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NCERT Solutions for Class 7 Maths Chapter 1: Integers

Chapter 1 Integers
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NCERT Solutions for Class 7 Maths Chapter 1 involves the study of Integers. In Class 6 we have learned about whole numbers and integers. Now, we will study more about integers, their properties, and operations. Likewise, the addition and the subtraction of the integers, properties of addition and subtraction of integers, multiplication, and division of integers, properties of multiplication and division of integers.

 

NCERT Solutions for Class 7 Maths Chapter 1 by Swiflearn are by far the best and most reliable NCERT Solutions that you can find on the internet. These NCERT Solutions for Class 7 Maths Chapter 1 are designed as per the CBSE Class 7 Maths Syllabus. These NCERT Solutions will surely make your learning convenient & fun. Students can also Download FREE PDF of NCERT Solutions Class 7 Chapter 1.

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NCERT Solutions for Class 7 Maths Chapter 1 Integers: EX 1.1 PDF

 

NCERT 7th Maths Chapter 1 Exercise 1.1 1

 

 

Exercise No: 1.1
Question 1:
Following number line shows the temperature in degree Celsius (℃) at
different places on a particular day.
(a)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 Lahulspiti 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):-

Solution (b):-
Observations from the line
Temperature at Hottest place i.e., Bengaluru is 22 ℃
Temperature at Coldest place i.e., Lahulspiti is -8 ℃
Difference in temperature between hottest and coldest place is
= 22 ℃ – (-8 ℃)
= 22 ℃ + 8℃
= 30 ℃
Hence, the answer is 30 ℃.

Solution (c):-
According to the reference image,
Temperature at Lahulspiti is -8 ℃
Temperature at Srinagar is -2 ℃
∴ Temperature difference between Lahulspiti and Srinagar is = -2℃ – (8℃)
= – 2℃ + 8℃
NCERT Solutions for Class 7th Maths Chapter 1
Integers
https://www.swiflearn.com/
= 6℃

Solution (d):-
According to the reference image,
Temperature at Srinagar = -2℃
Temperature at Shimla = 5℃
Temperature taken together = – 2℃ + 5℃ = 3 ℃
∴ 5℃ > 3℃
Hence, the temperature of Srinagar and Shimla taken together is less than the temperature at Shimla.
Now,
3℃ > -2℃
No, the temperature of Srinagar (-2℃) is less than the temperature of Srinagar and Shimla taken together (3℃)
Therefore, the statement is wrong.

 

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 2:-
According to the question,
Score by Jack in five successive rounds are 25, -5, -10, 15 and 10
The total score of Jack at the end of rounds
= 25 + (-5) + (-10) + 15 + 10
= 25 – 5 – 10 + 15 + 10
= 50 – 15
= 35
∴ Total Score by Jack = 35.

 

Question 3:
At Srinagar temperature was – 5℃ on Monday and then it dropped by
2℃ on Tuesday. What was the temperature of Srinagar on Tuesday? On
Wednesday, it rose by 4℃. What was the temperature on this day?

 

Solution 3:-
According to the question,
Monday’s temperature at Srinagar = -5℃
Drop in temperature on Tuesday = 2℃
∴ Temperature on Tuesday = Temperature of Monday – Temperature of Tuesday
= -5℃ – 2℃
= -7℃
Temperature on Wednesday = Temperature on Tuesday + Rise of 4℃
= -7℃ + 4℃
= -3℃
Thus, the temperature on Wednesday was -3℃

 

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 floating 1200 m below the sea level.
What is the vertical distance between them?

Solution 4:-
Plane is flying at the height (above the sea level) = 5000 m
Depth of Submarine (below the sea level) = -1200 m
The vertical distance between ‘plane and submarine’
= 5000 m – (- 1200) m
= 5000 m + 1200 m
= 6200 m

 

Question 5:
Mohan deposits ₹ 2,000 in his bank account and withdraws ₹ 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 5:-
Withdrawal is represented by a negative integer.
And, deposit of amount is represented by a positive integer.
According to the question,
Amount deposited in bank account = ₹ 2000
Amount withdrawn from the bank account = – ₹ 1642
Balance in account after the withdrawal
= ₹ 2000 + (-₹ 1642)
= ₹ 2000 – ₹ 1642
= ₹ 358
Hence, Balance after the withdrawal is ₹ 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 the
distance travelled towards west? By which integer will you represent her
final position from A?

