NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

NCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 1.3

In the NCERT solutions for class 7 Maths exercise 1.3 we get into the intriguing properties of integers as outlined in the CBSE syllabus. The closure property is a key focus, highlighting that the sum or difference of any two integers invariably results in another integer. This property is vital for understanding how integers interact under addition and subtraction. For instance, when we add or subtract integers like 17 + 23 or 7 – 9, the outcome is always an integer. The subtraction of integers, irrespective of their signs, consistently adheres to the closure property, ensuring that the difference between any two integers is always an integer. These foundational properties are crucial for the mathematical operations involving integers and are essential for further exploration of their characteristics and applications in the NCERT solutions for class 7 Maths exercise 1.3.

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

Class 7 Maths Ex 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)

Sol.

(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)

= (–15) × 0 × (–18)

= 0

(f) (–12) × (–11) × (10)

= 132 × 10 … [∵ (- × – = +)]

= 1320

(g) 9 × (–3) × (– 6)

= 9 × 18 … [∵ (- × – = +)]

= 162

(h) (–18) × (–5) × (– 4)

= 90 × -4 … [∵ (- × – = +)]

= – 360 … [∵ (+ × – = -)]

(i) (–1) × (–2) × (–3) × 4

= 2 × (-12) … [∵ (- × – = +), (- × + = -)]

= – 24

(j) (–3) × (–6) × (–2) × (–1)

= 18 × 2 … [∵ (- × – = +)

= 36

Class 7 Maths Ex 1.3 Question 2

Verify the following:

(a) 18 × [7 + (-3)] = [18 × 7] + [18 × (-3)]

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

Sol.

(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.

Class 7 Maths Ex 1.3 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

Sol.

(i) (-1) × a = -a

(ii) (-1) × 0 = 0 [∵ a × 0 = 0

Hence (c) 0 is the required integer.

Class 7 Maths Ex 1.3 Question 4

Starting from (-1) × 5, write various products showing some pattern to show

(-1) × (-1) = 1

Sol.

(-1) × 5 = -5

(-1) × 4 = -4 = (-5) + 1

(-1) × 3 = -3 = (-4) + 1

(-1) × 2 = -2 = (-3) + 1

(-1) × (1) = -1 = (-2) + 1

(-1) × 0 = 0 – (-1) + 1

(-1) × (-1) = 1 = 0+1

Class 7 Maths Ex 1.3 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

Sol.

(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

Class 7 Maths Ex 1.3 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?

Sol.

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

Ex 1.3 Class 7 Maths Question 7

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

(i) Mohan gets four correct and si× incorrect answers. What is his score?

(ii) Reshma gets five correct answers and five incorrect answers, what is her score?

(iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score?

Sol.

(i) Marks awarded to Mohan = 4 × 5

=20 for correct answers Marks awarded to Mohan = 6 × (-2)

= -12 for incorrect answers.

∴ Total marks obtained by Mohan

= 20 + (-12) = 20 – 12 = 8

(ii) Marks awarded to Reshma for correct answers

= 5 × 5 = 25

Marks awarded to Reshma for incorrect answers

= 5 × (-2) = -10

∴ Total marks obtained by Reshma

= 25 + (-10) = 25 – 10 = 15

(iii) Marks awarded to Heena for correct answers

= 2 × 5 = 10

Marks awarded to Heena for incorrect answer

= 5 × (-2) = -10

Number of question not attempted by Heena

= 10 – (2 + 5) = 10 – 7 = 3

Marks awarded to Heena for non-attempted questions

=3×0=0

∴ Total marks obtained by Heena

= 10 + (-10) + 0 = 10-10+ 0 = 0

Class 7 Maths Ex 1.3 Question 8

A cement company earns a profit of ₹ 8 per bag of white cement sold and a loss of ₹ 5 per bag of grey cement sold.

(a) The company sells 3,000 bags of white cement and sold 5,000 bags of grey cement in a month. What is its profit or loss?

