CBSE Class 7

These NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 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)]
Answer:
(a) (-30) ÷ 10 = – (30 ÷ 10) = – (3) = -3

(b) 50 ÷ (-5) = – (50 ÷ 5) = – (10) = -10

(c) (-36) ÷ (- 9) = + (36 ÷ 9) = + 4

(d) (-49) ÷ 49 = – (49 ÷ 49) = – (1) = -1

(e) 13 ÷ [(-2) + 1] = 13 ÷ (-2 + 1)
= 13 ÷ (-1) = – (13 ÷ 1) = -13

(f) 0 ÷ (-12) = – (0 ÷ 12) = 0

(g) (-31) ÷ [(-30) + (-1) = (-31) ÷ [-30 -1]
-31 ÷ (-31) = + (31 ÷ 31) = + 1

(h) [(-36) ÷ 12] ÷ 3 = [-36 ÷ 12] ÷ 3
= – (36 ÷ 12) ÷ 3 = – (3) ÷ 3
= -3 ÷ 3 = – 1

(i) [(-6) + 5] ÷ [(-2) + 1]
= (-6 + 5) ÷ (-2 + 1) = – 1 ÷ (-1)
= + (1 ÷ 1) = 1

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
Answer:
(a) Given a = 12, b = – 4 and c = 2
L.H.S = a + (b + c)
= 12 ÷ (-4 + 2) = 12 ÷ (- 2)
= – [12 ÷ 2] = – 6
R.H.S = (a ÷ b) + (a ÷ c)
= (12 + -4) + [12 ÷ 2]
= -(12 ÷ 4) + (6) = -3 + 6 = 3
∴ L.H.S ≠ R.H.S

(b) Here a = – 10, b = 1, c = 1
L.H.S = a ÷ (b + c)
= (-10) ÷ (1 + 1) = -10 ÷ 2
= -(10 ÷ 2) = -5
R.H.S = (a ÷ b) + (a ÷ c)
= (-10 ÷ 1) + (-10 ÷ 1)
= -(10 ÷ 1) + (-10) ÷ 1
= -10 -10 = -20
L.H.S ≠ R.H.S

Question 3.
Fill in the blanks :
Answer:
(a) 369 ÷ 1 = 369 (a ÷ 1 = a)
(b) (-75) ÷ 75 = -1 (-a ÷ a = -1)
(c) (-206) ÷ 206 = 1 (- a) ÷ (-a) = 1
(d) -87 ÷ -1 = 87 (-a ÷ – 1 = a)
(e) -87 ÷ = -87 [(-a) ÷ 1 = -a]
(f) -48 ÷ 48 = -1 (-a ÷ a = -1)
(g) 20 ÷ -10 = -2
(h) -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).
Answer:
Five pairs of integers (a, b) such that a ÷ b = -3
(1) (3, -1)
(2) (-3, 1)
(3) (9, -3)
(4) (-9, 3)
(5) (-12, 4)

Question 5.
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 mid-night?
Answer:
Temperature at 12 noon = + 10°C
Rate of change in temperature = -2°C per hour
Number of hours from 12 noon to midnight = 12
∴ Change in temperature in 12 hours = 12° × (-2) = -24°C
Temperatureat mid-night= 10°C + (-24°C)
(ie 12 hours after 12 noon) = -14°C
Temperature difference between + 10°C and -8°C = + 10°C – (-8°C)
= 10°C + 8°C = 18°C
Time taken = 18 ÷ 2 = 9 hours
Temperature change of 18°C will take place in 9 hours from 12 noon
Time after 9 hours from 12 noon = 9 P.M.

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?
Answer:
(i) Let ‘x’ be the number of incorrect question attempted by Radhika
According to the given question, we get
(+3) × 12 + x × (-2) = 20
36 – 2x = 20
2x = 36 – 20
2x = 16
x = 8
∴ Radhika attempted 8 incorrect questions.

