### NCERT Solutions for Class 7 Chapter 1 – Integers Exercise 1.3

**1. Find each of the following products:**

**(a) 3 × (–1)**

**Solution:-**

By the rule of Multiplication of integers,

= 3 × (-1)

= -3 … [∵ (+ × – = -)]

**(b) (–1) × 225**

**Solution:-**

By the rule of Multiplication of integers,

= (-1) × 225

= -225 … [∵ (- × + = -)]

**(c) (–21) × (–30)**

**Solution:-**

By the rule of Multiplication of integers,

= (-21) × (-30)

= 630 … [∵ (- × – = +)]

**(d) (–316) × (–1)**

**Solution:-**

By the rule of Multiplication of integers,

= (-316) × (-1)

= 316 … [∵ (- × – = +)]

**(e) (–15) × 0 × (–18)**

**Solution:-**

By the rule of Multiplication of integers,

= (–15) × 0 × (–18)

= 0

∵Any integer is multiplied with zero and the answer is zero itself.

**(f) (–12) × (–11) × (10)**

**Solution:-**

By the rule of Multiplication of integers,

= (–12) × (-11) × (10)

First multiply the two numbers having same sign,

= 132 × 10 … [∵ (- × – = +)]

= 1320

**(g) 9 × (–3) × (– 6)**

**Solution:-**

By the rule of Multiplication of integers,

= 9 × (-3) × (-6)

First multiply the two numbers having same sign,

= 9 × 18 … [∵ (- × – = +)]

= 162

**(h) (–18) × (–5) × (– 4)**

**Solution:-**

By the rule of Multiplication of integers,

= (-18) × (-5) × (-4)

First multiply the two numbers having same sign,

= 90 × -4 … [∵ (- × – = +)]

= – 360 … [∵ (+ × – = -)]

**(i) (–1) × (–2) × (–3) × 4**

**Solution:-**

By the rule of Multiplication of integers,

= [(–1) × (–2)] × [(–3) × 4]

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

= – 24

**(j) (–3) × (–6) × (–2) × (–1)**

**Solution:-**

By the rule of Multiplication of integers,

= [(–3) × (–6)] × [(–2) × (–1)]

First multiply the two numbers having same sign,

= 18 × 2 … [∵ (- × – = +)

= 36

**2. Verify the following:**

**(a) 18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]**

**Solution:-**

From the given equation,

Let us consider the Left Hand Side (LHS) first = 18 × [7 + (–3)]

= 18 × [7 – 3]

= 18 × 4

= 72

Now, consider the Right Hand Side (RHS) = [18 × 7] + [18 × (–3)]

= [126] + [-54]

= 126 – 54

= 72

By comparing LHS and RHS,

72 = 72

LHS = RHS

Hence, the given equation is verified.

**(b) (–21) × [(– 4) + (– 6)] = [(–21) × (– 4)] + [(–21) × (– 6)]**

**Solution:-**

From the given equation,

Let us consider the Left Hand Side (LHS) first = (–21) × [(– 4) + (– 6)]

= (-21) × [-4 – 6]

= (-21) × [-10]

= 210

Now, consider the Right Hand Side (RHS) = [(–21) × (– 4)] + [(–21) × (– 6)]

= [84] + [126]

= 210

By comparing LHS and RHS,

210 = 210

LHS = RHS

Hence, the given equation is verified.

3. (i) For any integer a, what is (–1) × a equal to?

Solution:-

= (-1) × a = -a

Because, when we multiplied any integer a with -1, then we get additive inverse of that integer.

**(ii). Determine the integer whose product with (–1) is**

**(a) –22**

**Solution:-**

Now, multiply -22 with (-1), we get

= -22 × (-1)

= 22

Because, when we multiplied integer -22 with -1, then we get additive inverse of that integer.

**(b) 37**

**Solution:-**

Now, multiply 37 with (-1), we get

= 37 × (-1)

= -37

Because, when we multiplied integer 37 with -1, then we get additive inverse of that integer.

**(c) 0**

**Solution:-**

Now, multiply 0 with (-1), we get

= 0 × (-1)

= 0

Because, the product of negative integers and zero give zero only.

**4. Starting from (–1) × 5, write various products showing some pattern to show**

**(–1) × (–1) = 1.**

**Solution:-**

The various products are,

= -1 × 5 = -5

= -1 × 4 = -4

= -1 × 3 = -3

= -1 × 2 = -2

= -1 × 1 = -1

= -1 × 0 = 0

= -1 × -1 = 1

We concluded that the product of one negative integer and one positive integer is negative integer. Then, the product of two negative integers is a positive integer.

**5. Find the product, using suitable properties:**

**(a) 26 × (– 48) + (– 48) × (–36)**

**Solution:-**

The given equation is in the form of Distributive law of Multiplication over Addition.

= a × (b + c) = (a × b) + (a × c)

Let, a = -48, b = 26, c = -36

Now,

= 26 × (– 48) + (– 48) × (–36)

= -48 × (26 + (-36)

= -48 × (26 – 36)

= -48 × (-10)

= 480 … [∵ (- × – = +)

**(b) 8 × 53 × (–125)**

**Solution:-**

The given equation is in the form of Commutative law of Multiplication.

= a × b = b × a

Then,

= 8 × [53 × (-125)]

= 8 × [(-125) × 53]

= [8 × (-125)] × 53

= [-1000] × 53

= – 53000

**(c) 15 × (–25) × (– 4) × (–10)**

**Solution:-**

The given equation is in the form of Commutative law of Multiplication.

= a × b = b × a

Then,

= 15 × [(–25) × (– 4)] × (–10)

= 15 × [100] × (–10)

= 15 × [-1000]

= – 15000

**(d) (– 41) × 102**

**Solution:-**

The given equation is in the form of Distributive law of Multiplication over Addition.

