 0

(iv) [13 + (β 12)] + (β¦β¦β¦β¦) = 13 + [(β12) + (β7)]

= [13 + (β 12)] + (x) = 13 + [(β12) + (β7)]

= [13 β 12] + (x) = 13 + [β12 β7]

= [1] + (x) = 13 + [-19]

= 1 + (x) = 13 β 19

= 1 + (x) = -6

By sending 1 from LHS to RHS, it becomes -1.

= x = -6 β 1

= [13 + (β 12)] + (-7) = 13 + [(β12) + (β7)] β¦ [This equation is in the form of Associative Property of Addition]

(v) (β 4) + [15 + (β3)] = [β 4 + 15] +β¦β¦β¦β¦

Let us assume the missing integer be x.

= (β 4) + [15 + (β3)] = [β 4 + 15] + x

= (β 4) + [15 β 3)] = [β 4 + 15] + x

= (-4) + [12] = [11] + x

= 8 = 11 + x

By sending 11 from RHS to LHS, it becomes -11,

= 8 β 11 = x

= (β 4) + [15 + (β3)] = [β 4 + 15] + -3 β¦ [This equation is in the form of Associative Property of Addition]

Exercise 1.3 Page: 21

1. Find each of the following products:

(a) 3 Γ (β1)

By the rule of Multiplication of integers,

= -3 β¦ [β΅ (+ Γ β = -)]

(b) (β1) Γ 225

= (-1) Γ 225

= -225 β¦ [β΅ (- Γ + = -)]

(c) (β21) Γ (β30)

= (-21) Γ (-30)

= 630 β¦ [β΅ (- Γ β = +)]

(d) (β316) Γ (β1)

= (-316) Γ (-1)

= 316 β¦ [β΅ (- Γ β = +)]

(e) (β15) Γ 0 Γ (β18)

= (β15) Γ 0 Γ (β18)

β΅ If any integer is multiplied by zero, the answer is zero itself.

(f) (β12) Γ (β11) Γ (10)

= (β12) Γ (-11) Γ (10)

First, multiply the two numbers having the same sign.

= 132 Γ 10 β¦ [β΅ (- Γ β = +)]

(g) 9 Γ (β3) Γ (β 6)

= 9 Γ (-3) Γ (-6)

= 9 Γ 18 β¦ [β΅ (- Γ β = +)]

(h) (β18) Γ (β5) Γ (β 4)

= (-18) Γ (-5) Γ (-4)

= 90 Γ -4 β¦ [β΅ (- Γ β = +)]

= β 360 β¦ [β΅ (+ Γ β = -)]

(i) (β1) Γ (β2) Γ (β3) Γ 4

= [(β1) Γ (β2)] Γ [(β3) Γ 4]

= 2 Γ (-12) β¦ [β΅ (- Γ β = +), (- Γ + = -)]

(j) (β3) Γ (β6) Γ (β2) Γ (β1)

= [(β3) Γ (β6)] Γ [(β2) Γ (β1)]

= 18 Γ 2 β¦ [β΅ (- Γ β = +)

2. Verify the following:

(a) 18 Γ [7 + (β3)] = [18 Γ 7] + [18 Γ (β3)]

From the given equation,

Let us consider the Left Hand Side (LHS) first = 18 Γ [7 + (β3)]

= 18 Γ [7 β 3]

Now, consider the Right Hand Side (RHS) = [18 Γ 7] + [18 Γ (β3)]

= [126] + [-54]

Hence, the given equation is verified.

(b) (β21) Γ [(β 4) + (β 6)] = [(β21) Γ (β 4)] + [(β21) Γ (β 6)]

Let us consider the Left Hand Side (LHS) first = (β21) Γ [(β 4) + (β 6)]

= (-21) Γ [-4 β 6]

= (-21) Γ [-10]

Now, consider the Right Hand Side (RHS) = [(β21) Γ (β 4)] + [(β21) Γ (β 6)]

= [84] + [126]

3. (i) For any integer a, what is (β1) Γ a equal to?

= (-1) Γ a = -a

When we multiply any integer a with -1, then we get the additive inverse of that integer.

(ii). Determine the integer whose product with (β1) is

Now, multiply -22 with (-1), and we get

= -22 Γ (-1)

When we multiply integer -22 with -1, then we get the additive inverse of that integer.

Now, multiply 37 with (-1), and we get

= 37 Γ (-1)

When we multiply integer 37 with -1, then we get the additive inverse of that integer.

Now, multiply 0 by (-1), and we get

Because the product of negative integers and zero gives zero only.

4. Starting from (β1) Γ 5, write various products showing some pattern to show

(β1) Γ (β1) = 1.

The various products are,

= -1 Γ 5 = -5

= -1 Γ 4 = -4

= -1 Γ 3 = -3

= -1 Γ 2 = -2

= -1 Γ 1 = -1

= -1 Γ 0 = 0

= -1 Γ -1 = 1

We concluded that the product of one negative integer and one positive integer is a negative integer. Then, the product of two negative integers is a positive integer.

5. Find the product using suitable properties:

(a) 26 Γ (β 48) + (β 48) Γ (β36)

The given equation is in the form of the Distributive Law of Multiplication over Addition.

= a Γ (b + c) = (a Γ b) + (a Γ c)

Let, a = -48, b = 26, c = -36

= 26 Γ (β 48) + (β 48) Γ (β36)

= -48 Γ (26 + (-36)

= -48 Γ (26 β 36)

= -48 Γ (-10)

= 480 β¦ [β΅ (- Γ β = +)

(b) 8 Γ 53 Γ (β125)

The given equation is in the form of the Commutative Law of Multiplication.

