Category: CBSE Class 8

RS Aggarwal Class 8 Solutions Chapter 25 Graphs Ex 25C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 25 Graphs Ex 25C.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 25 Graphs Ex 25A
• RS Aggarwal Solutions Class 8 Chapter 25 Graphs Ex 25B
• RS Aggarwal Solutions Class 8 Chapter 25 Graphs Ex 25C

OBJECTIVE QUESTIONS
Tick the correct answer in each of the following:

Question 1.
Solution:
Answer = (a)
∵ In point (3, 6), both x and y are positive.

Question 2.
Solution:
Answer = (c)
∵ In point ( – 7, – 1) both x and y are negative.

Question 3.
Solution:
Answer = (d)
∵ In point (2, – 3), x is positive and y is negative.

Question 4.
Solution:
Answer = (b)
∵ In point ( – 4, 1), x is negative and y is positive.

Question 5.
Solution:
Answer = (c)
∵ Abscissa is distance of a point from y- axis

Question 6.
Solution:
Answer = (d)
y = a is a line parallel to x-axis at a distance of ‘a’ units.

Question 7.
Solution:
Answer = (a)
The equation of the line y-axis, is x = 0

Hope given RS Aggarwal Solutions Class 8 Chapter 25 Graphs Ex 25C are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 25 Graphs Ex 25B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 25 Graphs Ex 25B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 25 Graphs Ex 25A
• RS Aggarwal Solutions Class 8 Chapter 25 Graphs Ex 25B
• RS Aggarwal Solutions Class 8 Chapter 25 Graphs Ex 25C

Question 1.
Solution:
(a) y = 3x
By giving some different values to x, we shall get corresponding values of y.
x = 1 then y = 3 x 1 = 3
if x = 2, then y = 3 x 2 = 6
if x = 0, then y = 3 x 0 = 0

Now plotting the points given above, and joining them.
(b) We get a line, From the graph.
(i) When x = 3, then y = 9
(ii) When x = 5, then y = 15
(iii) When x = 6, then y = 18 Ans.

Question 2.
Solution:
(a) P = 4x
By giving some different values to x, we get the corresponding values of y or P
If x = 1, then P = 4 x 1 = 4
if x = 2, then P = 4 x 2 = 8
if x = 0, then P = 4 x 0 = 0

Plot the points (1, 4), (2, 8) and 0, 0) on the graph and join then to get the graph of P = 4x as shown
(b) From the graph we see that
(i) When x = 3, then P = 12
(ii) When x = 4, then P = 16
(iii) When x = 6, then P = 24 Ans.

Question 3.
Solution:
A = x²
giving some values to x, we get corresponding values of y or A
If x = 1, then y or A = (1)² = 1
If x = 2, then y or A = (2)² = 4
If x = 0, then y or A = (0)² = 0

Now plot the point (1, 1), (2, 4), (0, 0) on the graph, and join them to get the graph of A = x2 as shown

(b) From the graph we see that
(i) When x = 2, then A = 4
(ii) When x = 3, then A = 9
(ii) When x = 4 then A = 16 Ans.

Hope given RS Aggarwal Solutions Class 8 Chapter 25 Graphs Ex 25B are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 25 Graphs Ex 25A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 25 Graphs Ex 25A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 25 Graphs Ex 25A
• RS Aggarwal Solutions Class 8 Chapter 25 Graphs Ex 25B
• RS Aggarwal Solutions Class 8 Chapter 25 Graphs Ex 25C

Question 1.
Solution:
Below is given the graph in which X’OX and YOY’ are the co-ordinate axes intersecting each other at O. Now the. points given have been plotted as shown on the graph.

