Table of contents  
Exercise 1.1  
Exercise 1.2  
Exercise 1.3  
Old Syllabus Questions 
(a) Sum is 7.
Ans:
= – 4 + (3)
= – 4 – 3 … [∵ (+) × (–) = (–)]
= –7
(b) Difference is 10.
Ans:
= 25 – (15)
= – 25 + 15 … [∵ (–) × (–) = (+)]
= 10
(c) Sum is 0.
Ans:
= 4 + (4)
= 4 – 4
= 0
Note: You can also think other combinations, it completely depends on you.
Q2. (a) Write a pair of negative integers whose difference gives 8.
Ans:
= ( –5) – ( –13)
= –5 + 13 … [∵ ( × – = +)]
= 8
(b) Write a negative integer and a positive integer whose sum is 5.
Ans:
= 25 + 20
= 5
(c) Write a negative integer and a positive integer whose difference is 3.
Ans: = – 2 – (1)
= – 2 – 1
= –3
Q3. In a quiz, team A scored 40, 10, 0 and team B scores 10, 0, 40 in three successive rounds. Which team scored more? Can we say that we can add integers in any order?
Ans:
 Team A scored 40, 10, 0
⇒ Total score of Team A = 40 + 10 + 0 =  30 Team B scored 10, 0, 40
⇒ Total score of Team B = 10 + 0+ (40)  10 + 0  40 = 30
Thus, the scores of both teams are same. Yes, we can add integers in any order due to the commutative property of addition of integers which says: A + B = B + A
Q4. Fill in the blanks to make the following statements true:
(i) (5) + (8) = (8) + (....)
Ans:
Let us assume the missing integer be x,
Then,
⇒ (–5) + (– 8) = (– 8) + (x)
⇒ – 5 – 8 = – 8 + x
⇒ – 13 = – 8 + x
By sending – 8 from RHS to LHS it becomes 8,
⇒ – 13 + 8 = x
⇒ x = – 5
Now substitute the x value in the blank place,
(–5) + (– 8) = (– 8) + ( 5) … [This equation is in the form of Commutative law of Addition]
(ii) 53 +... = 53
Ans:
Let us assume the missing integer be x,
Then,
⇒ –53 + x = –53
By sending – 53 from LHS to RHS it becomes 53,
⇒ x = 53 + 53 = x = 0
Now substitute the x value in the blank place,
⇒ –53 + 0 = –53 … [This equation is in the form of Closure property of Addition]
(iii) 17 +... = 0
Ans:
Let us assume the missing integer be x,
Then,
⇒17 + x = 0
By sending 17 from LHS to RHS it becomes 17,
⇒ x = 0 – 17
⇒ x = – 17
Now substitute the x value in the blank place,
⇒ 17 + (17) = 0 … [This equation is in the form of Closure property of Addition]
⇒ 17 – 17 = 0
(iv) [13 + (12)] + (....) = 13 + [(12) + (7)]
Ans:
Let us assume the missing integer be x,
Then,
⇒ [13 + (– 12)] + (x) = 13 + [(–12) + (–7)]
⇒ [13 – 12] + (x) = 13 + [–12 –7]
⇒ [1] + (x) = 13 + [19]
⇒ 1 + (x) = 13 – 19
⇒1 + (x) = 6
By sending 1 from LHS to RHS it becomes 1,
⇒ x = 6 – 1 = x = 7
Now substitute the x value in the blank place,
⇒ [13 + (– 12)] + (7) = 13 + [(–12) + (–7)] … [This equation is in the form of Associative property of Addition]
(v) (4) + [15 + (3)] = [4 + 15] + .....