= a Γ b = b Γ a

= 8 Γ [53 Γ (-125)]

= 8 Γ [(-125) Γ 53]

= [8 Γ (-125)] Γ 53

= [-1000] Γ 53

(c) 15 Γ (β25) Γ (β 4) Γ (β10)

= 15 Γ [(β25) Γ (β 4)] Γ (β10)

= 15 Γ [100] Γ (β10)

= 15 Γ [-1000]

(d) (β 41) Γ 102

= (-41) Γ (100 + 2)

= (-41) Γ 100 + (-41) Γ 2

= β 4100 β 82

(e) 625 Γ (β35) + (β 625) Γ 65

= 625 Γ [(-35) + (-65)]

= 625 Γ [-100]

(f) 7 Γ (50 β 2)

The given equation is in the form of the Distributive Law of Multiplication over Subtraction.

= a Γ (b β c) = (a Γ b) β (a Γ c)

= (7 Γ 50) β (7 Γ 2)

(g) (β17) Γ (β29)

= (-17) Γ [-30 + 1]

= [(-17) Γ (-30)] + [(-17) Γ 1]

= [510] + [-17]

(h) (β57) Γ (β19) + 57

= (57 Γ 19) + (57 Γ 1)

= 57 [19 + 1]

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

Let us take the lowered temperature as negative.

Initial temperature = 40 o C

Change in temperature per hour = -5 o C

Change in temperature after 10 hours = (-5) Γ 10 = -50 o C

β΄ The final room temperature after 10 hours of freezing process = 40 o C + (-50 o C)

7. In a class test containing 10 questions, 5 marks are awarded for every correct answer and (β2) 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?

Marks awarded for 1 correct answer = 5

Total marks awarded for 4 correct answer = 4 Γ 5 = 20

Marks awarded for 1 wrong answer = -2

Total marks awarded for 6 wrong answer = 6 Γ -2 = -12

β΄ Total score obtained by Mohan = 20 + (-12)

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

Total marks awarded for 5 correct answer = 5 Γ 5 = 25

Total marks awarded for 5 wrong answer = 5 Γ -2 = -10

β΄ Total score obtained by Reshma = 25 + (-10)

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

Total marks awarded for 2 correct answer = 2 Γ 5 = 10

Marks awarded for questions not attempted is = 0

β΄ Total score obtained by Heena = 10 + (-10)

8. A cement company earns a profit of βΉ 8 per bag of white cement sold and a loss of

βΉ 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?

We denote profit in a positive integer and loss in a negative integer,

The cement company earns a profit on selling 1 bag of white cement = βΉ 8 per bag

The cement company earns a profit on selling 3000 bags of white cement = 3000 Γ βΉ 8

Loss on selling 1 bag of grey cement = β βΉ 5 per bag

Loss on selling 5000 bags of grey cement = 5000 Γ β βΉ 5

= β βΉ 25000

Total loss or profit earned by the cement company = profit + loss

= 24000 + (-25000)

Thus, a loss of βΉ 1000 will be incurred by the company.

(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?

Let the number of white cement bags be x.

Cement company earns a profit on selling x bags of white cement = (x) Γ βΉ 8

Loss on selling 6400 bags of grey cement = 6400 Γ β βΉ 5

= β βΉ 32000

According to the question,

The company must sell to have neither profit nor loss.

= profit + loss = 0

= 8x + (-32000) =0

By sending -32000 from LHS to RHS, it becomes 32000

= 8x = 32000

= x = 32000/8

Hence, the 4000 bags of white cement have neither profit nor loss.

9. Replace the blank with an integer to make it a true statement.

(a) (β3) Γ _____ = 27

= (β3) Γ (x) = 27

= x = β (27/3)

Let us substitute the value of x in the place of blank,

= (β3) Γ (-9) = 27 β¦ [β΅ (- Γ β = +)]

(b) 5 Γ _____ = β35

= (5) Γ (x) = -35

= x = β (-35/5)

Let us substitute the value of x in the place of the blank.

= (5) Γ (-7) = -35 β¦ [β΅ (+ Γ β = -)]

(c) _____ Γ (β 8) = β56

= (x) Γ (-8) = -56

= x = (-56/-8)

= (7) Γ (-8) = -56 β¦ [β΅ (+ Γ β = -)]

(d) _____ Γ (β12) = 132

= (x) Γ (-12) = 132

= x = β (132/12)

= (β11) Γ (-12) = 132 β¦ [β΅ (- Γ β = +)]

Exercise 1.4 Page: 26

1. Evaluate each of the following.

(a) (β30) Γ· 10

= (β30) Γ· 10

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

(b) 50 Γ· (β5)

= (50) Γ· (-5)

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

(c) (β36) Γ· (β9)

= (-36) Γ· (-9)

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

(d) (β 49) Γ· (49)

= (β49) Γ· 49

(e) 13 Γ· [(β2) + 1]

= 13 Γ· [(β2) + 1]

= 13 Γ· (-1)

(f) 0 Γ· (β12)

= 0 Γ· (-12)

When we divide zero by a negative integer, it gives zero.

(g) (β31) Γ· [(β30) + (β1)]

= (β31) Γ· [(β30) + (β1)]

= (-31) Γ· [-30 β 1]

= (-31) Γ· (-31)

(h) [(β36) Γ· 12] Γ· 3

First, we have to solve the integers within the bracket.

= [(β36) Γ· 12]

= (β36) Γ· 12

(i) [(β 6) + 5)] Γ· [(β2) + 1]

The given question can be written as,

= [-1] Γ· [-1]

2. Verify that a Γ· (b + c) β (a Γ· b) + (a Γ· c) for each of the following values of a, b and c.

(a) a = 12, b = β 4, c = 2

From the question, a Γ· (b + c) β (a Γ· b) + (a Γ· c)

Given, a = 12, b = β 4, c = 2

Now, consider LHS = a Γ· (b + c)

= 12 Γ· (-4 + 2)

= 12 Γ· (-2)

Then, consider RHS = (a Γ· b) + (a Γ· c)

= (12 Γ· (-4)) + (12 Γ· 2)

= (-3) + (6)

= LHS β RHS

Hence, the given values are verified.

