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)

Answer:

(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)

Answer:

(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)

Answer:

(iv) 15 × (-25) × (-4) × (-10)

= 15 × [(–25) × (– 4)] × (–10)

= 15 × [100] × (–10)

= 15 × [-1000]

= – 15000

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

(i) (-4) × …… = 44
(ii) 7 × …… = -42
(iii) …… × (-13) = 143
(iv) (-5) × …… = 0

Answer:

(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?

Answer:

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

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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?

Answer:

Scoring System:
Correct Answer : +5 marks
Incorrect Answer : -2 marks
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 :–

Return to –  ML Aggarwal Maths Solutions for ICSE Class -7

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