(ii) Let ‘x’ be the number of incorrect question attempted by Mohini
According to the given question, we get
(+3) × 7 + x × (-2) = -5
21 – 2x = -5
2x = 21 + 5
2x = 26
x = \(\frac{26}{2}\) = 13
∴ Mohini attempted 13 incorrect questions.

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?
Answer:
Present position of the elevator is at 10 m above the ground level;
Distance moved by the elevator below the ground level = 350 m
∴ Total distance moved by the elevator = 350 m + 10 m = 360 m
∴ The rate of descend = 6 m/min
Time taken = \(\frac{360}{6}\) minutes = 60 minutes 6 = 1 hour
Required time = 1 hour

These NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes InText Questions

NCERT In-text Question Page No. 277

Question 1.
Match the shape with the name:


Answer:
(i) – (a) Cuboid
(ii) – (d) Cuboid
(iii) – (a) Cone
(iv) – (c) Sphere
(v) – (f) Cube
(vi) – (e) Pyramid

NCERT In-text Question Page No. 281

Question 1.
Here you find four nets. There are two correct nets among them to make a tetrahedron. See if you can work out which nets will make a tetrahedron.

Answer:
(i) and (iii) are correct nets to make a tetrahedron.

NCERT In-text Question Page No. 286

Question 1.
Try to guess the number of cubes in the following arrangements.

Answer:
(i) There are 24 cubes.
(ii) There are 8 cubes.
(ii) There are 9 cubes.

NCERT In-text Question Page No. 287

Question 1.
Two dice are placed side by side as shown: Can you say what the total would be on the face opposite to

(a) 5 + 6 (b) 4 + 3
(Remember that in a die sum of numbers on opposite faces is 7.)
Answer:
(a) The total opposite to face 5 + 6 is 2 + 1, i.e. 3.
(b) The total opposite to face 4 + 3 is 3 + 4, i.e. 7.

Question 2.
Three cubes each with 2 cm edge are placed side by side to form a cuboid. Try to make an oblique sketch and say what would be its length, breadth and height.

Answer:
Length of the cuboid
= 2 cm + 2 cm + 2 cm = 6 cm.
Breadth of the cuboid = 2 cm.
Height of the cuboid = 2 cm.

NCERT In-text Question Page No. 291

Question 1.
For each solid, the three views (1), (2), (3) are given. Identify for each solid the corresponding top, front and side views.
Answer:


(a) (1) → Front (2) → Side (3) → Top
(b) (1) → Top (2) → Side (3) → Front
(a) (1) → Side (2) → Front (3) → Top
(d) (1) → Side (2) → Top (3) → Front

Question 2.
Draw a view of each solid as seen from the direction indicated by the arrow.

Answer:

These NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Exercise 15.4

Question 1.
A bulb is kept burning just right above the following solids. Name the shape of the shadows obtained in each case. Attempt to give a rough sketch of the shadow. (You may try to experiment first and then answer these questions).

Answer:
When light falls just above the solid
(i) A ball, the shadow looks like a circle.
(ii) A cylinder pipe, the shadow looks like a rectangle.
(iii) A book, the shadow looks like nearly a rectangle.

Question 2.
Here are the shadow of some 3-D objects, when seen under the lamp of an overhead projector. Identify the solid (s) that match each shadow. (There may be multiple answers for these!)

(i) The given shadow corresponds to a
sphere.
(ii) The given shadow corresponds to a cube.
(iii) The given shadow corresponds to a cone.
(iv) The given shadow corresponds to a cuboid or a cylinder.

Question 3.
Examine if the following are true statements.
(i) The cube can cast a shadow in the shape of a rectangle.
(ii) The cube can cast a shadow in the shape of a hexagon.
Answer:
(i) The given statement is true.
(ii) The given statement is false.

These NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Exercise 15.1

Question 1.
Identify the nets which can be used to make cubes (cut out copies of the nets and try it):


Answer:
(ii), (iii), (iv) and (vi) can be used to make cubes.

Question 2:
Dice are cubes with dots on each face. Opposite faces of a die always have a total of seven dots on them.
Here are two nets to make dice (cubes); the numbers inserted in each square indicate the number of dots in that box.

