NCERT Solutions for Class 7 Maths - Integers (Exercise 1.1, 1.2 and 1.3)

Exercise 1.1

Q1. Following number line shows the temperature in degree celsius (°C) at different places on a particular day.

(a) Observe this number line and write the temperature of the places marked on it.

Ans: By observing the number line, we can find the temperature of the cities as follows,
The temperature at Lahulspiti is -8^oC
The temperature at Srinagar is -2^oC
The temperature at Shimla is 5^oC
The temperature at Ooty is 14^oC
The temperature at Bengaluru is 22^oC

(b) What is the temperature difference between the hottest and the coldest places among the above cities?

Ans: From the number line, we observe that,
The temperature at the hottest place, i.e. Bengaluru, is 22°C
The temperature at the coldest place, i.e., Lahulspiti, is -8°C
Temperature difference between hottest and coldest place is = 22°C – (-8°C)
= 22°C + 8°C
= 30°C
Hence, the temperature difference between the hottest and the coldest place is 30°C.

(c) What is the temperature difference between Lahulspiti and Srinagar?

Ans: From the given number line,
The temperature at Lahulspiti is -8°C
The temperature at Srinagar is -2°C
∴ The temperature difference between Lahulspiti and Srinagar is = -2°C – (8°C)
= – 2°C + 8°C
= 6°C

(d) Can we say the temperature of Srinagar and Shimla, taken together, is less than the temperature at Shimla? Is it also less than the temperature at Srinagar?

Ans: From the given number line,
The temperature at Srinagar =-2°C
The temperature at Shimla = 5°C
The temperature of Srinagar and Shimla, taken together, is = – 2°C + 5°C
= 3°C
∴ 5°C > 3°C
So, the temperature of Srinagar and Shimla, taken together, is less than the temperature at Shimla.
Then,
3° > -2°
Hence, the temperature of Srinagar and Shimla, taken together, is not less than the temperature of Srinagar.

Q2. In a quiz, positive marks are given for correct answers and negative marks are given for incorrect answers. If Jack’s scores in five successive rounds were 25, – 5, – 10, 15 and 10, what was his total at the end?

Ans: From the question,
Jack’s scores in five successive rounds are 25, -5, -10, 15 and 10
The total score of Jack at the end will be = 25 + (-5) + (-10) + 15 + 10
= 25 – 5 – 10 + 15 + 10
= 50 – 15
= 35
∴ Jack’s total score at the end is 35.

Q3. At Srinagar temperature was – 5°C on Monday, and then it dropped by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day?

Ans: From the question,
The temperature on Monday at Srinagar = -5°C
The temperature on Tuesday at Srinagar has dropped by 2°C = Temperature on Monday –2°C
= -5°C – 2°C
= -7°C
The temperature on Wednesday at Srinagar rose by 4°C = Temperature on Tuesday + 4°C
= -7°C + 4°C
= -3°C
Thus, the temperature on Tuesday and Wednesday was -7°C and -3°C, respectively.

Q4. A plane is flying at a height of 5000 m above sea level. At a particular point, it is exactly above a submarine floating 1200 m below the sea level. What is the vertical distance between them?

Ans: From the question,
The plane is flying at the height = 5000 m
Depth of Submarine = -1200 m
The vertical distance between plane and submarine = 5000 m – (- 1200) m
= 5000 m + 1200 m
= 6200 m

Q5. Mohan deposits ₹ 2,000 in his bank account and withdraws ₹ 1,642 from it the next day. If the withdrawal of the amount from the account is represented by a negative integer, then how will you represent the amount deposited? Find the balance in Mohan’s account after the withdrawal.