 

Solution 6:-
According to the question, we can see that
Distance travelled to the east is shown by a Positive (+ve) Integer.
And, Distance towards the west will be represented by a Negative Integer
Distance in east direction = 20 km
Distance in west direction = – 30 km
∴Distance travelled from A = 20 + (- 30)
= 20 – 30
= -10 km
Hence, a negative integer will represent the distance travelled by Rita from point A, i.e. – 10 km

 

Question 7:
In a magic square each row, column and diagonal have the same sum.
Check which of the following is a magic square.

 

Solution 7:-
First for square (i)
By adding the numbers in each rows,
R1 = 5 + (- 1) + (- 4)
= 5 – 1 – 4
= 5 – 5
= 0
R2 = -5 + (-2) + 7
= – 5 – 2 + 7
= -7 + 7
= 0
R3 = 0 + 3 + (-3)
= 3 – 3
= 0
By adding the numbers in each columns,
C1 = 5 + (- 5) + 0
= 5 – 5
= 0
C2 = (-1) + (-2) + 3
= -1 – 2 + 3
= -3 + 3
= 0
C3 = -4 + 7 + (-3)
= -4 + 7 – 3
= -7 + 7
= 0
By adding the numbers in diagonals,
D1 = 5 + (-2) + (-3)
= 5 – 2 – 3
= 5 – 5
= 0
D2 = -4 + (-2) + 0
= – 4 – 2
= -6
As sum of one diagonal is not equals to zero like the other values,
∴, Square (i) is not a magic square
Now, For Square (ii)
By adding the numbers in each rows,
R1= 1 + (-10) + 0
= 1 – 10 + 0
= -9
R2= (-4) + (-3) + (-2)
= -4 – 3 – 2
= -9
R3= (-6) + 4 + (-7)
= -6 + 4 – 7

= -13 + 4
= -9
By adding the numbers in each columns,
C1= 1 + (-4) + (-6)
= 1 – 4 – 6
= 1 – 10
= -9
C2= (-10) + (-3) + 4
= -10 – 3 + 4
= -13 + 4
= -9
C3= 0 + (-2) + (-7)
= 0 – 2 – 7
= -9
By adding the numbers in diagonals,
D1= 1 + (-3) + (-7)
= 1 – 3 – 7
= 1 – 10
= -9
D2= 0 + (-3) + (-6)
= 0 – 3 – 6
= -9
Thus, square (ii) is a magic square, because sum of all row, column and diagonal is equal to – 9.

 

Question 8:
Verify a – (– b) = a + b for the following values of a and b.
a = 21, b = 18
a = 118, b = 125
a = 75, b = 84
a = 28, b = 11

Solution 8 (i):-
According to the question,
a = 21 and b = 18
To verify a – (- b) = a + b
Taking Left Hand Side as = a – (- b)
= 21 – (- 18)
= 21 + 18
= 39
Now, Putting values in Right Hand Side = a + b
= 21 + 18
= 39
By comparing LHS and RHS
LHS = RHS
39 = 39
Hence, verified.

Solution 8 (ii):-
According to the question,
a = 118 and b = 125
Taking Left Hand Side (LHS) as = a – (- b)
= 118 – (- 125)
= 118 + 125
= 243
Now, Putting values in Right Hand Side (RHS) = a + b
= 118 + 125
= 243
Comparing LHS and RHS
LHS = RHS
243 = 243
Hence, verified.