Ans:
Let us assume the missing integer be x,
Then,
⇒ (– 4) + [15 + (–3)] = [– 4 + 15] + x
⇒(– 4) + [15 – 3)] = [– 4 + 15] + x
⇒ (4) + [12] = [11] + x
⇒ 8 = 11 + x
By sending 11 from RHS to LHS it becomes 11,
⇒ 8 – 11 = x
⇒ x = 3
Now substitute the x value in the blank place,
⇒ (– 4) + [15 + (–3)] = [– 4 + 15] + 3 … [This equation is in the form of Associative property of Addition]
Q1. Find the each of the following products
(a) 3 x (1)
Ans:
By the rule of Multiplication of integers,
= 3 × (1)
= 3 … [∵ (+ × – = )]
(b) (1) x 225
Ans:
By the rule of Multiplication of integers,
= (1) × 225
= 225 … [∵ ( × + = )]
(c) (21) x (30)
Ans:
By the rule of Multiplication of integers,
= (21) × (30)
= 630 … [∵ ( × – = +)]
(d) (316) x (1)
Ans:
By the rule of Multiplication of integers,
= (316) × (1)
= 316 … [∵ ( × – = +)]
(e) (15) x 0 x (18)
Ans:
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) x (11) x (10)
Ans:
By the rule of Multiplication of integers,
= (–12) × (11) × (10)
First multiply the two numbers having same sign,
= 132 × 10 … [∵ ( × – = +)]
= 1320
(g) 9 x (3) x (6)
Ans:
By the rule of Multiplication of integers,
= 9 × (3) × (6)
First multiply the two numbers having same sign,
= 9 × 18 … [∵ ( × – = +)]
= 162
(h) (18) x (5) x (4)
Ans:
By the rule of Multiplication of integers,
= (18) × (5) × (4)
First multiply the two numbers having same sign,
= 90 × 4 … [∵ ( × – = +)]
= – 360 … [∵ (+ × – = )]
(i) (1) x (2) x (3) x 4
Ans:
By the rule of Multiplication of integers,
= [(–1) × (–2)] × [(–3) × 4]
= 2 × (12) … [∵ ( × – = +), ( × + = )]
= – 24
(j) (3) x (6) x (2) x (1)
Ans:
By the rule of Multiplication of integers,
= [(–3) × (–6)] × [(–2) × (–1)]
First multiply the two numbers having same sign,
= 18 × 2 … [∵ ( × – = +) = 36
Q2. Verify the following
(a) 18 x [7 + (3)] = [18 x 7] + [18 x (3)]
Ans:
18 x [7 + (3)] = [18 x 7] + [18 x (3)]
⇒ 18 x 4 = 126 + (54)
⇒ 72 = 72
⇒ L.H.S. = R.H.S.
Hence, Verified.
(b)(21) x [(4) + (6)] = [(21) x (4)] + [(21) x (6)]
Ans:
(21) x [(4) + (6)] = [(21)x (4)] + [(21) x (6)]
⇒ (21) x (10) = 84 + 126
⇒ 210 = 210
⇒ L.H.S. = R.H.S.
Hence, Verified.
Q3. (i) For any integer a, what is (1) x a equal to?
Ans: = (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
Ans:
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
Ans:
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
Ans:Now, multiply 0 with (1), we get
= 0 × (1)
= 0
Because, the product of negative integers and zero give zero only.
Q4. Starting from (1) x 5, write various products showing some patterns to show (1) x (1) = 1
Ans:
According to the pattern,
(−1) × (5) = −5
(−1) × (4) = −4
(−1) × (3) = −3
(−1) × (2) = −2
(−1) × (1) = −1
(−1) × (0) = 0
(−1) × (−1) = 1
Thus, we can conclude that this pattern shows the product of one negative integer and one positive integer is a negative integer whereas the product of two negative integers is a positive integer.
Q1. Evaluate each of the following
(a) (30) ÷ 10
Ans:
When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put minus sign () before the quotient.
= (–30) ÷ 10
= – 3
(b) 50 ÷ (5)
Ans:
When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put minus sign () before the quotient.
= (50) ÷ (5)
= – 10
(c) (36) ÷ (9)
Ans:
When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put positive sign (+) before the quotient.
= (36) ÷ (9)
= 4
(d) (49) ÷ 49
Ans:
When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put minus sign () before the quotient.
= (–49) ÷ 49
= – 1
(e) 13 + [(2) + 1]
Ans:
When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put minus sign () before the quotient.
= 13 ÷ [(–2) + 1]
= 13 ÷ (1)
= – 13
(f) 0 ÷ (12)
Ans:
When we divide zero by a negative integer gives zero.
= 0 ÷ (12)
= 0
(g) (31) ÷ [(30) ÷ (1)]
Ans:
When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put positive sign (+) before the quotient.
= (–31) ÷ [(–30) + (–1)]
= (31) ÷ [30 – 1]
= (31) ÷ (31)
= 1
(h) [(36) ÷ 12] ÷ 3
Ans:
When we divide a negative integer by a positive integer, we first divide them as whole numbers and then put minus sign () before the quotient.
First we have to solve the integers with in the bracket,
= [(–36) ÷ 12]
= (–36) ÷ 12
= – 3
Then, = (3) ÷ 3 = 1
(i) [(6) + 5] ÷ [(2) + 1]
Ans:
When we divide a negative integer by a negative integer, we first divide them as whole numbers and then put positive sign (+) before the quotient.
