1.3 MULTIPLICATION OF INTEGERS:
While multiplying a positive integer and a negative integer,we multiply them as whole numbers.
Then we put a minus or plus sign as the case may be;
Case I: We put a positive sign if both the numbers are positive.
Case II; We put a negative sign if both the numbers are negative
Case III :We put a negative sign even if if one of the number is negative.
EXERCISE 1.3
1) Find 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)
Solution:
(a) 3 x (– 1) = – (3 x 1) = –3
(b) (–1) x 225 = – (1 x 225) = –225
(c) (–21) x (– 30) = (21 x 30)
= [(20 + 1) x 30]
= 20 x 30 + 1 x 30
= 600 + 30 = 630
(d) (–316 x –1) = + (316 x 1) = 316
(e) (–15) x 0 (–18) = + [(–15) x 0] x (–18)
= 0 x (–18) = 0
(f) (–12) x (–11) x 10 = + (12 x 11 x 10)
= (132 x 10) = 1320
Product of even number of negative integers is positive.
(g) 9 x (–3) x (–6) = + [9 x 3 x 6] = 162
Product of even number of negative integers is positive.
(h) (–18) x (–5) x (–4) = – [18 x 5 x 4]
= – [18 x 20] = –360
Product of odd number of negative integers is negative.
(i) (–1) x (–2) x (–3) x 4 = – [1 x 2 x 3 x 4]
= – [2 x 12] = –24
Product of odd number of negative integers is negative.
(j) (–3) x (–6) x (–2) x (–1) = + [3 x 6 x 2 x 1]
= + [36 x 1 ] = 36
Product of even number of negative integers is positive.
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)]
Solution:-
a) L.H.S = 18 x [ 7 + (–3)]
= 18 x [7 – 3] = 18 × 4 = 72
R.H.S = [18 x 7] + [18 x (–3)]
= 126 + [–154] = 126 – 54 = 72
∴ L.H.S = R.H.S
∴ 18 x [7 + (–3)] = [18 x 7] + [18 x (–3)]
b) L.H.S = (–21) x [(– 4) + (– 6)]
= (–21) x [(– 10) = + ( 21 x 10) = 210
R.H.S = [(–21) x (– 4)] + [(–21) x (– 6)]
= [(+84)] + [(+126)] = 84 + 126 = 210
∴ L.H.S = R.H.S
∴ [(–21) x (– 4)] + (– 6) = [(–21) x (– 4)] + [(–21) x (– 6)]
3. i) For any integer a, what is (–1) equal to?
ii) Determine the integer whose product with −1 is
a) –22 b) 37 c) 0
Solution:
i) (–1) x a = –a
ii) (a) ∴ (–1) x [any integer]
∴ –1 x 22 = –22
22 is the additive inverse of the integer−22.
b) (–1) x (–37) = 37
−37 is the additive inverse of the integer 37.
c) (–1) x 0 = 0
0 is the additive inverse of the integer 0.
Product of a negative integer and zero is zero.
4. Starting from(–1) x 5 write various products showing some pattern to show (–1) x.(–1) = 1
Solution:
(–1) x 5 = –5
(–1) x 4 = –4 = (–5) + 1
(–1) x 3 = –3 = (–4) + 1
(–1) x 2 = –2 = (–3) + 1
(–1) x 1 = –1 = (–2) + 1
(–1) x 0 = 0 = (–1) + 1
(–1) x (–1) = 1 = 0 + 1
5. Find the product, using suitable properties:
(a) 26 x (–48) + (–48) x (–36)
(b) 8 x 53 x (–125)
(c) 15 x (–25) x (4) x (–10)
(d) (–41) x 102
(e) 625 x (–35) + (–625) x 65
(f) 7 x (50– 2)
(g) (–17) x (–29)
(h) (–57) x (–19) + 57
Solution:
(a) 26 x (–48) + (–48) x (–36):
[Using distributive property]
26 x (–48) + (–48) x (–36)
= (–48) x [26 + (36)]
= (–48) x [10]
= (–48) x (10) = 480
(b) 8 x 53 x (–125):
8 x 53 x (–125) [8 x (–125)] x 53
[Using associative property]
= [–1000] x 53
= – [1000 x 53] = – 53000
(c) 15 x (–25) x (–4) (–10)
[Using associative property ]
15 x (–25) x (–4) x (–10)
= [(–25) x (–4)] x [(–10) x 15]
= [100] x [–150]
= – (100 x 150) = –15000
(d) (–41) x 102 = (–41) x [100 + 2]
[Using distributive property]
= (–41) x 100 + (–41)
a (b – c) = a x b = – a x c
= – 4100 + (–82) = 4182
(e) 625 x (–35) + (–625) x 65
Using Commutative property:
= 625 x [(–35) + (–65)]
= 625 x [–100]
= – [625 x 100] = – 62500
(f) 7 x (50 – 2) = 7 x 50 − 7 x 2
Using Distributive property:
= 350 – 14 = 336
(g) (–17) x (–29) = (17 x 29)
Using Commutative property:
= 17 x (30 – 1)
= 17 x 30 – 17 1
a (b – c) = a x b = – a x c
510 – 17 = 493
(h) (–57) x (–19) + 57
Using Distributive property:
= (–57) x (–19) + [(–57) x (–1)]
[∴ 57 = (–1) x (–57)]
= (–57) x [(–19) + (–1)]
= (–57) x (–20)
= + [57 x 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:
Room temperature = 40°C
Change in temperature per hours = –5°C
Change in temperature in 10 hours = 10 x (–5°C) = –50°C
∴ Room temperature after 10 hours = 40°C + (–50°C) = –10°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 not attempted.
