Class-7 Subject-Maths Chapter-1 Exercise1.3 INTEGERS

 
EVENTS CONVENT HIGH SCHOOL
10/08/2021          CLASS-7         SESSION2021-22(SLOT-1)
Maths
Chapter-1
INTEGERS EXERCISE 1.3
______________________________________

Question 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)

Solution:

(a) 3 × (-1) = -3 × 1 = -3

(b) (-1) × 225 = -1 × 225 = -225

(c) (-21) × (-30) = (-) × (-) × 21 × 30 = 630

(d) (-316) × (-1) = (-) × (-) × 316 × 1 = 316

(e) (-15) × 0 × (-18) = 0 [∵ a × 0 = a]

(f) (-12) × (-11) × (10)

= (-) × (-) × 12 × 11 × 10 = 1320

(g) 9 × (-3) × (-6) = (-3) × (-6) × 9

= (—) × (-) × 3 × 6 × 9 = 162

(h) (-18) × (-5) × (-4)

= (-) × (-) × (-) × 18 × 5 × 4 = -360

(i) (-1) × (-2) × (-3) × 4

= (-) × (-) × (-) × 1 × 2 × 3 × 4 = -24

(j) (-3) × (-6) × (-2) × (-1)

= (-) × (-) × (-) × (-) × 3 × 6 × 2 × 1 = 36

 

Question 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)]

LHS = 18 × [7 + (-3)] = 18 × 4 = 72

RHS = [18 × 7] + [18 × (-3)] = 126 + (-54)

= 126 – 54 = 72

LHS = RHS

Hence, verified.

 

 (b) (-21) × [(-4) + (-6)] = [(-21) × (-4)] + [(-21) × (-6)]

LHS = (-21) × [(-4) + (-6)]

= (-21) × (-10)

= (-) × (-) × 21 × 10 = 210

RHS = [(-21) × (-4)] + [(-21) × (-6)]

= (84) + (126) = 84 + 126 = 210

LHS = RHS

Hence, verified.

 

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

(ii) Determine the integer whose product with (-1) is 0.

(a) -22

(b) 37

(c) 0

Solution:

(i) (-1) × a = -a

(ii) (-1) × 0 = 0 [∵ a × 0 = 0]

Hence (c) 0 is the required integer.

 

Question 5.Find the product, using suitable properties:

(a) 26 × (-48) + (-48) × (-36)

(b) 8 × 53 × (-125)

(c) 15 × (-25) × (-4) × (-10)

(d) (-41) × 102

(e) 625 × (-35) + (-625) × 65

(f) 7 × (50 – 2)

(g) (-17) × (-29)

(h) (-57) × (-19) + 57

Solution:

(a) 26 × (-48) + (-48) × (-36)

= -48 × [26 + (-36)] = -48 × [26 – 36] = -48 × -10 = 480 [Distributive property of multiplication over

addition]

 

(b) 8 × 53 × (-125) = 53 × [8 × (-125)]

[Associative property of multiplication] = 53 × (-1000) = -53000

 

(c) 15 × (-25) × (-4) × (-10)

= [(-25) × (-4)] × [15 × (-10)]

[Regrouping the terms] = 100 × (-150) = -15000

 

(d) (-41) × 102 = (-41) × [100 + 2]

= (-41) × 100 + (-41) × 2

[Distributive property of multiplication over addition] = -4100 – 82 = -4182

 

(e) 625 × (-35) + (-625) × 65

= 625 × [(-35) + (-65)]

[Distributive property of multiplication over addition]

= 625 × (-100) = -62500

 

(f) 7 × (50 – 2) = 7 × 48 = 336 or

7 × (50 – 2) = 7 × 50 -7 × 2 = 350 – 14 = 336 [Distributive property of multiplication over addition]

 

(g) (-17) × (-29) = (-17) × [30 + (-1)]

= (-17) × 30 + (-17) × (-1)

= -510 + 17 = -493

[Distributive property of multiplication over addition]

 

(h) (-57) × (-19) + 57 = 57 × 19 + 57

= 57 × 19 + 57 × 1 [Y (-) × (-) = (+)] [Distributive property of multiplication over addition]

= 57 × (19 + 1) = 57 × 20 = 1140

 

Question 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:

Temperature of the room in the beginning = 40°C

Temperature after 1 hour

= 40°C – 1 × 5°C = 40°C – 5°C – 35°C

Similarly, temperature of the room after 10 hours

= 40°C – 10 × 5°C = 40°C – 50°C = -10°C

 

Class-7 Subject-Maths Chapter-1 Exercise1.2 INTEGERS

 
EVENTS CONVENT HIGH SCHOOL
05/08/2021          CLASS-           SESSION2021-22(SLOT-1)
Maths
Chapter-1
INTEGERS EXERCISE 1.2
______________________________________

Que-1 Explanation

Question 1.Write down a pair of integers whose:

(a) sum is -7

(b) difference is -10

(c) sum is 0.

Solution:

(a) Let us take a pair of integers -3 and -4.

∴ (-3) + (-4) = -3 – 4 = -7

(b) Let us take a pair of integers -12 and -2

∴ (-12) – (-2) = -12 + 2 = -10

(c) Let us take a pair of integers -3 and 3

∴ (-3) + (3) = -3 + 3 = 0
 

Que-2 Explanation

Question 2.