Solution 8 (iii):-
According to the question,
a = 75 and b = 84
Taking Left Hand Side (LHS) as = a – (- b)
= 75 – (- 84)
= 75 + 84
= 159
Now, Putting values in Right Hand Side (RHS) = a + b
= 75 + 84
= 159
Comparing LHS and RHS
LHS = RHS
159 = 159
Hence, verified.

Solution 8 (iv):-
According to the question,
a = 28 and b = 11
Taking Left Hand Side (LHS) as = a – (- b)
= 28 – (- 11)
= 28 + 11
= 39
Now, Putting values in Right Hand Side (RHS) = a + b
= 28 + 11
= 39
Comparing LHS and RHS
LHS = RHS
39 = 39
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 9 (a):-
Left Hand Side (LHS) = (-8) + (-4)
= -8 – 4
= -12
Right Hand Side (RHS) = (-8) – (-4)
= -8 + 4
= -4
Comparing LHS and RHS
LHS < RHS
-12 < -4
∴ (-8) + (-4) [<] (-8) – (-4)

Solution 9 (b):-
Left Hand Side (LHS) = (-3) + 7 – 19
= -3 + 7 – 19
= -22 + 7
= -15
Right Hand Side (RHS) = 15 – 8 + (-9)
= 15 – 8 – 9
= 15 – 17
= -2
Comparing LHS and RHS
LHS < RHS
-15 < -2
∴ (-3) + 7 – (19) [<] 15 – 8 + (-9)

Solution 9 (c):-
Left Hand Side (LHS) = 23 – 41 + 11
= 34 – 41
= – 7
Right Hand Side (RHS) = 23 – 41 – 11
= 23 – 52
= – 29
Comparing LHS and RHS
LHS > RHS
– 7 > -29
∴ 23 – 41 + 11 [>] 23 – 41 – 11

Solution 9 (d):-
Left Hand Side (LHS) = 39 + (-24) – 15
= 39 – 24 – 15
= 39 – 39
= 0
Right Hand Side (RHS) = 36 + (-52) – (- 36)
= 36 – 52 + 36
= 72 – 52
= 20
Comparing LHS and RHS
LHS < RHS
0 < 20
∴ 39 + (-24) – (15) [<] 36 + (-52) – (- 36)

Solution 9 (e):-
Left Hand Side (LHS) = – 231 + 79 + 51
= – 231 + 130
= -101
Right Hand Side (RHS) = – 399 + 159 + 81
= – 399 + 240
= – 159
Comparing LHS and RHS
LHS > RHS
-101 > -159
∴ – 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 moves in part (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?

 

Solution 10 (i):-
Considering steps moved down are represented by positive integers and then, steps moved up
are represented by negative integers.
Initially monkey is sitting on the top most step i.e., 1st step
In 1st jump = 1 + 3 = 4th step
In 2nd jump = 4 + (-2) = 4 – 2 = 2nd step
In 3rd jump = 2 + 3 = 5th step
In 4th jump = 5 + (-2) = 5 – 2 = 3rd step
In 5th jump = 3 + 3 = 6th step
In 6th jump = 6 + (-2) = 6 – 2 = 4th step
In 7th jump = 4 + 3 = 7th Step
In 8th jump = 7 + (-2) = 7 – 2 = 5th step
In 9th jump = 5 + 3 = 8th step
In 10th jump = 8 + (-2) = 8 – 2 = 6th step
In 11th jump = 6 + 3 = 9th step
∴A total of 11 Jumps (i.e., Up to the 9th step) was taken by the monkey to reach the water level.

 

Solution 10 (ii):-
Initial Position of the monkey: 9th Step (Water Level)
In 1st jump = 9 + (-4)
= 9 – 4
= 5th step
In 2nd jump = 5 + 2 = 7th step
In 3rd jump = 7 + (-4) = 7 – 4 = 3rd step
In 4th jump = 3 + 2 = 5th step
In 5th jump = 5 + (-4) = 5 – 4 = 1st step
∴It took 5 jumps by the monkey to reach back at the top (i.e., first step.)