 

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RS Aggarwal Class 8 Solutions Chapter 24 Probability Ex 24B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 24 Probability Ex 24B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 24 Probability Ex 24A
• RS Aggarwal Solutions Class 8 Chapter 24 Probability Ex 24B

Tick the correct answer in each of the following :

Question 1.
Solution:
A spinning wheel has 3 white and 5 green sectors
Possible out come = 3 + 5 = 8
It is spinners, then
Probability of getting a green sector = \(\\ \frac { 5 }{ 8 } \) (b)

Question 2.
Solution:
8 cards are numbered 1, 2, 3, 4, 5, 6, 7, 8
They are mixed and kept in a box One card is chosen at random, then Probability of card having 9 number less than 4 = \(\\ \frac { 3 }{ 8 } \) (c)

Question 3.
Solution:
Two coins are tossed simultaneously, then Possible outcomes = 4
Now probability of getting one head and one tail = \(\\ \frac { 2 }{ 4 } \) = \(\\ \frac { 1 }{ 2 } \) (b)

Question 4.
Solution:
. In a bag, there are 3 white and 2 red balls
Possible outcomes = 3 + 2 = 5
Now probability of a red ball drawn
= \(\\ \frac { 2 }{ 5 } \) (d)

Question 5.
Solution:
A die is thrown then
Possible outcomes = 6
Now probability of getting 6 is \(\\ \frac { 1 }{ 6 } \) (b)

Question 6.
Solution:
A die is thrown
Possible outcomes = 6
Now probability of getting an even number
which are 2, 4, 6 = \(\\ \frac { 3 }{ 6 } \) = \(\\ \frac { 1 }{ 2 } \) (a)

Question 7.
Solution:
One card is drawn from a well shuffled deck of 52 cards, possible out comes = 52
The probability of card which is a queen = \(\\ \frac { 4 }{ 52 } \)
= \(\\ \frac { 1 }{ 13 } \) (c)

Question 8.
Solution:
One card is drawn from a well-shuffled deck of 52 card, possible out comes = 52 Probability of a card being a black 6
(which are two) = \(\\ \frac { 2 }{ 52 } \) = \(\\ \frac { 1 }{ 26 } \) (b)

Hope given RS Aggarwal Solutions Class 8 Chapter 24 Probability Ex 24B are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 24 Probability Ex 24A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 24 Probability Ex 24A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 24 Probability Ex 24A
• RS Aggarwal Solutions Class 8 Chapter 24 Probability Ex 24B

Question 1.
Solution:
(i) When a coin is tossed, we get outcomes 2 as H or T (Head or Tail)
(ii) When two coins are tossed together, we get possible four outcomes as HH, HT, TH, TT
(iii) A die is thrown, we get possible outcomes as 1,2, 3, 4, 5, 6
(iv) From a well – shuffled deck of 52 cards, 0ne card is at random drawn, we get the possible outcomes is 52

Question 2.
Solution:
Possible outcomes = 2
In a single throw of a coin, we get
probability of getting a tail = \(\\ \frac { 1 }{ 2 } \)

Question 3.
Solution:
In a single throw of two coins, possible outcomes = 4
(i) Probability of getting both tails = \(\\ \frac { 1 }{ 4 } \)
(ii) Probability of getting at least one tail = \(\\ \frac { 3 }{ 4 } \)
(iii) Probability of getting at the most one tail = \(\\ \frac { 2 }{ 4 } \) = \(\\ \frac { 1 }{ 2 } \)

Question 4.
Solution:
In a bag, there are 4 white and 5 blue balls ,
Possible outcomes = 4 + 5 = 9
One ball is drawn at random, then
(i) the probability of a white ball = \(\\ \frac { 4 }{ 9 } \)
(ii) the probability of a blue ball = \(\\ \frac { 5 }{ 9 } \)

Question 5.
Solution:
In a bag, there are 5 white, 6 red and 4 green balls
Possible outcome is 5 + 6 + 4 = 15
One ball is drawn at random, then
(i) Probability of a green ball = \(\\ \frac { 4 }{ 15 } \)
(ii) Probability of a white ball = \(\\ \frac { 5 }{ 15 } \) = \(\\ \frac { 1 }{ 3 } \)
(iii) Probability of a non-red ball = \(\\ \frac { 5+4 }{ 15 } \)
= \(\\ \frac { 9 }{ 15 } \)
= \(\\ \frac { 3 }{ 5 } \)
(5 white and 4 green balls are non-red balls)

Question 6.
Solution:
In a lottery, there are 10 prizes and 20 blanks
Possible outcomes = 10 + 20 = 30
A ticket is chosen at random, then
probability of getting a prize = \(\\ \frac { 10 }{ 30 } \) = \(\\ \frac { 1 }{ 3 } \)

