Integers
Exercise 1.3
Questions with (*) sign are open-ended questions.
1). Evaluate each of the following:
(a) (–30) ÷ 10
(b) 50 ÷ (–5)
(c) (–36) ÷ (–9)
(d) (– 49) ÷ (49)
(e) 13 ÷ [(–2) + 1]
(f) 0 ÷ (–12)
(g) (–31) ÷ [(–30) + (–1)]
(h) [(–36) ÷ 12] ÷ 3
(i) [(– 6) + 5)] ÷ [(–2) + 1]
Answers:
(a) (–30) ÷ 10
= (–30) ÷ 10 = -3
when we divide a negative integer by a positive integer, we first divide them as whole numbers and then put a negative sign (-).
(b) 50 ÷ (–5)
50 ÷ (–5) = -5
when we divide a positive integer by a negative integer, we first divide them as whole numbers and then put a minus sign (–) before the quotient.
(c) (–36) ÷ (–9)
(–36) ÷ (–9) = 4
when we divide a negative integer by a negative integer, we first divide them as whole numbers and then put a positive sign (+).
(d) (– 49) ÷ (49)
(– 49) ÷ (49) = -1
when we divide a negative integer by a positive integer, we first divide them as whole numbers and then put a negative sign (-).
(e) 13 ÷ [(–2) + 1]
13 ÷ [(–2) + 1] = 13 ÷ [–2 + 1]
=13 ÷ (-1)
= -13
when we divide a positive integer by a negative integer, we first divide them as whole numbers and then put a negative sign (-).
(f) 0 ÷ (–12)
0 ÷ (–12)
When 0 is divided by any number it gives 0
(g) (–31) ÷ [(–30) + (–1)]
(–31) ÷ [(–30) + (–1)] = (–31) ÷ [–30 +( –1)]
= (–31) ÷ [–30 –1]
= (–31) ÷ (–31)
= 1
when we divide a negative integer by a negative integer, we first divide them as whole numbers and then put a positive sign (+).
(h) [(–36) ÷ 12] ÷ 3
[(–36) ÷ 12] ÷ 3 = [(–36) ÷ 12] ÷ 3
= (-3) ÷ 3
when we divide a negative integer by a positive integer, we first divide them as whole numbers and then put a negative sign (-).
= = (-3) ÷ 3 = -1
when we divide a negative integer by a positive integer, we first divide them as whole numbers and then put a negative sign (-).
(i) [(– 6) + 5)] ÷ [(–2) + 1]
[(– 6) + 5)] ÷ [(–2) + 1] = (-6+5) ÷ (-2+1)
= (-1) ÷ (-1)
= 1
when we divide a negative integer by a negative integer, we first divide them as whole numbers and then put a positive sign (+).
2). Verify that a ÷ (b + c) ¹ (a ÷ b) + (a ÷ c) for each of the following values of a, b and c.
(a) a = 12, b = – 4, c = 2
a ÷ (b + c) = 12 ÷ [ (-4) + 2]
= 12 ÷ (-4 + 2)
= 12 ÷ (-2)
= -6
(a ÷ b) + (a ÷ c) = [12 ÷ (-4)] + [12 ÷ 2]
= (-3) + 6
= -3 + 6 = 3
a ÷ (b + c) ¹ (a ÷ b) + (a ÷ c)
(b) a = (–10), b = 1, c = 1
a ÷ (b + c) = (-10) ÷ [ 1 + 1]
= (-10)÷ (2)
= -5
(a ÷ b) + (a ÷ c) = [(-10) ÷ (1)] + [(-10) ÷ (1)]
= (-10) + (-10)
= -10 – 10
= -20
a ÷ (b + c) ¹ (a ÷ b) + (a ÷ c)
3). Fill in the blanks:
(a) 369 ÷ _____ = 369
Ans: 369 ÷ 1 = 369
(b) (–75) ÷ _____ = –1
Ans: (–75) ÷ 1= –1
(c) (–206) ÷ _____ = 1
Ans: (–206) ÷ (-1) = 1
(d) – 87 ÷ _____ = 87
Ans: – 87 ÷ 1 = 87
(e) _____ ÷ 1 = – 87
Ans: (-87) ÷ 1 = – 87
(f) _____ ÷ 48 = –1
Ans: (-48) ÷ 48 = –1
(g) 20 ÷ _____ = –2
Ans: 20 ÷ (-10) = –2
(h) _____ ÷ (4) = –3
Ans: (-12) ÷ (4) = –3
4). Write five pairs of integers (a, b) such that a ÷ b = –3. One such pair is (6, –2) because 6 ÷ (–2) = (–3).
(12, –3), (-6, 2), (-15, 5), (3, –1), (18, –6)
5). The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at mid-night?