Ans: Withdrawal of the amount from the account is represented by a negative integer.
Then, the deposit of the amount to the account is represented by a positive integer.
From the question,
Total amount deposited in bank account by the Mohan = ₹ 2000
The total amount is withdrawn from the bank account by the Mohan = – ₹ 1642
Balance in Mohan’s account after the withdrawal = amount deposited + amount withdrawn
= ₹ 2000 + (-₹ 1642)
= ₹ 2000 – ₹ 1642
= ₹ 358
Hence, the balance in Mohan’s account after the withdrawal is ₹ 358

Q6. Rita goes 20 km towards the east from point A to point B. From B, she moves 30 km towards the west along the same road. If the distance towards the east is represented by a positive integer, then how will you represent the distance travelled towards the west? By which integer will you represent her final position from A?

Ans: From the question, it is given that
A positive integer represents the distance towards the east.
Then, the distance travelled towards the west will be represented by a negative integer.
Rita travels a distance in the east direction = 20 km
Rita travels a distance in the west direction = – 30 km
∴ Distance travelled from A = 20 + (- 30)
= 20 – 30
= -10 km
Hence, we will represent the distance travelled by Rita from point A by a negative integer, i.e., –10 km

Q7. In a magic square, each row, column and diagonal have the same sum. Check which of the following is a magic square.

Ans: First, we consider the square (i)
By adding the numbers in each row, we get
= 5 + (- 1) + (- 4) = 5 – 1 – 4 = 5 – 5 = 0
= -5 + (-2) + 7 = – 5 – 2 + 7 = -7 + 7 = 0
= 0 + 3 + (-3) = 3 – 3 = 0
By adding the numbers in each column, we get
= 5 + (- 5) + 0 = 5 – 5 = 0
= (-1) + (-2) + 3 = -1 – 2 + 3 = -3 + 3 = 0
= -4 + 7 + (-3) = -4 + 7 – 3 = -7 + 7 = 0
By adding the numbers in diagonals, we get
= 5 + (-2) + (-3) = 5 – 2 – 3 = 5 – 5 = 0
= -4 + (-2) + 0 = – 4 – 2 = -6
Because the sum of one diagonal is not equal to zero,
(i) is not a magic square
Now, we consider the square (ii)
By adding the numbers in each row, we get
= 1 + (-10) + 0 = 1 – 10 + 0 = -9
= (-4) + (-3) + (-2) = -4 – 3 – 2 = -9
= (-6) + 4 + (-7) = -6 + 4 – 7 = -13 + 4 = -9
By adding the numbers in each column, we get
= 1 + (-4) + (-6) = 1 – 4 – 6 = 1 – 10 = -9
= (-10) + (-3) + 4 = -10 – 3 + 4 = -13 + 4
= 0 + (-2) + (-7) = 0 – 2 – 7 = -9
By adding the numbers in diagonals, we get
= 1 + (-3) + (-7) = 1 – 3 – 7 = 1 – 10 = -9
= 0 + (-3) + (-6) = 0 – 3 – 6 = -9
This (ii) square is a magic square because the sum of each row, each column and the diagonal is equal to -9.

Q8. Verify a – (– b) = a + b for the following values of a and b.
(i) a = 21, b = 18

Ans: From the question,
a = 21 and b = 18
To verify a – (- b) = a + b
Let us take Left Hand Side (LHS) = a – (- b)
= 21 – (- 18)
= 21 + 18
= 39
Now, Right Hand Side (RHS) = a + b
= 21 + 18
= 39
By comparing LHS and RHS
LHS = RHS
39 = 39
Hence, the value of a and b is verified.

(ii) a = 118, b = 125

Ans: From the question,
a = 118 and b = 125
To verify a – (- b) = a + b
Let us take Left Hand Side (LHS) = a – (- b)
= 118 – (- 125)
= 118 + 125
= 243
Now, Right Hand Side (RHS) = a + b
= 118 + 125
= 243
By comparing LHS and RHS
LHS = RHS
243 = 243
Hence, the value of a and b is verified.

(iii) a = 75, b = 84

Ans: From the question,
a = 75 and b = 84
To verify a – (- b) = a + b
Let us take Left Hand Side (LHS) = a – (- b)
= 75 – (- 84)
= 75 + 84
= 159
Now, Right Hand Side (RHS) = a + b
= 75 + 84
= 159
By comparing LHS and RHS
LHS = RHS
159 = 159
Hence, the value of a and b is verified.