Question 7.
Solution:
In a ,box of 100 electric bulb, 8 are defective
Then non-defective bulbs = 100 – 8 = 92
Now possible outcomes = 100
(i) Probability of a drawn bulb, which is defective = \(\\ \frac { 8 }{ 100 } \) = \(\\ \frac { 2 }{ 25 } \)
(ii) Probability of a drawn bulb which is non defective = \(\\ \frac { 92 }{ 100 } \) = \(\\ \frac { 23 }{ 25 } \)

Question 8.
Solution:
A die is thrown, then
Possible outcomes = 6
(i) Now probability of getting 2 = \(\\ \frac { 1 }{ 6 } \)
(ii) Probability of a number less than 3 (which are 1 and 2) = \(\\ \frac { 2 }{ 6 } \) = \(\\ \frac { 1 }{ 3 } \)
(iii) Probability of a composite number (a composite number is a number which is not a prime number which are 4, 6) = \(\\ \frac { 2 }{ 6 } \) = \(\\ \frac { 1 }{ 3 } \)
(iv) Probability of a number not less than 4 (which are 5, 6) = \(\\ \frac { 2 }{ 6 } \) = \(\\ \frac { 1 }{ 3 } \)

Question 9.
Solution:
Total number of ladies = 200
Those who like coffee = 82
Those who dislike coffee = 118
Possible number of outcomes = 200
One lady is chosen at random, then
(i) Probability of a lady who dislikes coffee = \(\\ \frac { 118 }{ 200 } \)
= \(\\ \frac { 59 }{ 100 } \)

Question 10.
Solution:
19 ball bearing numbers, 1, 2, 3,…19
possible outcomes = 19
A ball is drawn at random from the box, then
(i) Probability of a ball which bears a prime numbers which are 2, 3, 5, 7, 11, 13, 17 and 19 = 8 = \(\\ \frac { 8 }{ 19 } \)
(ii) Probability of a ball which bears an even number which are 2, 4, 6, 8, 10, 12, 14, 16, 18 = 9 = \(\\ \frac { 9 }{ 19 } \)
(iii) Probability of a number which bears a number divisible by 3 which are 3, 6, 9, 12, 15, 18 = 6 = \(\\ \frac { 6 }{ 19 } \)

Question 11.
Solution:
A card’s drawn at random from a deck
of well-shuffled deck of 52 cards Probability = 52
(i) Probability of a card being a king = \(\\ \frac { 4 }{ 52 } \) = \(\\ \frac { 1 }{ 13 } \)
(ii) Probability of a card being spade = \(\\ \frac { 13 }{ 52 } \) = \(\\ \frac { 1 }{ 4 } \)
(iii) Probability of a card being a red queen = \(\\ \frac { 2 }{ 52 } \) = \(\\ \frac { 1 }{ 26 } \)
(iv) Probability of a card being a black 8 = \(\\ \frac { 2 }{ 52 } \) = \(\\ \frac { 1 }{ 26 } \)

Question 12.
Solution:
One card is drawn at random from a deck of well shuffled deck of 52 cards
Possible outcomes = 52
(i) Probability of a card being a 4 = \(\\ \frac { 4 }{ 52 } \) = \(\\ \frac { 1 }{ 13 } \)
(ii) Probability of a card being a queen = \(\\ \frac { 4 }{ 52 } \) = \(\\ \frac { 1 }{ 13 } \)
(iii) Probability of a card being a black card = \(\\ \frac { 26 }{ 52 } \) = \(\\ \frac { 1 }{ 2 } \)

 

Hope given RS Aggarwal Solutions Class 8 Chapter 24 Probability Ex 24A are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 23 Pie Charts Ex 23B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 23 Pie Charts Ex 23B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 23 Pie Charts Ex 23A
• RS Aggarwal Solutions Class 8 Chapter 23 Pie Charts Ex 23B

OBJECTIVE QUESTIONS :
Tick the correct answer in each of the following:

