Class 7 Maths Chapter 1 Question Answers - Integers

Q1: Mohan deposits ₹ 2,000 in his bank account and then withdraws ₹ 1,642 from it the following day. Now, if the withdrawal of the amount from the account is represented by a negative integer, then how will you represent the total amount deposited? Also, Find the balance in Mohan’s account after the withdrawal.
Ans: Withdrawal of these amounts 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 above question,
The total amount that is deposited in the bank account by the Mohan = ₹ 2000
The total amount that is withdrawn from the bank account by the Mohan is = – ₹ 1642
Final Balance in Mohan’s account after the withdrawal = amount deposited + amount is withdrawn
= ₹ 2000 + (-₹ 1642)
= ₹ 2000 – ₹ 1642
= ₹ 358
Hence, the total balance in Mohan’s account after the withdrawal is ₹ 358

Q2: In the city of Srinagar, temperature was – 5°C on Monday, and then it dropped by two °C on Tuesday. What was the temperature of the city of Srinagar on Tuesday? On Wednesday, the temperature rose by 4°C. What was the temperature on this day?
Ans: From the above question,
The temperature on Monday at Srinagar is = -5C
The temperature on Tuesday at the city of Srinagar is dropped by 2C = Temperature on Monday – 2C
= -5C – 2C
= -7 celsius
The temperature on Wednesday at the city Srinagar rose by 4C = Temperature on Tuesday + 4C.
= -7C + 4C
= -3 celsius
Thus, the temperature on days Tuesday and Wednesday was found to be -7C and -3C, respectively.

Q3: Verify a – (– b) is equal to a + b for the following values of alphabets a and b.
(i) a = 21, b = 18
Ans:
From the above question,
a = 21 and b = 18
So To verify a – (- b) is equal to a + b
Let us take the Left Hand Side (LHS) = a – (- b)
= 21 – (- 18)
= 21 + 18
= 39
Now, lets take Right Hand Side (RHS) = a + b
= 21 + 18
= 39
By comparing both the LHS and the RHS.
LHS = RHS
39 = 39
Hence, the value of a and b are verified.
(ii) a = 118, b = 125
From the above question,
a = 118 and b = 125
To verify this a – (- b) = a + b
Let us take the Left Hand Side (LHS) = a – (- b)
= 118 – (- 125)
= 118 + 125
= 243
Now, take the Right Hand Side (RHS) = a + b
= 118 + 125
= 243
By comparing both the LHS and the RHS
LHS = RHS
243 = 243
Hence, the values of a and b are verified.
(iii) a = 75, b = 84
From the above question,
a = 75 and b = 84
To verify that the a – (- b) = a + b
Let us take the Left Hand Side (LHS) = a – (- b)
= 75 – (- 84)
= 75 + 84
= 159
Now, the Right Hand Side (RHS) = a + b
= 75 + 84
= 159
By comparing both LHS and RHS, we find that,
LHS = RHS
159 = 159
Hence, the value of a and b is verified as.
(iv) a = 28, b = 11
From the above question,
a = 28 and b = 11
To verify that a – (- b) = a + b
Let us now take Left Hand Side (LHS) = a – (- b)
= 28 – (- 11)
= 28 + 11
= 39
Now, Right Hand Side (RHS) = a + b
= 28 + 11
= 39
By comparing both the LHS and the RHS
LHS = RHS
39 = 39
Hence, the value of a and b are verified.