(iv) a = 28, b = 11

Ans: From the question,
a = 28 and b = 11
To verify a – (- b) = a + b
Let us take Left Hand Side (LHS) = a – (- b)
= 28 – (- 11)
= 28 + 11
= 39
Now, Right Hand Side (RHS) = a + b
= 28 + 11
= 39
By comparing LHS and RHS
LHS = RHS
39 = 39
Hence, the value of a and b is verified.

Q9. Use the sign of >, < or = in the box to make the statements true.
(a) (-8) + (-4) [ ] (-8) – (-4)

Ans: Let us take Left Hand Side (LHS) = (-8) + (-4)
= -8 – 4
= -12
Now, Right Hand Side (RHS) = (-8) – (-4)
= -8 + 4
= -4
By comparing LHS and RHS
LHS < RHS
-12 < -4
∴ (-8) + (-4) [<] (-8) – (-4)

(b) (-3) + 7 – (19) [ ] 15 – 8 + (-9)

Ans: Let us take Left Hand Side (LHS) = (-3) + 7 – 19
= -3 + 7 – 19
= -22 + 7
= -15
Now, Right Hand Side (RHS) = 15 – 8 + (-9)
= 15 – 8 – 9
= 15 – 17
= -2
By comparing LHS and RHS
LHS < RHS
-15 < -2
∴ (-3) + 7 – (19) [<] 15 – 8 + (-9)

(c) 23 – 41 + 11 [ ] 23 – 41 – 11

Ans: Let us take Left Hand Side (LHS) = 23 – 41 + 11
= 34 – 41
= – 7
Now, Right Hand Side (RHS) = 23 – 41 – 11
= 23 – 52
= – 29
By comparing LHS and RHS
LHS > RHS
– 7 > -29
∴ 23 – 41 + 11 [>] 23 – 41 – 11

(d) 39 + (-24) – (15) [ ] 36 + (-52) – (- 36)

Ans: Let us take Left Hand Side (LHS) = 39 + (-24) – 15
= 39 – 24 – 15
= 39 – 39
= 0
Now, Right Hand Side (RHS) = 36 + (-52) – (- 36)
= 36 – 52 + 36
= 72 – 52
= 20
By comparing LHS and RHS
LHS < RHS
0 < 20
∴ 39 + (-24) – (15) [<] 36 + (-52) – (- 36)

(e) – 231 + 79 + 51 [ ] -399 + 159 + 81

Ans: Let us take Left Hand Side (LHS) = – 231 + 79 + 51
= – 231 + 130
= -101
Now, Right Hand Side (RHS) = – 399 + 159 + 81
= – 399 + 240
= – 159
By comparing LHS and RHS
LHS > RHS
-101 > -159
∴ – 231 + 79 + 51 [>] -399 + 159 + 81

Q10. A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the ninth step.

(i) He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level?

Ans: Let us consider steps moved down are represented by positive integers, and steps moved up are represented by negative integers.
Initially, the monkey is sitting on the topmost step, i.e., the first step
In the 1^st jump the monkey will be at step = 1 + 3 = 4 steps
In the 2^nd jump the monkey will be at step = 4 + (-2) = 4 – 2 = 2 steps
In the 3^rd jump the monkey will be at step = 2 + 3 = 5 steps
In the 4^th jump the monkey will be at step = 5 + (-2) = 5 – 2 = 3 steps
In the 5^th jump the monkey will be at step = 3 + 3 = 6 steps
In the 6^th jump the monkey will be at step = 6 + (-2) = 6 – 2 = 4 steps
In the 7^th jump the monkey will be at step = 4 + 3 = 7 steps
In the 8^th jump the monkey will be at step = 7 + (-2) = 7 – 2 = 5 steps
In the 9^th jump the monkey will be at step = 5 + 3 = 8 steps
In the 10^th jump the monkey will be at step = 8 + (-2) = 8 – 2 = 6 steps
In the 11^th jump the monkey will be at step = 6 + 3 = 9 steps
∴ The monkey took 11 jumps (i.e., the 9^th step) to reach the water level.