(b) a = (β10), b = 1, c = 1

Given, a = (-10), b = 1, c = 1

= (-10) Γ· (1 + 1)

= (-10) Γ· (2)

= ((-10) Γ· (1)) + ((-10) Γ· 1)

= (-10) + (-10)

3. Fill in the blanks:

(a) 369 Γ· _____ = 369

= 369 Γ· x = 369

= x = (369/369)

Now, put the valve of x in the blank.

= 369 Γ· 1 = 369

(b) (β75) Γ· _____ = β1

= (-75) Γ· x = -1

= x = (-75/-1)

= (-75) Γ· 75 = -1

(c) (β206) Γ· _____ = 1

= (-206) Γ· x = 1

= x = (-206/1)

= (-206) Γ· (-206) = 1

(d) β 87 Γ· _____ = 87

= (-87) Γ· x = 87

= x = (-87)/87

= (-87) Γ· (-1) = 87

(e) _____ Γ· 1 = β 87

= (x) Γ· 1 = -87

= x = (-87) Γ 1

= (-87) Γ· 1 = -87

(f) _____ Γ· 48 = β1

= (x) Γ· 48 = -1

= x = (-1) Γ 48

= (-48) Γ· 48 = -1

(g) 20 Γ· _____ = β2

= 20 Γ· x = -2

= x = (20)/ (-2)

= (20) Γ· (-10) = -2

(h) _____ Γ· (4) = β3

= (x) Γ· 4 = -3

= x = (-3) Γ 4

= (-12) Γ· 4 = -3

4. Write five pairs of integers (a, b) such that a Γ· b = β3. One such pair is (6, β2) because 6 Γ· (β2) = (β3).

(i) (15, -5)

Because, 15 Γ· (β5) = (β3)

(ii) (-15, 5)

Because, (-15) Γ· (5) = (β3)

(iii) (18, -6)

Because, 18 Γ· (β6) = (β3)

(iv) (-18, 6)

Because, (-18) Γ· 6 = (β3)

(v) (21, -7)

Because, 21 Γ· (β7) = (β3)

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

From the question, it is given,

The temperature at the beginning, i.e., at 12 noon = 10 o C

Rate of change of temperature = β 2 o C per hour

Temperature at 1 PM = 10 + (-2) = 10 β 2 = 8 o C

Temperature at 2 PM = 8 + (-2) = 8 β 2 = 6 o C

Temperature at 3 PM = 6 + (-2) = 6 β 2 = 4 o C

Temperature at 4 PM = 4 + (-2) = 4 β 2 = 2 o C

Temperature at 5 PM = 2 + (-2) = 2 β 2 = 0 o C

Temperature at 6 PM = 0 + (-2) = 0 β 2 = -2 o C

Temperature at 7 PM = -2 + (-2) = -2 -2 = -4 o C

Temperature at 8 PM = -4 + (-2) = -4 β 2 = -6 o C

Temperature at 9 PM = -6 + (-2) = -6 β 2 = -8 o C

β΄ At 9 PM, the temperature will be 8 o C below zero.

The temperature at midnight, i.e., at 12 AM

Change in temperature in 12 hours = -2 o C Γ 12 = β 24 o C

So, at midnight temperature will be = 10 + (-24)

So, at midnight, the temperature will be 14 o C below 0.

6. In a class test, (+ 3) marks are given for every correct answer and (β2) 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 scored β5 marks on this test, though she got 7 correct answers. How many questions has she attempted incorrectly?

Marks awarded for 1 correct answer = + 3

(i) Radhika scored 20 marks.

Total marks awarded for 12 correct answers = 12 Γ 3 = 36

Marks awarded for incorrect answers = Total score β Total marks awarded for 12 correct

So, the number of incorrect answers made by Radhika = (-16) Γ· (-2)

(ii) Mohini scored -5 marks.

Total marks awarded for 7 correct answers = 7 Γ 3 = 21

So, the number of incorrect answers made by Mohini = (-26) Γ· (-2)

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 β 350 m?

The initial height of the elevator = 10 m

The final depth of the elevator = β 350 m β¦ [β΅distance descended is denoted by a negative

The total distance to descended by the elevator = (-350) β (10)

Time taken by the elevator to descend -6 m = 1 min

So, the time taken by the elevator to descend β 360 m = (-360) Γ· (-6)

= 60 minutes

Disclaimer:

Dropped Topics βΒ Introduction, Recall, 1.4.3 Product of three or more negative numbers and 1.5.7 Making multiplication easier.

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- Important Questions for CBSE Class 7 Maths Chapter 1 - Integers (Free PDF Download)

## Important Practice Problems for CBSE Class 7 Maths Chapter 1: Integers

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## Study Important Questions for Class 7 Maths Chapter 1- Integers

A.Very short answer question β 1 marks

1. Define Integers.

Ans: The numbers range from negative infinity to positive infinity including zero. They are denoted by I i.e. $ \text{I=}\left\{ \left. .....\text{-3,-2,-1,0,1,2,3}..... \right\} \right. $ .

2. We move to the left in the number line when we__________ or __________.

Ans: We move to the left in the number line when we add a negative integer or subtract a positive integer .

3. Additive inverse of $ \text{-25} $ is ____.

Ans: $ 25 $ .