Q4: Fill in the blanks to make the following statements true:
(i) (–5) + (– 8) = (– 8) + (______)
Ans:
Let us assume that the missing integer is x,
So,
= (–5) + (– 8) which equals to (– 8) + (x)
= – 5 – 8 = – 8 + x
= – 13 = – 8 + x
By sending – 8 from the RHS to the LHS, it becomes 8,
= – 13 + 8 = x
= x = – 5
Now substitute the x value in the place of the blank place present,
(–5) + (– 8) = (– 8) + (- 5) … [This following equation is present in the form of the Commutative law of Addition]
(ii) –53 + ________ = –53
Let us assume that the missing integer is x,
So,
= –53 + x = –53
By sending – 53 from the LHS to the RHS, it becomes 53,
= x = -53 + 53
= x = 0
Now substitute the following x value in the blank place,
= –53 + 0 = –53 … [This equation is present in the form of Closure property of Addition]
(iii) 17 + _______ = 0
Let us assume that the missing integer is x,
So,
= 17 + x = 0
By sending 17 from the LHS to the RHS, it becomes -17,
= x = 0 – 17
= x = – 17
Now substitute this x value in the blank place,
= 17 + (-17) = 0 … [This equation is present in the form of Closure property of Addition]
= 17 – 17 = 0
(iv) [13 + (– 12)] + (______) = 13 + [(–12) + (–7)]
Let us assume that the missing integer is x,
So,
= [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 one from the LHS to the RHS, it becomes -1,
= x = -6 – 1
= x = -7
Now substitute the following x value in the blank place value,
= [13 + (– 12)] + (-7) equals to 13 + [(–12) + (–7)] … [This equation is present in the form of the Associative Property of Addition]
(v) (– 4) + [15 + (–3)] equals to [– 4 + 15] +_____
Let us assume that the missing integer is x,
So,
= (– 4) + [15 + (–3)] is equal to [– 4 + 15] + x
= (– 4) + [15 – 3)] equals to [– 4 + 15] + x
= (-4) + [12] = [11] + x
= 8 = 11 + x
Now, By sending 11 from the RHS to the LHS, it becomes -11,
= 8 – 11 = x
= x = -3
Now substitute the x value in the place of the blank place,
= (– 4) + [15 + (–3)] equals to [– 4 + 15] + -3 … [The following equation is in the form of the Associative property of the Addition]

Q5: A certain freezing process requires that the room temperature be lowered from 40°C at the rate of 5°C every hour. What will be the final room temperature 10 hours after the actual process begins?
Ans: 
From the above question, it is given that
Let us take the lowered temperature as a negative integer,
Initial temperature will be = 40^oC
Change in temperature per hour is = -5^oC
Change in temperature after 10 hours will be = (-5) × 10 = -50^oC
∴The final room temperature after the 10 hours of freezing process = 40^oC + (-50^oC)
= -10^oC

Q6: Following number line given below shows the temperature present in degree celsius at different places on a particular day.

(i) Observe the number line and write down the temperature of the places marked on it.

Ans:
By observing the above number line, we can find out the temperature of the cities as follows,
The temperature in the city of Lahulspiti is -8°C.
The temperature in the city of Srinagar is -2°C
The temperature in the city of Shimla is 5°C.
The temperature in the city of Ooty is 14°C.
The temperature in the city of Bengaluru is 22°C.
(ii) What is the temperature difference between the hottest and the coldest places among the cities stated above?
From the above number line, we can observe that,
The temperature at the given hottest place, that is, Bengaluru, is 22°C.
The temperature at the given coldest place, that is, Lahulspiti, is -8°C
The temperature difference between the hottest and the coldest place is given as = 22°C – (-8°C)
= 22°C + 8°C
= 30° Celsius
Hence, the total temperature difference between the hottest and the coldest place is 30oC
(iii) What is the temperature difference between the cities of Lahulspiti and Srinagar?
From the above-given number line,
The temperature in the city of Lahulspiti is -8°C.
The temperature in the city of Srinagar is -2°C
∴ The temperature difference between the cities Lahulspiti and Srinagar is = -2oC – (8oC)
= – 2°C + 8°C
= 6°C
(iv) Can we say that the temperature of Srinagar and Shimla taken together is less than the temperature present at Shimla? Is it also less than the temperature present at Srinagar?
From the above-given number line,
The temperature in the city of Srinagar =-2°C
The temperature in the city of Shimla = 5°C
The temperature of the cities Srinagar and Shimla taken together becomes = – 2°C + 5°C
= 3° degree C
5°C > 3°C
Hence, the temperature of the cities Srinagar and Shimla taken together is indeed less than the temperature present at Shimla.
Then,
3° > -2°
And No, the temperature of the cities Srinagar and Shimla taken together is not less than the temperature of the city Srinagar.