(ii) After drinking water, the monkey wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down with every move. In how many jumps will he reach back to the top step?

Ans: Let us consider steps moved down are represented by positive integers, and then steps moved up are represented by negative integers.
Initially, the monkey is sitting on the ninth step, i.e., at the water level
In the 1^st jump the monkey will be at step = 9 + (-4) = 9 – 4 = 5 steps
In the 2^nd jump the monkey will be at step = 5 + 2 = 7 steps
In the 3^rd jump the monkey will be at step = 7 + (-4) = 7 – 4 = 3 steps
In the 4^th jump the monkey will be at step = 3 + 2 = 5 steps
In the 5^th jump the monkey will be at step = 5 + (-4) = 5 – 4 = 1 step
∴ The monkey took 5 jumps to reach back to the top step, i.e., the first step.

(iii) If the number of steps moved down is represented by negative integers, and the number of steps moved up by positive integers, represent his moves in parts (i) and (ii) by completing the following; (a) – 3 + 2 – … = – 8 (b) 4 – 2 + … = 8. In (a), the sum (– 8) represents going down by eight steps. So, what will the sum 8 in (b) represent?

Ans: From the question, it is given that
If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers,
Monkey moves in part (i)
= – 3 + 2 – ……….. = – 8
Then LHS = – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3
= – 18 + 10
= – 8
RHS = -8
∴ Moves in part (i) represent that the monkey is going down 8 steps because of a negative integer.
Now,
Monkey moves in part (ii)
= 4 – 2 + ……….. = 8
Then LHS = 4 – 2 + 4 – 2 + 4
= 12 – 4
= 8
RHS = 8
∴ Moves in part (ii) represent that the monkey is going up 8 steps because of a positive integer.

Exercise 1.2

Q1. Write down a pair of integers whose

(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]

Exercise 1.3 

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.

Q5. Find the product using suitable properties:
(a) 26 × (– 48) + (– 48) × (–36)

Ans: The given equation is in the form of the 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)

Ans: The given equation is in the form of the 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)

Ans: The given equation is in the form of the Commutative Law of Multiplication.
= a × b = b × a
Then,
= 15 × [(–25) × (– 4)] × (–10)
= 15 × [100] × (–10)
= 15 × [-1000]
= – 15000

(d) (– 41) × 102

Ans: The given equation is in the form of the 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

Ans: The given equation is in the form of the Distributive Law of Multiplication over Addition.
= a × (b + c) = (a × b) + (a × c)
= 625 × [(-35) + (-65)]
= 625 × [-100]
= – 62500

(f) 7 × (50 – 2)

Ans: The given equation is in the form of the Distributive Law of Multiplication over Subtraction.
= a × (b – c) = (a × b) – (a × c)
= (7 × 50) – (7 × 2)
= 350 – 14
= 336

(g) (–17) × (–29)

Ans: The given equation is in the form of the 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

Ans: The given equation is in the form of the Distributive Law of Multiplication over Addition.
= a × (b + c) = (a × b) + (a × c)
= (57 × 19) + (57 × 1)
= 57 [19 + 1]
= 57 × 20
= 1140

Q6. 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?

Ans: From the question, it is given that
Let us take the lowered temperature as negative,
Initial temperature = 40^oC
Change in temperature per hour = -5^oC
Change in temperature after 10 hours = (-5) × 10 = -50^oC
∴ The final room temperature after 10 hours of freezing process = 40^oC + (-50^oC)
= -10^oC

Q7. 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: 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?

Ans: 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?

Ans: 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

Q8. 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 5,000 bags of grey cement in a month. What is its profit or loss?