(b) What is the number of white cement bags it must sell to have neither profit nor loss, if the number of grey bags sold is 6,400 bags.

Sol.

(a) Profit on one white cement bag = ₹ 8

loss on one grey cement bag = ₹ – 5

Profit on 3,000 bags of white cement

= ₹ (8 × 3,000) = ₹ 24,000

Loss on 5,000 bags of grey cement

= ₹ (-5 × 5000) = – ₹ 25,000

Total loss = – ₹ 25,000 + ₹ 24,000

= – ₹ 1000 i.e. ₹ 1000

(b) Selling price of grey bags at a loss of ₹ 5

= ₹ (5 × 6,400) – ₹ 32,000

For no profit and no loss, the selling price of white bags = ₹ 32,000

Rate of selling price of white bags at a profit of ₹ 8 per bag.

∴ Number of white cement bags sold

= 5 × 6400 / 8

Hence, the required number of bags = 4,000

Class 7 Maths Ex 1.3 Question 9

Replace the blank with an integer to make it a true statement.

(a) (-3) × __ = 27

(b) 5 × __ = -35

(c) __ × (-8) = -56

(d) __ × (-12) = 132

Sol.

(a) (-3) × __ = 27 = (-3) × (-9) = 27 [∵ (-) × (-) = (+)]

(b) 5 × __ = -35 = 5 × (-7) = -35 [∵ (+) × (-) = (-)]

(c) __ × (-8) = -56 = 7 × (-8) = -56 [∵ (+) × (-) = (-)]

(d) __ × (-12) = 132 = (-11) × (-12) = 132 [∵ (-) × (-) = (+)]

NCERT Solutions for Class 7 Maths Chapter 1 Integers Other Exercise

• NCERT Solutions for Class 7 Maths Ex 1.1
• NCERT Solutions for Class 7 Maths Ex 1.2
• NCERT Solutions for Class 7 Maths Ex 1.4

NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3 PDF Download

Infinity learn’s easy-to-understand NCERT solutions for class 7 Maths chapter 1 Integers Ex 1.3 pdf download. Learn about the basic rules of adding and subtracting numbers, as explained in the CBSE maths syllabus for class 7. Our NCERT solutions are written in a simple and clear way to help you understand the ideas well. These NCERT solutions are great for students who want to do well in their studies and learn the basics of numbers. Don’t miss this chance to improve your math skills. Download the PDF from Infinity Learn and start a learning journey that will make you stand out in your studies.

NCERT Solutions for Class 7 Maths Other Chapters

• Chapter 2 Fractions and Decimals
• Chapter 3 Data Handling
• Chapter 4 Simple Equations
• Chapter 5 Lines and Angles
• Chapter 6 The Triangles and Its Properties
• Chapter 7 Congruence of Triangles
• Chapter 8 Comparing Quantities
• Chapter 9 Rational Numbers
• Chapter 10 Practical Geometry
• Chapter 11 Perimeter and Area
• Chapter 12 Algebraic Expressions
• Chapter 13 Exponents and Powers
• Chapter 14 Symmetry
• Chapter 15 Visualizing Solid Shapes

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FAQs on NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3

What is the focus of Exercise 1.3 in Class 7 Maths Chapter 1 Integers?

Exercise 1.3 focuses on understanding the properties of addition and subtraction of integers, including the closure property.

What is the closure property as discussed in Exercise 1.3?

The closure property states that the sum or difference of any two integers will always result in another integer.

Are integers closed under subtraction according to Exercise 1.3?

Yes, integers are closed under subtraction, which means that the difference between any two integers is always an integer.

Can I find solutions to Exercise 1.3 online?

Yes, solutions to Exercise 1.3 are available online on various educational platforms, including Infinity Learn which is best among all.

Will practicing Exercise 1.3 help me in my exams?

Yes, practicing Exercise 1.3 will help you understand the properties of integers, which is a fundamental concept in mathematics and will be useful in your exams.