# NCERT Solutions for Class 7 Maths chapter 1 Integers | integers class 7

NCERT Solutions for Class 7 Maths chapter 1 Integers Class 7th maths chapter 1 Class 7 maths. Here We learn what is in Class 7 maths solution chapter 1 Integers and how to solve questions with easiest method. In this chapter we solve the question of NCERT Solutions for integers class 7 Exercise 1.1, Class 7 maths chapter 1 Exercise 1.2, Class 7 maths chapter 1 Exercise 1.3 and Class 7 maths chapter 1 Exercise 1.4. chapter 1 Class 7 maths class 7th maths chapter 1

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# ncert solutions for Class 7 maths chapter 1 Integers

integers class 7th maths chapter 1

## Exercise – 1

integers class 7

### ncert solutions for Class 7 maths chapter 1 Integers

Class 7th maths solution ex 1.1

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

(a) Observe this number line and 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 temperature of Srinagar and Shimla taken together is less than the temperature at Shimla? Is it also less than the temperature at Srinagar?

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

Places |
Temperature |

Lahulspiti | тИТ 8┬░C |

Srinagar | тИТ 2┬░C |

Shimla | + 5┬░C |

Ooty | + 15┬░C |

Bengaluru | + 22┬░C |

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

The hottest places Bengaluru temperature = + 22┬░C

The coldest places Lahulspiti temperature = тИТ 8┬░C

The temperature difference between the hottest and the coldest places is = + 22 тИТ (тИТ 8)┬а

= + 22 + 8

= + 30┬░C

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

the temperature of Lahulspiti = тИТ 8┬░C

the temperature of Srinagar = тИТ 2┬░C

the temperature difference between Lahulspiti and Srinagar is = тИТ 2 тИТ (тИТ 8)

= тИТ 2 + 8

= 6┬░C

**(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?**

the temperature at Shimla is = 5┬░C

the temperature at Srinagar = тИТ 2┬░C

The sum of the temperature at Shimla and Srinagar = 5 + (тИТ 2)

= 5 тИТ 2

= 3┬░C

The sum of the temperature at Shimla and Srinagar < the temperature at Shimla

3┬░C < 5┬░C

Yes, We can say that temperature of Srinagar and Shimla taken together is less than the temperature at Shimla

The sum of the temperature at Shimla and Srinagar < the temperature at Srinagar

3┬░C > тИТ 2┬░C

No, The sum of the temperature at Shimla and Srinagar is not less than Srinagar temperature.

**2. In a quiz, positive marks are given for correct answers and negative marks are given for incorrect answers. If JackтАЩs scores in five successive rounds were 25, тАУ 5, тАУ 10, 15 and 10, what was his total at the end?****Solution :**

JackтАЩs scores = 25, тАУ 5, тАУ 10, 15 and 10

The sum of Jack’s scores = 25 + (тИТ 5) + (тИТ 10) + 15 + 10

= 25 тИТ 5 тИТ 10 + 15 + 10

= 25 + 15 + 10 тИТ 5 тИТ 10

= 50 тИТ 15

= 35 Marks

**3. At Srinagar temperature was тАУ 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 :**

Srinagar temperature on Monday = тИТ 5┬░C

Temperature dropped at Tuesday = тИТ 2┬░C

So, Tuesday temperature is = тИТ 5 + (тИТ 2)

= тИТ 5 тИТ 2

= тИТ 7┬░C

Srinagar temperature on Monday = тИТ 7┬░C

Temperature rose on Wednesday = + 4┬░C (Temperature rise show in positive sign)

Temperature on Wednesday = тИТ 7 + 4

= тИТ 3┬░C

**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 :**

Plane height from the sea level = + 5000 m

submarine below the sea level = тИТ 1200 m

the vertical distance between them = 5000 тИТ (тИТ 1200)

= 5000 + 1200

= + 6200 m

**5. Mohan deposits rupees 2,000 in his bank account and withdraws rupees 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 MohanтАЩs account after the withdrawal.****Solution :**

Deposits money in Mohan bank account = rupees + 2000

Withdraws money from Mohan bank account = rupees тИТ 1642

We will represent the deposited amount in positive form.