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

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

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

Solution:

(a) Let us have -2 and -10

∴ Difference = (-2) – (-10) = -2 + 10 = 8

(b) Let us have -7 and 2

∴ (-7) + (2) = -7 + 2 = -5

(c) Let us have -2 and 1

∴ (-2) – (1) = – 2 – 1 = -3

Que-3 Explanation 

 

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

table

Solution:Total score of team

A = (-40) + (10) + (0) – -40 + 10 + 0 = -30

Total score of team

B = 10 + 0 + (-40) = 10 + 0 – 40 – -30

∴ The scores of both the teams are same i.e. -30.

Yes, we can add the integers in any order.

 

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

 (i) -5 + (-8) – (-8) + (-5) [Commutative law of additions]

(ii) -53 + 0 = -53 [Additive Identity]

[Adding 0 to any integer, it gives the same value]

(iii) 17 + (-17) = 0 [Additive inverse]

(iv) [13 + (-12)] + (-7) =13 + [(-12) + (-7)] [Associative law of addition]

(v) (-4) + [15 + (-3)] – [-4 + 15] + (-3) [Associative law of addition]

 

Class-7 Subject-Maths Chapter-1 Slot-1 INTEGERS(Ex1.1)

 

EVENTS CONVENT HIGH SCHOOL
31/07/2021          CLASS-7         SESSION2021-22(SLOT-1)
Maths
Chapter-1(EX.1.1)
INTEGERS
______________________________________

INTRODUCTION INTEGERS 

Que-1 Explain
Que-2 Explain
Que-3 Explain
Que-4 Explain
Que-5 Explain
Que-8 Explain
Que-9 Explain

Question 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 Lahulspriti and Srinagar?

 

Cities                            Temperature

Lahulspriti -                           8°C

Srinagar     -                           2°C

Shimla                                    5°C

 (b) The temperature of the hottest place = 22°C

The temperature of the coldest place = -8°C

Difference = 22°C – (-8°C)

= 22°C + 8°C = 30°C

(c) Temperature of Lahulspriti = -8°C

Temperature of Srinagar = -2°C

∴ Difference = -2°C – (-8°C)

= -2°C + 8°C = 6°C

 

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

Solution:

Given scores are 25, -5, -10, 15, 10

Marks given for correct answers

= 25 + 15 + 10 = 50

Marks given for incorrect answers

= (-5) + (-10) = -15

∴ Total marks given at the end

= 50 + (-15) = 50 – 15 = 35

 

 

Question 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:

Initial temperature of Srinagar on Monday = -5°C

Temperature on Tuesday = -5°C – 2°C = -7°C

Temperature was increased by 4°C on Wednesday.

∴ Temperature on Wednesday

= -7°C + 4°C = -3°C

Hence, the required temperature on Tuesday = -7°C

and the temperature on Wednesday = -3°C

 

Question 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 flowing 1200 m below the sea level. What is the vertical distance between them?

Solution:

Height of the flying plane = 5000 m

Depth of the submarine = -1200 m

∴ Distance between them

= + 5000 m – (-1200 m)

= 5000 m + 1200 m = 6200 m

Hence, the vertical distance = 6200 m

 

 

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

Solution:

The deposited amount will be represented by a positive integer i.e., ₹ 2000.

Amount withdrawn = ₹ 1,642

∴ Balance in the account

= ₹ 2,000 – ₹ 1,642 = ₹ 358

Hence, the balance in Mohan’s account after the withdrawal

= ₹ 358

 

Question 8.Verify a – (-b) = a + b for the following values of a and 6.

(i) a = 21, b = 18

(ii) a = 118, b = 125

(iii) a = 75, b = 84

(iv) a= 28, 6 = 11

Solution:

(i) a – (-b) = a + b

LHS = 21 – (-18) = 21 + 18 = 39

RHS = 21 + 18 = 39

LHS = RHS Hence, verified.

 

(ii) a – (-b) = a + b

LHS = 118 – (-125) = 118 + 125 = 243

RHS = 118 + 125 = 243

LHS = RHS Hence, verified.

 

(iii) a – (-b) = a + b

LHS = 75 – (-84) = 75 + 84 = 159

RHS = 75 + 84 = 159

LHS = RHS Hence, verified.

 

 Question 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

Solution:

(a) (-8) + (-4) □ (-8) – (-4)

LHS = (-8) + (-4) = -8 – 4 = – 12

RHS = (-8) – (-4) = -8 + 4 = -4

Here – 12 < -4

Hence, (-8) + (-4) [<] (-8) – (-4)

 

(b) (-3) + 7 – (19) □ 15 – 8 + (-9)

LHS = (-3) + 7 – (19) =-3 + 7-19

= -3 – 19 + 7

= -22 + 1 = -15

RHS = 15 – 8 + (-9)

= 15-8-9

= 15 – 17 = -2

Here -15 < -2

Hence, (-3) + 7 – (19) [<] 15 – 8 + (-9)

 

(c) 23 – 41 + 11 □ 23 – 41 – 11

LHS = 23 – 41 + 11 = 23 + 11 – 41 = 34 – 41 = -7

RHS = 23 – 41 – 11 = 23 – 52 = -29 Here, -7 > -29

Hence, 23 – 41 + 11 [>] 23 – 41 – 11