NCERT Solutions Maths ch-1 Integers for Class 7th
Integers are a fundamental concept in mathematics, representing whole numbers that can be positive, negative, or zero. They are typically denoted on a number line, where positive integers lie to the right of zero and negative integers lie to the left. Integers include numbers like -3, -2, -1, 0, 1, 2, and 3, extending infinitely in both directions. Operations with integers, such as addition, subtraction, multiplication, and division, follow specific rules. For instance, adding two negative integers results in a more negative integer, while the product of two integers with opposite signs is always negative. Understanding integers is crucial for solving real-world problems, as they are commonly used to represent temperatures, elevations, and financial transactions, making them an essential part of the mathematical toolkit.
NCERT Solutions of Class 7th Chapter 1 Integers Exercise 1.1, 1.2, 1.3 and 1.4
We try to teach you all Questions in easy way. We solve all chapter wise sums of maths textbook. In every chapter include NCERT solutions. For solutions of Exercise 1.1, 1.2, 1.3 and 1.4 click on Tabs :
NCERT Solutions of class 7th maths Chapter 1 Integers Exercise 1.1
Question 1.
Following number line shows the temperature in degree Celsius (°C) at different places on a particular day.
- Observe this number line and write the temperature of the places marked on it.
(b) What is the temperature difference between the hottest and the coldest places among the above?
(c) What is the temperature difference between Lahulspriti and Srinagar?
(d) Can we say temperature of Srinagar and Shimla taken together is less than the temperature at Shimla? Is it also less than the temperature at Srinagar?
Solution:
(a) From the given number line, we observe the following temperatures.
Cities | Temperature |
Lahulspriti | -8°C |
Srinagar | -2°C |
Shimla | 5°C |
Ooty | 14°C |
Bengaluru | 22°C |
(b) The temperature of the hottest place = 22°C
The temperature of the coldest place = -8°C
Difference = 22°C – (-8°C)
= 22°C + 8°C = 30°C
(c) Temperature of Lahulspriti = -8°C
Temperature of Srinagar = -2°C
∴ Difference = -2°C – (-8°C)
= -2°C + 8°C = 6°C
(d) Temperature of Srinagar = -2°C
Temperature of Shimla = 5°C
∴ Temperature of the above cities taken together
= -2°C + 5°C = 3°C
Temperature of Shimla = 5°C
Hence, the temperature of Srinagar and Shimla taken together is less than that of Shimla by 2°C.
i.e., (5°C – 3°C) = 2°C
Question 2.
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?
Solution:
Given scores are 25, -5, -10, 15, 10
Marks given for correct answers
= 25 + 15 + 10 = 50
Marks given for incorrect answers
= (-5) + (-10) = -15
∴ Total marks given at the end
= 50 + (-15) = 50 – 15 = 35
Question 3.
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?
Solution:
Initial temperature of Srinagar on Monday = -5°C
Temperature on Tuesday = -5°C – 2°C = -7°C
Temperature was increased by 4°C on Wednesday.
∴ Temperature on Wednesday
= -7°C + 4°C = -3°C
Hence, the required temperature on Tuesday = -7°C
and the temperature on Wednesday = -3°C
Question 4.
A plane is flying at the height of 5000 m above the sea level. At a particular point, it is exactly above a submarine flowing 1200 m below the sea level. What is the vertical distance between them?
Solution:
Height of the flying plane = 5000 m
Depth of the submarine = -1200 m
∴ Distance between them
= + 5000 m – (-1200 m)
= 5000 m + 1200 m = 6200 m
Hence, the vertical distance = 6200 m
Question 5.
Mohan deposits Rs. 2,000 in a bank account and withdraws Rs. 1,642 from it, the next day. If withdrawal of 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.
Solution:
The deposited amount will be represented by a positive integer i.e., Rs. 2000.
Amount withdrawn = Rs. 1,642
∴ Balance in the account
= Rs. 2,000 – Rs. 1,642 = Rs. 358
Hence, the balance in Mohan’s account after the withdrawal
= Rs. 358
Question 6.