Balance in Mohan’s account after withdrawal = 2000 тИТ 1642

= rupees 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?****Solution :**

Distance covered by Rita from point A towards east to point B = 20 km (positive integer)

Distance covered along the road from point B to the west = тИТ 30 km

Note : If the distance covered towards East is represented by a positive integer, then we will express the distance covered towards West as a negative integer.

its final position from point A = 20 + (тИТ 30)

= 20 тИТ 30

= тИТ 10┬а

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

In square (i) sum of first row = 5 + (тИТ 1) + (тИТ 4) = 5 тИТ 1 тИТ 4 = 5 тИТ 5 = 0

In square (i) sum of second row = тИТ 5 + (тИТ 2) + 7 = тИТ 5 тИТ 2 + 7 = тИТ 7 + 7 = 0

In square (i) sum of third row = 0 + 3 + (тИТ 3) = 0 + 3 тИТ 3 = 3 тИТ 3 = 0

In square (i) sum of first column = 5 + (тИТ 5) + 0 = 5 тИТ 5 = 0

In square (i) sum of second column = тИТ 5 + (тИТ 5) + 0 = тИТ 5 тИТ 5 + 0 = 0

In square (i) sum of third column = тИТ 1 + (тИТ 2) + 3 = тИТ 1 тИТ 2 + 3 = тИТ 3 + 3 = 0

In square (i) sum of first diagonal = 5 + (тИТ 2) + (тИТ 3) = 5 тИТ 2 тИТ 3 = 5 тИТ 5 = 0

In square (i) sum of second diagonal = тИТ 4 + (тИТ 2) + 0 = тИТ 4 тИТ 2 + 0 = тИТ 6

In square (i) each row, column and diagonal have not the same sum. So, this is not a magical square.

In square (ii) sum of first row = 1 + (тИТ 10) + 0 = 1 тИТ 10 = тИТ 9

In square (ii) sum of second row = тИТ 4 + (тИТ 3) + (тИТ 2) = тИТ 4 тИТ 3 тИТ 2 = тИТ 9

In square (ii) sum of third row = тИТ 6 + 4 + (тИТ 7) = тИТ 6 + 4 тИТ 7 = тИТ 13 + 4 = тИТ 9

In square (ii) sum of first column = 1 + (тИТ 4) + (тИТ 6) = 1 тИТ 4 тИТ 6 = 1 тИТ 10 = тИТ 9

In square (ii) sum of second column = тИТ 4 + (тИТ 3) + (тИТ 2) = тИТ 4 тИТ 3 тИТ 2 = тИТ 9┬а

In square (ii) sum of third column = тИТ 6 + 4 + (тИТ 7) = тИТ 6 + 4 тИТ 7 = тИТ 13 + 4 = тИТ 9

In square (ii) sum of first diagonal = 1 + (тИТ 3) + (тИТ 7) = 1 тИТ 3 тИТ 7 = 1 тИТ 10 = тИТ 9

In square (ii) sum of second diagonal = 0 + (тИТ 3) + (тИТ 6) = тИТ 3 тИТ 6 = тИТ 9

In square (ii) each row, column and diagonal have the same sum. So, this is a magical square.