Insert suitable numbers in the blanks, remembering that the number on the opposite faces should total to 7.

Answer:

Question 3.
Can this be a net for a die? Explain your answer.

Answer:
No, because one pair of opposite faces will have 1 and 4 on them whose total is not equal to 7. Another faces are having 3 and 6 on them whose total is also equal to 7.

Question 4.
Here is an incomplete net for making a cube. Complete it in at least two different ways. Remember that a cube has six faces. How many are there in the net here? (Give two separate diagrams. If you like, you may use a squared sheet for easy manipulation.)

Answer:

Question 5.
Match the nets with appropriate solids:


Answer:
(a) → (ii): because, the net (ii) can be folded as square.
(b) → (iii): because, the net (iii) can be folded as cylinder.
(c) → (iv): because, the net (iv) can be folded as cone.
(d) → (i): because, the net (i) can be folded as pyramid.

These NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 1 Integers Exercise 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)
Answer:
(a) 3 × (-1) = – (3 × 1) = -3
(b) (-1) × 225 = – (1 × 225) = -225
(c) (-21) × (- 30) = 21 × 30 = 630
(d) (-316) × (-1) = 316 × 1 = 316
(e) (-15) × 0 × (-18) = (15 × 0) × 18
= 0 × 18 = 0

(f) (-12) × (-11) × 10
= [(-12) × (-11)] × 10
= (12 × 11) × 10 = 132 × 10 = 1320

(g) 9 × (-3) × (-6) = 9 × [(-3) × (-6)]
= 9 × (3 × 6) = 9 × 18 = 162

(h) (-18) × (-5) × (-4)
= [(-18) × (-5)] × (-4)
= (18 × 5) × (-4) = 90 × (-4)
= – (90 × 4) = -360

(i) (-1) × (-2) × (-3) × 4
= [(-1) × (-2)] × [(-3) × 4]
= (1 × 2) × [- (3 × 4)]
= 2 × [12] = 2 × (-12) = -24

(j) (-3) × (-6) × (-2) × (-1)
= [(-3) × (-6)] × [(- 2) × (-1)]
= (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)]
Answer:
(a) L.H.S = 18 × [7 + (-3)]
= 18 × [7-3] = 18 × 4 = 72
R.H.S = [18 × 7] + [18 × (-3)]
= 126 + (-54) = 126 – 54 = 72
∴ L.H.S = R.H.S

(b) L.H.S = (-21) × [(-4) + (-6)]
= (-21) × [-4 -6]
= (-21) × (-10) = 210
R.H.S = [(-21) × (-4)] + [(-21) × (-6)]
= (21 × 4) + (21 × 6)
= 84 + 126 = 210
∴ L.H.S = R.H.S

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
Answer:
(i) (-1) × a = – a
∴ (- 1) × (any integer)
= additive inverse of the integer

(ii) (a) – 22 = (-1) × 22
(b) 37 = (-1) × (- 37)
(c) 0 = (-1) × 0

Question 4.
Starting from (-1) × 5, write various products showing some pattern to show
(-1) × (-1) = 1.
Answer:
(-1) × 5 = -5
(-1) × 4 = -4
(-1) × 3 = -3
(-1) × 2 = -2
(-1) × 1 = -1
(-1) × 0 = 0
(-1) × (-1) = 1

Question 5.
Find the product, using suitable properties:
(a) 26 × (-48) + (-48) × (-36)
(b) 8 × 53 × (-125)
(c) 15 × (-25) × (-4) × (-10)
(d) (-41) × 102
(e) 625 × (-35) + (-625) × 65
(f) 7 × (50 – 2)
(g) (-17) × (-29)
(h) (-57) × (-19) + 57
Answer:
(a) 26 × (- 48) + (-48) × (-36)
= (-48) × [26 + (-36)]
(using distributive property)
= (-48) × (26 – 36)
= (-48) × (-10) = 480

(b) 8 × 53 × (-125) = (8 × 53) × (-125)
(using associative property)
= 424 × (-125) = -424 × 125 = – 53000