(i) Mohan gets four correct and six incorrect answer. What is his score?
(ii) Reshma gets five correct answer and five incorrect answer. What is her score?
(iii) Heena gets two correct and five incorrect answer out of seven questions she attempts. What’s her score?
Solution:
Total number of questions = 10
Marks for correct answer = 5
Marks for incorrect answer = (–2)
Marks for unanswered questions = 0
(i) Mohan’s score = 4 x (5) + 6 x (–2)
= 20 + (–12)= 8
(ii) Reshma’s score = 5 x (5) + 5 x (–2)
= 25 + (–10)= 15
(iii) Heena’s score = 2 x (5) + 5 x (–2) + 3 x (0)
= 10 + (–10) + 0 = 0
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 5000 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?
Solution:
(a) Profit of ₹ 8 is earned on a bag of white cement and a loss of ₹ 5 on a bag of grey cement
Number of white cement bags sold = 3,000
Number of grey cement bags sold = 5,000
Profit = 3000 x ₹ 8 = ₹ 24.000
Loss = 5000 x ₹ 5 = ₹ 25,000
Here Loss > Profit
∴ Loss = ₹ (25,000 – 24.000) = ₹ 1000
(b) Number of grey cement bags sold = 6400
Total loss = ₹ 5 x 6400 = ₹ 32000
For no profit and no loss, there should be a profit of ₹ 32000
Number of white cement bags sold to earn a profit of ₹ 32000 = 32000 ÷ 8 = 4000 bags
9. Replace the blank with an integer to make it a true statement:
(a) (–3) x ___= 27
(b) 5 x ___= –35
(c) ____ x (–8) = –56
d) ___ x (–12) = 132
Solution:
(a) –3 x ___= 27
= –3 x (–9) = 27
[∴ 3 x 9 = 27]
(b) _____ x = – 35
= 5 x (–7) = –35
[ ∴ 5 x 7 = 35]
(c) ___ x (–8) = –56
7 x (–8) = –56
[∴ 7 x 8 = 56]
(d) ____ x (–12) = 132
= (–11) x (–12) = 132
[∴ 11 x 12 = 132]
Try these: (page 22)
Find:
a) (–100) ÷ 5 b) (–81) ÷ 9
c) (–75) ÷ 5 d) (–32) ÷ 2
Solution:
We know that to divide a negative integer by a positive integer, we divide then as whole numbers and then put minus sign before the quotient.
a) ∴ 100 ÷ 5 = 20 ∴ (–100) ÷ 5 = –20
b) ∴ 81 ÷ 9 = 9 ∴ (–81) ÷ 9 = –9
c) ∴ 75 ÷ 5 = 15 ∴ (–75) ÷ 5 = –15
d) ∴ 32 ÷ 2 = 16 ∴ (–32) ÷ 2 = –16
Try these: (page 23)
Find:
(a) 125 ÷ (–25) (b) 80 ÷ (–5) (c) 64 ÷ (–16)
Solution:
We know that to divide a positive integer by a negative integer, we first divide them as whole numbers and then put a minus sign (–) before the quotient.
a) ∴ 125 ÷ 25 = 5 b) 80 ÷ (5) = 16
∴ 125 ÷ (–25) = –5 ∴ 80 ÷ (–5) = –16
c) ∴ 64 ÷ 16 = 4
∴ 64 ÷ (–16) = –4
Find:
(a) (–36) ÷ (–4) (b) (–201) ÷ (–3) (c) (–325) ÷ (–13)
Solution:
To divide a negative integer by a negative integer, we first divide them as a whole number and put a positive sign before the quotient.
a) ∴ 36 ÷ 4 = 9 b) ∴ 201 ÷ 3 = 67 c) ∴ 325 ÷ 13 = 25
∴ (–36) ÷ (–4) = 9 ∴ (–201) ÷ (–3) = 67 ∴ (–325) ÷ (–13) = 25
Find: (i) 1 ÷ a = 1 and (ii) a ÷ (–1) = –a for any integer a?
Take different values of and check.
Solution:
(i) Let us take a = –1, 1, 2, 3, …
For a = –1
L.H.S = 1 ÷ (–1) = –1 [∴ 1 ÷ 1 = 1]
R.H.S = 1
i.e. L.H. S. ≠ R.H.S.
For a = 1
L.H.S = 1 ÷ a = 1 ÷ 1 = 1
R.H.S = 1
i.e. L.H.S = R.H.S
For a = 2
L.H.S = 1 ÷ a = 1 ÷ 2 = 1/2 ≠ R.H.S
For a = 3
L.H.S = 1 ÷ a = 1 ÷3 = 1/3 ≠ R.H.S
Thus, 1 ÷ a = 1 is true only for a = 1
(ii) Let us take a = 1, 2, 3,….
For a = 1,
L.H.S = a ÷ (–1) = 1 ÷ (–1) = –2
R.H.S = –a = –1
i.e L.H.S = R.H.S.
For a = 2
L.H.S = a ÷ (–1) = 2 ÷ (–1) = –2
R.H.S = a = –2
i.e. L.H. S = R.H.S.
For a = 3
L.H.S = a ÷ (–1) = 3 ÷ (–1) = –3
R.H.S = –a = –3
i.e. L.H. S = R.H.S.
For a = 7
L.H.S = a ÷ (–1) = 7 ÷ (–1) = –7
R.H.S = –a = –7
i.e. L.H. S = R.H.S.
For every integer, we have a ÷ (–1) = –a