## 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

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%

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

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

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.

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

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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 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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