4. Fill the blanks for $ \text{-228+96+125} $ ___ $ \text{-451+197+76}\left( \text{use,,=} \right) $

Ans: $ > $

5. What would come in place of ? in $ \text{-11+0=?} $

Ans: $ \text{-11} $

6. Fill the blanks for $ \text{-22 }\!\!\times\!\!\text{ -13 }\!\!\times\!\!\text{ 5=} $ _____

Ans: $ \text{1430} $

7.Fill the blanks for $ \text{-3 }\!\!\times\!\!\text{ 125=} $ ___

Ans: $ \text{-375} $

B. Short Answer Questions β 2 marks

8. Verify $ \text{a-}\left( \text{-b} \right)\text{=a+b} $ for the following values of $ \text{a} $ and $ \text{b} $

a. $ \text{a=25,b=12} $

Ans: Substituting value of $ \text{a} $ and $ \text{b} $ in given equation

$ \text{a-}\left( \text{-b} \right)\text{=a+b} $

$ \text{25-}\left( \text{-12} \right)\text{=25+12} $

$ \,\text{25+12=25+12} $

$ \text{37=37} $

Hence, verified.

b. $ \text{a=113,b=16} $

$ \text{a-}\left( \text{-b} \right)\text{=a+b} $

$ \text{113-}\left( \text{-16} \right)\text{=113+16} $

$ \text{113+16=113+16} $

$ \text{129=129} $

9. Use $ \text{,} $ or $ \text{=} $ sign for the below statements to make it true

a. $ \left( \text{-9} \right)\text{+}\left( \text{-28} \right) $ ____ $ \left( \text{-9} \right)\text{-}\left( \text{-28} \right) $

Ans: Solving both sides-

$ \left( \text{-9} \right)\text{+}\left( \text{-28} \right)=-37 $

$ \left( \text{-9} \right)\text{-}\left( \text{-28} \right)=19 $

Thus, $ \left( \text{-9} \right)\text{+}\left( \text{-28} \right) < \left( \text{-9} \right)\text{-}\left( \text{-28} \right) $

b. $ \text{25+}\left( \text{-14} \right)\text{-18} $ ____ $ \text{25+}\left( \text{-14} \right)\text{-}\left( \text{-18} \right) $

$ \text{2}5+\left( -14 \right)-18=11-18 $

$ =\text{-7} $

$ \text{25+}\left( \text{-14} \right)\text{-}\left( \text{-18} \right)=11+18 $

$ =29 $

Thus, $ \text{2}5+\left( -14 \right)-18 < \text{25+}\left( \text{-14} \right)\text{-}\left( \text{-18} \right) $.

10. Write down a pair of integers for the following

a. Sum gives $ \text{-9} $

Ans: A pair of integers that gives sum $ \text{-9} $ is $ \left( -6,-3 \right) $.

b. Difference gives $ \text{-11} $

Ans: A pair of integers that gives sum $ \text{-11} $ is $ \left( -14,3 \right) $.

11. a. Write a positive and negative integer whose sum is $ \text{-4} $ .

Ans: $ \left( 4,-8 \right) $ is a positive and negative integer whose sum is $ \text{-4} $ .

b.Write a negative integer and a positive integer whose difference is $ \text{-2} $ .

Ans: $ \left( -1,1 \right) $ is a positive and negative integer whose sum is $ \text{-2} $ .

12. Fill in the blanks

a. $ \left( \text{-4} \right)\text{+}\left( \text{-11} \right)\text{=}\left( \text{-11} \right)\text{+ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ } $

Ans: $ \left( \text{-4} \right)\text{+}\left( \text{-11} \right)\text{=}\left( \text{-11} \right)\text{+ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ } $

\[\Rightarrow \left( \text{-4} \right)\text{+}\left( \text{-11} \right)\text{+11}\]

\[\Rightarrow -4\]

Thus, $ \left( \text{-4} \right)\text{+}\left( \text{-11} \right)\text{=}\left( \text{-11} \right)\text{+-4} $

b. $ \left[ \text{22+}\left( \text{-9} \right) \right]\text{+}\left( \text{-2} \right)\text{=22+}\left[ \text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ +}\left( \text{-2} \right) \right] $

Ans: \[\left[ \text{22+}\left( \text{-9} \right) \right]\text{+}\left( \text{-2} \right)\text{=22+}\left[ \text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ +}\left( \text{-2} \right) \right]\]

\[\Rightarrow \text{13-2=22+}\left[ \text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ +}\left( \text{-2} \right) \right]\]

\[\Rightarrow \text{11-22=}\left[ \text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ +}\left( \text{-2} \right) \right]\]

\[\Rightarrow \text{-11= }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ +}\left( \text{-2} \right)\]

\[\Rightarrow \text{-11+2}\]

\[\Rightarrow \text{-9}\]

Thus, \[\left[ \text{22+}\left( \text{-9} \right) \right]\text{+}\left( \text{-2} \right)\text{=22+}\left[ \text{-9+}\left( \text{-2} \right) \right]\]

13. Verify $ \text{7 }\!\!\times\!\!\text{ }\left[ \left( \text{22} \right)\text{+}\left( \text{-9} \right) \right]\text{=}\left[ \left( \text{7} \right)\text{ }\!\!\times\!\!\text{ 22} \right]\text{+}\left[ \text{7 }\!\!\times\!\!\text{ -9} \right] $

Ans: On solving both sides

$ \text{7 }\!\!\times\!\!\text{ }\left[ \left( \text{22} \right)\text{+}\left( \text{-9} \right) \right]\text{=}\left[ \left( \text{7} \right)\text{ }\!\!\times\!\!\text{ 22} \right]\text{+}\left[ \text{7 }\!\!\times\!\!\text{ -9} \right] $

$ 7\times \left[ 13 \right]=154-63 $

$ 91=91 $

14.Find the product of

a. $ \text{63 }\!\!\times\!\!\text{ 0 }\!\!\times\!\!\text{ -7} $

Ans: The product of $ \text{63 }\!\!\times\!\!\text{ 0 }\!\!\times\!\!\text{ -7} $ is $ 0 $ .