**8. Verify a тАУ (тАУ 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 |

**Solution :┬а****(i) a = 21, b = 18**

Put the a and b value in a тИТ (тИТ b) = a + b

21 тИТ (тИТ 18) = 21 + 18

21 + 18 = 21 + 18

39 = 39

L.H.S = R.H.S

**(ii) a = 118, b = 125**

Put the a and b value in a тИТ (тИТ b) = a + b

118 тИТ (тИТ 125) = 118 + 125

118 + 125 = 118 + 125

243 = 243

L.H.S = R.H.S┬а

**(iii) a = 75, b = 84**

Put the a and b value in a тИТ (тИТ b) = a + b

75 тИТ (тИТ 84) = 75 + 84

75 + 84 = 75 + 84

159 = 159

L.H.S = R.H.S┬а

**(iv) a = 28, b = 11**

Put the a and b value in a тИТ (тИТ b) = a + b

28 тИТ (тИТ 11) = 28 + 11

28 + 11 = 28 + 11

39 = 39

L.H.S = R.H.S┬а

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

(a) | (тИТ 8) + (тИТ 4) | . . . . | (тИТ 8) тИТ (тИТ 4) |

(b) | (тИТ 3) + 7 тИТ (19) | . . . .┬а | 15 тИТ 8 + (тИТ 19) |

(c) | 23 тИТ 41 + 11 | . . . . | 23 тИТ 41 тИТ 11 |

(d) | 39 + (тИТ 24) тИТ (15) | . . . . | 23 тИТ 41 тИТ 11 |

(e) | 39 + (тИТ 24) тИТ (15) | . . . . | 36 + (тИТ 52) тИТ (тИТ 36) |

**Solution :****(a) (тИТ 8) + (тИТ 4) . . . . (тИТ 8) тИТ (тИТ 4)**

тИТ 8 тИТ 4 . . . . тИТ 8 + 4

тИТ 12 **<** тИТ 4

**(b) (тИТ 3) + 7 тИТ (19) . . . . 15 тИТ 8 + (тИТ 19)**

тИТ 3 + 7 тИТ 19 . . . . 15 тИТ 8 тИТ 19

тИТ 22 + 7 . . . . 15 тИТ 27

тИТ 15 **<** тИТ 12

**(c) 23 тИТ 41 + 11 . . . . 23 тИТ 41 тИТ 11**

23 + 11 тИТ 41 . . . . 23 тИТ 52

34 тИТ 41 . . . . тИТ 29

тИТ 7 **>** тИТ 29

**(d) 39 + (тИТ 24) тИТ (15) . . . . 36 + (тИТ 52) тИТ (тИТ 36)**

39 тИТ 24 тИТ 15 . . . . 36 тИТ 52 + 36

39 тИТ 39 . . . . 36 + 36 тИТ 52

0 . . . . 72 тИТ 52

0 **<** 20

**(e) тИТ 231 + 79 + 51 . . . . тИТ 399 + 159 + 81**

тИТ 231 + 130 . . . . тИТ 399 + 240

тИТ 101 **>** тИТ 159

**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?

(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) тАУ 3 + 2 тАУ … = тАУ 8

(b) 4 тАУ 2 + … = 8. In (a) the sum (тАУ 8) represents going down by eight steps. So, what will the sum 8 in (b) represent?**Solution :****(i) He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level?**

monkey is sitting on the topmost step

The position of the monkey after the jumps will be as follows:

In the first jump he will reach the step = 4 step

In the second jump he will reach the step = 4 тИТ 2 = 2 step

In the third jump he will reach the step = 2 + 3 = 5 step

In the four jump he will reach the step = 5 тИТ 2 = 3 step

In the five jump he will reach the step = 3 + 3 = 6 step

In the six jump he will reach the step = 6 тИТ 2 = 4 step

In the seven jump he will reach the step = 4 + 3 = 7 step

In the eight jump he will reach the step = 7 тИТ 2 = 5 step

In the nine jump he will reach the step = 5 + 3 = 8 step

In the ten jump he will reach the step = 8 тАУ 2 = 6 step

In the eleven jump he will reach the step = 6 + 3 = 9 step

Hence, the ninth step which is at the water level will be able to reach in 11 jumps.

**(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?**

Here, the monkey is at the water level i.e. the ninth step.

After the jumps, the position of the monkey from the top step will be as follows:

In the first jump he will reach the step = 9 тАУ 4 = 5 steps

In the second jump he will reach the step = 5 + 2 = 7 steps

In the third jump he will reach the step = 7 тАУ 4 = 3 steps

In the fourth jump he will reach the step = 3 + 2 = 5 steps

In the fifth jump he will reach the step = 5 тАУ 4 = 1 step

Number of jumps required = 5

**(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) тАУ 3 + 2 тАУ … = тАУ 8**

(a) тИТ 3 + 2 + . . . = тИТ 8 (b) 4 тИТ 2 + . . . = 8.

In (a) the sum (тИТ 8) represents going down eight steps, then

(b) What does the sum 8 represent?

solution :**(a) тИТ 3 + 2 + . . . = тИТ 8**

тИТ 3 + 2 тИТ 3 + 2 тИТ 3 + 2 тИТ 3 + 2 тИТ 3 + 2 тИТ 3 = тИТ 8

**(b) 4 тИТ 2 + 4 тИТ 2 + 4 = 8**

The number 8 in (b) represents climbing 8 steps uphill.