The given question can be written as,
= [1] ÷ [1]
= 1
Q2. Verify that a ÷ (b + c) ≠ (a + b) + (a ÷ c) for each of the following values of a, b and c.
(a) a = 12, b = 4,c = 2
Ans:
From the question,
a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
Given that, a = 12, b = – 4, c = 2
Now, consider
LHS = a ÷ (b + c)When we divide a positive integer by a negative integer, we first divide them as whole numbers and then put minus sign () before the quotient.
= 12 ÷ (4 + 2)
= 12 ÷ (2) = 6
Then, consider
RHS = (a ÷ b) + (a ÷ c)
= (12 ÷ (4)) + (12 ÷ 2)
= (3) + (6) = 3
By comparing LHS and RHS = 6 ≠ 3
= LHS ≠ RHS
Hence, the given values are verified.
(b) a = (10), b = 1 c = 1
Ans:
From the question,
a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
Given, a = (10), b = 1, c = 1
Now, considerWhen we divide a negative integer by a positive integer, we first divide them as whole numbers and then put minus sign () before the quotient.
LHS = a ÷ (b + c)= (10) ÷ (1 + 1)
= (10) ÷ (2) = 5
Then, consider
RHS = (a ÷ b) + (a ÷ c)
= ((10) ÷ (1)) + ((10) ÷ 1)
= (10) + (10)
= 10 – 10 = 20
By comparing LHS and RHS
= 5 ≠ 20
= LHS ≠ RHS
Hence, the given values are verified.
Q3. Fill in the blanks
(a) 369 ÷ _____ = 369
Ans:
Let us assume the missing integer be x,
Then,
= 369 ÷ x = 369
= x = (369/369)
= x = 1
Now, put the valve of x in the blank.
= 369 ÷ 1 = 369
(b) (75) ÷ _____ = (1)
Ans:
Let us assume the missing integer be x,
Then,
= (75) ÷ x = 1
= x = (75/1)
= x = 75
Now, put the valve of x in the blank.
= (75) ÷ 75 = 1
(c) (206) ÷ _____ = 1
Ans:
Let us assume the missing integer be x,
Then,
= (206) ÷ x = 1
= x = (206/1)
= x = 206
Now, put the valve of x in the blank.
= (206) ÷ (206) = 1
(d) (87) ÷ _____ = 87
Ans:
Let us assume the missing integer be x,
Then,
= (87) ÷ x = 87
= x = (87)/87
= x = 1
Now, put the valve of x in the blank.
= (87) ÷ (1) = 87
(e) _____ ÷1 = 87
Ans:
Let us assume the missing integer be x,
Then, = (x) ÷ 1 = 87
= x = (87) × 1
= x = 87
Now, put the valve of x in the blank.
= (87) ÷ 1 = 87
(f) _____ ÷ 48 = 1
Ans:
Let us assume the missing integer be x,
Then,
= (x) ÷ 48 = 1
= x = (1) × 48
= x = 48
Now, put the valve of x in the blank.
= (48) ÷ 48 = 1
(g) 20 ÷ _____ = 2
Ans:
Let us assume the missing integer be x,
Then,
= 20 ÷ x = 2
= x = (20) / (2)
= x = 10
Now, put the valve of x in the blank.
= (20) ÷ (10) = 2
(h) _____ ÷ (4) = 3
Ans:
Let us assume the missing integer be x,
Then,
= (x) ÷ 4 = 3
= x = (3) × 4
= x = 12
Now, put the valve of x in the blank.
= (12) ÷ 4 = 3
Q4. Write five pairs of integers (a, b) such that a ÷ b = 3. One such pair is (6,2) because 6 ÷ (2) = (3)
Ans:
(i) (6) ÷ 2 = 3
(ii) 9 ÷ (3) = 3
(iii) 12 ÷ (4) = 3
(iv) (9) ÷ 3 = 3
(v) (15) ÷ 5 = 3
Q5. The temperature at noon was 10ºC above zero. If it decreases at the rate of 2ºC per hour until midnight, at what time would the temperature be 8ºC below zero? What would be the temperature at midnight?
Ans:
The following number line is representing the temperature:
Temperature in (^{0}C)The temperature decreases 2ºC = 1 hour
The temperature decreases 1ºC = 1/2 hour
The temperature decreases
Total time = 12 noon + 9 hours = 21 hours = 9 pm
Thus, at 9 pm the temperature would be 8ºC below 0ºC.
Q6. 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?