Rita goes 20 km towards east from a point A to the point B. From B, she moves 30 km towards west along the same road. If the distance towards east is represented by a positive integer, then how will you represent her final position from A?
Solution:
Distances travelled towards east from point A will be represented by positive integer i.e. +20 km.
Distance travelled towards the west from point B will be represented by negative integer, i.e., —30 km.
Final position of Rita from A
= 20 km – 30 km = – 10 km
Hence, the required position of Rita will be presented by a negative number, i.e., -10.
Question 7.
In a magic square each row, column and the diagonal have the same sum. Check which of the following is a magic square?
Solution:
(i) Row one R1 = 5 + (-1) + (—4)
=5 – 1 – 4 = 5 – 5 = 0
Row two R2 = (-5) + (-2) + 7
= -5 – 2 + 7 = -7 + 7 = 0
Row three R3 = 0 + 3 + (-3)
= 0 + 3- 3 = 0
Column one C1t = 5 + (-5) + 0
= 5 – 5 + 0 = 0
Column two C2 = (-1) + (-2) + (3)
=-1 – 2 + 3 = -3 + 3 = 0
Column three C3 = (-4) + 7 + (-3)
= -4 + 7 – 3 = 7 – 7 = 0
Diagonal d12 = 5 + (-2) + (-3)
= 5 – 2- 3 = 5 – 5 = 0
Diagonal d2 = (-4) + (-2) + 0
= -4 – 2 + 0 = -6 + 0 = -6
Here, the sum of the integers of diagonal d2 is different from the others.
Hence, it is not a magic square.
(ii) Row one R1 = 1 + (-10) + 0
= 1 – 10 + 0 = -9
Row two R2 = (-4) + (-3) + (-2)
= -4 – 3 – 2 = -9
Row three R3 = (-6) + (4) + (-7)
= -6 + 4 – 7 = -9
Column one C3 = 1 + (-4) + (-6)
= 1 – 4 – 6 = -9
Column two C2 = (-10) + (-3) + 4
= -10 – 3 + 4 = -9
Column three C3 = 0 + (-2) + (-7)
= 0 – 2 -7 = -9
Diagonal d1 = 1 + (-3) + (-7)
= 1 – 3 – 7 = 1 – 10 = -9
Diagonal d2 = 0 + (-3) + (-6)
= 0 – 3- 6 = -9
Here, sum of the integers column wise, row wise and diagonally is same i.e. -9.
Hence, (ii) is a magic square.
Question 8.
Verify a – (-b) = a + b for the following values of a and 6.
(i) a = 21, b = 18
(ii) a = 118, b = 125
(iii) a = 75, b = 84
(iv) a= 28, 6 = 11
Solution:
(i) a – (-b) = a + b
LHS = 21 – (-18) = 21 + 18 = 39
RHS = 21 + 18 = 39
LHS = RHS Hence, verified.
(ii) a – (-b) = a + b
LHS = 118 – (-125) = 118 + 125 = 243
RHS = 118 + 125 = 243
(iii) a – (-b) = a + b
LHS = 75 – (-84) = 75 + 84 = 159
RHS = 75 + 84 = 159
LHS = RHS Hence, verified.
(iv) a – (-b) = a + b
LHS = 28 – (-11) = 28 + 11 = 39
RHS = 28 + 11 = 28 + 11 = 39
LHS = RHS Hence, verified.
Question 9.
Use the sign of >, < or = in the box to make the statements true.