### ncert solutions for Class 7 maths chapter 1 Integers

Class 7th maths solution ex 1.2

1. Write down a pair of integers whose:

(a) sum is тИТ 7 | (b) difference is тИТ 10 | (c) sum is 0 |

Solution :

(a) sum is тИТ 7

тИТ 4 + (тИТ 3) = тИТ 7

тИТ 10 + 3 = тИТ 7

(b) difference is тИТ 10

тИТ 6 тИТ (+4) = тИТ 10

+ 10 тИТ 20 = тИТ 10

(c) sum is 0

2 + 2 = 0

тИТ 4 + 4 = 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 тАУ 5.

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

Solution :

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

тИТ 4 тИТ (тИТ 12) = тИТ 4 + 12 = 8

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

6 + (тИТ 11) = 6 тИТ 11 = тИТ 5

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

9 тИТ (+12) = 9 тИТ 12 = тИТ 3

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

Solution :

team A scored = тИТ 40, 10, 0

sum of team A scored = тИТ 40 + 10 + 0

= тИТ 30

team A scored = 10, 0, тИТ 40

sum of team B scored = 10 + 0 + (тИТ 40)

= 10 + 0 тИТ 40

= тИТ 30

So both the teams got equal marks тИТ 30.

Yes, integers can be added in any order.

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

(i) (тАУ5) + (тАУ 8) = (тАУ 8) + (…………)

(ii) тАУ53 + ………… = тАУ53

(iii) 17 + ………… = 0

(iv) [13 + (тАУ 12)] + (…………) = 13 + [(тАУ12) + (тАУ7)]

(v) (тАУ 4) + [15 + (тАУ3)] = [тАУ 4 + 15] + …………

Solution :

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

(тИТ 5) + (тИТ 8) = (тИТ 8) + (. . . . . . . .)

тИТ 5 тИТ 8 = (тИТ 8) + (. . . . . . . .)

тИТ 13 = (тИТ 8) + (тИТ 5)

So adding тИТ 5 to тИТ 8 will make both sides equal.

(ii) тАУ 53 + . . . . . . . . = тИТ 53

тИТ 53 + 0 = тИТ 53

So adding 0 to тИТ 53 will make both sides equal.

(iii) 17 + . . . . . . . . . = 0

17 + (тИТ 17) = 0

0 = 0

So adding тИТ 17 to 17 will make both the sides equal.

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

13 тИТ 12 + . . . . . . . = 13 тИТ 12 тИТ 7

1 + . . . . . . . = 13 тИТ 19

1 + . . . . . . . = тИТ 6

1 + (тИТ 7) = тИТ 6

So adding тИТ 7 to 1 will make both sides equal.

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

тИТ 4 + 15 тИТ 3 = тИТ 4 + 15 + . . . . . . . . .

тИТ 7 + 15 = 11 + . . . . . . .┬а

8 = 11 + (тИТ 3)

8 = 8

So adding тИТ 3 to 11 will make both sides equal.

### ncert solutions for Class 7 maths chapter 1 Integers

Class 7th maths solution ex 1.3

**1. Find each of the following products:**

(a) 3 ├Ч (тАУ 1) | (b) (тАУ 1) ├Ч 225 |

(c) (тАУ 21) ├Ч (тАУ 30) | (d) (тАУ 316) ├Ч (тАУ 1) |

(e) (тАУ 15) ├Ч 0 ├Ч (тАУ 18) | (f) (тАУ 12) ├Ч (тАУ 11) ├Ч (10) |

(g) 9 ├Ч (тАУ 3) ├Ч (тАУ 6) | (h) (тАУ 18) ├Ч (тАУ 5) ├Ч (тАУ 4) |

(i) (тАУ 1) ├Ч (тАУ 2) ├Ч (тАУ 3) ├Ч 4 | (j) (тАУ 3) ├Ч (тАУ 6) ├Ч (тАУ 2) ├Ч (тАУ 1) |

**Solution**** :****(a) 3 ├Ч (тАУ 1)**

Since the product of + and тИТ is тИТ .

= тИТ 3

**(b) (тАУ 1) ├Ч 225**

Since the product of + and тИТ is тИТ .

= тИТ 225

**(c) (тАУ 21) ├Ч (тАУ 30)**

Since the product of тИТ and тИТ is + .

= 630

**(d) (тАУ 316) ├Ч (тАУ 1)**

Since the product of тИТ and тИТ is + .