Question 1.
Solution:
Answer = (b)
Central angles = \(\\ \frac { 250 }{ 2400 } \) x 360°
= \(\\ \frac { 75 }{ 2 } \)
= \(37{ \frac { 1 }{ 2 } }^{ o } \)

Question 2.
Solution:
Answer (c)
Central angle = \(\\ \frac { 35 }{ 100 } \) x 360°
= 126°

Question 3.
Solution:
Answer = (a)
Total number of strength = 1650
Arts stream’s central angle = 48°
No. of students of Arts stream
= \(\\ \frac { 48 }{ 360 } \) x 1650
= 220

Question 4.
Solution:
Answer = (c)
Central angle of students reading novels = 81°
\(\\ \frac { 81 }{ 360 } \) x 100
= \(\\ \frac { 45 }{ 2 } \)
= \(22{ \frac { 1 }{ 2 } } \)%

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RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22.

Question 1.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph draw one horizontal line OX and other vertical line OY representing x-axis and y-axis respectively.
(ii) Along x-axis, write the name of subjects taken at uniform gaps.
(iii) Choose the scale = 1 small division = 1 mark
(vi) Then the heights of various bars will be drawn as shown on the graph.

Question 2.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph paper, draw a horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along the x-axis write the years taken on uniform gaps.
(iii) Choose scale : 1 small division = 20 students
(iv) Then the heights of various bars will be drawn as shown on the graph.

Question 3.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph paper, draw a horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along the x-axis, write the names of sports taken on uniform gaps.
(iii) Choose the scale : 1 small division = 1 student
(iv) Then the heights of various bars will be drawn as shown on the graph

Question 4.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph, draw a horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along x-axis, write cities with a uniform gaps.
(iii) Choose the scale : 1 small division = 200 km
(iv) Then we shall draw the heights of various bars as shown on the graph.

Question 5.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph, draw a horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along x-axis, write countries
(iii) Choose the scale : 1 small division = 10 year
(iv) Then we shall draw the heights of various bars as shown on the graph.

Question 6.
Solution:
We can draw a bar graph by the following steps :
(i) On the graph, draw a horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along x-axis, write modes of transport with uniform gaps.
(iii) Choose the scale : 1 small division = 100 Students
(iv) Then we shall draw the heights of bars as shown on the graph.

Question 7.
Solution:
(i) Draw a horizontal line OX and a vertical line OY which represent x-axis and y-axis respectively on the graph.
(ii) Along OX, write years and along OY, number of motorcycles.
(iii) Choose 1 division = 300
(iv) Now draw bars of different heights according to give data as shown on the graph.

Question 8.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph, draw horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along x-axis, write the names of States given at uniform gaps.
(iii) Choose scale : 1 small division = 200 lakhs
(iv) Then we shall draw the heights of various bars as shown on the graph.

Question 9.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph, draw horizontal line OX and another vertical line OY representing x-axis
and y-axis respectively.
(ii) Along x-axis, write the names of animals given at uniform gaps.
(iii) Choose scale : 1 small division = 200 lakhs
(iv) Then we shall draw the heights of various bars as shown on the graph.

Question 10.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph paper, draw a horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along the x-axis write the years taken on uniform gaps.
(iii) Choose scale : 1 small division = 20 export earnings
(iv) Then the heights of various bars will be drawn as shown on the graph.

Question 11.
Solution:
We can draw the bar graph by the following steps :
(i) On the graph, draw horizontal line OX and another vertical line OY representing x-axis and y-axis respectively.
(ii) Along x-axis, write the names of years given at uniform gaps.
(iii) Choose scale 1 small division = 200 lakhs
(iv) Then we shall draw the heights of various bars as shown on the graph.

Question 12.
Solution:
(i) The bar graph shows the number of members in each of the 100 families of a village.
(ii) 90
(iii) 65
(iv) 5

Question 13.
Solution:
(i) The given bar graph shows the marks obtained by a student in an examination in each of the five subjec ts.
(ii) English.
(iii) From the given graph,
(iv) Mathematics.