(a) (-8) +(-4) □(-8)-(-4)
(b) (-3) + 7 – (19) □ 15 – 8 + (-9)
(c) 23 – 41 + 11 □ 23 – 41 – 11
(d) 39 + (-24) – (15) □ 36 + (-52) – (-36)
(e) -231 + 79 + 51 □ -399 + 159 + 81
Solution:
(a) (-8) + (-4) □ (-8) – (-4)
LHS = (-8) + (-4) = -8 – 4 = – 12
RHS = (-8) – (-4) = -8 + 4 = -4
Here – 12 < -4
Hence, (-8) + (-4) [<] (-8) – (-4)
(b) (-3) + 7 – (19) □ 15 – 8 + (-9)
LHS = (-3) + 7 – (19) =-3 + 7-19
= -3 – 19 + 7
= -22 + 1 = -15
RHS = 15 – 8 + (-9)
= 15-8-9
= 15 – 17 = -2
Here -15 < -2
Hence, (-3) + 7 – (19) [<] 15 – 8 + (-9)
(c) 23 – 41 + 11 □ 23 – 41 – 11
LHS = 23 – 41 + 11 = 23 + 11 – 41 = 34 – 41 = -7
RHS = 23 – 41 – 11 = 23 – 52 = -29 Here, -7 > -29
Hence, 23 – 41 + 11 [>] 23 – 41 – 11
(d) 39 + (-24) – (15) □ 36 + (-52) – (-36)
LHS = 39 + (-24) – (15)
= 39 – 24 – 15
= 39 – 39 = 0
RHS = 36 + (-52) – (-36) = 36 – 52 + 36
= 36 + 36 – 52
= 72 – 52 = 20
Here 0 < 20
Hence, 39 + (-24) – (15) [<] 36 + (-52) – (-36)
(e) -231 + 79 + 51 □ -399 + 159 + 81
LHS = -231 + 79 + 51 = -231 + 130 = -101
RHS = -399 + 159 + 81 = -399 + 240 = -159
Here, -101 > -159
Hence, -231 + 79 + 51 [>] -399 + 159 + 81
Question 10.
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?
(ii) After drinking water, he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. In how many jumps will he reach back the top 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 move in part (t) 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?
Solution:
(i) The position of monkey after the
1st jump J1 is at 4th step ↓
2nd jump J2 is at 2nd step ↑
3rd jump J3 is at 5th step ↓
4th jump J4 is at 3rd step ↑
5th jump J5 is at 6th step ↓
6th jump J6 is at 4th step ↑
7th jump J7 is at 7th step ↓
8th jump J8 is at 5th step ↑
9th jump J9 is at 8th step ↓
10th jump J10 is at 6th step ↑
11th jump J11 is at9th step ↓ (Water level)
Hence the required number of jumps = 11.
(ii) Monkey’s position after the
1st jump J1 is at 5th step ↑
2nd jump J2 is at 7th step ↓
3rd jump J3 is at 3rd step ↑
4th jump J4 is at 5th step ↓
5th jump J5 is at 1st step ↑
Hence, the required number of jumps = 5.
(iii) According to the given conditions we have the following tables
Therefore (a) Total number of steps
=-3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3
= -8 which represents the monkey goes down by 8 steps.
In case (ii), we get
Jumps | J1 | J2 | J3 | J4 | J5 |
Number of steps | +4 | -2 | +4 | -2 | +4 |
Therefore (b) Total number of steps.
= +4 – 2 + 4 – 2 + 4 = 8
NCERT Solutions of class 7th maths Chapter 1 Integers Exercise 1.2
Question 1.
Write down a pair of integers whose:
(a) sum is -7
(b) difference is -10
(c) sum is 0.
Solution:
(a) Let us take a pair of integers -3 and -4.
∴ (-3) + (-4) = -3 – 4 = -7
(b) Let us take a pair of integers -12 and -2
∴ (-12) – (-2) = -12 + 2 = -10
(c) Let us take a pair of integers -3 and 3
∴ (-3) + (3) = -3 + 3 = 0
Question 2.
(a) Write a pair of negative integers whose difference gives 8.
(b) Write a negative integer and positive integer whose sum is -5.
(c) Write a negative integer and a positive integer whose difference is -3.
