## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 1 Integers Ex 1.3

Question 1.

Find the following products:

(i) 7 × (-35)

(ii) (-13) × (-15)

(iii) (-12) × (-11) × (-10)

(iv) (-13) × 0 × (-24)

(v) (-1) × (-2) × (-3) × 4

(vi) (-3) × (-6) × (-2) × (-1)

Solution:

(i) 7 × (-35) = -245

{∵ (+) × (-) = -; (-) × (-) = +; (-) × (-) × (-) = -}

(ii) (-13) × (-15) = +195

(iii) (-12) × (-11) × (-10) = 132 × (-10) = -1320

(iv) (-13) × 0 × (-24) = -13 × 0 = 0 × (-24) = 0 (Multiplication by 0, result is always zero)

(v) (-1) × (-2) × (-3) × 4 = (2) (-3) × 4 = -6 × 4 = -24

(vi) (-3) × (-6) × (-2) × (-1) = +(18) × (+2) = 36

Question 2.

Verify the following:

(i) 37 × [6 + (-3)] = 37 × 6 + 37 × (-3)

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

Solution:

(i) 37 × [6 + (-3)] = 37 × 6 + 37 × (-3)

LHS = 37 × [6 – 3] = 37 × 3 = 111

RHS = 37 × 6 + 37 × (-3) = 222 – 111 = 111

LHS = RHS

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

LHS = (-21) × [(-6) + (-4)] = -21 × [-6 – 4] = (-21) × (-10) = +210

RHS = (-21) × (-6) + (-21) × (-4) = 126 + 84 = 210

LHS = RHS

Question 3.

Using suitable properties, evaluate the following:

(i) 8 × 53 × (-125)

(ii) (-8) × (-2) × 3 × (-5)

(iii) (-6) × 2 × (-8) × 5

(iv) 15 × (-25) × (-4) × (-10)

(v) 26 × (-48) + (-48) × (-36)

(vi) 724 × (-56) + (-724) × 44

(vii) (-47) × 102

(viii)(-39) × (-97)

Solution:

(i) 8 × 53 × (-125)

Using associative property

= 8 × (-125) × 53 = -1000 × 53 = -53000

(ii) (-8) × (-2) × 3 × (-5)

Using associative property

= (-2) × (-5) × (-8) × 3

= +10 × (-8) × 3

= -80 × 3

= -240

(iii) (-6) × 2 × (-8) × 5

Using associative property

= 2 × 5 × (-6) × (-8)

= 10 × (+48)

= 480

(iv) 15 × (-25) × (-4) × (-10)

= (-25) × (-4) × (-10) × 15

= 100 × (-150)

= -15000

(v) 26 × (-48) + (-48) × (-36)

Using distributive law of mutliplication

= (-48) [26 + (-36)]

= (-48) × (-10)

= 480

(vi) 724 × (-56) + (-724) × 44

Using distributive law of mutliplication

= 724 × (-56 – 44)

= 724 × 100

= 72400

(vii) (-47) × 102

= (-47) × (100 + 2)

Using distributive law of mutliplication

= (-47) × 100 + (-47 × 2)

= -4700 – 94

= -4794

(viii) (-39) × (-97)

= -39 × (-100 + 3)

= (-39) × (-100) + (-39) (3)

= 3900 – 117

= 3783

Question 4.

Fill in the blanks to make the following true statements:

(i) (-4) × …… = 44

(ii) 7 × …… = -42

(iii) …… × (-13) = 143

(iv) (-5) × …… = 0

Solution:

(i) (-4) × …… = 44

⇒ (-4) × (41) = 44 (∵ 44 ÷ -4 = -11)

(ii) 7 × …… = 42

7 × (-6) = -42 (∵ -42 ÷ 7 = -6)

(iii) ….. × (-13) = 143

(-11) × (-13) = 143 (∵ -143 ÷ -13 = -11)

(iv) (-5) × ….. = 0

(-5) × 0 = 0 (∵ 0 ÷ -5 = 0)

Question 5.

A certain freezing process requires that room temperature be lowered from 32°C at the rate of 5°C every hour. What will be the room temperature 8 hours after the freezing process begins?

Solution:

Original temperature = 32°C

Rate of lowering the temperature = 5°C per hour

After 8 hours, the freezing process begins = 32°C – (5°C × 8) = 32°C – 40°C = -8°C

Question 6.

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

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

(ii) Seema gets 5 correct and 5 incorrect answers. What is her score?

(iii) Ritu attempted 7 questions and gets only 2 correct answers. What is her score?

Solution:

Total number of questions in a test = 10

Award for a correct answer = 5 marks

Deduction for an incorrect answer = 2 marks

Answer for not attempting the question = 0 mark

(i) Rohit award for 4 correct answers and 6 incorrect answers = 54 – 6 × 2 = 20 – 12 = 8 marks

(ii) Seema’s award for 5 correct answers and 5 incorrect answers = 5 × 5 – 5 × 2 = 25 – 10 = 15 marks

(iii) Ritu’s award for attempting 7 questions 3 questions not attempted, 2 correct answers

= (10 – 7) = 3 not attempted = 0 marks and 5 incorrect answers

= 2 × 5 + 3 × 0 – 5 × 2

= 10 + 0 – 10

= 0 marks

Question 7.

(i) Find a pair of integers whose product is -15 and whose difference is 8.

(ii) Find a pair of integers whose product is -36 and whose difference is 15.

Solution:

(i) A pair of integers whose product = -15

and difference = 8

These integers can be -3, 5 as

-3 × 5 = -15

5 – (-3) = 5 + 3 = 8

as -5, 3 is another pair as

3 × (-5) = -15

and 3 – (-5) = 8

(ii) A pair of integer whose product is -36

and differnce = 15

There integers can be -12 and 3

as -12 × 3 = -36

and 3 – (-12) = 3 + 12 = 15

Other pair can be -3, 12 as

-3 + 12 = -36

and 12 – (-3)

= 12 + 3 = 15

## Leave a Reply