= 316

**(e) (тАУ 15) ├Ч 0 ├Ч (тАУ 18)**

Since multiplying zero by any number gives the product zero.

= 0

**(f) (тАУ 12) ├Ч (тАУ 11) ├Ч (10)**

Since the product of тИТ and тИТ is + .

= 132 ├Ч 10

= 1320

**(g) 9 ├Ч (тАУ 3) ├Ч (тАУ 6)**

Since the product of + and тИТ is тИТ .

= тИТ 27 ├Ч (тИТ 6)

Since the product of тИТ and тИТ is + .

= 162

**(h) (тАУ 18) ├Ч (тАУ 5) ├Ч (тАУ 4)**

Since the product of тИТ and тИТ is + .

= 90 ├Ч (тИТ 4)

Since the product of + and тИТ is тИТ .

= тИТ 360

**(i) (тАУ 1) ├Ч (тАУ 2) ├Ч (тАУ 3) ├Ч 4**

Since the product of тИТ and тИТ is + .

= 2 ├Ч (тИТ 3) ├Ч 4

Since the product of тИТ and + is тИТ .

= тИТ 6 ├Ч 4

= тИТ 24

**(j) (тАУ 3) ├Ч (тАУ 6) ├Ч (тАУ 2) ├Ч (тАУ 1)**

Since the product of тИТ and тИТ is + .

= 18 ├Ч 2

= 36

**2. Verify the following:**

(a) 18 ├Ч [7 + (тАУ3)] = [18 ├Ч 7] + [18 ├Ч (тАУ3)]

(b) (тАУ21) ├Ч [(тАУ 4) + (тАУ 6)] = [(тАУ21) ├Ч (тАУ 4)] + [(тАУ21) ├Ч (тАУ 6)]**Solution :****(a) 18 ├Ч [7 + (тАУ 3)] = [18 ├Ч 7] + [18 ├Ч (тАУ 3)]**

18 ├Ч [7 тИТ 3] = 126 + (тИТ 54)

18 ├Ч 4 = 126 тИТ 54

72 = 72

L.H.S. = R.H.S.┬а

**(b) (тАУ 21) ├Ч [(тАУ 4) + (тАУ 6)] = [(тАУ 21) ├Ч (тАУ 4)] + [(тАУ 21) ├Ч (тАУ 6)]**

тИТ 21 ├Ч [тИТ 4 тИТ 6] = 84 + 126

тИТ 21 ├Ч (тИТ 10) = 210

210 = 210

L.H.S. = R.H.S.

**3.** (i) For any integer a, what is (тАУ 1) ├Ч a equal to?

(ii) Determine the integer whose product with (тАУ1) is

(a) тИТ 22 | (b) 37 | (c) 0 |

**Solution :****(i) For any integer a, what is (тАУ 1) ├Ч a equal to?**

(тИТ 1) ├Ч a

= тИТ a

**(ii) Determine the integer whose product with (тАУ 1) is****(a) тИТ 22**

тИТ 22 ├Ч (тИТ 1)

= 22

**(b) 37**

37 ├Ч (тИТ 1)

= тИТ 37

**(c) 0**

0 ├Ч (тИТ 1)

= 0

**4. Starting from (тАУ 1) ├Ч 5, write various products showing some pattern to show (тАУ 1) ├Ч (тАУ 1) = 1.****Solution :**

(тАУ 1) ├Ч 5 = тИТ 5┬а

(тАУ 1) ├Ч 4 = тИТ 4

(тАУ 1) ├Ч 3 = тИТ 3

(тАУ 1) ├Ч 2 = тИТ 2

(тАУ 1) ├Ч 1 = тИТ 1

(тАУ 1) ├Ч 0 = 0

(тАУ1) ├Ч (тИТ 1) = 1

**5. Find the product, using suitable properties:**

(a) 26 ├Ч (тАУ 48) + (тАУ 48) ├Ч (тАУ36) | (b) 8 ├Ч 53 ├Ч (тАУ125) |

(c) 15 ├Ч (тАУ25) ├Ч (тАУ 4) ├Ч (тАУ10) | (d) (тАУ 41) ├Ч 102 |

(e) 625 ├Ч (тАУ35) + (тАУ 625) ├Ч 65 | (f) 7 ├Ч (50 тАУ 2) |

(g) (тИТ 17) ├Ч (тИТ 29) | (h) (тИТ 57) ├Ч (тИТ 19) + 57 |

**Solution :****(a) 26 ├Ч (тАУ 48) + (тАУ 48) ├Ч (тАУ 36)**

= (тИТ 48) (26 тИТ 36)