Question 14.
Solution:
(i) Mount Everest is the heighest peak and its heights is 8800 m.
(ii) Highest peak is Mount Everest and lowest peak is Annapurna and their heights are 8800 m and 6000 m respectively.
Ratio = 8800 : 6000 => 22 : 15
(iii) Heights of peaks in ascending order is 6000 m, 7500 m, 8000 m, 8200 m and 8800 m.
(iv) Kanchenjunga peak differ by 600 meter from Mount Everest.

 

Hope given RS Aggarwal Class 8 Solutions Chapter 22 Constructing and Interpreting Bar Graphs Ex 22 are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 21 Data Handling Ex 21B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 21 Data Handling Ex 21A
• RS Aggarwal Solutions Class 8 Chapter 21 Data Handling Ex 21B

Question 1.
Solution:
Frequency distribution table is given below:

Question 2.
Solution:
Arranging the given data in increasing order:
312, 324, 356, 365, 378, 400, 435, 472, 506, 548, 565, 570, 584, 596, 617, 630, 674, 685, 700, 736, 745, 754, 763, 776, 780.
Now frequency distribution table is given below :

Question 3.
Solution:
Frequency Distribution table is given below:

Question 4.
Solution:
Frequency distribution table is given below

Question 5.
Solution:
Frequency table is given below :

Question 6.
Solution:

Hope given RS Aggarwal Solutions Class 8 Chapter 21 Data Handling Ex 21B are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 21 Data Handling Ex 21A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 21 Data Handling Ex 21A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 21 Data Handling Ex 21A
• RS Aggarwal Solutions Class 8 Chapter 21 Data Handling Ex 21B

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

 

Hope given RS Aggarwal Solutions Class 8 Chapter 21 Data Handling Ex 21A are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20C.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20A
• RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20B
• RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20C

Tick the correct answer in each of the following:

Question 1.
Solution:
Answer = (b)
Length (l) = 12 cm
Breadth (b) = 9cm
height (h) = 8 cm

Question 2.
Solution:
Total surface area of cube = 150 cm²
Side = \( \sqrt { \frac { 150 }{ 6 } } \)
= √25
= 5 cm
Volume = (side)³
= (5)³
= 125 cm³ (b)

Question 3.
Solution:
Volume of cube = 343 cm²
Side = \( \sqrt [ 3 ]{ 343 } =\sqrt [ 3 ]{ 7\times 7\times 7 } \)
= 7 cm
Total surface area = 6 (side)²
= 6 x (7)²
= 6 x 49 cm²
= 294 cm² (c)

Question 4.
Solution:
Rate of painting = 10 paise per cm²
Total cost = Rs. 264.60

Question 5.
Solution:
Answer = (c)
Length of wall (l) = 8m = 800 cm
Breadth (b) = 22.5 cm
Height (h) = 6 m
= 600 cm

Question 6.
Solution:
Answer = (c)
Edge of cube = 10 cm
Volume = a³ = (10)³ = 1000 cm³
Edge of box = 1 m = 100 cm

Question 7.
Solution:
Answer = (a)
Ratio in sides of a cuboid = 1 : 2 : 3
Surface area = 88 cm²

Question 8.
Solution:
Ratio in the two volumes = 1 : 27
Let volume of first volume = x³
and volume of second volume = 27x³
Side of first cube = x

Question 9.
Solution:
Surface area of a brick of measure 10 cm x 4 cm x 3 cm
= 2 (l x b + b x h + h x l)
= 2 [10 x 4 + 4 x 3 + 3 x 10] cm²
= 2 [40 + 12 + 30]
= 82 x 2
= 164 cm² (c)

Question 10.
Solution:
Length of beam (l) = 9 m

Question 11.
Solution:
Water in rectangular reservoir = 42000
Volume = \(\\ \frac { 42000 }{ 1000 } \) = 42 m³
Length (l) = 6 m
Breadth (b) = 3.5 m
Depth = \(\\ \frac { volume }{ l\times b } \)
= \(\\ \frac { 42 }{ 6\times 3.5 } \)
= 2 m (c)

Question 12.
Solution:
Dimensions of a room are 10 m, 8 m, 3.3 m
Volume of air in it = lbh
= 10 x 8 x 3.3 = 264 m³
Air required for one man = 3 m³
No. of men = \(\\ \frac { 264 }{ 3 } \)
= 88 (b)