Solution:
(a) Let us have -2 and -10
∴ Difference = (-2) – (-10) = -2 + 10 = 8
(b) Let us have -7 and 2
∴ (-7) + (2) = -7 + 2 = -5
(c) Let us have -2 and 1
∴ (-2) – (1) = – 2 – 1 = -3
Question 3.
In a quiz, team A scored -40, 10, 0 and team B scored 10, 0, -40 in three successive rounds. Which team scored more? Can you say that we can add integers in any order?
table
Solution:
Total score of team
A = (-40) + (10) + (0) – -40 + 10 + 0 = -30
Total score of team
B = 10 + 0 + (-40) = 10 + 0 – 40 – -30
∴ The scores of both the teams are same i.e. -30.
Yes, we can add the integers in any order.
Question 4.
Fill in the blanks to make the following statements true:
(i) (-5) + (-8) = (-8) + (…)
(ii) -53 + … = -53
(iii) 17 + … = 0
(iv) [13 + (-12)] + (…) = 13 + [(-12) + (-7)]
(v) (-4) + [15 + (-3)] = [-4 + 15] + …
Solution:
(i) -5 + (-8) – (-8) + (-5) [Commutative law of additions]
(ii) -53 + 0 = -53 [Additive Identity]
[Adding 0 to any integer, it gives the same value]
(iii) 17 + (-17) = 0 [Additive inverse]
(iv) [13 + (-12)] + (-7) – 13 + [(-12) + (-7)] [Associative law of addition]
(v) (-4) + [15 + (-3)] – [-4 + 15] + (-3) [Associative law of addition]
NCERT Solutions of class 7th maths Chapter 1 Integers Exercise 1.3
1. Find each of the following products:
(a) 3 × (–1)
Solution:-
By the rule of Multiplication of integers,
= 3 × (-1)
= -3 … [∵ (+ × – = -)]
(b) (–1) × 225
Solution:-
By the rule of Multiplication of integers,
= (-1) × 225
= -225 … [∵ (- × + = -)]
(c) (–21) × (–30)
Solution:-
By the rule of Multiplication of integers,
= (-21) × (-30)
= 630 … [∵ (- × – = +)]
(d) (–316) × (–1)
Solution:-
By the rule of Multiplication of integers,
= (-316) × (-1)
= 316 … [∵ (- × – = +)]
(e) (–15) × 0 × (–18)
Solution:-
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) × (–11) × (10)
Solution:-
By the rule of Multiplication of integers,
= (–12) × (-11) × (10)
First multiply the two numbers having same sign,
= 132 × 10 … [∵ (- × – = +)]
= 1320
(g) 9 × (–3) × (– 6)
Solution:-
By the rule of Multiplication of integers,
= 9 × (-3) × (-6)
First multiply the two numbers having same sign,
= 9 × 18 … [∵ (- × – = +)]
= 162
(h) (–18) × (–5) × (– 4)
Solution:-
By the rule of Multiplication of integers,
= (-18) × (-5) × (-4)
First multiply the two numbers having same sign,
= 90 × -4 … [∵ (- × – = +)]
= – 360 … [∵ (+ × – = -)]
(i) (–1) × (–2) × (–3) × 4
Solution:-
By the rule of Multiplication of integers,
= [(–1) × (–2)] × [(–3) × 4]
= 2 × (-12) … [∵ (- × – = +), (- × + = -)]
= – 24
(j) (–3) × (–6) × (–2) × (–1)
Solution:-
By the rule of Multiplication of integers,
= [(–3) × (–6)] × [(–2) × (–1)]
First multiply the two numbers having same sign,
= 18 × 2 … [∵ (- × – = +)
= 36
2. Verify the following:
(a) 18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]
Solution:-
From the given equation,
Let us consider the Left Hand Side (LHS) first = 18 × [7 + (–3)]
= 18 × [7 – 3]
= 18 × 4
= 72
Now, consider the Right Hand Side (RHS) = [18 × 7] + [18 × (–3)]
= [126] + [-54]
= 126 – 54
= 72
By comparing LHS and RHS,
72 = 72
LHS = RHS
Hence, the given equation is verified.