= a × (b + c) = (a × b) + (a × c)

= (-41) × (100 + 2)

= (-41) × 100 + (-41) × 2

= – 4100 – 82

= – 4182

**(e) 625 × (–35) + (– 625) × 65**

**Solution:-**

The given equation is in the form of Distributive law of Multiplication over Addition.

= a × (b + c) = (a × b) + (a × c)

= 625 × [(-35) + (-65)]

= 625 × [-100]

= – 62500

**(f) 7 × (50 – 2)**

**Solution:-**

The given equation is in the form of Distributive law of Multiplication over Subtraction.

= a × (b – c) = (a × b) – (a × c)

= (7 × 50) – (7 × 2)

= 350 – 14

= 336

**(g) (–17) × (–29)**

**Solution:-**

The given equation is in the form of Distributive law of Multiplication over Addition.

= a × (b + c) = (a × b) + (a × c)

= (-17) × [-30 + 1]

= [(-17) × (-30)] + [(-17) × 1]

= [510] + [-17]

= 493

**(h) (–57) × (–19) + 57**

**Solution:-**

The given equation is in the form of Distributive law of Multiplication over Addition.

= a × (b + c) = (a × b) + (a × c)

= (57 × 19) + (57 × 1)

= 57 [19 + 1]

= 57 × 20

= 1140

**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?**

**Solution:-**

From the question, it is given that

Let us take the lowered temperature as negative,

Initial temperature = 40^{o}C

Change in temperature per hour = -5^{o}C

Change in temperature after 10 hours = (-5) × 10 = -50^{o}C

∴The final room temperature after 10 hours of freezing process = 40^{o}C + (-50^{o}C)

= -10^{o}C

**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 six incorrect answers. What is his score?**

**Solution:-**

From the question,

Marks awarded for 1 correct answer = 5

Then,

Total marks awarded for 4 correct answer = 4 × 5 = 20

Marks awarded for 1 wrong answer = -2

Then,

Total marks awarded for 6 wrong answer = 6 × -2 = -12

∴Total score obtained by Mohan = 20 + (-12)

= 20 – 12

= 8

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

**Solution:-**

From the question,

Marks awarded for 1 correct answer = 5

Then,

Total marks awarded for 5 correct answer = 5 × 5 = 25

Marks awarded for 1 wrong answer = -2

Then,

Total marks awarded for 5 wrong answer = 5 × -2 = -10

∴Total score obtained by Reshma = 25 + (-10)

= 25 – 10

= 15

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

**Solution:-**

From the question,

Marks awarded for 1 correct answer = 5

Then,

Total marks awarded for 2 correct answer = 2 × 5 = 10

Marks awarded for 1 wrong answer = -2

Then,

Total marks awarded for 5 wrong answer = 5 × -2 = -10

Marks awarded for questions not attempted is = 0

∴Total score obtained by Heena = 10 + (-10)

= 10 – 10

= 0

**8. A cement company earns a profit of Rs 8 per bag of white cement sold and a loss of**

**Rs 5 per bag of grey cement sold.**

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

**Solution:-**

We denote profit in positive integer and loss in negative integer,

From the question,

Cement company earns a profit on selling 1 bag of white cement = Rs 8 per bag

Then,

Cement company earns a profit on selling 3000 bags of white cement = 3000 × Rs 8

= Rs 24000

Loss on selling 1 bag of grey cement = – Rs 5 per bag

Then,

Loss on selling 5000 bags of grey cement = 5000 × – Rs 5

= – Rs 25000

Total loss or profit earned by the cement company = profit + loss

= 24000 + (-25000)

= – Rs1000

Thus, a loss of Rs 1000 will be incurred by the company.

**(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.**

**Solution:-**

We denote profit in positive integer and loss in negative integer,

From the question,

Cement company earns a profit on selling 1 bag of white cement = Rs 8 per bag

Let the number of white cement bags be x.

Then,

Cement company earns a profit on selling x bags of white cement = (x) × Rs 8

= Rs 8x

Loss on selling 1 bag of grey cement = – Rs 5 per bag

Then,

Loss on selling 6400 bags of grey cement = 6400 × – Rs 5

= – Rs 32000

According to the question,

Company must sell to have neither profit nor loss.

= Profit + loss = 0

= 8x + (-32000) =0

By sending -32000 from LHS to RHS it becomes 32000

= 8x = 32000

= x = 32000/8

= x = 4000

Hence, the 4000 bags of white cement have neither profit nor loss.

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

**(a) (–3) × _____ = 27**

**Solution:-**

Let us assume the missing integer be x,

Then,

= (–3) × (x) = 27

= x = – (27/3)

= x = -9

Let us substitute the value of x in the place of blank,

= (–3) × (-9) = 27 … [∵ (- × – = +)]

**(b) 5 × _____ = –35**

**Solution:-**

Let us assume the missing integer be x,

Then,

= (5) × (x) = -35

= x = – (-35/5)

= x = -7

Let us substitute the value of x in the place of blank,

= (5) × (-7) = -35 … [∵ (+ × – = -)]

**(c) _____ × (– 8) = –56**

**Solution:-**

Let us assume the missing integer be x,

Then,

= (x) × (-8) = -56

= x = (-56/-8)

= x = 7

Let us substitute the value of x in the place of blank,

= (7) × (-8) = -56 … [∵ (+ × – = -)]

**(d) _____ × (–12) = 132**

**Solution:-**

Let us assume the missing integer be x,

Then,

= (x) × (-12) = 132

= x = – (132/12)

= x = – 11

Let us substitute the value of x in the place of blank,

= (–11) × (-12) = 132 … [∵ (- × – = +)]