b. $ \text{5 }\!\!\times\!\!\text{ }\left( \text{-3} \right)\text{ }\!\!\times\!\!\text{ -2} $

Ans: So, $ 5\times \left( -3 \right)\times -2=5\times 6 $

$ =30 $

The product of $ 5\times \left( -3 \right)\times -2 $ is $ 30 $ .

a. $ \text{-2 }\!\!\times\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ =14} $

Ans: So, $ \text{-2 }\!\!\times\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ =14} $

$ \Rightarrow \dfrac{\text{14}}{\text{-2}} $

$ \Rightarrow \text{-7} $

Hence, $ \text{-2 }\!\!\times\!\!\text{ 7 =14} $ .

b. \[\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\times\!\!\text{ -8=-32}\]

Ans: So, \[\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\times\!\!\text{ -8=-32}\]

\[\Rightarrow \dfrac{-32}{-8}\]

\[\Rightarrow 4\]

Hence, \[\text{4 }\!\!\times\!\!\text{ -8=-32}\]

16. Evaluate

a. $ \text{-39 }\!\!\div\!\!\text{ 13} $

Ans: $ -39\div 13 $

$ \Rightarrow \dfrac{-39}{13} $

$ \Rightarrow -3 $

Hence, $ -39\div 13=-3 $

b. $ \text{-64 }\!\!\div\!\!\text{ }\left[ \text{-8 }\!\!\times\!\!\text{ -8} \right] $

Ans: $ -64\div \left[ -8\times -8 \right] $

$ \Rightarrow \dfrac{-64}{\left[ -8\times -8 \right]} $

$ \Rightarrow \dfrac{-64}{64} $

$ \Rightarrow -1 $

Hence, $ -64\div \left[ -8\times -8 \right]=-1 $

17. Write two pairs of integers such that $ \text{a }\!\!\div\!\!\text{ b=-5} $

Ans: The two pairs of integers such that $ \text{a }\!\!\div\!\!\text{ b=-5} $ are:

> $ \left( 10,-2 \right) $

> $ \left( -70,14 \right) $

C. Short answer questions β 3 marks

18. Manvita deposits Rs. $ \text{5000} $ in her bank account after two days. She withdraws Rs. $ \text{3748} $ from it. If the amount deposited is a positive integer. How will you represent the amount withdrawn and also find the balance amount in the account?

Ans: The amount withdrawn should always be represented as a negative integer.

Thus, it would be $ -3748 $.

Since, Total balance $ = $ Amount deposited $ - $ Amount withdrawn

Total balance $ =5000-3748 $

$ \text{=Rs}\text{. 1252} $ .

Hence, the amount withdrawn would be negative integer i.e., $ -3748 $ and the balance amount in the account is $ \text{Rs}\text{. 1252} $ .

19. In a game Mishala scored $ \text{20,}\,\text{-40,}\,\text{10} $ and Meera scored $ \text{-40,10,}\,\text{20} $ . Who scored more and can we add scores (integers) in any order?

Ans: Since, Mishala scored $ \text{20,}\,\text{-40,}\,\text{10} $ .

Therefore, total score of Mishala is

$ \text{=20-40+}\,\text{10} $

$ \text{=-20+10} $

$ \text{=-10} $

And since, Meera scored $ \text{-40,}\,1\text{0,}\,2\text{0} $ .

Therefore, total score of Meera is

$ \text{=-40+}\,\text{10+20} $

Hence, both scored the same points in a game but in a different order.

Yes, we can add integers in any order.

20. Find the product with suitable properties for the following-

a. $ \text{16 }\!\!\times\!\!\text{ }\left( \text{-34} \right)\text{+}\left( \text{-34} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{-18} \right) $

Ans: Given

$ \text{16 }\!\!\times\!\!\text{ }\left( \text{-34} \right)\text{+}\left( \text{-34} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{-18} \right) $

By distributive property-

$ \text{a }\!\!\times\!\!\text{ b+a }\!\!\times\!\!\text{ c=a}\left[ \text{b+c} \right] $

Thus,

$ \text{=-34}\left[ \text{16-18} \right] $

$ \text{=-34 }\!\!\times\!\!\text{ -2} $

$ \text{=68} $

Hence, $ \text{16 }\!\!\times\!\!\text{ }\left( \text{-34} \right)\text{+}\left( \text{-34} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{-18} \right)=68 $ .

b. $ \text{23 }\!\!\times\!\!\text{ -36 }\!\!\times\!\!\text{ 10} $

$ 23\times -36\times 10 $

By commutative property-

$ \left( \text{a }\!\!\times\!\!\text{ b} \right)\text{ }\!\!\times\!\!\text{ c=a }\!\!\times\!\!\text{ }\left( \text{b }\!\!\times\!\!\text{ c} \right) $

$ =23\times \left[ -36\times 10 \right] $

$ =23\times -360 $

$ =-8280 $

21. A fruit merchant earns a profit of Rs. $ \text{6} $ per bag of orange sold and a loss of Rs. $ \text{4} $ per bag of grapes sold.

a. Merchant sells $ \text{1800} $ bags of orange and $ \text{2500} $ bags of grapes. What is the profit or loss?

Ans: Since, profit is denoted by a positive integer and a loss is denoted by a positive integer.

