Integers Ex-1.3

NCERT solutions for Class 7 Integers Maths 

NCERT Solutions for Class 7 Maths Integers

Class –VII Mathematics (Ex. 1.3)

Question 1.Find the each of the following products:

(a) 3 x (–1)

(b) (–1) x 225

(c) (–21) x (–30)

(d) (–316) x (–1)

(e) (–15) x 0 x (–18)

(f) (–12) x (–11) x (10)

(g) 9 x (–3) x (–6)

(h) (–18) x (–5) x (–4)

(i) (–1) x (–2) x (–3) x 4

(j) (–3) x (–6) x (2) x (–1)

Answer:

(a) 3 x (–1) = –3

(b) (–1) x 225 = –225

(c) (–21) x (–30) = 630

(d) (–316) x (–1) = 316

(e) (–15) x 0 x (–18) = 0

(f) (–12) x (–11) x (10) = 132 x 10 = 1320

(g) 9 x (–3) x (–6) = 9 x 18 = 162

(h) (–18) x (–5) x (–4) = 90 x (–4) = –360

(i) (–1) x (–2) x (–3) x 4 = (–6 x 4) = –24

(j) (–3) x (–6) x (2) x (–1) = (–18) x (–2) = 36

NCERT Solutions for Class 7 Maths Exercise 1.3

Question 2.Verify the following:

(a) 18 x [7 + (–3)] = [18 x 7] + [18 x (–3)]

(b) (–21) x [(–4) + (–6)] = [(–21) x (-4)] + [(–21) x (–6)]

Answer:

(a) 18 x [7 + (–3)] = [18 x 7] + [18 x (–3)]

$\Rightarrow$ 18 x 4 = 126 + (–54)

$\Rightarrow$ 72 = 72

$\Rightarrow$ L.H.S. = R.H.S. Hence verified.

(b) (–21) x [(–4) + (–6)] = [(–21) x (–4)] + [(–21) x (–6)]

$\Rightarrow$ (–21) x (–10) = 84 + 126

$\Rightarrow$ 210 = 210

$\Rightarrow$ L.H.S. = R.H.S. Hence verified.

NCERT Solutions for Class 7 Maths Exercise 1.3

Question 3.

(i)For any integer $a,$ what is $\left( { – 1} \right) \times a$ equal to?

(ii)Determine the integer whose product with $\left( { – 1} \right)$ is:

(a) –22

(b) 37

(c) 0

Answer:

(i) $\left( { – 1} \right) \times a = – a,$ where $a$ is an integer.

(ii) (a) $\left( { – 1} \right) \times \left( { – 22} \right) = 22$

(b) $\left( { – 1} \right) \times 37 = – 37$

(c) $\left( { – 1} \right) \times 0 = 0$

NCERT Solutions for Class 7 Maths Exercise 1.3

Question 4.Starting from $\left( { – 1} \right) \times 5,$ write various products showing some patterns to show $\left( { – 1} \right) \times \left( { – 1} \right) = 1.$

Answer:

$\left( { – 1} \right) \times 5 = – 5$ $\left( { – 1} \right) \times 4 = – 4$

$\left( { – 1} \right) \times 3 = – 3$ $\left( { – 1} \right) \times 2 = – 2$

$\left( { – 1} \right) \times 1 = – 1$ $\left( { – 1} \right) \times 0 = 0$

$\left( { – 1} \right) \times \left( { – 1} \right) = 1$

Thus, we can conclude that this pattern shows the product of one negative integer and one positive integer is negative integer whereas the product of two negative integers is a positive integer.

NCERT Solutions for Class 7 Maths Exercise 1.3

Question 5.Find the product, using suitable properties:

(a) $26 \times \left( { – 48} \right) + \left( { – 48} \right) \times \left( { – 36} \right)$

(b) $8 \times 53 \times \left( { – 125} \right)$

(c) $15 \times \left( { – 25} \right) \times \left( { – 4} \right) \times \left( { – 10} \right)$

(d) $\left( { – 41} \right) \times \left( {102} \right)$

(e) $625 \times \left( { – 35} \right) + \left( { – 625} \right) \times 65$

(f) $7 \times \left( {50 – 2} \right)$

(g) $\left( { – 17} \right) \times \left( { – 29} \right)$

(h) $\left( { – 57} \right) \times \left( { – 19} \right) + 57$

Answer:

(a) $26 \times \left( { – 48} \right) + \left( { – 48} \right) \times \left( { – 36} \right)$

$\Rightarrow$ $\left( { – 48} \right) \times \left[ {26 + \left( { – 36} \right)} \right]$ [Distributive property]

$\Rightarrow$ $\left( { – 48} \right) \times \left( { – 10} \right)$

$\Rightarrow$ 480

(b) $8 \times 53 \times \left( { – 125} \right)$

$\Rightarrow$ $53 \times \left[ {8 \times \left( { – 125} \right)} \right]$ [Commutative property]

$\Rightarrow$ $53 \times \left( { – 1000} \right)$

$\Rightarrow$ $– 53000$

(c) $15 \times \left( { – 25} \right) \times \left( { – 4} \right) \times \left( { – 10} \right)$

$\Rightarrow$ $15 \times \left[ {\left( { – 25} \right) \times \left( { – 4} \right) \times \left( { – 10} \right)} \right]$ [Commutative property]

