NCERT SOLUTIONS FOR CLASS 7th MATHS CHAPTER 1,INTEGERS SOLUTIONS

CBSE SOLUTIONS FOR CLASS 7th MATHS CHAPTER 1,SOLUTIONS FOR INTEGERS

Important points

1. Integers are closed under addition. Integers are also closed under subtraction.

2. Addition is commutative for integers. Subtraction is not commutative for integers.

3. Addition is associative for integers. Multiplication is commutative for integers. Multiplication is associative for integers.

Here you can also see ncert solutions for class 7

Chapter 2 – fractions and decimals

Chapter 4 – simple equations solutions

Chapter 9 – Rational numbers solutions

Maths solutions class 10 real numbers
INTEGERS CHAPTER 1

EXERCISE 1.1

Problem 1

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

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

b. What is the temperature difference between the hottest and the coolest places among the above ?

c. What is the temperature difference between Lahulspiti and Srinagar ?

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

Solutions :

a. By observing the number line, the temperature of the cities as follows

Lahulspiti : – 8°C

Srinagar : – 2°C

Shimla : – 5°C

Ooty : – 14°C

Bengaluru : – 22°C

b. The temperature at the hottest place – Bengaluru = 22°C

The temperature at the coolest place – Lahulstipti = – 8°C

The temperature difference between the hottest and coolest place = 22° C – ( – 8°C ) = 22°C + 8°C = 30°C

c. The temperature at the Lahulspiti = – 8°C

The temperature at the Srinagar = – 2°C

The temperature between Srinagar and Lahulspiti = – 2°C – ( – 8°C ) = – 2°C + 8°C = 6°C

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

No, it is more than the temperature at Srinagar.

Problem 2

2. In a quiz, positive marks are given for correct answers and negative marks are given for incorrect answered. If Jack’s scores in five successive rounds were 25, – 5, – 10, 15 and 10, what was his total at the end ?

Solution :

Jack’s scores in successive rounds are 25, – 5, – 10, 15 and 10.

The total Jack’s score at the end = 25 + ( – 5 ) + ( – 10 ) + 15 + 10

= 25 – 5 – 10 + 15 + 10 = 35

Therefore, Jack’s score at the end = 35

Problem 3

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 ?

Solution :

The temperature on Monday = – 5°C

The temperature on Tuesday = Temperature on Monday – 2°C = – 5°C – 2°C = – 7°C

The temperature on Wednesday = Temperature on Tuesday + 4°C = – 7°C + 4°C = – 3°C

Problem 4

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

Solution :

The height of the plane = 5000 m

The depth of the submarine = 5000 m – ( – 1200 m ) = 5000 m + 1200 m = 6200 m

Therefore, the vertical distance between plane and submarine = 6200 m

Problem 5

5. Mohan deposits Rs. 2000 in his bank account and withdraws Rs. 1642 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.

Solution :

From the problem, the amount deposited is represented by a positive integer and the amount withdrawal is represented by a negative integer.

Mohan deposited the amount = Rs. 2,000

Mohan withdrawal the amount = – Rs. 1,600

Balance in account = amount deposited + amount withdrawal = Rs. 2,000 + ( – Rs. 1,642 ) = Rs. 358

Therefore, the balance in Mohan’s account after withdrawal = Rs. 358

Problem 6

6. Rita goes 20 km towards east from a point A to the point 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 traveled towards west ? By which integer will you represent her final position from A ?

Solution :

From the problem, the distance traveled towards is represented by positive integer and the distance traveled towards west is represented by negative integer.

The distance traveled by Rita in east direction = 20 km

The distance traveled by Rita in west direction = – 30 km

The distance traveled from the point A = 20 km + ( – 30 km ) = 20 km – 30 km = – 10 km

The distance traveled by Rita from the point A is represented by negative integer = – 10

Therefore, Now Rita is in west direction.

Problem 7

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

Solution :

From the squares, it can be observe that in square ( i ),

The sum of the integers in every row = 0

The sum of the integers in every column = 0

The sum of one of its diagonal is not zero i.e – 4 – 2 = – 6

In square ( ii ),

The sum of the integers in each row = – 9

The sum of the integers in each column = – 9

The sum of the integers in its each diagonal = – 9

Therefore, the square ( ii ) is a magic figure.