= (тИТ 48) ├Ч (тИТ 10)

= 480

**(b) 8 ├Ч 53 ├Ч (тАУ125)**

= 8 ├Ч (тИТ 125) ├Ч 53

= тИТ 1000 ├Ч 53

= тИТ 53000

**(c) 15 ├Ч (тАУ 25) ├Ч (тАУ 4) ├Ч (тАУ 10)**

= 15 ├Ч (тАУ 10) ├Ч (тАУ 25) ├Ч (тАУ 4)

= тИТ 150 ├Ч 100

= тИТ 15000

**(d) (тАУ 41) ├Ч 102**

= (тАУ 41) ├Ч (100 + 2)

= (тИТ 41) ├Ч 100 + (тИТ 41) ├Ч 2

= тИТ 4100 + (тИТ 82)

= тИТ 4100 тИТ 82

= тИТ 4182

**(e) 625 ├Ч (тАУ 35) + (тАУ 625) ├Ч 65**

= 625 ├Ч (тИТ 35 тИТ 65)

= 625 ├Ч (тИТ 100)

= тИТ 62500

**(f) 7 ├Ч (50 тАУ 2)**

= 7 ├Ч 50 тИТ 7 ├Ч 2

= 350 тИТ 14

= 336

**(g) (тИТ 17) ├Ч (тИТ 29)**

= (тИТ 17) ├Ч (тИТ 30 + 1)

= (тИТ 17) ├Ч (тИТ 30) + (тИТ 17) ├Ч 1

= 510 тИТ 17

= 493

**(h) (тИТ 57) ├Ч (тИТ 19) + 57**

= тИТ 57 ├Ч (тИТ 19 тИТ 1)

= тИТ 57 ├Ч (тИТ 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?****Solution :**

Room temperature = 40┬░C

Decrease per hour = 5C

Decrease in temperature after 10 hours = 10 ├Ч (тИТ 5)

= тИТ 50┬░C

Hence temperature after 10 hours = 40 + (тИТ 50)

= 40 тИТ 50

= тИТ 10┬░C

**7. In a class test containing 10 questions, 5 marks are awarded for every correct answer and (тАУ2) 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 correct answers and five incorrect answers, what is her score?

(iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score?**Solution :****(i) Mohan gets four correct and six incorrect answers. What is his score?**

Sum of 4 correct questions attempted by Mohan = 4 ├Ч 5 = 20

Sum of 6 wrong questions attempted by Mohan = 6 ├Ч (тИТ 2) = тИТ 12

Total marks obtained by Mohan = 20 тИТ 12 = 8

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

Sum of 5 correct questions attempted by Reshma = 5 ├Ч 5 = 25

Sum of 5 wrong questions scored by Reshma = 5 ├Ч (тИТ 2) = тИТ 10

Total marks obtained by Reshma = 25 тИТ 10 = 15

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

Sum of 2 correct questions attempted by Heena = 2 ├Ч 5 = 10

Sum of 5 wrong questions attempted by Heena = 5 ├Ч (тИТ 2) = тИТ 10

Total marks obtained by Heena = 10 тИТ 10 = 0

**8. A cement company earns a profit of rupees 8 per bag of white cement sold and a loss of rupees 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 :****(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?**

Profit on selling one white bag of cement = Rs.8

Profit on selling 3000 white bags of cement = 8 ├Ч 3000 = Rs 24000

Loss on selling one gray bag of cement = Rs 5

Loss on selling 5000 gray bags of cement = 5 ├Ч 5000 = Rs 25000

Loss = 24000 тИТ 25000

= тИТ 1000

Hence he will have a loss of Rs 1000.

