Tag: RS Aggarwal Class 8 Maths Solutions

RS Aggarwal Class 8 Solutions Chapter 10 Profit and Loss Ex 10A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10A
• RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10B
• RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10C
• RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10D

Question 1.
Solution:
(i) C.P. = Rs. 620
S.P. = Rs. 713
Gain = S.P. – C.P. = Rs. 713 – Rs. 620
= Rs. 93

Question 2.
Solution:
(i) C.P. = Rs. 1650
Gain% = 4%

You can also Download NCERT Solutions for Class 8 Maths to help you to revise complete Syllabus and score more marks in your examinations.

Question 3.
Solution:
(i)S.P.= Rs 1596
Gain % = 12%

Question 4.
Solution:
C.P. of iron safe = Rs 12160
Paid for its transportation = Rs 340
Total cost = Rs 12160 + Rs 340
= Rs 12500

Question 5.
Solution:
C.P. of car = Rs 73500
Overhead charges = Rs 10300 + 2600
= Rs 12900

Question 6.
Solution:
C.P. of 20 kg @ Rs 36 per kg.
= 20 x 36 = Rs 720
C.P. of 25 kg @ Rs32 per kg.

Question 7.
Solution:
Ratio of the mixture = 5 : 2
Let 5 kg of coffee is mixed with 2 kg of chicory

Question 8.
Solution:
Let CR of 17 water bottles = Rs. 17
Then C.P. of one bottle

Question 9.
Solution:
Let C.P. of 12 candIes =Rs. 12
C.P of 1 candle = Re. 1

Question 10.
Solution:
S.P. of 125 cassettes – gain = C.P. of 125 cassettes
=> S.P. of 125 cassettes – S.P. of 5 cassettes = C.P. of 125 cassettes
=> S.P. of 120 cassettes = C.P. of 125 cassettes

Question 11.
Solution:
S.P. of 45 lemons = C.P. of 45 lemons – loss

Question 12.
Solution:
CP. of 6 oranges = Rs 20
and C.P of 1 orange = Rs \(\\ \frac { 20 }{ 6 } \) = Rs \(\\ \frac { 10 }{ 3 } \)

Question 13.
Solution:
C.P. of 12 bananas = Rs 40
and C.P of 1 banana = Rs \(\\ \frac { 40 }{ 12 } \) = Rs \(\\ \frac { 10 }{ 3 } \)

Question 14.
Solution:
C.P. of 10 apples = Rs 75
and C.P of 1 apple = Rs \(\\ \frac { 75 }{ 10 } \) = Rs \(\\ \frac { 15 }{ 2 } \)

Question 15.
Solution:
Let eggs purchased were = 3 x 16 = 48

Question 16.
Solution:
S.P. of camera = Rs 1080
C.P. of camera + gain = S.P. of camera
=> C.P. of camera + \(\\ \frac { 1 }{ 8 } \) of C.P. = S.P. of camera
= \(\\ \frac { 9 }{ 8 } \) x C.P. of camera = S.P. of camera = Rs 1080

Question 17.
Solution:
S.P. of a pen = Rs 54
Loss = \(\\ \frac { 1 }{ 10 } \) of her outlay

Question 18.
Solution:
Let C.P. of table = Rs 100
In first case, loss = 10%
S.P. = Rs 100 – 10 = Rs 90
and in second case, gain = 10%

Question 19.
Solution:
Let C.P. of chair = Rs 100
In first case, gain = 15%
then S.P. = Rs 100 + 15 = Rs 115
and in second case, gain = 8%

Question 20.
Solution:
Let the C.P. of cycle = Rs 100
In first case, gain = 10%
then S.P. = Rs 100 + 10 = Rs 110
In second case, gain = 14%
Difference between their S.P.s = Rs 114 – Rs 110 = Rs 4

Question 21.
Solution:
Cost price of 40 kg @ Rs 12.50 per kg.
= 40 x 12.50 = Rs 500
and cost price of 30 kg @ of Rs 14 per kg
= 30 x 14 = Rs 420

Question 22.
Solution:
C.P. of first bat = Rs 840
Gain% = 15%

Question 23.
Solution:
C.P. of first jean = Rs 1450
Gain% = 8%

Question 24.
Solution:
Total quantity of rice = 200 kg.
C.P. of 200 kg @ Rs 25 per kg.
= Rs 200 x 25 = Rs 5000
Gain% on total = 8%

Question 25.
Solution:
Let C.P. = Rs 100
then S.P. = \(\\ \frac { 6 }{ 5 } \) of Rs 100

Question 26.
Solution:
Let C.P. of flower vase = Rs 100
then S.P. \(\\ \frac { 5 }{ 6 } \) of C.P. = \(\\ \frac { 5 }{ 6 } \) of Rs 100
= Rs \(\\ \frac { 500 }{ 6 } \)

Question 27.
Solution:
S.P. of a bouquet = Rs 322
Gain = 15%

Question 28.
Solution:
In first case,
S.P. of an umbrella = Rs 336
Loss% = 4%

Question 29.
Solution:
S.P. of Radio = Rs 3120
Loss% = 4%

Question 30.
Solution:
S.P. Of first sarees = Rs 1980
Loss% = 10%

Question 31.
Solution:
S.P. of first fan = Rs 1140
Gain% = 14%

Question 32.
Solution:
C.P. for Manoj = Rs 3990
or S.P. for Arun = Rs 3990
loss% = 5%
C.P for Arun
\(=\frac { S.P\times 100 }{ 100-losspercent} =\frac { 3990\times 100 }{ 100-5 } \)

Question 33.
Solution:
C.P. of plot of land = Rs 480000
C.P. of \(\\ \frac { 2 }{ 5 } \)th part = Rs 480000 x \(\\ \frac { 2 }{ 5 } \)
= Rs 192000

Question 34.
Solution:
C.P. of sugar = Rs 4500
C.P. of \(\\ \frac { 1 }{ 3 } \) of sugar

 

Hope given RS Aggarwal Solutions Class 8 Chapter 10 Profit and Loss Ex 10A are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 6 Operations on Algebraic Expressions Ex 6C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6C.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6D
• RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6E

Question 1.
Solution:
(i) 24x²y³ ÷ 3xy

Question 2.
Solution:
(i)(5m³ – 30m² + 45m) ÷ 5m

Write the quotient and remainder when we divide:

Question 3.
Solution:

Question 4.
Solution:

Quotient = x – 2
Remainder = 0

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Hope given RS Aggarwal Solutions Class 8 Chapter 6 Operations on Algebraic Expressions Ex 6C are helpful to complete your math homework.

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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
= \(\\ \frac { 45 }{ 2 } \)
= \(22{ \frac { 1 }{ 2 } } \)%

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 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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.