(b) (–21) × [(– 4) + (– 6)] = [(–21) × (– 4)] + [(–21) × (– 6)]
Solution:-
From the given equation,
Let us consider the Left Hand Side (LHS) first = (–21) × [(– 4) + (– 6)]
= (-21) × [-4 – 6]
= (-21) × [-10]
= 210
Now, consider the Right Hand Side (RHS) = [(–21) × (– 4)] + [(–21) × (– 6)]
= [84] + [126]
= 210
By comparing LHS and RHS,
210 = 210
LHS = RHS
Hence, the given equation is verified.
3. (i) For any integer a, what is (–1) × a equal to?
Solution:-
= (-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
Solution:-
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
Solution:-
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
Solution:-
Now, multiply 0 with (-1), we get
= 0 × (-1)
= 0
Because, the product of negative integers and zero give zero only.
4. Starting from (–1) × 5, write various products showing some pattern to show
(–1) × (–1) = 1.
Solution:-
The various products are,
= -1 × 5 = -5
= -1 × 4 = -4
= -1 × 3 = -3
= -1 × 2 = -2
= -1 × 1 = -1
= -1 × 0 = 0
= -1 × -1 = 1
We concluded that the product of one negative integer and one positive integer is negative integer. Then, the product of two negative integers is a positive integer.
5. Find the product, using suitable properties:
(a) 26 × (– 48) + (– 48) × (–36)
Solution:-
The given equation is in the form of 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)
Solution:-
The given equation is in the form of 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)
Solution:-
The given equation is in the form of Commutative law of Multiplication.
= a × b = b × a
Then,
= 15 × [(–25) × (– 4)] × (–10)
= 15 × [100] × (–10)
= 15 × [-1000]
= – 15000
(d) (– 41) × 102
Solution:-
The given equation is in the form of 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
Solution:-
The given equation is in the form of Distributive law of Multiplication over Addition.
= a × (b + c) = (a × b) + (a × c)
= 625 × [(-35) + (-65)]
= 625 × [-100]
= – 62500
(f) 7 × (50 – 2)
Solution:-
The given equation is in the form of Distributive law of Multiplication over Subtraction.
= a × (b – c) = (a × b) – (a × c)
= (7 × 50) – (7 × 2)
= 350 – 14
= 336
(g) (–17) × (–29)
Solution:-
The given equation is in the form of 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
Solution:-
The given equation is in the form of Distributive law of Multiplication over Addition.
= a × (b + c) = (a × b) + (a × c)
= (57 × 19) + (57 × 1)
= 57 [19 + 1]
= 57 × 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:-
From the question, it is given that
Let us take the lowered temperature as negative,
Initial temperature = 40oC
Change in temperature per hour = -5oC
Change in temperature after 10 hours = (-5) × 10 = -50oC
∴The final room temperature after 10 hours of freezing process = 40oC + (-50oC)
= -10oC
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 questions not attempted.
(i) Mohan gets four correct and six incorrect answers. What is his score?
Solution:-
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?
Solution:-
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?
Solution:-
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
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 5,000 bags of grey cement in a month. What is its profit or loss?
Solution:-
We denote profit in positive integer and loss in negative integer,
From the question,
Cement company earns a profit on selling 1 bag of white cement = ₹ 8 per bag
Then,
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.
Solution:-
We denote profit in positive integer and loss in negative integer,
From the question,
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,
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,
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.
9. Replace the blank with an integer to make it a true statement.
(a) (–3) × _____ = 27
Solution:-
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
Solution:-
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
Solution:-
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
Solution:-
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 … [∵ (- × – = +)]
NCERT Solutions of class 7th maths Chapter 1 Integers Exercise 1.4
1. Evaluate each of the following:
(a) (–30) ÷ 10
Solution:-
= (–30) ÷ 10
= – 3
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.