Therefore, profit earned by selling $ \text{1} $ bag of orange is Rs. $ 6 $

Profit earned by selling $ \text{1800} $ bags or orange is

$ \text{6 }\!\!\times\!\!\text{ 1800} $

$ \text{=Rs}\text{. 10,800} $

Loss incurred by selling $ 1 $ bag of grapes is Rs. $ -4 $

Loss incurred by selling $ 2500 $ bags of grapes is

$ =-4\times 2500 $

$ =10,000 $

Total profit or loss earned $ = $ Profit $ + $ Loss

$ =10,800+10,000 $

$ =800 $

Hence, a profit of Rs. $ 800 $ will be earned by a merchant.

b. What is the number of bags of oranges to be sold to have neither profit nor loss if the number of grapes bags are sold is $ \text{900} $ bags?

Ans: Since profit is denoted by a positive integer and a loss is denoted by a positive integer.

Therefore, Loss incurred while selling $ 1 $ bag of grapes $ \text{=-Rs}\text{.4} $

Loss incurred while selling 900 bags of grapes be

$ =-4\times 900 $

$ =-3600 $

Let the number of bags of oranges to be sold $ \text{=x} $

Profit earned when $ 1 $ bag of orange is sold $ \text{=Rs}\text{.6} $

Profit earned while selling x bags of orange $ \text{=6x} $

Condition for no profit, no loss

Profit earned $ + $ Loss incurred $ =0 $

$ \text{6x-3600=0} $

$ \text{6x=3600} $

$ \text{x=}\dfrac{\text{3600}}{\text{6}} $

\[\text{x=600}\]

Hence, to have neither profit nor loss \[\text{600}\] number of bags of oranges to be sold.

22. Verify that \[\text{a }\!\!\div\!\!\text{ }\left( \text{b+c} \right)\ne \left( \text{a }\!\!\div\!\!\text{ b} \right)\text{+}\left( \text{a }\!\!\div\!\!\text{ c} \right)\] for each of the following values of $ \text{a,b} $ and $ \text{c} $ .

a. $ \text{a=8,}\,\text{b=4,}\,\text{c=2} $

Ans: For equation \[\text{a }\!\!\div\!\!\text{ }\left( \text{b+c} \right)\ne \left( \text{a }\!\!\div\!\!\text{ b} \right)\text{+}\left( \text{a }\!\!\div\!\!\text{ c} \right)\].

L.H.S \[\text{=a }\!\!\div\!\!\text{ }\left( \text{b+c} \right)\]

\[\text{=8 }\!\!\div\!\!\text{ }\left( \text{-4+2} \right)\]

\[\text{=8 }\!\!\div\!\!\text{ }\left( \text{-2} \right)\]

\[\text{=-4}\]

R.H.S \[=\left( \text{a }\!\!\div\!\!\text{ b} \right)\text{+}\left( \text{a }\!\!\div\!\!\text{ c} \right)\]

$ \text{=}\left( \text{8 }\!\!\div\!\!\text{ -4} \right)\text{+}\left( \text{8 }\!\!\div\!\!\text{ 2} \right) $

$ \text{=-2+4} $

$ \text{=2} $

Hence, $ \text{L}\text{.H}\text{.S}\ne \text{R}\text{.H}\text{.S} $ .

Thus, \[\text{a }\!\!\div\!\!\text{ }\left( \text{b+c} \right)\ne \left( \text{a }\!\!\div\!\!\text{ b} \right)\text{+}\left( \text{a }\!\!\div\!\!\text{ c} \right)\] for $ \text{a=8,}\,\text{b=4,}\,\text{c=2} $ .

b. $ \text{a=-15,}\,\text{b=2,}\,\text{c=1} $

$ \text{=-15 }\!\!\div\!\!\text{ }\left( \text{2+1} \right) $

$ \text{=-15 }\!\!\div\!\!\text{ 3} $

$ \text{=-5} $

$ \text{=}\left( \text{-15 }\!\!\div\!\!\text{ 2} \right)\text{+}\left( \text{-15 }\!\!\div\!\!\text{ 1} \right) $

$ \text{=-7}\text{.5+}\left( \text{-15} \right) $

$ \text{=-22}\text{.5} $

Hence, $ \text{L}\text{.H}\text{.S}\ne \text{R}\text{.H}\text{.S} $

Thus, \[\text{a }\!\!\div\!\!\text{ }\left( \text{b+c} \right)\ne \left( \text{a }\!\!\div\!\!\text{ b} \right)\text{+}\left( \text{a }\!\!\div\!\!\text{ c} \right)\] for $ \text{a=-15,}\,\text{b=2,}\,\text{c=1} $ .

23. In a CET Examination $ \left( \text{+2} \right) $ marks are given for every current answer and $ \left( \text{-0}\text{.5} \right) $ marks are given for every wrong answer and $ 0 $ for non-attempting any question.

a. Likitha scores $ \text{30} $ marks. If she got $ \text{20} $ correct answers, how many questions she has attempted incorrectly?

Ans: Marks obtained for $ 1 $ correct answer $ \text{=+2} $

Marks obtained for $ 1 $ wrong answer $ \text{=-0}\text{.5} $

So, Marks scored by Likitha = $ 30 $

Marks obtained by $ 20 $ correct answers $ =20\times 2=40 $

Marks obtained for incorrect answer $ = $ Total score $ - $ Marks obtained by $ 20 $ correct answer

$ =30-40 $

$ =-10 $

Marks obtained for $ 1 $ wrong answer $ =-0.5 $

$ \therefore $ The number of incorrect answers $ =\dfrac{-10}{-0.5} $

$ =20 $

Hence, she attempted $ 20 $ questions wrongly.

b. Saara scores $ \text{-4} $ marks if she got $ \text{3} $ correct answers. How many were incorrect?

So, Marks scored by Saara $ =-4 $

Marks obtained for 3 correct answers $ =3\times 2=6 $

Marks obtained for incorrect answers $ = $ Total score $ - $ Marks obtained for $ 3 $ correct answer

$ =-4-6=-10 $

Marks obtained for 1 wrong answer $ =-0.5 $

$ \therefore $ The number of incorrect questions $ =\dfrac{-10}{-0.5} $

Hence, $ 20 $ questions were incorrect.