**(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.**

Loss on selling one gray bag of cement = Rs 5

Loss on selling 6400 gray bags of cement = 5 ├Ч 6400 = Rs 32000

Since profit of Rs.8 = 1 bag

Hence profit of Rs 1 = 1/8 on the bag

Number of white bag sold for a profit of Rs 32000 = 32000/8

= 4000 bags

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

(a) (тАУ 3) ├Ч _____ = 27 | (b) 5 ├Ч _____ = тИТ 35 |

(c) _____ ├Ч (тАУ 8) = тАУ 56 | (d) _____ ├Ч (тАУ 12) = 132 |

**Solution :**(a) (тАУ 3) ├Ч _____ = 27

(тИТ 3) ├Ч

**(тИТ 9)**= 27

(b) 5 ├Ч _____ = тАУ 35

5 ├Ч **(тИТ 7)** = тИТ 35

(c) _____ ├Ч (тАУ 8) = тАУ 56**7** ├Ч (тАУ 8) = тАУ 56

(d) _____ ├Ч (тАУ 12) = 132**(тИТ 11)** ├Ч (тАУ 12) = 132

**Ncert class 7 maths chapter 1**

class 7th math

Ex 1.4

class 7th math

Ex 1.4

**1. Evaluate each of the following:**

(a) (тАУ30) ├╖ 10 | (b) 50 ├╖ (тАУ5) | (c) (тАУ36) ├╖ (тАУ9) |

(d) (тАУ 49) ├╖ (49) | (e) 13 ├╖ [(тАУ2) + 1] | (f) 0 ├╖ (тАУ12) |

(g) (тАУ31) ├╖ [(тАУ30) + (тАУ1)] | ||

(h) [(тАУ36) ├╖ 12] ├╖ 3 | (i) [(тАУ 6) + 5)] ├╖ [(тАУ2) + 1] | ┬а |

Solution :

**(a) (тАУ30) ├╖ 10**

= \(\displaystyle \frac{-30}{10}\)= тИТ 3

**(b) 50 ├╖ (тАУ 5)**

= \(\displaystyle \frac{{50}}{{-5}}\)= тИТ 10

**(c) (тАУ36) ├╖ (тАУ9)**

= \(\displaystyle \frac{{-36}}{{-9}}\)= 4

**(d) (тАУ 49) ├╖ (49)**

= \(\displaystyle \frac{{-49}}{{49}}\)= тИТ 1

**(e) 13 ├╖ [(тАУ2) + 1]**

= \(\displaystyle \frac{{13}}{{-1}}\)= тИТ 13

**(g) (тАУ 31) ├╖ [(тАУ 30) + (тАУ 1)]**

= \(\displaystyle \frac{{-31}}{{(-30-1)}}\)= \(\displaystyle \frac{{-31}}{{-31}}\)= 1

**(h) [(тАУ 36) ├╖ 12] ├╖ 3**

= \(\displaystyle \left( {\frac{{-36}}{{12}}} \right)\div 3\)= \(\displaystyle \frac{{-3}}{{3}}\)

= тИТ 1

**(i) [(тАУ 6) + 5)] ├╖ [(тАУ2) + 1]**

= \(\displaystyle \frac{{-6+5}}{{-2+1}}\)= \(\displaystyle \frac{{-1}}{{-1}}\)

= 1

**2. Verify that a ├╖ (b + c) тЙа (a ├╖ b) + (a ├╖ c) for each of the following values of a, b and c.**

(a) a = 12, b = тАУ 4, c = 2 | (b) a = (тАУ 10), b = 1, c = 1 |

**Solution :**

**(a) a = 12, b = тАУ 4, c = 2**

put the value of a, b and c in a ├╖ (b + c) тЙа (a ├╖ b) + (a ├╖ c)

12 ├╖ (тИТ 4 + 2) тЙа 12 ├╖ (тИТ 4) + (12 ├╖ 2)

12 ├╖ (тИТ 2) тЙа тИТ 3 + 6

тИТ 6 тЙа 3

L.H.S тЙа R.H.S.

**(b) a = (тАУ 10), b = 1, c = 1**

put the value of a, b and c in a ├╖ (b + c) тЙа (a ├╖ b) + (a ├╖ c)

(тАУ 10) ├╖ (1 + 1) тЙа (тИТ 10 ├╖ 1) + (тИТ 10 ├╖ 1)

(тАУ 10) ├╖ 2 тЙа тИТ 10 + (тИТ 10)