Problem 8

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

Solutions :

i. a = 21, b = 18

a – ( – b ) = 21 – ( – 18 ) = 21 + 18 = 39

a + b = 21 + 18 = 39

Therefore, a – ( – b ) = a + b

ii. a = 118, b = 125

a – ( – b ) = 118 – ( – 125 ) = 118 + 125 = 143

a + b = 118 + 125 = 143

Therefore, a – ( – b ) = a + b

iii. a = 75, b = 84

a – ( – b ) = 75 – ( – 84 ) = 74 + 84 = 159

a + b = 75 + 84 = 159

Therefore, a – ( – b ) = a + b

iv. a = 28, b = 11

a – ( – b ) = 28 – ( – 11 ) = 28 + 11 = 39

a + b = 28 + 11 = 39

Therefore, a – ( – b ) = a + b

Problem 9

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

a. ( – 8 ) + ( – 4 ) ……… ( – 8 ) – ( – 4 )

b. ( – 3 ) + 7 – ( 19 ) …….. 15 – 8 + ( – 9 )

c. 23 – 41 + 11 ………… 23 – 41 – 11

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

e. – 231 + 79 + 51 ……… – 399 + 159 + 81

Solutions :

a. ( – 8 ) + ( – 4 )

– 8 – 4 = – 12

( – 8 ) – ( – 4 )

– 8 + 4 = – 4

– 12 < – 4

Therefore, ( – 8 ) + ( – 4 ) < ( – 8 ) – ( – 4 )

b. ( – 3 ) + 7 – ( 19 ) …….. 15 – 8 + ( – 9 )
( – 3 ) + 7 – ( 19 )

= – 3 + 7 – 19 = – 15

15 – 8 + ( – 9 )

= 15 – 8 – 9 = – 2

– 15 < – 2

Therefore , ( – 3 ) + 7 – ( 19 ) < 15 – 8 + ( – 9 )

c. 23 – 41 + 11 ………… 23 – 41 – 11
23 – 41 + 11 ………… 23 – 41 – 11
34 – 41 ………………… 23 – 52

– 7 ……………………… – 29
– 7 > – 19

Therefore, 23 – 41 + 11 > 23 – 41 – 11
d. 39 + ( – 24 ) – ( 15 ) …….. 36 + ( – 52 ) – ( – 36)
39 – 24 – 15 ………………….. 36 – 52 + 36
39 – 39 …………………………. 72 – 52
0 ……………………………………. 20

0 < 20

Therefore, 39 + ( – 24 ) – ( 15 ) < 36 + ( – 52 ) – ( – 36)
e. – 231 + 79 + 51 ……… – 399 + 159 + 81
– 231 + 79 + 51 ……… – 399 + 240
– 231 + 130 ……… – 399 + 159 + 81
– 101 …………………… – 159

Therefore, – 231 + 79 + 51 > – 399 + 159 + 81

Problem 10

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 2 steps up. In how many jumps will he 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 in every move. In how many jumps will he reach back the top 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 part ( i ) and ( ii ) by completing the following ; ( a ) – 3 + 2 – …. = – 8, ( b ) 4 – 2 + …. = 8. In ( a ) the sun ( – 8 ) represents going down by eight steps. S, what will the sum 8 in ( b ) represent ?

Solution :

From the problem, the steps moved down is represented by negative integer and the steps moved up by positive integer.

The monkey is at the topmost or first step.

The water level is at the ninth step.

i. Let the monkey will be at step = 1

After first jump, the monkey will be at step = 1 + 3 = 4

After second jump, the monkey will be at step = 4 + ( – 2 ) = 2

After third jump, the monkey will be at step = 2 + 3 = 5

After 4th jump, the monkey will be at step = 5 + ( – 2 ) = 3

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

After 6th jump, the monkey will be at step = 6 + ( – 2 ) = 4

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

After 8th jump, the monkey will be at step = 7 + ( – 2 ) = 5

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

After 10th jump, the monkey will be at step = 8 + ( – 2 ) = 6

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

Therefore, the monkey will be at water level after 11 jumps.

ii. Let he monkey will be at step = 9

After first jump, the monkey will be at step = 9 + ( – 4 ) = 5

After second jump, the monkey will be at step = 5 + 2 = 7

After third jump, the monkey will be at step = 7 + ( – 4 ) = 3

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

After 5th jump, the monkey will be at step = 5 + ( – 4 ) = 1

Therefore, the monkey will reach back at first step after five jumps.

iii. a. The monkey moves in part ( i ) = – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 = – 8

b. The monkey moves in part ( ii ) = 4 – 2 + 4 – 2 + 4 = 8

The sum 8 in ( b ) represent the monkey going up 8 steps.