Question 13.
Solution:
Length of water tank (l) = 3 m
Width (b) = 2 m
and height (h) = 5 m
Volume = lbh = 3 x 2 x 5 = 30 m³
Water in it = 30 x 1000
= 30000 (a)

Question 14.
Solution:
Size of box = 25 cm, 15 cm, 8 cm
Surface area = (lb + bh + hl)
= 2 ( 25 x 15 + 15 x 8 + 8 x 25) cm²
= 2 (375 + 120 + 200) cm²
= 2(695)
= 1390 cm² (b)

Question 15.
Solution:
Diagonal of cube = 4√3
Side = \( \frac { 4\sqrt { 3 } }{ \sqrt { 3 } } \)
= 4 cm
Volume = a³ = (4)³
= 64 cm³ (d)

Question 16.
Solution:
Diagonal of cube = 9√3 cm
Side = \( \frac { 9\sqrt { 3 } }{ \sqrt { 3 } } \)
= 9 cm
Surface area = 6a²
= 6 (9)² = 6 x 81 cm²
= 486 cm² (b)

Question 17.
Solution:
Let side of cube in first case = a
Then volume = a³
If side of cube is doubled, then side = 2a
Volume (2a)³ = 8a³
Becomes 8 times (d)

Question 18.
Solution:
Let side of cube in first case = a
Then surface area = 6a²
and side of second cube = 2a
Surface area = 6 (2a)² = 6 x 4a² = 24a²
Ratio = \(\frac { { 24a }^{ 2 } }{ { 6a }^{ 2 } } \) = 4
Becomes 4 times (b)

Question 19.
Solution:
Sides (edges) of 3 cubes are 6 cm, 8 cm, and 10 cm respectively
Volume of first cube = (6)³ = 216 cm³
Volume of second cube = (8)³ = 512 cm³
and volume of third cube
= (10)³ = 1000 cm³
Sum of volumes of 3 cubes = 216 + 512 + 1000
= 1728 cm³
Volume of new single cube = 1728 cm³
Edge = \(\sqrt [ 3 ]{ 1728 } \)
\(\sqrt [ 3 ]{ { \left( 12 \right) }^{ 3 } } \)
= 12 cm (a)

Question 20.
Solution:
Each edge of 5 cubes = 5 cm
Placing than adjacent to each other
Length of new cuboid (l)
= 5 x 5 = 25 cm
Breadth (b) = 5 cm
and height (h) = 5 cm
Volume of new cuboid = lbh
= 25 x 5 x 5 cm³
= 625 cm³ (d)

Question 21.
Solution:
Diameter of circular well = 2n
Radius = \(\\ \frac { 2 }{ 2 } \) = 1 m
Depth(h) = 14 m
Volume of earth dug out = πr²h
= \(\\ \frac { 22 }{ 7 } \) x 1 x 1 x 14
= 44 m (d)

Question 22.
Solution:
Capacity of cylindrical tank = 1848 m³
Diameter = 14 m

Question 23.
Solution:
Radius of a cylinder (r) = 20 cm
and height (h) = 60 cm

Question 24.
Solution:
Radius of each coin (r) = 0.75 cm
and thickness (h) = 0.2 cm

Question 25.
Solution:
Volume of silver = 66 cm³
Diameter of wire = 1 mm = \(\\ \frac { 1 }{ 10 } \)

Question 26.
Solution:
Diameter of cylinder = 10 cm
Radius (r) = \(\\ \frac { 10 }{ 2 } \) = 5 cm

Question 27.
Solution:
Diameter of cylinder = 7 cm
Radius (r) = \(\\ \frac { 7 }{ 2 } \) cm

Question 28.
Solution:
Curved surface area of a cylinder = 264 cm³
Height (h) = 14 cm

Question 29.
Solution:
Diameter of cylinder = 14 cm
Radius (r) = 7 cm
Curved surface area = 220 cm²

Question 30.
Solution:
Ratio in radii of two cylinder = 2 : 3
and ratio in their height = 5 : 3
Let radii of two cylinder = 2x and 3x
and corresponding heights = 5y, 3y

Hope given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20C are helpful to complete your math homework.

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