## Important Questions for CBSE Class 7 Maths Chapter 1 - Integers

Students who have not developed a solid base feel that Maths is hard to excel in and may perceive the subject as a burden. As they have not worked on the subject or failed to grasp its theories and logic, they must start well in class 7 to lead a strong foundation in Mathematics, to conquer good marks in the examination. The magic wand of a good Mathematician is practice. Therefore, to secure a good score in the examination, the students must practice questions based on every theory and formula of Maths.

They may begin with class 7 Maths Chapter 1 important questions to start their practice from the first chapter. If the students find the topics challenging or cannot deal with complex issues, they may use efficient study material to dismantle diplomatic problems into manageable solutions. The critical questions for Class 7 Maths Chapter 1 are available in text and PDF formats online, which is accessible to them for free.

## Details of Class 7 Maths Chapter 1 Integers

In the first chapter of class 7 Maths, the students get a sneak peek into the integer number system. The students must ensure a deep understanding of the topic to proceed with the upcoming chapters. To grasp the topicβs in-depth knowledge, they must practice the important questions for Class 7 Maths Chapter 1. Now, we will highlight some of the important concepts that are included in chapter 1:

An integer is one of the fundamental parts of Mathematics. It can be quoted as a number that can be depicted without any fractional components. For instance, 3, 61, 70, 5 all are integers, while 6.54, 5.89 are non-integer numbers.

We can easily blend out an integer from a series of counting numbers. Let's make it clear with an example, suppose if a counting number is subtracted from itself, the result will be zero. If a larger counting number is removed from a smaller whole number, the output becomes a negative integer. When we subtract the smaller number from the larger whole number, it results in a positive integer. Applying this methodology, we can derive many integers ranging from negative to positive. A set of integers are depicted by 'Z.'

Z = {β¦.., -4,-3,-2,-1,0,1,2,3,4,β¦..n}.

## Properties of Integers

Some of the properties of an integer are as follows:

Commutative Property

The commutative properties of an integer depict that if we perform any operation like multiplication or some numbers, the numbersβ position can be swapped without differing in the output.

Let's make the property clear with an example:

Suppose X and Y are two non-zero integers,

Therefore, the commutative property of addition is X + Y = Y + X.

And, the commutative property of multiplication is X x Y = Y x X.

Associative Property

The associative property of integers depicts that if we perform an addition or multiplication operation on any set of numbers, the result will be identical, irrespective of the grouping of the multiplicands or addends. Some of the traits of associative properties are mentioned below:

The associative property of integers involves a minimum of 3 numbers.

Generally, the integers are grouped using parenthesis.

The numbers defined within the parenthesis are depicted by one unit.

The associative property can only be implemented for addition and multiplication operations and not for division or subtraction.

Let's take an example to make the associative property clear. According to the property, 7 + (8 + 2) = 2 + (7 + 8).

Distributive Property

The distributive property depicts that if two or more numbers are added and multiplied with another number, it will be identical to the current output if each addend is individually multiplied and then added together.

Here's an example to clear the distributive property of integers:

(7 + 1 + 2) x 5

This equation can be simplified to 10 x 5 = 50,

While if we dismantle the equation as 7 x 5 + 1 x 5 + 2 x 5, the result will be equal, i.e., 50.

## Arithmetical Operations Using Integers

Addition of integers.

There are a set of rules to add integer with same and different signs:

During the addition of two integer numbers with the same sign, the output generated also depicts the same sign. Example: 7 + 8 = 15.

For the addition of two integers, one with positive and one with negative signs, the result must retain the largest integer sign. The operation must be performed by subtracting the two integers. Example: 8 + (-14) = -6.

## Subtraction of Integers

The rules to perform subtraction of integers are as follows:

If the subtraction is to be performed between two integers with different signs, i.e., one negative and one positive, the output will retain the largest integer's sign. Example: 9 - 5 = 4.

If the subtraction is performed between two negative integers, the result can be obtained by adding the same number with the opposite sign. Example: -7 - (-9) = -7 + 9 = 2.

## Multiplication of Integers

Multiplication of a positive and a negative integer - The result of the multiplication of a positive and a negative integer can be generated simply by multiplying both the numbers and denoting the output with a minus(-) sign. For example: -7 x 6 = -42.

Multiplication of Two Negative Integers - The product of two negative integers is always a positive integer. For example: -2 x -7 = 14.

Multiplication of Three or More Negative Integers - If the total integers to be multiplied is even, then the output will carry plus (+) sign. The total number of integers to be multiplied is odd. The result will carry the minus(+) sign.

## Division of Integers

Division of a Negative Integer by a Positive Integer - When a negative number is divided by a positive number, the quotient comes with a negative sign.

Division of a Negative Integer by Another Negative Integer - when a negative number is divided by another negative number, the quotient comes with a positive sign.

To get in-depth details about the topics, students can practice the important questions in class 7 Maths chapter 1.

## Important Questions for Class 7 Maths Chapter 1 Integers

Thorough practicing of the important questions for Class 7 Maths integers may help the students grip the topics efficiently. Here we present you the vital questions on the chapter:

What is an Integer? Explain with examples.

What are the dissimilarities between Integers and Real numbers?

What is the commutative property of integers? Explain with examples.

Define the additive identity of an integer.

What is the distributive property of an integer?

Explain the multiplication properties of integers.

What are the division properties of the integers?

A lift is moving towards a mine craft at a rate of 8m/min. How much time will it take to reach -300m if the descent began from 15m above the ground level?

Explain the associative property of integers with an example.

What are the five integer operations?