INTEGERS SOLUTIONS FOR 7TH CLASS CHAPTER 1,SOLUTIONS FOR INTEGERS

CBSE SOLUTIONS FOR CLASS 7TH MATHS CHAPTER 1,INTEGERS SOLUTIONS

EXERCISE 1. 2

1. Write down a pair of integers whose:

a. Sum is – 7

b. Difference is – 10

c. Sum is 0

Solutions :

a. Sum is – 7

( – 3 ) + ( – 4 ) = – 3 – 4 = – 7 or

( -5 ) + ( – 2 ) = – 5 – 2 = – 7

b. Difference is – 10

( – 15 ) – ( – 5 ) = – 15 + 5 = – 10 or

( – 18 ) – ( – 8 ) = – 18 + 8 = – 10

c. Sum is 0

( – 6 ) + 6 = 0 or

( – 3 ) + 3 = 0

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.

Solutions :

a. ( – 4 ) – ( – 12 ) = – 4 + 12 = 8

b. ( – 9 ) + 4 = – 9 + 4 = – 5

c. ( – 2 ) – ( + ) = – 2 – 1 = – 3

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

Solution :

The scores of team A = – 40, 10 and 0

The scores of team B = 10, 0 and – 40

The total score of team A = – 40 + 10 + 0 = – 30

The total score of team B = 10 + 0 + ( – 40 ) = – 30

The scores of both the teams are equal.

Yes, we observe that the scores got by both teams in successive rounds are numerically equal. But they are different in order.

Problem :

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

i. ( – 5 ) + ( – 8 ) = ( – 8 ) + ( ……… )

ii. – 53 + …………. = – 53

iii. 17 + ………. = 0

iv. [ 13 + ( – 12 ) ] + ( ………… ) = 13 + [ ( – 12 ) + ( – 7 ) ]

v. ( – 4 ) + [ 15 + ( – 3 ) ] = [ – 4 + 15 ] + ……..

Solutions :

i. ( – 5 ) + ( – 8 ) = ( – 8 ) + ( – 5 )

ii. – 53 + 0 = – 53

iii. 17 + ( – 17 ) = 0

iv. [ 13 + ( – 12 ) ] + ( – 7 ) = 13 + [ ( – 12 ) + ( – 7 ) ]

v. ( – 4 ) + [ 15 + ( – 3 ) ] = [ – 4 + 15 ] + ( – 3 )

NCERT SOLUTIONS FOR MATHS CLASS 7TH CHAPTER 1

EXERCISE 1 . 3

Problems :

1. Find each of the following products :

a. 3 × ( -1 )

b. ( – 1 ) × 225

c. ( – 21 ) × ( – 30 )

d. ( – 316 ) × ( – 1 )

e. ( – 15 ) × 0 × ( – 18 )

f. ( – 12 ) × ( – 11 ) × ( 10 )

g. 9 × ( – 3 ) × ( – 6 )

h. ( – 18 ) × ( – 5 ) × ( – 4 )

i. ( – 1 ) × ( – 2 ) × ( – 3 ) × 4

j. ( – 3 ) × ( – 6 ) × ( – 2 ) × ( – 1 )

Solutions :

a. 3 × ( – 1 ) = – 3

b. ( – 1 ) × 225 = – 225

c. ( – 21 ) × ( – 30 ) = 630

d. ( – 316 ) × ( – 1 ) = 316

e. ( – 15 ) × 0 × ( – 18 ) = 0

f. ( – 12 ) × ( – 11 ) × ( 10 ) = 1320

g. 9 × ( – 3 ) × ( – 6 ) = 9 × 18 = 162

h. ( – 18 ) × ( – 5 ) × ( – 4 )

( – 18 ) × [ ( – 5 ) × ( – 4 ) ]

= – 18 × 20 = – 360

i. ( – 1 ) × ( – 2 ) × ( – 3 ) × 4

[ ( – 1 ) × ( – 2 ) ] × [ ( – 3 ) × 4 ]

= 2 × ( – 12 ) = – 24

j. ( – 3 ) × ( – 6 ) × ( – 2 ) × ( – 1 )

[ ( – 3 ) × ( – 6 ) ] × [ ( – 2 ) × ( – 1 ) ]

= 18 × 2 = 36

Problem :

2. Verify the following :

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

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

Solution :

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

18 × [ 7 + ( – 3 ) ] = 18 × ( 7 – 3 ) = 18 × 4 = 72

[ 18 × 7 ] + [ 18 × ( – 3 )] = ( 126 ) + ( – 54 ) = 126 – 54 = 72

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

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

( – 21 ) × [ ( – 4 ) + ( – 6 ) ] = ( – 21 ) × [ – 4 – 6 ] = ( – 21 ) × ( – 6 ) = 210

[ ( – 21 ) × ( – 4 ) ] + [ ( – 21 ) × ( – 6 ) ] = [ 84 ] + [ 126 ] = 210

Problems :

3. i. For any integer a, what is ( – 1 ) × a equal to ?