(c) 15 × (-25) × (-4) × (-10)
= [(-25) × (- 4)] × [(-10) × 15]
(using associative property)
= (100) × (-150) = -100 × 150 = -15000

(d) (-41) × 102 = -(41 × 102)
= – [41 × (100 + 2)]
(Distributive property)
= – [41 × 100 + 41 × 2]
= – [4100 + 82] = -4182

(e) 625 × (-35) + (-625) × 65
= 625 × [-35 + (-1) (65)]
(Distributive property)
= 625 × [-35 -65]
= 625 × (-100) = -62500

(f) 7 × (50 – 2) = 7 × 50 + 7 × (-2)
= 350 – 14 = 336
(using distributive property)

(g) (-17) × (-29) = 17 × 29 = 17 × (30 – 1)
(using distributive property)
= 17 × 30 + 17 × (-1) = 510 – 17 = 493

(h) (-57) × (-19) + 57 = 57 × 19 + 57
= 57 × (19 + 1)
(using distributive property)
= 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?
Answer:
Room temperature = 40°C
Change of temperature = – 5°C In 10 hours,
Change of temperature = 40°C + 10 x (-5)°C
= 40°C + (-50°C) = 40°C – 50°C = -10°C

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?
Answer:
Total number of questions = 10
Marks for a correct answer = 5
Marks for an incorrect answer = -2
Marks for un-answered question = 0
(i) Mohan’s score = 4 × (5) + 6 × (-2)
= 20 + (-1)(6 × 2) = 20 – 12 = 8

(ii) Reshma’s score = 5 × (5) + 5 × (-2)
= 25 – 10 = 15

(iii) Heena’sscore = 2 × (5) + 5 × (-2) + 3 × 0
= 10 + (-10) + 0= 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.
Answer:
(a) The company sells 3,000 bags of white cement
∴ It’s profit = ₹ 3000 × 8 = ₹ 24000
Also, the company sells 5000 bags of grey cement
∴ It’s loss = ₹ 5000 × 5 = ₹ 25000
Since ₹ 25000 > 24,000
∴ The company is in loss
Loss amount = 25000 – 24,000
= ₹ 1000

(b) Let the number of white cement bags sold be ‘x‘
Given : No profit and No loss
Number of grey bags sold = 6400
Number of white bags × profit
= Number of grey bags × loss
x × 8 = 6400 × 5
x = \(\frac{6400 \times 5}{8}\) = 800 × 5 = 4000
∴ Number of white cement bags = 4000

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
Answer:
(a) -3 × = 27
(-3) × (-9) = 27

(b) 5 × = -35
5 × (-7) = -35

(c) ………….. × (-8) = -56
7 × (-8) = -56

(d) …………. × (-12) = 132
(-11) × (-12) = 132

These NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 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
Answer:
(a) – 15 and 8
sum = – 15 + 8 = – 7

(b) – 30 and – 20
Difference = – 30 – (- 20)
= -30 + 20 = – 10

(c) – 12 and + 12
Difference = – 12 + (12)
= -12 + 12 = 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.
Answer:
(a) – 13 and – 21
Difference = -13 – (- 21)
= – 13 + 21 = 8

(b) – 15 and 10
sum = – 15 + 10 = – 5
(c) – 1 and 2
Difference = – 1 – (+ 2)
= – 1 – 2 = – 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?
Answer:
Total score of team A = (- 40) + 10 + 0
= – 40 + 10 + 0 = – 30
Total score of team B=10 + 0 + (- 40)
= 10 + 0 – 40 = 10 – 40 = – 30
So, both the team scored equal. Yes, we can say that we can add 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] + …………….
Answer:
(i) (- 5) + (- 8) = (- 8) + (- 5)
(ii) – 53 + 0 = – 53
(iii) 17 + (- 17) = 0
(iv) [13 + (- 12)] + (- 7) = 13 + [(-12) + (- 7)]
(v) (-4)+ [15 +(-3)] = [-4 + 15] + (- 3)