Ans:
Marks are given for one correct answer = 3
 Marks given for 12 correct answers = 3 x 12 = 36
Radhika scored 20 marks. Therefore, Marks obtained for incorrect answers = 20  36 = 16
Now, marks given for one incorrect answer = 2
Therefore, number of incorrect answers = (16) ÷(2) = 8
Thus, Radhika has attempted 8 incorrect questions.
(ii) Mohini scores (5) marks in this test, though she has got 7 correct answers.
How many questions has she attempted incorrectly?
Ans:
Marks given for seven correct answers = 3 x 7 = 21
 Mohini scores = 5
 Marks obtained for incorrect answers = 5 21 = 26
 Now, marks given for one incorrect answer = 2
 Therefore, number of incorrect answers = (26) ÷ (2) = 13
Thus, Mohini has attempted 13 incorrect questions.
Q7. An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 above the ground level, how long will it take to reach 350 m?
Ans:
The starting position of mine shaft is 10 m above the ground but it moves in the opposite direction so it travels the distance (350) m below the ground.
 So total distance covered by mineshaft = 10 m  (350) m = 10 + 350 = 360 m
 Now, time taken to cover a distance of 6 m by it = 1 minute
 So, time taken to cover a distance of 1 m by it = 1/6 minute
 Therefore, time taken to cover a distance of 360 m
= 60 minutes = 1 hour
[Since 60 minutes = 1 hour]
Thus, in one hour the mine shaft reaches 350m below the ground.
Q1. Find the product, using suitable properties
(a) 26 x (48) + (48) x (36)
Ans:
⇒ (48) x [26 + (36)] [Distributive property]
⇒ (48) x (10)
⇒ 480
(b) 8 x 53 x (125)
Ans:
⇒ 53 x [8 x (125)] [Commutative property]
⇒ 53 x (1000)
⇒ 53000
(c) 15 x (25) x (4) x (10)
Ans:
⇒ 15 x [(25) x (4) x (10)] [Commutative property]
⇒ 15 x (1000)
⇒ 15000
(d) (41) x (102)
Ans:
⇒ 41 x [100 + 2] [Distributive property]
⇒ [(41) x 100] + [(41) x 2]
⇒ 4100 + (82)
⇒ 4182
(e) 625 x (35) + (625) x 65
Ans:
⇒ 625 x [(35) + (65)] [Distributive property]
⇒ 625 x (100)
⇒ 62500
(f) 7 x (502)
Ans:
⇒ 7 x 50  7 x 2 [Distributive property]
⇒ 350  14 = 336
(g) (17) x (29)
Ans:
⇒ (17) x [(30) + 1] [Distributive property]
⇒ (17) x (30) + ( 17) x 1
⇒ 510 + (17)
⇒ 493
(h) (57) x (19) + 57
Ans:
⇒ (57) x (19) + 57 x 1
⇒ 57 x 19 + 57 x 1
⇒ 57 x (19 + 1) [Distributive property]
⇒ 57 x 20 = 1140
Q2. A certain freezing process requires that room temperature be lowered from 40^{0}C at the rate of 5ºC every hour. What will be the room temperature 10 hours after the process begins?
Ans:
Given: Present room temperature = 40^{0}C
► Decreasing the temperature every hour = 5^{0}C
► Room temperature after 10 hours = 40^{0}C + 10 x (5^{0}C)
= 40^{0}C  50^{0}C
=  10^{0}C
Thus, the room temperature after 10 hours is  10^{0}C after the process begins.
Q3. 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?
Ans:
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 five correct answers and five incorrect answers, what is her score?
Ans:
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 two correct and five incorrect answers out of seven questions she attempts. What is her score?
Ans:
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.
Q4. 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?
Ans:
Given: Profit of 1 bag of white cement = Rs 8
And Loss of 1 bag of grey cement = Rs 5
Profit on selling 3000 bags of white cement = 3000 x Rs 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
= Rs 25,000  Rs 24,000
= Rs 1,000
Thus, he has lost Rs 1,000 on selling the grey cement bags.
(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.
Ans:
Given: Profit of 1 bag of white cement = Rs 8
And Loss of 1 bag of grey cement = Rs 5
Let the number of bags of white cement be x.
According to question,
Thus, he must sell 4000 white cement bags to have neither profit nor loss.
Q5. Replace the blank with an integer to make it a true statement.
(a) (3) x _____= 27
Ans:
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 x _____ = 35
Ans:
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) _____ x (8) = 56
Ans:
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) _____ x (12) = 132
Ans:
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 … [∵ ( × – = +)]
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