Solution 10 (iii):-
According to the question,
The number of steps moved up is given by positive integers and number of steps moved down by negative integers.
Monkey moves in part (i)
= – 3 + 2 – ……….. = – 8
LHS = – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3
= – 18 + 10
= – 8
RHS = -8
{Negative integer shows downward movement of 8 steps}
Now,
Monkey moves in part (ii)
= 4 – 2 + ……….. = 8
Then LHS = 4 – 2 + 4 – 2 + 4
= 12 – 4
= 8
RHS = 8
{Positive integer shows upward movement of 8 steps}.

 

NCERT Solutions for Class 7 Maths Chapter 1 Integers: EX 1.2 PDF

 

NCERT 7th Maths Chapter 1 Exercise 1.2 1

 

Exercise No: 1.2
Question 1:
Write down a pair of integers whose:
(a) Sum is -7
(b) Difference is – 10
(c) Sum is 0

 

Solution 1 (a):-
(-3) + (-4)
= -3 – 4
= – 7

(b):-
(-12) – (-2)
= -12 + 2
= -10

(c):-
(-3) + (3)
= -3 + 3
= 0

 

Question 2:
(a)Write a pair of negative integers whose difference gives 8
(b)Write a negative integer and a positive integer whose sum is – 5.
(c) Write a negative integer and a positive integer whose difference is – 3.

 

Solution 2
(a):-
(-2) – (-10)
= – 2 + 10
= 8

(b):-
= -25 + 20
= -5

(c):-
(-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 we say that we can add
integers in any order?

 

Solution:-
According to the question, it is given that
Score of team A = -40, 10, 0
Sum of score obtained by team A = – 40 + 10 + 0
= – 30
Sum of Score obtained by team B
= 10, 0, -40
Total score of team B
= 10 + 0 + (-40)
= 10 + 0 – 40
= – 30
Both team scored the same score 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 4 (i):-
-5 + (-8) – (-8) + (-5)
{Commutative law of additions}

(ii):-
53 + 0 = -53
{Additive Identity} {Adding 0 to any integer, 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 for Class 7 Maths Chapter 1 Integers: EX 1.3 PDF

 

NCERT 7th Maths Chapter 1 Exercise 1.3 1

 

Exercise No: 1.3
Question 1:
Find each of the following products:
(a) 3 × (–1)
(b) (–1) × 225
(c) (–21) × (–30)
(d) (–316) × (–1)
(e) (–15) × 0 × (–18)
(f) (–12) × (–11) × (10)
(g) 9 × (–3) × (– 6)
(h) (–18) × (–5) × (– 4)
(i) (–1) × (–2) × (–3) × 4
(j) (–3) × (–6) × (–2) × (–1)

Solution:
(a) Applying the rule of Multiplication of integers.
3 × (-1) = -3 [∵ (+ × – = -)]
(b) Applying the rule of Multiplication of integers.
(-1) × 225= -225 [∵ (- × + = -)]
(c) Applying the rule of Multiplication of integers.
(-21) × (-30) = 630 [∵ (- × – = +)]
(d) Applying the rule of Multiplication of integers.
(-316) × (-1)= 316
(e) Applying the rule of Multiplication of integers.
(–15) × 0 × (–18)= 0
Note: If an integer is multiplied with zero then the answer comes out as zero.
(f) Applying the rule of Multiplication of integers.
(–12) × (-11) × (10)= 132 × 10
= 1320
(g) Applying the rule of Multiplication of integers.
9 × (-3) × (-6) = 9 × 18
= 162
(h) Applying the rule of Multiplication of integers.
(-18) × (-5) × (-4) = 90 × -4
= – 360
[∵ (+ × – = -)]
(i) Applying the rule of Multiplication of integers.
[(–1) × (–2)] × [(–3) × 4] = 2 × (-12) {∵ (- × – = +), (- × + = -)}
= – 24
(j) Applying the rule of Multiplication of integers.
[(–3) × (–6)] × [(–2) × (–1)] = 18 × 2
= 36

 