## Benefits of Important Questions for Class 7 Maths Chapter 1

Students deserve the best study material to guide them on the topics. Therefore, they can unleash the actual benefits of the Class 7 Maths Chapter 1 important questions by thoroughly practicing to secure good marks in the examination. Here are some of the fruitfulness that the students can enjoy from the questions:

The questions are carefully designed under the strict guidelines of CBSE and are reviewed by top-notch professionals.

The students must practise these important questions to understand every concept and theory of the chapter.

In case the students find any difficulty to solve problems or understand any concept, they may refer to the in-detailed solutions. Vedantu, with its team of expert educators, has framed every important question of Class 7 Maths Integers after efficient analysis of the past yearβs question papers.

Students who find it challenging to study Mathematics can build their logic to score well in the examination. To kick start the class 7 academic journey, the students must efficiently practice the Important Questions of Class 7 Maths Chapter 1 Integers, which they can access for free on Vedantuβs website. Students just have to download the PDF and then they can solve the extra questions from the comfort of their homes anytime.

For more study materials related to different chapters of Class 7 Mathematics, students can visit Vedantuβs website and explore the huge collection of free resources available with us.

## FAQs on Important Questions for CBSE Class 7 Maths Chapter 1 - Integers (Free PDF Download)

1. What are the various types of integers?

The various types of Integers are as follows:

Positive Numbers - are those numbers which have a plus sign (+). They are always found on the right side of zero on the number line. Most of the time positive numbers are represented simply as a whole number without the plus sign (+). Every positive number is greater than zero as well as negative numbers.

Example: 1,2, 234, 5667,99999999, etc.

Negative Numbers

Negative numbers are numbers symbolized with a minus sign (-). Negative numbers are represented to the left of zero on a number line.

Example: β¦., -99, -158, -110, -1.

Zero is neither a positive nor a negative integer. It is a neutral number i.e. zero has no sign (+ or -).

2: What are the operations that can be done on Integers? What are its main properties?

The various operations that are performed on Integers are as follows:Β

Subtraction

Multiplication

The main properties of integers are as follows:Β

Closure Property

Identity Property

3: What do you mean by commutative property?

The commutative property of addition and multiplication states that irrespective of the order of terms the result will not change. Swapping of terms will have no effect on the sum or product.

Ex:Β x and y are any two integers, then

β x + y = y + x

β x Γ y = y Γ x

14 + (β6) = 8 = (β6) +1 4;

10 Γ (β3) = β30 = (β3) Γ 10

Note: Subtraction (x β y β y β x) and division (x Γ· y β y Γ· x) are not commutative for integers and whole numbers.

4: Fill in the blanks to make the following statements true:

(β5) + (β 8) = (β 8) + (β¦β¦β¦β¦)

A4: Let us assume the missing integer be x,

= (β5) + (β 8) = (β 8) + (x)

= β 5 β 8 = β 8 + x

= β 13 = β 8 + x

By sending β 8 from RHS to LHS it becomes 8,

= β 13 + 8 = x

Now substitute the x value in the blank place,

(β5) + (β 8) = (β 8) + (- 5) β¦ (This equation is in the form of Commutative law of Addition)

5. What is Chapter 1 of Class 7 Maths about?

The first Chapter of Class 7 Maths begins with a review of the fundamentals of integers that were covered in the previous classes. These recollected ideas are used in the first exercise in the Class 7 Mathematics Chapter 1 answers. You will learn more about integers, their characteristics, and operations in Chapter 1 of Class 7 Mathematics. Expert teachers prepare the important questions to assist students. For a student, the Important Questions of Chapter 1 of Class 7 Mathematics provided by Vedantu are very useful.

6. How many questions are there in Chapter 1 -Integer of Class 7 Maths?

There are four exercises in Chapter 1 βIntegersβ of Class 7 Maths. These four exercises contain a total of 30 questions. For more practice of Chapter 1 of Class 7 Maths, students can refer to the Important Questions of Chapter 1 of Class 7 Maths, prepared by experts at Vedantu for the benefit of the students.

7. Do I need to practice all questions provided in Important Questions for Chapter 1 of Class 7 Maths?

Yes, you need to practise all the questions. All the questions in this chapter deal with a unique concept and in order to get clarity on all the concepts, you must practice every sum. A short review of integers and their representation on the number line, operations such as addition, subtraction, and others that may be done on them, additive inverse, and negative integers are just a few of the subjects covered at the beginning. For important questions of Chapter 1 of Class 7 Maths, visit Vedantu website or mobile app and download the PDF free of cost.

8. What is the associative property of integers?

The associative property of integers states that no matter how the multiplicands or addends are grouped, the result of an addition or multiplication operation on any set of numbers will be the same. The following are some of the characteristics of associative properties:

In general, parentheses are used to arrange the numbers.

One unit represents the numbers defined within the parentheses.

Integers have an associative characteristic that requires a minimum of three numbers.

Q9. What is Associative Property?

Ans: The associative property of integers states that no matter how the multiplicands or addends are grouped, the result of an addition or multiplication operation on any set of numbers will be the same. The following are some of the characteristics of associative properties:

## Chapterwise Important Questions for CBSE Class 7 Maths

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## 7th grade (Illustrative Mathematics)

Unit 1: scale drawings, unit 2: introducing proportional relationships, unit 3: measuring circles, unit 4: proportional relationships and percentages, unit 5: rational number arithmetic, unit 6: expressions, equations, and inequalities, unit 7: angles, triangles, and prisms, unit 8: probability and sampling.

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If you have a scale factor that is less than 1, your scaled copy will be

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Quadrilateral EFGH is a scaled copy of quadrilateral ABCD. Select the 2 statements that are true:

Segment EF is twice as long as segment AB

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The measure of angle HEF is twice the measure of angle DAB

The length of segment EH is 16 units

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The scale of a map says that 1cm represents 5km.

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