тИТ 5 тЙа 1

**3. Fill in the blanks:**

(a) 369 ├╖ _____ = 369 | (b) (тАУ75) ├╖ _____ = тАУ1 |

(c) (тАУ206) ├╖ _____ = 1 | (d) тАУ 87 ├╖ _____ = 87 |

(e) _____ ├╖ 1 = тАУ 87 | (f) _____ ├╖ 48 = тАУ1 |

(g) 20 ├╖ _____ = тАУ2 | (h) _____ ├╖ (4) = тИТ 3 |

**Solution :**

**(a) 369 ├╖ _____ = 369**

369 ├╖

**1**= 369

**(b) (тАУ 75) ├╖ _____ = тАУ 1**

(тАУ 75) ├╖ **75** = тАУ 1

**(c) (тАУ 206) ├╖ _____ = 1**

(тАУ 206) ├╖ **(тАУ 206)** = 1

**(d) тАУ 87 ├╖ _____ = 87**

тАУ 87 ├╖ **тИТ 1** = 87

**(e) _____ ├╖ 1 = тАУ 87****тИТ 87** ├╖ 1 = тАУ 87

**(f) _____ ├╖ 48 = тАУ 1****тИТ 48** ├╖ 48 = тАУ 1

**(g) 20 ├╖ _____ = тАУ 2**

20 ├╖ **(тИТ 10)** = тАУ 2

**(h) _____ ├╖ (4) = тАУ 3****тИТ 12** ├╖ (4) = тАУ 3

**4. Write five pairs of integers (a, b) such that a ├╖ b = тАУ 3. One such pair is (6, тАУ 2) because 6 ├╖ (тАУ 2) = (тАУ 3).┬а****Solution :****(тАУ 6, 2)**

because 6 ├╖ (тАУ2) = (тАУ 3)

**(тАУ 12, 4)**

because тИТ 12 ├╖ 4 = (тАУ 3)

**(12, тАУ 4)**

because ├╖ (тАУ 4) = (тАУ 3)

**(9, тАУ 3)**

because 9 ├╖ (тАУ 3) = (тАУ 3)

**(тАУ 9, 3)**

because тИТ 9 ├╖ 3 = (тАУ 3)

**(тИТ 45, 15)**

because тИТ 45 ├╖ 15 = (тАУ 3)

**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?****Solution :**Temperature at 12 noon = + 10┬░C

Decrease in temperature till 12 noon = тИТ 2┬░C

Decrease in temperature to be тИТ 8 ┬░C = тИТ 8 тИТ (+ 10)

= тИТ 18┬░C

Time taken to lower the temperature by тИТ 2 ┬░C = 1 hour

Time taken for the temperature to drop to тИТ18┬░C = 18/2 = 9 hours

Decrease in temperature in 12 hours = 12 ├Ч (тИТ 2)

= тИТ 24┬░C Decrease in temperature from 12 noon to 12 noon = тИТ 24 + 10

= тИТ 14┬░C

So at 12 o’clock the temperature will be тИТ14┬░C.

**6. In a class test (+ 3) marks are given for every correct answer and (тАУ2) marks are given for every incorrect answer and no marks for not attempting any question. (i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly? (ii) Mohini scores тАУ5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?Solution :**

**(i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?**

Marks obtained for correct answer = + 3

Correct answer of Radhika = 12

Total marks scored by Radhika for correct answer = 12 ├Ч 3

= 36 points

Marks obtained by Radhika = 20 marks

Marks of wrong answer = 36 тИТ 20 = 16 marks

Marks for one wrong answer = тИТ 2

Hence 16 marks will be deducted = 16/2 = 8 Answer

**(ii) Mohini scores тАУ5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?**

Marks obtained by Mohini = (тИТ 5) marks

Correct answer given by Mohini = 7

Marks of correct answer given by Mohini = 7 ├Ч 3 = 21 marks

Marks of wrong answer obtained by Mohini = тИТ 5 тИТ (+ 21)

= тИТ 5 тИТ 21 = тИТ 26 points

Hence, wrong answer given for тИТ 26 marks = \(\displaystyle \frac{{26}}{2}\) = 13 answer

**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 тАУ 350 m.****Solution :**

Total distance = тИТ 350 тИТ (+ 10)

= тИТ 350 тИТ 10

= тИТ 360 m

Since time taken to walk 6 m = 1 minute

Time taken to walk 1 m =┬а\(\displaystyle \frac{{1}}{6}\)┬аminute

Time taken to walk 360 m = ┬а\(\displaystyle \frac{{1}}{6}\) ├Ч 360

= 60 minutes or 1 hour