$\Rightarrow$ $15 \times \left( { – 1000} \right)$

$\Rightarrow$ $– 15000$

(d) $\left( { – 41} \right) \times \left( {102} \right)$

$\Rightarrow$ $– 41 \times \left[ {100 + 2} \right]$ [Distributive property]

$\Rightarrow$ $\left[ {\left( { – 41} \right) \times 100} \right] + \left[ {\left( { – 41} \right) \times 2} \right]$ $\Rightarrow$ $– 4100 + \left( { – 82} \right)$

$\Rightarrow$ $– 4182$

(e) $625 \times \left( { – 35} \right) + \left( { – 625} \right) \times 65$

$\Rightarrow$ $625 \times \left[ {\left( { – 35} \right) + \left( { – 65} \right)} \right]$ [Distributive property]

$\Rightarrow$ $625 \times \left( { – 100} \right)$

$\Rightarrow$ $– 62500$

(f) $7 \times \left( {50 – 2} \right)$

$\Rightarrow$ $7 \times 50 – 7 \times 2$ [Distributive property]

$\Rightarrow$ $350 – 14 = 336$

(g)$\left( { – 17} \right) \times \left( { – 29} \right)$$\Rightarrow$

$\left( { – 17} \right) \times \left[ {\left( { – 30} \right) + 1} \right]$ [Distributive property]

$\Rightarrow$ $\left( { – 17} \right) \times \left( {30} \right) + \left( { – 17} \right) \times 1$ $\Rightarrow$ $510 + \left( { – 17} \right)$

$\Rightarrow$ 493

(h)$\left( { – 57} \right) \times \left( { – 19} \right) + 57$

$\Rightarrow$ $\left( { – 57} \right) \times \left( { – 19} \right) + 57 \times 1$

$\Rightarrow$ 57 x 19 + 57 x 1

$\Rightarrow$ 57 x (19 + 1) [Distributive property]

$\Rightarrow$ 57 x 20 = 1140

NCERT Solutions for Class 7 Maths Exercise 1.3

Question 6.A certain freezing process requires that room temperature be low^ered from 40oC at the rate of 5oC every hour. What will be the room temperature 10 hours after the process begins?

Answer:

Given: Present room temperature = 40oC

Decreasing the temperature every hour = 5oC

Room temperature after 10 hours = 40oC + 10 x (–5oC )

= 40oC – 50oC

= – 10oC

Thus, the room temperature after 10 hours is – 10oC after the process begins.

NCERT Solutions for Class 7 Maths Exercise 1.3

Question 7.In a class test containing 10 questions, 5 marks are awarded for e^very correct answer and $\left( { – 2} \right)$ 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?

(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?

Answer:

(i) Mohan gets marks for four correct questions = 4 x 5 = 20

He gets marks for six incorrect questions = 6 x (–2) = –12

Therefore, total scores of Mohan = (4 x 5) + [6 x (–2)]

= 20 – 12 = 8

Thus, Mohan gets 8 marks in a class test.

(ii) Reshma gets marks for five correct questions = 5 x 5 = 25

She gets marks for five incorrect questions = 5 x (–2) = –10

Therefore, total score of Resham = 25 + (–10) = 15

Thus, Reshma gets 15 marks in a class test.

(iii) Heena gets marks for two correct questions = 2 x 5 = 10

She gets marks for five incorrect questions = 5 x (–2) = –10

Therefore, total score of Resham = 10 + (–10) = 0

Thus, Reshma gets 0 marks in a class test.

NCERT Solutions for Class 7 Maths Exercise 1.3

Question 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?

(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.

Answer:

Given: Profit of 1 bag of white cement = Rs. 8

And Loss of 1 bag of grey cement = Rs. 5

(a) Profit on selling 3000 bags of white cement = 3000 x 8 = Rs. 24,000

Loss of selling 5000 bags of grey cement = 5000 x Rs. 5 = Rs. 25,000

Since Profit < Loss

Therefore, his total loss on selling the grey cement bags = Loss – Profit

= 25,000 – 24,000

= Rs. 1,000

Thus, he has lost of Rs. 1,000 on selling the grey cement bags.

(b)Let the number of bags of white cement be $x.$

According to question, Loss = Profit

$\therefore$ 5 x 6,400 = $x$ x 8

$\Rightarrow$ $x = \frac{{5 \times 6400}}{8}$ = 5000 bags

Thus, he must sell 4000 white cement bags to have neither profit nor loss.

NCERT Solutions for Class 7 Maths Exercise 1.3

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

(a)$\left( { – 3} \right) \times \_\_\_\_\_\_\_ = 27$

(b) $5 \times \_\_\_\_\_\_\_ = – 35$

(c) $\_\_\_\_\_\_\_ \times \left( { – 8} \right) = – 56$

(d) $\_\_\_\_\_\_\_ \times \left( { – 12} \right) = 132$

Answer:

(a) $\left( { – 3} \right) \times \underline {\left( { – 9} \right)} = 27$

(b) $5 \times \underline {\left( { – 7} \right)} = – 35$

(c) $\underline 7 \times \left( { – 8} \right) = – 56$

(d) $\underline {\left( { – 11} \right)} \times \left( { – 12} \right) = 132$