Question 2.
Verify the following:
(a) 18 × [7 + (–3)] = [18 × 7] + [18 × (–3)]
(b) (–21) × [(– 4) + (– 6)] = [(–21) × (– 4)] + [(–21) × (– 6)]

 

Solution:-
(a) From the given question
Considering (LHS) first = 18 × [7 + (–3)]
= 18 × [7 – 3]
= 18 × 4
= 72
Now, (RHS) = [18 × 7] + [18 × (–3)]
= [126] + [-54]
= 126 – 54
= 72
Comparing LHS and RHS,
72 = 72
LHS = RHS
Hence, Verified
(b) From the given question,
Considering (LHS) first = (–21) × [(– 4) + (– 6)]
= (-21) × [-4 – 6]
= (-21) × [-10]
= 210
Now, (RHS) = [(–21) × (– 4)] + [(–21) × (– 6)]
= [84] + [126]
= 210
Comparing LHS and RHS,
210 = 210
LHS = RHS
Hence, Verified

 

 

Question 3:
(i) For any integer a, what is (–1) × a equal to?
(ii) Determine the integer whose product with (–1) is
(a) –22
(b) 37
(c) 0

 

Solution:-
(i) (-1) × a = -a {Additive inverse of an integer property}
(ii)
(a) Multiplying -22 with (-1),
-22 × (-1) = 22 {Additive inverse of an integer property}
(b) Multiplying 37 with (-1),
37 × (-1) = -37 {Additive inverse of an integer property}
(c) Multiplying 0 with (-1),
0 × (-1) = 0 As the product of negative integers and zero gives zero only.

 

Question 4:
Starting from (–1) × 5, write various products showing some pattern to
show (–1) × (–1) = 1.

 

Solution 4:-
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
This pattern concludes 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.

 

Question 5:
Find the product, using suitable properties:
(a) 26 × (– 48) + (– 48) × (–36)
(b) 8 × 53 × (–125)
(c) (c) 15 × (–25) × (– 4) × (–10)
(d)(d) (– 41) × 102
(e) 625 × (–35) + (– 625) × 65
(f) 7 × (50 – 2)
(g) (–17) × (–29)
(h)(–57) × (–19) + 57

 

Solution:-
(a) a × (b + c) = (a × b) + (a × c)
{Distributive law of Multiplication over Addition}
Assuming, a = -48, b = 26, c = -36
Now,
= 26 × (– 48) + (– 48) × (–36)
= -48 × (26 + (-36)
= -48 × (26 – 36)
= -48 × (-10)
= 480
{∵ (- × – = +)}
(b) a × b = b × a
{Commutative law of Multiplication}
Now,
= 8 × [53 × (-125)]
= 8 × [(-125) × 53]
= [8 × (-125)] × 53
= [-1000] × 53
= – 53000
(c) a × b = b × a
{Commutative law of Multiplication}
Now,
= 15 × [(–25) × (– 4)] × (–10)
= 15 × [100] × (–10)
= 15 × [-1000]
= – 15000
(d) a × (b + c) = (a × b) + (a × c)
{Distributive law of Multiplication over Addition}
= (-41) × (100 + 2)
=(-41) × 100 + (-41) × 2
= – 4100 – 82
= – 4182
(e) a × (b + c) = (a × b) + (a × c)
{Distributive law of Multiplication over Addition}
= 625 × [(-35) + (-65)]
= 625 × [-100]
= – 62500

(f) a × (b – c) = (a × b) – (a × c)
{Distributive law of Multiplication over Subtraction}
= (7 × 50) – (7 × 2)
= 350 – 14
= 336
(g) a × (b + c) = (a × b) + (a × c)
{Distributive law of Multiplication over Addition}
= (-17) × [-30 + 1]
= [(-17) × (-30)] + [(-17) × 1]
= [510] + [-17]
= 493
(h) a × (b + c) = (a × b) + (a × c)
{Distributive law of Multiplication over Addition}
= (57 × 19) + (57 × 1)
= 57 [19 + 1]
= 57 × 20
= 1140

 

Question 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 6:-
According to the question,
Taking lowered temperature as negative,
Initially temperature was = 40oC
Per hour change= -5 oC
Change after 10 hours = (-5) × 10 = -50oC
∴The final temperature after freezing process = 40oC + (-50oC)
= -10oC

 

Question 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?
(ii) Reshma gets five correct answers and five incorrect answers, what
is her score?
(iii) Heena gets two correct and five incorrect answers out of seven
questions she attempts. What is her score?