The temperature at 12 noon = 10°C
Rate of temperature decrease = 2°C per hour
The decrease in temperature = 10 – (-8)
= 10 + 8
= 18
Time taken = 18 ÷ 2 = 9 hours
The temperature is 8°C below zero at 9 pm
The temperature at midnight = 12 X 2 = 24
The temperature at midnight = 10 – (24) = 10 – 24
= -14°C
6). In a class test (+ 3) marks are given for every correct answer and (–2) marks are given for every incorrect answer and no marks for not attempting any question.
(i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?
Score of Radhika = 20
Number of correct answers = 12
Marks for correct answers = 12 X 3 = 36
Difference in marks = 36 – 20 = 16
Number of incorrect answers = 16 ÷ 2 = 8
(ii) Mohini scores –5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
Score of Mohini = -5
Number of correct answers = 7
Marks for correct answers = 7 X 3 = 21
Difference in marks = 21 – (-5) = 21 + 5 = 26
Number of incorrect answers = 26 ÷ 2 = 13
7). An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 m.
Rate of descending = 6 m/min
Start of the descend = 10 m
End point of the descend = -350 m
Length of the descend = 10- (-350)
= 10 + 350
= 360 m
Time taken to descend = 360 ÷ 6 = 60 min = 1 hour
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Additional Examples for Practice
1). Find the product, using suitable properties:
(a) 26 × (– 48) + (– 48) × (–36)
26 × (– 48) + (– 48) × (–36)
= (-48) X [ 26 + (-36)]
= (-48) X ( 26 -36)
= (-48) X ( -10)
= 480
Ans: 26 × (– 48) + (– 48) × (–36) = 480
(b) 8 X 53 X (–125)
8 X 53 X (–125)
= 8 X (-125) X 53
= -1000 X 53
= -53000
Ans: 8 X 53 X (–125) = -53000
(c) 15 X (–25) X (– 4) X (–10)
15 X (–25) X (– 4) X (–10)
= 15 X 100 X (-10)
= 1500 X (-10)
= -15000
Ans: 15 X (–25) X (– 4) X (–10) = -15000
(d) (– 41) × 102
(– 41) × 102 = (-41) X (100 + 2)
= (-41) X 100 + (-41) X 2
= -4100 + (-82)
= -4100 -82
= -4182
Ans: (– 41) × 102 = -4182
(e) 625 × (–35) + (– 625) × 65
625 × (–35) + (– 625) × 65
= 625 × (–35) – 625 × 65
= 625 [-35 – 65]
= 625 (-100)
= -62500
Ans: 625 × (–35) + (– 625) × 65 = -62500
(f) 7 × (50 – 2)
7 × (50 – 2) = 7 X 50 – 7 X 2
= 350 – 14
= 336
Ans: 7 × (50 – 2) = 336
(g) (–17) × (–29)
(–17) × (–29) = -17 X (1 – 30)
= -17 X 1 + 17 X 30
= -17 + 510
= 493
Ans: (–17) × (–29) = 493
(h) (–57) × (–19) + 57
(–57) × (–19) + 57 = 57 X 19 + 57
= 57 X (19 + 1)
= 57 X 20
= 1140
Ans: (–57) × (–19) + 57 = 1140
2). 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?
The initial temperature of the room = 400C
Rate of lowering temperature = -50 C per hour
Decrease in temperature after 10 hours = 10 X (-5)
= -50
Room temperature after 10 hours = 40 + (-50)
= 40 – 50
= -10
Room temperature after 10 hours = -100C
3). 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?
Marks awarded for every correct answer = 5
Marks awarded for every incorrect answer = -2
Score of Mohan = 4 X 5 + 6 X (-2)
= 20 – 12
= 8
Score of Mohan = 8 marks
(ii) Reshma gets five correct answers and five incorrect answers, what is her score?
Marks awarded for every correct answer = 5
Marks awarded for every incorrect answer = -2
Score of Reshma = 5 X 5 + 5 X (-2)
= 25 – 10
= 15
Score of Reshma = 15 marks
(iii) Heena gets two correct and five incorrect answers out of seven questions she attempts. What is her score?
Marks awarded for every correct answer = 5
Marks awarded for every incorrect answer = -2
Score of Heena = 2 X 5 + 5 X (-2)
= 10 – 10
= 0
Score of Heena = 0 marks
4). 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?
Profit of 1 bag of white cement = Rs 8
Loss on 1 bag of grey cement = Rs 5
Profit or Loss made by the company is = 3000 X 8 + 5000 X (-5)
= 24000 – 25000
= -1000
Negative mark indicates loss.
Loss made by the company = Rs 1000
(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.
Loss made by the company by selling grey cement = 6400 X 5
= 32000
Number of white cement bags to sell to make neither profit nor loss
= 32000 ÷ 8
= 4000.
5). Replace the blank with an integer to make it a true statement.
(a) (–3) × _____ = 27
(–3) × 9 = 27
(b) 5 × _____ = –35
5 × (-7) = –35
(c) _____ × (– 8) = –56
7 × (– 8) = –56
(d) _____ × (–12) = 132
11 × (–12) = 132