Ans: We denote profit in positive integers and loss in negative integers,
From the question,
The cement company earns a profit on selling 1 bag of white cement = ₹ 8 per bag
Then,
The cement company earns a profit on selling 3000 bags of white cement = 3000 × ₹ 8
= ₹ 24000
Loss on selling 1 bag of grey cement = – ₹ 5 per bag
Then,
Loss on selling 5000 bags of grey cement = 5000 × – ₹ 5
= – ₹ 25000
Total loss or profit earned by the cement company = profit + loss
= 24000 + (-25000)
= – ₹1000
Thus, a loss of ₹ 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

Ans: We denote profit in positive integers and loss in negative integers,
From the question,
The cement company earns a profit on selling 1 bag of white cement = ₹ 8 per bag
Let the number of white cement bags be x.
Then,
The cement company earns a profit on selling x bags of white cement = (x) × ₹ 8
= ₹ 8x
Loss on selling 1 bag of grey cement = – ₹ 5 per bag
Then,
Loss on selling 6400 bags of grey cement = 6400 × – ₹ 5
= – ₹ 32000
According to the question,
The 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.

Q9. Replace the blank with an integer to make it a true statement.
(a) (–3) × _____ = 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 × _____ = –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) _____ × (– 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) _____ × (–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 … [∵ (- × – = +)]

Exercise 1.4

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, consider

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.
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 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 midnight?

Ans: From the question, given that,
The temperature at the beginning, i.e. at 12 noon = 10°C
Rate of change of temperature = – 2°C per hour
Then,
Temperature at 1 p.m. = 10 + (-2) = 10 – 2 = 8°C
Temperature at 2 p.m. = 8 + (-2) = 8 – 2 = 6°C
Temperature at 3 p.m. = 6 + (-2) = 6 – 2 = 4°C
Temperature at 4 p.m. = 4 + (-2) = 4 – 2 = 2°C
Temperature at 5 p.m. = 2 + (-2) = 2 – 2 = 0°C
Temperature at 6 p.m. = 0 + (-2) = 0 – 2 = -2°C
Temperature at 7 p.m. = -2 + (-2) = -2 -2 = -4°C
Temperature at 8 p.m. = -4 + (-2) = -4 – 2 = -6°C
Temperature at 9 p.m. = -6 + (-2) = -6 – 2 = -8°C
∴ At 9 p.m., the temperature will be 8°C below zero
Then,
The temperature at midnight, i.e. at 12 a.m.
Change in temperature in 12 hours = -2°C × 12 = – 24°C
So, at midnight temperature will be = 10 + (-24)
= – 14°C
So, at midnight temperature will be 14°C below 0.

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? (ii) Mohini scored –5 marks in this test, though she got 7 correct answers. How many questions has she attempted incorrectly?

Ans: From the question,
Marks awarded for 1 correct answer = +3
Marks awarded for 1 wrong answer = -2
(i) Radhika scored 20 marks
Then,
Total marks awarded for 12 correct answers = 12 × 3 = 36
Marks awarded for incorrect answers = Total score – Total marks awarded for 12 correct
Answers
= 20 – 36
= – 16
So, the number of incorrect answers made by Radhika = (-16) ÷ (-2)
= 8
(ii) Mohini scored -5 marks
Then,
Total marks awarded for 7 correct answers = 7 × 3 = 21
Marks awarded for incorrect answers = Total score – Total marks awarded for 12 correct
Answers
= – 5 – 21
= – 26
So, the number of incorrect answers made by Mohini = (-26) ÷ (-2)
= 13

Q7. 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?

Ans: From the question,
The initial height of the elevator = 10 m
The final depth of the elevator = – 350 m … [∵distance descended is denoted by a negative integer]
The total distance to descended by the elevator = (-350) – (10)
= – 360 m
Then,
Time taken by the elevator to descend -6 m = 1 min
So, the time taken by the elevator to descend – 360 m = (-360) ÷ (-6)
= 60 minutes
= 1 hour