Maths RD Sharma Solutions Class 8

RD Sharma Class 8 Solutions Chapter 24 Data Handling II (Graphical Representation of Data as Histograms) Ex 24.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1

Question 1.
Given below is the frequency distribution of the heights of 50 students of a class :

Draw a histogram representing the above data.
Solution:
We represent class intervals along x-axis and frequency along y-axis. Taking suitable intervals along x-axis and y-axis we construct the rectangles as shown in the figure. This is the required histogram.

Question 2.
Draw a histogram of the following data :

Solution:
We represent class-intervals along x-axis and frequency along y-axis. Taking suitable intervals along x-axis andy-axis, we construct rectangles as shown in the figure. This is the required histogram.

Question 3.
Number of workshops organized by a school in different areas during the last five years is as follows :

Draw a histogram representing the above data.
Solution:
We represent years along x-axis and number of workshops along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.

Question 4.
In a hypothetical sample of 20 people the amounts of money with them were found to be as follows :
114, 108,100, 98, 101,109,117,119, 126, 131, 136, 143, 156, 169, 182, 195, 207, 219, 235, 118.
Draw the histogram of the frequency distribution (taking one of the class intervals as 50-100).
Solution:
Highest sample = 235
Lowest sample = 98
Range = 235-98 = 137
Now frequency distribution table will be as under:

We represent class intervals along x-axis and frequency along j’-axis. Taking suitable intervals, we construct a rectangles as shown in the figure. This is the required histogram.

Question 5.
Construct a histogram for the following data:

Solution:
We represent monthly school fee (in Rs) along x-axis and number of schools along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.

Question 6.
Draw a histogram for the daily earnings of 30 drug stores in the following table :

Solution:
We represent daily earnings (in Rs) along x-axis and number of stores along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.

Question 7.
Draw a histogram to represent the following data:


Solution:
We represent monthly salary (in Rs) along x-axis and number of teachers along y-axis. Taking suitable intervals we construct rectangles as shown in the figure. This is the required histogram

Question 8.
The following histogram shows the number of literate females in the age group of 10 to 40 years in a town :
(i) Write the age group in which the number of literate female is highest.
(ii) What is the class width ?

(iii) What is the lowest frequency ?
(iv) What are the class marks of the classes ?
(v) In which age group literate females are least ?
Solution:
(i) The age group in which the number of literate females is 15-20.
(ii) The class width is 5.
(iii) Lowest frequency is 320.
(iv) The class marks of the classes are
\(\frac { 10+15 }{ 2 }\) = \(\frac { 25 }{ 2 }\) =12.5, similarly other class marks will be 17.5,22.5,27.5,32.5,37.5
(v) The least literate females is in the class 10-15

Question 9.
The following histogram shows the monthly wages (in Rs) of workers in a factory:

(i) In which wage-group largest number of workers are being kept ? What is their number ?
(ii) What wages are the least number of workers getting ? What is the number of such workers ?
(iii) What is the total number of workers ?
(iv) What is the factory size ?
Solution:
(i) The largest number of workers are in wage group 950-1000 and is 8.
(ii) The least number of workers are in the wage group 900-950 and is 2.
(iii) Total number of workers is 40 (3 + 7 + 5 + 4 + 2 + 8 + 6 + 5)
(iv) The factory size is 50.

Question 10.
Below is the histogram depicting marks obtained by 43 students of a class :
(i) Write the number of students getting highest marks.
(ii) What is the class size ?

Solution:
(i) The number of students getting highest marks is 3.
(ii) The class size is 10.

Question 11.
The following histogram shows the frequency distribution of the ages of 22 teachers in a school:

(i) What is the number of eldest and youngest teachers in the school ?
(ii) Which age group teachers are more in the school and which least ?
(iii) What is the size of the classes ?
(iv) What are the class marks of the classes?
Solution:
(i) The number of eldest teacher is 1 and the number of youngest teacher is 2.
(ii) The teachers in age group 35-40 is most.
(iii) Size of classes is 5.
(iv) Class marks of class 20-25 is \(\frac { 20+25 }{ 2 }\)= \(\frac { 45 }{ 2 }\) = 22.5
and similarly others will be 27.5, 32.5, 37.5, 42.5, 47.5, 52.5.