(b) 50 ÷ (–5)
Solution:-
= (50) ÷ (-5)
= – 10
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.
(c) (–36) ÷ (–9)
Solution:-
= (-36) ÷ (-9)
= 4
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.
(d) (– 49) ÷ (49)
Solution:-
= (–49) ÷ 49
= – 1
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.
(e) 13 ÷ [(–2) + 1]
Solution:-
= 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 minus sign (-) before the quotient.
(f) 0 ÷ (–12)
Solution:-
= 0 ÷ (-12)
= 0
When we divide zero by a negative integer gives zero.
(g) (–31) ÷ [(–30) + (–1)]
Solution:-
= (–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 positive sign (+) before the quotient.
(h) [(–36) ÷ 12] ÷ 3
Solution:-
First we have to solve the integers with in the bracket,
= [(–36) ÷ 12]
= (–36) ÷ 12
= – 3
Then,
= (-3) ÷ 3
= -1
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.
(i) [(– 6) + 5)] ÷ [(–2) + 1]
Solution:-
The given question can be written as,
= [-1] ÷ [-1]
= 1
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.
2. 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
Solution:-
From the question, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
Given, a = 12, b = – 4, c = 2
Now, consider LHS = a ÷ (b + c)
= 12 ÷ (-4 + 2)
= 12 ÷ (-2)
= -6
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.
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
Solution:-
From the question, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
Given, a = (-10), b = 1, c = 1
Now, consider LHS = a ÷ (b + c)
= (-10) ÷ (1 + 1)
= (-10) ÷ (2)
= -5
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.
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.
3. Fill in the blanks:
(a) 369 ÷ _____ = 369
Solution:-
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
Solution:-
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
Solution:-
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
Solution:-
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
Solution:-
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
Solution:-
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
Solution:-
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
Solution:-
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
4. Write five pairs of integers (a, b) such that a ÷ b = –3. One such pair is (6, –2) because 6 ÷ (–2) = (–3).
Solution:-
(i) (15, -5)
Because, 15 ÷ (–5) = (–3)
(ii) (-15, 5)
Because, (-15) ÷ (5) = (–3)
(iii) (18, -6)
Because, 18 ÷ (–6) = (–3)
(iv) (-18, 6)
Because, (-18) ÷ 6 = (–3)
(v) (21, -7)
Because, 21 ÷ (–7) = (–3)
5. The temperature at 12 noon was 10oC above zero. If it decreases at the rate of 2oC per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at mid-night?
Solution:-
From the question is given that,
Temperature at the beginning i.e., at 12 noon = 10oC
Rate of change of temperature = – 2oC per hour
Then,
Temperature at 1 PM = 10 + (-2) = 10 – 2 = 8oC
Temperature at 2 PM = 8 + (-2) = 8 – 2 = 6oC
Temperature at 3 PM = 6 + (-2) = 6 – 2 = 4oC
Temperature at 4 PM = 4 + (-2) = 4 – 2 = 2oC
Temperature at 5 PM = 2 + (-2) = 2 – 2 = 0oC
Temperature at 6 PM = 0 + (-2) = 0 – 2 = -2oC
Temperature at 7 PM = -2 + (-2) = -2 -2 = -4oC
Temperature at 8 PM = -4 + (-2) = -4 – 2 = -6oC
Temperature at 9 PM = -6 + (-2) = -6 – 2 = -8oC
∴At 9 PM the temperature will be 8oC below zero
Then,
The temperature at mid-night i.e., at 12 AM
Change in temperature in 12 hours = -2oC × 12 = – 24oC
So, at midnight temperature will be = 10 + (-24)
= – 14oC
So, at midnight temperature will be 14oC below 0.
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? (ii) Mohini scores –5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
Solution:-
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
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.
Solution:-
From the question,
The initial height of the elevator = 10 m
Final depth of 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, time taken by the elevator to descend – 360 m = (-360) ÷ (-60)
= 60 minutes
= 1 hour