## 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

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

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

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

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

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

Question 5.
Solution:
∵ Abscissa is distance of a point from y- axis

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

Question 7.
Solution:
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

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

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

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

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

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

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

Question 2.
Solution:
Central angle = $$\\ \frac { 35 }{ 100 }$$ x 360°
= 126°

Question 3.
Solution:
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:
Central angle of students reading novels = 81°
$$\\ \frac { 81 }{ 360 }$$ x 100
= $$\\ \frac { 45 }{ 2 }$$
= $$22{ \frac { 1 }{ 2 } }$$%

Hope given RS Aggarwal Solutions Class 8 Chapter 23 Pie Charts Ex 23B are helpful to complete your math homework.

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## RS Aggarwal Class 8 Solutions Chapter 23 Pie Charts Ex 23A

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

Other Exercises

Question 1.
Solution:
Total expenditure = Rs. 4000 + 5400 + 2800 + 1800 + 400 = Rs. 14400

Construction of pie chart :
1. Draw a circle of any convenient radius.
3. Staring from this radius, draw sectors of central angle 100°, 135°, 70°, 45° and 10° respectively.
4. Shade these sectors with different colors or designs as shown in the figure.
This is the required pie chart.

Question 2.
Solution:
Total number of creatures 900

(i) Draw a circle with a suitable radius.
(iii) Starting from this radius, draw sectors whose central angles are 60°, 160°, 70°, 50° and 20° respectively.
(iv) Now shade each sector with different colours or designs as shown in the figure.

Question 3.
Solution:
Total number of students = 350 + 245 + 210 + 175 + 280 = 1260

(i) Draw a circle with a suitable radius.
(iii) Starting from this radius draw sectors whose central angles are 100°, 70°, 60°, 50° and 80° respectively.
(iv) Now shade each sector with different colours or designs as shown in the figure.

Question 4.
Solution:

(i) Draw a circle with a suitable radius.
(iii) Starting from this radius, draw sectors whose actual angles are 105°, 60°, 30°, 120° and 45° respectively.
(iv) Now shade each sector with different colours or design as shown in the figure.

Question 5.
Solution:
Here total number of workers = 1080

Now (i) Draw a circle with a suitable radius
(iii) Starting from this radius, draw sectors whose central angle are 150°, 90°, 85°, 35° respectively.
(iv) Now shade the sectors with different colours or designs as shown in the figure.

Question 6.
Solution:
Total marks obtained by Sudhir
= 105 + 75 + 150 + 120 + 90 = 540

(i) Draw a circle with a suitable radius
(iii) Starting from this radius, draw sectors whose central angles are 70°, 50°, 100°, 80° and 60° respectively
(iv) Now shade these sectors with different colours or designs as shown in the figure.

Question 7.
Solution:
Total number of fruits = 26 + 30 + 21 + 5 + 8 = 90

(i) Draw a circle with a suitable radius.
(iii) Starting from this radius, draw sectors of central angles 104°, 120°, 84°, 20° and 32° respectively.
(iv) Shade these sectors with different colours or designs as shown in the figure.

Question 8.
Solution:
Total number of million of tonnes of food grains = 57 + 76 + 38 + 19 = 190 million of tonnes

(i) Draw a circle with suitable radius.
(iii) Starting with this radius, draw sectors of central angles 108°, 144°, 72° and 366 respectively.
(iv) Shade these sectors with different colours or designs as shown in the figure.

Question 9.
Solution:
Total percentage = 25 + 45 + 20 + 10 = 100%

(i) Draw a circle with a suitable radius.
(iii) Starting from this radius, draw sectors of central angles 90°, 162°, 72° and 36° respectively.
(iv) Shade these sectors with different colours or designs as shown in the figure.

Question 10.
Solution:
Total percentage = 20 + 40 + 25 + 15 = 100%

(i) Draw a circle with a suitable radius.
(iii) Starting from this radius, draw sectors of central angles 72°, 144°, 90° and 54° respectively.
(iv) Now shade these sectors with different colours or designs as shown in the figure.

Hope given RS Aggarwal Solutions Class 8 Chapter 23 Pie Charts Ex 23A are helpful to complete your math homework.

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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

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

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

Tick the correct answer in each of the following:

Question 1.
Solution:
Length (l) = 12 cm
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:
Length of wall (l) = 8m = 800 cm
Height (h) = 6 m
= 600 cm

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

Question 7.
Solution:
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
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 cm(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
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