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

a. – 22

b. 37

c. 0

Solutions :

i. For any integer a, ( – 1 ) × a = – a

ii. a. ( – 1 ) × ( – 22 ) = 22

b. ( – 1 ) × ( – 37 ) = – 37

c. ( – 1 ) × ( 0 ) = 0

Problem :

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

Solution :

( – 1 ) × 5 = – 5

( – 1 ) × 4 = – 4 = – 5 + 1

( – 1 ) × 3 = – 3 = – 4 + 1

( – 1 ) × 2 = – 2 = – 3 + 1

( – 1 ) × 1 = – 1 = – 2 + 1

( – 1 ) × 0 = – 0 = – 1 + 1

So, ( – 1 ) × ( – 1 ) = 1 = 0 + 1

5. 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

Solutions :

a. 26 × ( – 48 ) + ( – 48 ) × ( – 36 )

[ ( – 48 ) × 26 ] + [ ( – 48 ) × ( – 36 ) ]

( a × b ) + ( a × c ) = a × ( b + c )

– 48 × [ 26 + ( – 36 ) ]

= – 48 × ( 26 – 36 )

= – 48 × ( – 10 ) = 480

b. 8 × 53 × ( – 125 )

a × b × c = ( a × b ) × c

8 × 53 × ( – 125 ) = ( 8 × 53 ) × ( – 125 )

424 × ( – 125 ) = – 53000

c. 15 × ( – 25 ) × ( – 4 ) × ( – 10 )

[ 15 × ( – 25 ) ] × [ ( – 4 ) × ( – 10 ) ]

= ( – 375 ) × ( 40 ) = – 15000

d. ( – 41 ) × ( 102 )

= – 41 × ( 100 + 2 )

= ( – 41 × 100 ) + ( – 41 × 2 )

= – 4100 + ( – 82 ) = – 4182

e. 625 × ( – 35 ) + ( – 625 ) × ( 65 )

= 625 × ( – 35 ) + ( – 625 ) × ( 65 )

= 625 × [ ( – 35 ) + ( – 65 ) ]

= 625 × ( – 100 ) = – 62500

f. 7 × ( 50 – 2 )

= ( 7 × 50 ) – ( 7 × 2 )

= 350 – 14 = 336

g. ( – 17 ) × ( – 29 )

( – 17 ) × ( – 20 – 9 )

= ( – 17 × – 20 ) × ( – 17 × 9 )

= – 340 + 153 = 493

h. ( – 57 ) × ( – 19 ) + 57

( – 57 ) × ( – 19 – 1 )

= – 57 × ( – 20 ) = 1140

Problem :

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 ?

Solution :

The initial temperature = 40° C

Change in temperature per one hour = – 5°C

Change in temperature per 10 hours = 10 × ( – 5 ) = – 50°C

The temperature of the room at the end ° 40°C + ( – 50°C ) = 40°C – 50°C = – 10°C

Problem :

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

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

ii. Reshma gets five answers and five incorrect answers, what is her score ?

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

Solution :

Marks given for one correct answer = 5

Marks given for one incorrect answer = – 2

i. For Mohan, marks given for four correct answers = 4 × 5 = 20

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

The score got by Mohan = 20 + ( – 12 ) = 20 – 12 = 8

ii. For Reshma, marks given for five correct answers = 5 × 5 = 25

Marks given for five incorrect answers = 5 × ( – 2 ) = – 10

The score got by Reshma = 25 + ( – 10 ) = 25 – 10 = 15

iii. For Hema, marks given for two correct answers = 2 × 5 = 10

Marks given for five incorrect answers = 5 × ( – 2 ) = – 10

The score got by Mohan = 10 + ( – 10 ) = 00 – 10 = 0

Problem

8. A cement company earns 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 ?

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.

Solution :

Profit is represented by a positive integer and loss is represented by a negative integer.

