# 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:**

**(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

**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

**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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