Question 12.
The weekly wages of 30 workers in a factory are given:
830,835,890,810,835,836,869,845,898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806, 840
Mark a frequency table with intervals as 800-810,810-820 and so on, using tally marks.
Also, draw a histogram and answer the following questions:
(i) Which group has the maximum number of workers ?
(ii) How many workers earn Rs 850 and more ?
(iii) How many workers earn less than Rs 850?
Solution:
The frequency table will be as given below:

We represent wages (in Rs) along x-axis and number of workers along y-axis. Taking suitable intervals, we construct rectangles as shown in the figure. This is the required histogram.
(i) Maximum workers are in the wage group 830-840.
(ii) Number of workers getting Rs 850 and more are 1 + 3 + 1 + 1 + 4 = 10.
(iii) Number of workers getting less than Rs 850 are 3 + 2 + 1 + 9 + 5 = 20

Hope given RD Sharma Class 8 Solutions Chapter 24 Data Handling II Ex 24.1 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.

RD Sharma Class 8 Solutions Chapter 25 Data Handling III (Pictorial Representation of Data as Pie Charts or Circle Graphs) Ex 25.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2

Other Exercises

• RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.1
• RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2

Question 1.
The pie-chart given in figure represents the expenditure on different items in constructing a flat in Delhi. If the expenditure incurred on cement is Rs 1,12,500, find the following:

(i) Total cost of the flat,
(ii) Expenditure incurred on labour.
Solution:
Expenditure on cement = Rs 1,12,500
and its central angle = 75°

Question 2.
The pie-chart given in the figure shows the annual agricultural production of an Indian state. If the total production of all the commodities is 81000 tonnes, find the production (in tonnes) of:

(i) Wheat
(ii) Sugar
(iii) Rice
(iv) Maize
(v) Gram
Solution:
Total production = 81000 tonnes

Question 3.
The following pie-chart shows the number of students admitted in different faculties of a college. If 1000 students are admitted in Science answer the following:

(i) What is the total number of students ?
(ii) What is the ratio of students in science and arts ?
Solution:
Students admitted in science = 1000
Central angle = 100°
(i) Total number of students

∴ Ratio in science and arts = 1000 : 1200 = 5:6

Question 4.
In the figure, the pie-chart shows the marks obtained by a student in an examination. If the student secures 440 marks in all, calculate his marks in each of the given subjects.

Solution:
Total marks secured by a student = 440

Question 5.
In the figure, the pie-charts shows the marks obtained by a student in various subjects. If the student scored 135 marks in mathematics, find the total marks in all the subjects. Also, find his score in individual subjects.

Solution:
Marks obtained in mathematics =135
Central angle = 90°

Question 6.
The following pie-chart shows the monthly expenditure of Shikha on various items. If she spends Rs 16,000 per month, answer the following questions:

(i) How much does she spend on rent ?
(ii) How much does she spend on education ?
(iii) What is the ratio of expenses on food and rent ?
Solution:
Total expenditure per month = Rs 16,000

Question 7.
The pie-chart (as shown in the figure) represents the amount spent on different sports by a sports club in a year. If the total money spent by the club on sports is Rs 1,08,000, find the amount spent on each sport.

Solution:
total amount spent on sports = Rs 1,08,000

Hope given RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2 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.

RD Sharma Class 8 Solutions Chapter 25 Data Handling III (Pictorial Representation of Data as Pie Charts or Circle Graphs) Ex 25.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.1

Other Exercises

• RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.1
• RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.2

Question 1.
The number of hours, spent by a school boy on different activities in a working day, is given below:

Present the information in the form of a pie-chart .
Solution:
Total = 24

Now we draw a circle and divide it in the sectors having above central angles as shown in the figure.

Question 2.
Employees of a company have been categorized according to their religions as given below:

Draw a pie-chart to represent the above information.
Solution:
Total =1080

Now draw a circle and divided it into sectors having the above central angles as shown in the figure.

Question 3.
In one day the sales (in rupees) of different items of a baker’s shop are given below :

Draw pie-chart representing the above sales.
Solution:
Total = 480


Now draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 4.
The following data shows the expenditure of a person on different items during a month. Represent the data by a pie-chart.

Solution:
Total = 10800

Now we draw a circle and divide it into sector having the above central angles as shown in figure.

Question 5.
The percentages of various categories of workers in a state are given in the following table:

Present the information in the form of a pie-chart.
Solution:
Total = 100

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 6.
The following table shows the expenditure incurred by a publisher in publishing a book:

Present the above data in the form of a pie-chart.
Solution:
Total = 100% = 100

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure

Question 7.
Percentage of the different products of a village in a particular district are given below. Draw a pie-chart representing this information.