The profit earned while selling one bag of white cement = Rs. 8

The loss incurred while selling one bag of grey cement = Rs. – 5

i. The profit earned while selling 3,000 bag of white cement = Rs. 8 × 3,000 = Rs. 24,000
The loss incurred while selling 5,000 bag of grey cement = Rs. – 5 × 5,000 = Rs. – 25,000
The total profit or loss earned = Rs 24,000 + ( – 25,000 ) = Rs. – 1,000

Therefore, a loss of Rs. 1,000 incurred by the company.

ii. Let the number of bags of white cement to be sold = x

The profit earned while selling x bags of white cement = x × Rs. 8 = Rs. 8x

The loss incurred while selling 6,400 bag of grey cement = Rs. – 5 × 6,400 = Rs. – 32,000
In the condition no profit and no loss

Profit earned + loss incurred = 0

Rs. 8x + ( Rs. – 32,000 ) = 0

Rs. 8x – Rs. 32,000 = 0

Rs. 8x = Rs. 32,000

x = 32,000 ÷ 8 = 4,000

Therefore, 4,000 bags of white cement must be sold the company.

Problem

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

a. ( – 3 ) × …….. = 27

b. 5 × …….. = – 35

c. …….. × ( – 8 ) = – 56

d. ………. × ( – 12 ) = 132

Solutions :

a. ( – 3 ) × – 9 = 27

b. 5 × ( – 7 ) = – 35

c. 7 × ( – 8 ) = – 56

d. 11 × ( – 12 ) = -132

NCERT SOLUTIONS FOR MATHS CLASS 7TH CHAPTER 1,INTEGERS SOLUTIONS

NCERT SOLUTIONS FOR CLASS 7 MATHS/ CHAPTER 1/SOLUTIONS FOR INTEGERS

SOLUTIONS FOR MATHS CLASS 7/ CHAPTER 1 /INTEGERS

EXERCISE 1 . 4

Problem 1

1. Evaluate each of the following :

a. ( – 30 ) ÷ 10

b. 50 ÷ ( – 5 )

c. ( – 36 ) ÷ ( – 9 )

d. ( – 49 ) ÷ ( 49 )

e. 13 ÷ [ ( – 2 ) + 1 ]

f. 0 ÷ ( – 12 )

g. ( – 31 ) ÷ [ ( – 30 ) + ( – 1 ) ]

h. [ ( – 36 ) ÷ 12 ] ÷ 3

i. [ ( – 6 + 5 ) ] ÷ ( – 2 ) + 1 ]

Solutions :

a. ( – 30 ) ÷ 10 = – 3

b. 50 ÷ ( – 5 ) = – 10

c. ( – 36 ) ÷ ( – 9 ) = 4

d. ( – 49 ) ÷ ( 49 ) – 1

e. 13 ÷ [ ( – 2 ) + 1 ] = 13 ÷ ( – 1 ) = – 13

f. 0 ÷ ( – 12 ) = 0

g. ( – 31 ) ÷ [ ( – 30 ) + ( – 1 ) ] = – 31 ÷ ( – 31 ) = 1

h. [ ( – 36 ) ÷ 12 ] ÷ 3 = ( – 3 ) ÷ 3 = – 1

i. [ ( – 6 + 5 ) ] ÷ [ ( – 2 ) + 1 ] = ( – 1 ) ÷ ( – 1 ) = 1

Problem 2

2. Verify that a ÷ ( b + c ) is not equal to ( a ÷ b ) + ( a ÷ c ) for each of the following values of a, b and c.

a. a = 12, b = – 4, c = 2

b, a = ( – 10 ), b = 1, c = 1

Solutions :

a. a = 12, b = – 4, c = 2

a ÷ ( b + c ) = 12 ÷ [ ( – 4 ) + 2 ]

= 12 ÷ ( – 2 ) = – 6

( a ÷ b ) + ( a ÷ c ) = [ 12 ÷ ( – 4 ) ] + [ 12 ÷ 2 ]

= ( – 3 ) + 6 = 3

Therefore, a ÷ ( b + c ) is not equal to ( a ÷ b ) + ( a ÷ c )

b, a = ( – 10 ), b = 1, c = 1

a ÷ ( b + c ) = – 10 ÷ ( 1 + 1 )

= – 10 ÷ 2 = – 5

( a ÷ b ) + ( a ÷ c ) = [( – 10 ) ÷ 1 ] + [ ( – 10 ) ÷ 1 ]

= ( – 10 ) + ( – 1 ) = – 10 – 10 = – 20

Therefore, a ÷ ( b + c ) is not equal to ( a ÷ b ) + ( a ÷ c )

