# ML Aggarwal Integers Exe-1.3 Class 7 ICSE Maths Solutions

ML Aggarwal Integers Exe-1.3 Class 7 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-1.3 Questions for Integers as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-7.

## ML Aggarwal Integers Exe-1.3 Class 7 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 7th Chapter-1 Integers Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-1.3 Questions Edition 2023-2024

### Integers Exe-1.3

ML Aggarwal Class 7 ICSE Maths Solutions

Page-17

#### 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)

(i) 7 × (-35)

= -245

(ii) (-13) × (-15)

= 195

(iii) (-12) × (-11) × (-10)

= 132 × (-10)

= -1320

(iv) (-13) × 0 × (-24)

= 0 × (-24)

= 0

(v) (-1) × (-2) × (-3) × 4

= 2 × (-12)

= -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)

(i) 37×[6+(-3)] = 37×6+37×(-3)

LHS = 37×[6+(-3)]

= 37×3

= 111

RHS = 37×6+37×(-3)

= 37×6+(-111)

= 222 + (-111)

= 111

Therefore, LHS=RHS

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

-21 × [-6 + -4] = (-21)×(-6)+(-21) + -4
-21 × -10 = 21×-6 + (-21) + -4
-210 = -126 – 21 × -4
-210 = -126 – 84
-210 = -210​

Therefore, 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)

(i) 8 × 53 × −125

= 8×−125 × 53

= −1000 × 53 = -53000

(ii) (-8) × (-2) × 3 × (-5)

=(-8) × 3 ×( -2 ) × (-5)

= -24 × 10

= -240

= -240

(iii) (−6) × 2 × 8 × 5

=(−6 × 8) × 2 × 5

=(−6 × 8) × 10

=−48 × 10=−480

(iv) 15 × (-25) × (-4) × (-10)

= 15 × [(–25) × (– 4)] × (–10)

= 15 × [100] × (–10)

= 15 × [-1000]

= – 15000

(v) 26 × −48 + (−48) × −36

=−48 (26 − 36)

= −48 × −10 = 480

(vi) 724 × (-56) + (-724) × 44

Using distributive law of multiplication

= 724 × (-56 – 44)

= 724 × 100

= 72400

(vii) (-47) × 102

(-47) × (100 + 2)

Using distributive law of multiplication

= (-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

(i) (-4) × …… = 44

(-4) × -11 = 44

(ii) 7 × …… = -42

7 × -6 = -42

(iii) …… × (-13) = 143

-11 × (-13) = 143

(iv) (-5) × …… = 0

(-5) × 0 = 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?

Current temperature of room = 32°C

rate of lowering of temperature -5°C per hour

so, in 8 hours temperature will be = (8×-5)°C = -40°C

therefore, the room temperature will be -40 + 32 = -8°C

### Integers Exe-1.3

ML Aggarwal Class 7 ICSE Maths Solutions

Page-18

#### 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?

Scoring System:
Not attempted : 0 marks

(i) Rohit gives 4 correct and 6 incorrect answers.

∴ Rohit’s Score = 4*(5) + 6*(-2)
= 20 – 12
8 marks

(ii) Seema gives 5 correct and 5 incorrect answers.

∴ Seema ‘s score = 5*(5) + 5*(-2)
= 25 – 10
15 marks

(iii) Ritu gives 2 correct and 5 incorrect answers.

∴ Ritu’s score = 2*(5) + 5*(-2)
= 10 – 10
0 marks

—  : End of ML Aggarwal Integers Exe-1.3 Class 7 ICSE Maths Solutions :–

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