 

Solution:-
(i)
According to the question,
Marks for 1 correct answer = 5
Then,
Total marks for 4 accurate answer = 4 × 5 = 20
Marks for 1 incorrect answer = -2
Then,
Total marks for 6 incorrect answer = 6 × -2 = -12
∴Total score of Mohan = 20 + (-12)
= 20 – 12
= 8
(ii)
According to the question,
Marks for 1 correct answer = 5
Then,
Marks for 5 correct answer = 5 × 5 = 25
Marks for 1 wrong answer = -2
Then,
Total marks for 5 wrong answer = 5 × -2 = -10
∴Total score of Reshma = 25 + (-10)
= 25 – 10
= 15
(iii)
According to the question,
Marks for 1 correct answer = 5
Then,
Total marks for 2 correct answer = 2 × 5 = 10
Marks for 1 wrong answer = -2
Then,
Total marks for 5 wrong answer = 5 × -2 = -10
Marks awarded for questions not attempted is = 0
∴Total score of Heena = 10 + (-10)
= 10 – 10
= 0

 

Question 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?
(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:-
(a)
Profit is denoted by a positive integer and loss by a negative integer,
According to the question,
Profit by selling 1 bag of white cement is ₹ 8 per bag
Then,
Profit by selling 3000 bags of white cement is 3000 × ₹ 8
= ₹ 24000
Loss to company by selling 1 bag of grey cement = – ₹ 5 per bag
Then,
Loss to company by selling 5000 bags of grey cement = 5000 × – ₹ 5
= – ₹ 25000
Total loss or profit earned by the company = profit + loss
= 24000 + (-25000)
= – ₹1000
Therefore, a total loss of ₹ 1000 will be incurred.
(b)
Profit is denoted by a positive integer and loss by a negative integer,
According to the question,
Loss by selling of 1 grey bag = ₹ 5
= ₹ (5 × 6,400) – ₹ 32,000
For no profit and no loss, the price of white bags should be ₹ 32,000
Rate of selling price of white bags for a profit of ₹ 8 per bag.
∴ Number of white cement bags sold
32000/8 =4000
Therefore, the required bags = 4,000

 

Question 9:
Replace the blank with an integer to make it a true statement.
(a) (–3) × _____ = 27
(b) 5 × _____ = –35
(c) _____ × (– 8) = –56
(d) _____ × (–12) = 132

 

Solution:-
(a)
Assuming the missing integer be x,
Now,
(–3) × (x) = 27
x = – (27/3)
x = – 9
Substituting the value of x,
(–3) × (-9) = 27
[∵ (- × – = +)]
(b)
Assuming the missing integer be x,
Now,
(5) × (x) = -35
x = – (-35/5)
x = -7
Substituting the value of x,
(5) × (-7) = -35
[∵ (+ × – = -)]
(c)
Assuming the missing integer be x,
Now,
(x) × (-8) = -56
x = (-56/-8)
x = 7
Substituting the value of x,
(7) × (-8) = -56
[∵ (+ × – = -)]
(d)
Assuming the missing integer be x,
Now,
(x) × (-12) = 132
x = – (132/12)
x = – 11
Substituting the value of x,
(–11) × (-12) = 132
[∵ (- × – = +)]

 

 

NCERT Solutions for Class 7 Maths Chapter 1 Integers: EX 1.4 PDF

 

NCERT 7th Maths Chapter 1 Exercise 1.4 1

 

Exercise No: 1.4
Question 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]

 