Q7: In the following quiz, positive marks are given for every correct answer and negative marks are given for each incorrect answer. If Jack’s scores in the quiz for five successive rounds were 25, – 5, – 10, 10, and 15 so, what was his total at the end?
Ans:
From the above question,
Jack’s scores in the five successive rounds are 25, -5, -10, 15 and 10
Hence, Their total score of Jack at the end will be = 25 + (-5) + (-10) + 15 + 10
= 25 – 5 – 10 + 15 + 10
= 50 – 15
= 35 marks
∴ Now, Jack’s total score at the end is 35.

Q8: In a magic square, every row, column and diagonal has the same sum. Check which of these following is a magic square. 

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

Q9: A water tank has stepped inside it. A monkey is sitting on the utter topmost step (which is the first step). The water level is present at the ninth step.
(i) He jumps three steps down the stairs and then successively jumps back two steps upwards. In how many jumps will the Monkey reach the following water level?
Ans:
Let us consider the steps moved down are represented by a positive integer, and then the steps moved up are represented by a negative integer.
Initially, the Monkey is sitting on the topmost step, which is the first step.
In the 1^st jump monkey will be at the step = 1 + 3 = 4 steps
In the 2^nd jump monkey will be at the step = 4 + (-2) = 4 – 2 = 2 steps
In the 3^rd jump monkey will be at the step = 2 + 3 = 5 steps
In the 4^th jump monkey will be at the step = 5 + (-2) = 5 – 2 = 3 steps
In the 5^th jump monkey will be at the step = 3 + 3 = 6 steps
In the 6^th jump monkey will be at the step = 6 + (-2) = 6 – 2 = 4 steps
In the 7^th jump monkey will be at the step = 4 + 3 = 7 steps
In the 8^th jump monkey will be at the step = 7 + (-2) = 7 – 2 = 5 steps
In the 9^th jump monkey will be at the step = 5 + 3 = 8 steps
In the 10^th jump monkey will be at the step = 8 + (-2) = 8 – 2 = 6 steps
In the 11^th jump monkey will be at the step = 6 + 3 = 9 steps
∴ Monkey took a total of 11 jumps (i.e., 9th step) to reach the water level.
(ii) After drinking water, the Monkey wants to go back. For this, the Monkey jumps four steps up and then successively jumps back two steps down in his every move. In how many total jumps will he reach back to the top step?
Let us consider the steps moved down are represented by the positive integers, and then the steps moved up are represented by the negative integers.
Initially, the Monkey is sitting on the ninth step, i.e., at the water level.
In the 1^st jump monkey will be at the step = 9 + (-4) = 9 – 4 = 5 steps
In the 2^nd jump monkey will be at the step = 5 + 2 = 7 steps
In the 3^rd jump monkey will be at the step = 7 + (-4) = 7 – 4 = 3 steps
In the 4^th jump monkey will be at the step = 3 + 2 = 5 steps
In the 5^th jump monkey will be at the step = 5 + (-4) = 5 – 4 = 1 step
∴ Hence the Monkey took five jumps to reach back to the top step, i.e., the first step.

Q10: Find the product using the suitable properties:
(i) 26 × (– 48) + (– 48) × (–36)
Ans: This given equation is in the form of the Distributive law of the  Multiplication property over Addition.
= a × (b + c) becomes equal to (a × b) + (a × c)
Let, a = -48, b = 26, c = -36
So,
= 26 × (– 48) + (– 48) × (–36)
= -48 × (26 + (-36)
= -48 × (26 – 36)
= -48 × (-10)
= 480 … [∵ (- × – = +)
(ii) 8 × 53 × (–125)
The given equation is present 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
(iii) 15 × (–25) × (– 4) × (–10)
This given equation is in the form of the Commutative law of the Multiplication property.
= a × b = b × a
So,
= 15 × [(–25) × (– 4)] × (–10)
= 15 × [100] × (–10)
= 15 × [-1000]
= – 15000
(iv) (– 41) × 102
This given equation is in the form of a Distributive law of the Multiplication property over Addition.
= a × (b + c) = (a × b) + (a × c)
= (-41) × (100 + 2)
= (-41) × 100 + (-41) × 2
= – 4100 – 82
= – 4182
(v) 625 × (–35) + (– 625) × 65
This 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