Problem : 3

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

Solutions :

a. 369 ÷ 1 = 369

b. ( – 75 ) ÷ 75 = – 1

c. ( – 206 ) ÷ ( – 206 ) = 1

d. – 87 ÷ ( – 1 ) = 87

e. – 87 ÷ 1 = – 87

f. – 48 ÷ 48 = – 1

g. 20 ÷ ( – 10 ) = – 2

h. – 12 ÷ ( 4 ) = – 3

Problem : 4

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

Solutions :

1. ( a, b )= ( 24, – 8 )

a ÷ b = 24 ÷ ( – 8 ) = – 3

2. ( a, b )= ( – 15, 5 )

a ÷ b = – 15 ÷ ( 5 ) = – 3

3. ( a, b )= ( 9, – 3 )

a ÷ b = 9 ÷ ( – 3 ) = – 3

4. ( a, b )= ( – 12, 4 )

a ÷ b = – 12 ÷ 4 = – 3

5. ( a, b )= ( 21, – 7 )

a ÷ b = 21 ÷ ( – 7 ) = – 3

Problem : 5

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 be the temperature at mid – night ?

Solution :

Initial temperature at noon, 12 : 00 = 8° C

Change in temperature per hour = – 2° C

The temperature at 1 : 00PM = = 10° C – 2°C = 8°C

The temperature at 2 : 00 PM = = 8° C – 2°C = 6°C

The temperature at 3 : 00 PM = = 6° C – 2°C = 4°C

The temperature at 4 : 00 PM = = 4° C – 2°C = 2°C

The temperature at 5 : 00 PM= = 2° C – 2°C = 0°C

The temperature at 6 : 00 PM = = 0° C – 2°C = – 2°C

The temperature at 7 : 00 PM = – 2° C – 2°C = – 4°C

The temperature at 8 : 00 PM = = – 4° C – 2°C = – 6°C

The temperature at 9 : 00 PM = = – 6 ° C – 2°C = – 8°C

Therefore, the temperature will be 8°C below 0 at 9 : 00 PM.

The temperature at 10 : 00 PM = = – 8° C – 2°C = – 10°C

The temperature at 11 : 00 PM = = – 10° C – 2°C = – 12°C

The temperature at 12 : 00, midnight = – 12° C – 2°C = – 14°C

Therefore, the temperature at midnight will be 14°C below zero .

Problem : 6

6. In a class test ( + 3 ) marks are given for every correct answer and ( – 2 ) marks are given for every incorrect answer and no marks for not attempting any question. i. Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly ? ii. Mohan scores – 5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly ?

Solution :

Marks obtained for one right answer = 3

Marks obtained for one wrong answer = – 2

i. For Radhika

The marks scored by Radhika = 20

Marks obtained for 12 correct answers = 12 × 3 = 36

Marks obtained for incorrect answers = total score – marks obtained for correct answers = 20 – 36 = – 16

The number of incorrect answers = – 16 ÷ ( – 2 ) = 8

Therefore, she attempted 8 questions incorrectly.

ii. For Mohini

Marks scored by Mohini = – 5

Marks obtained for 7 correct answers = 7 × 3 = 21

Marks obtained for incorrect answers = total score – marks obtained for correct answers = – 5 – 21 = – 26
The number of incorrect answers = – 26 ÷ ( – 2 ) = 13

Therefore, she attempted 13 questions incorrectly.

Problem : 7

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

Initial height = 10 m

Final depth – – 350

The total distance descended by the elevator = ( – 360 ) – ( + 10 ) = – 360

Time taken by the elevator to descend 6 m = 1 minute

Time taken by the elevator to descend ( – 360 ) = – 369 ÷ ( – 6 ) = 60 minutes = 1 hour.

7TH CLASS MATHS SOLUTIONS CHAPTER 1 CBSE

Note : Observe the solutions of integers  once and try them in your own methods.

1. Integers

2. Fractions and decimals

3. Simple equations

4. Rational numbers

You can also see

Ncert solutions for maths class 6 some chapters

Ncert solutions for maths class 7 some chapters

Ncert solutions for maths class 8 some chapters

Intermediate maths solutions for some chapters

Ncert solutions for class 8 chapter 12

Inter first year maths 1b some chapters

1Locus

2. Transformation of axes

3. The straight lines

4. Three dimensional coordinates

5. Direction cosines and direction ratios

6. The plane

7. Errors and approximations

8. Tangent and normal

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