Solution:-
(a)
(–30) ÷ 10 = – 3
Note: When a negative integer is divided by a positive integer, at first it is divided as whole numbers and then, minus sign (-) is added before the quotient.
(b)
(50) ÷ (-5) = – 10
Note: When a positive integer is divided by a negative integer, at first it is divided as whole numbers and then, minus sign (-) is added before the quotient.
(c)
(-36) ÷ (-9) = 4
Note: When a negative integer is divided by a negative integer, at first it is divided as whole numbers and then, positive sign (+) is added before the quotient.
(d)
(–49) ÷ 49 = – 1
Note: When a positive integer is divided by a negative integer, at first it is divided as whole numbers and then, minus sign (-) is added before the quotient.
(e)
13 ÷ [(–2) + 1] = 13 ÷ (-1)
= – 13
Note: When a positive integer is divided by a negative integer, at first it is divided as whole numbers and then, minus sign (-) is added before the quotient.
(f)
0 ÷ (-12)= 0
Note: Zero divided by a negative integer gives zero as result.
(g)
(–31) ÷ [(–30) + (–1)] = (-31) ÷ [-30 – 1]
= (-31) ÷ (-31)
= 1
Note: When a negative integer is divided by a negative integer, at first it is divided as whole numbers and then, positive sign (+) is added before the quotient.
(h)
Applying BODMAS Rule
[(–36) ÷ 12]= (–36) ÷ 12
= – 3
Now, = (-3) ÷ 3
= -1
Note: When a negative integer is divided by a positive integer, at first it is divided as whole numbers and then, minus sign (-) is added before the quotient.
(i)
Simplifying the equation,
[-1] ÷ [-1]= 1
Note: When a negative integer is divided by a negative integer, at first it is divided as whole numbers and then, positive sign (+) is added before the quotient

 

Question 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
(b) a = (–10), b = 1, c = 1

 

Solution:-
(a)
According to the question,
a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
Given values, a = 12, b = – 4, c = 2
Considering LHS = a ÷ (b + c)
12 ÷ (-4 + 2)
12 ÷ (-2)
-6
Note: When a positive integer is divided by a negative integer, at first it is divided as whole
numbers and then, minus sign (-) is added before the quotient.
Now,
Considering RHS = (a ÷ b) + (a ÷ c)
(12 ÷ (-4)) + (12 ÷ 2)
(-3) + (6)
3
Comparing LHS and RHS
-6 ≠ 3
LHS ≠ RHS
Hence, Verified
(b)
According to the question, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
Given values, a = (-10), b = 1, c = 1
Considering LHS = a ÷ (b + c)
(-10) ÷ (1 + 1)
(-10) ÷ (2)
-5
Note: When a negative integer is divided by a positive integer, at first it is divided as whole
numbers and then, minus sign (-) is added before the quotient
Considering RHS = (a ÷ b) + (a ÷ c)
((-10) ÷ (1)) + ((-10) ÷ 1)
(-10) + (-10)
-10 – 10
-20
Comparing LHS and RHS
-5 ≠ -20
LHS ≠ RHS
Hence, verified.

 

Question 3:
Fill in the blanks:
(a) 369 ÷ _____ = 369
(b)(–75) ÷ _____ = –1
(c) (–206) ÷ _____ = 1
(d) – 87 ÷ _____ = 87
(e) _____ ÷ 1 = – 87
(f) _____ ÷ 48 = –1
(g) 20 ÷ _____ = –2
(h) _____ ÷ (4) = –3

 

 

Solution:-
(a)
Assuming the missing integer be x
Now,
369 ÷ x = 369
x = (369/369)
x = 1
Replacing the value of x
369 ÷ 1 = 369
(b)
Assuming the missing integer be x
Now,
(-75) ÷ x = -1
x = (-75/-1)
= x = 75
Replacing the value of x
(-75) ÷ 75
-1
(c)
Assuming the missing integer be x
Now,
(-206) ÷ x = 1
x = (-206/1)
x = -206
Replacing the value of x,
(-206) ÷ (-206)
1
(d)
Assuming the missing integer be x
Now,
(-87) ÷ x = 87
x = (-87)/87
x = -1
Replacing the value of x
(-87) ÷ (-1)
87
(e)
Assuming the missing integer be x
Now,
(x) ÷ 1= -87
x = (-87) × 1
x = -87
Replacing the value of x
(-87) ÷ 1
-87
(f)
Assuming the missing integer be x
Now,
(x) ÷ 48 = -1
x = (-1) × 48
x = -48
Replacing the value of x
(-48) ÷ 48
-1
(g)
Assuming the missing integer be x
Now,
20 ÷ x = -2
x = (20)/ (-2)
x = -10
Replacing the value of x
(20) ÷ (-10) = -2
(h)
Solution:-
Assuming the missing integer be x
Now,
(x) ÷ 4 = -3
x = (-3) × 4
x = -12
Replacing the value of x
(-12) ÷ 4
-3

 

Question 4:
Write five pairs of integers (a, b) such that a ÷ b = –3. One such pair is (6, –
2) because 6 ÷ (–2) = (–3).

 

Solution 4:-
The five pairs are as followed,
(a) (24, -8)
Because 24 ÷ (-8) = -3
(b) (-12, 4)
Because (-12) ÷ 4 = -3
(c) (15, -5)
Because 15 ÷ (-5) = -3
(d) (18, -6)
Because 18 ÷ (-6) = -3
(e) (60, -20)
Because 60 ÷ (-20) = -3

 

Question 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:-
According to the question,
Temperature at 12 noon = 10oC
Rate of change = – 2 oC per hour
Now,
*Change in temperature at 1 PM =
10 + (-2) = 10 – 2 = 8oC
*Change in temperature at 2 PM =
8 + (-2) = 8 – 2 = 6oC
*Change in temperature at 3 PM =
6 + (-2) = 6 – 2 = 4oC
*Change in temperature at 4 PM =
4 + (-2) = 4 – 2 = 2oC
*Change in temperature at 5 PM =
2 + (-2) = 2 – 2 = 0oC
*Change in temperature at 6 PM =
0 + (-2) = 0 – 2 = -2
oC
*Change in temperature at 7 PM =
-2 + (-2) = -2 -2 = -4
oC
*Change in temperature at 8 PM =
-4 + (-2) = -4 – 2 = -6
oC
*Change in temperature at 9 PM =
-6 + (-2) = -6 – 2 = -8
oC
Therefore, at 9pm temperature will reach 8oC below zero
Now,
Temperature at 12 AM i.e. {Mid-night}
∴ Change in temperature in 12 hours
= 12 × (-2°C) = -24°C
So, temperature at 12 AM {i.e.mid-night} will be
= 10 + (-24)
= – 14℃

 

Question 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 6:-
According to the question,
Marks for 1 correct answer = + 3
Marks for 1 wrong answer = -2
(i)
Score by Radhika 20 marks
Now,
Total marks for 12 correct answers = 12 × 3 = 36
Marks deducted for incorrect answers = Total score – Total marks for 12 correct Answers
= 20 – 36
= – 16
So, the number of wrong answers made by Radhika
= (-16) ÷ (-2)
= 8
(ii) Score by Mohini -5 marks
Now,
Total marks for 7 right answers
= 7 × 3 = 21
Marks deducted for incorrect answers = Total score – Total marks for 12 correct Answers
= – 5 – 21
= – 26
So, the number of wrong answers made by Mohini
= (-26) ÷ (-2)
= 13

 

Question 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:-
According to the question,
Initial position of the elevator = 10 m
Final distance descended by elevator = – 350 m
[∵Distance descended is denoted by a negative integer]
The total distance descended by the elevator = (-350) – (10)
= – 360 m
Now,
Time taken by the elevator to go down by -6 m
= 1 min
So, time taken by the elevator to go down by -360 m
= (-360) ÷ (-60) = 60 minutes = 1 hour

 

 

 

 

 

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