Solution:
Total = 100

Now, we draw a circle and divided it into sectors having the above central angles as shown in the figure.

Question 8.
Draw a pie-diagram for the following data of expenditure pattern in a family.

Solution:
Total =100


Now we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 9.
Draw a pie-diagram of the areas of continents cf the world given in the following table :

Solution:
Total = 133.3

Now, we draw a circle and divide it into sectors having the above given central angles as shown in the figure.

Question 10.
The following data gives the amount spent of the construction of a house. Draw a pie diagram.

Solution:
Total = 300 (in thousands)

Now, we draw a circle and divide it into sectors having the above given central angles as shown in the figure.

Question 11.
The following table shows how a student spends his pocket money during the course of a month. Represent it by a pie-diagram.

Solution:
Total expenditure = 100


Now, we draw a circle and divide it into sectors having the above given central angles as shown in the figure.

Question 12.
Represent the following data by a pie-diagram :

Solution:
1. For family A
Total = 10000

Now, we draw a circle and divide it in sectors having the above central angles as shown in the figure.
(2) For family B
Total = 11680

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 13.
Following data gives the break up of the cost of production of a book:

Draw a pie – diagram depicting the above information.
Solution:
Total = 100

Now, we draw a circle and divide it into sectors of above central angles as shown in the figure.

Question 14.
Represent the following data with the help of pie-diagram.

Solution:
Total = 6000 tons

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 15.
Draw a pie-diagram representing the relative frequencies (expressed as percentage) of the eight classes as given below :
12.6,18.2,17.5,20.3,2.8,4.2,9.8,14.7
Solution:
Total =100

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure given.

Question 16.
Following is the break up of the expenditure of a family on different items of consumption :

Draw a a pie – diagram to represent the above data.
Solution:
Total = Rs. 3000

Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Question 17.
Draw a pie-diagram for the following data of the investment pattern in five year plan :

Solution:
Total = 100


Now, we draw a circle and divide it into sectors having the above central angles as shown in the figure.

Hope given RD Sharma Class 8 Solutions Chapter 25 Data Handling III Ex 25.1 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.

RD Sharma Class 8 Solutions Chapter 26 Data Handling IV (probability) Ex 26.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 26 Data Handling IV (probability) Ex 26.1

Question 1.
The probability that it will rain to morrow is 0.85. What is the probability that it will not rain tomorrow ?
Solution:
Total number of possible events = 1
∴ P (\(\bar { A }\) ) = 0.85
∴ P (\(\bar { A }\) ) = 1-0.85 = 0.15

Question 2.
A die is thrown. Find the probability of getting (i) a prime number (ii) 2 or 4 (iii) a multiple of 2 or 3.
Solution:
Total number of possible events = 6 (1 to 6)
(i) Let A be the favourable occurrence which are prime number i.e., 2,3,5
∴ P(A) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
(ii) Let B be the favourable occurrence which are 2 or 4
∴ P(B) = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)
(iii) Let C be the favourable occurrence which are multiple of 2 or 3 i.e., 2, 3, 4, 6.
∴ P(C) = \(\frac { 4 }{ 6 }\) = \(\frac { 2 }{ 3 }\)

Question 3.
In a simultaneous throw of a pair of dice, find the probability of getting:
(i) 8 as the sum
(ii) a doublet
(iii) a doublet, of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(viii) neither 9 nor 11 as the sum of the numbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) at least once
(xiii) a number other than 5 on any dice.
Solution:
By throwing of a pair of dice, total number of possible events = 6 × 6 = 36
(i) Let A be the occurrence of favourable events whose sum is 8 i.e. (2,6), (3,5), (4,4), (5,3) , (6,2) which are 5
∴ P(A) = \(\frac { 5 }{ 36 }\)
(ii) Let B be the occurrence of favourable events which are doublets i.e. (1, 1), (2, 2), (3, 3), (4,4), (5, 5) and (6, 6).
∴ P(B) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)
(iii) Let C be the occurrence of favourable events which are doublet of prime numbers which are (2, 2), (3,3), (5, 5)
∴ P(C) = \(\frac { 3 }{ 36 }\) = \(\frac { 1 }{ 12 }\)
(iv) Let D be the occurrence of favourable events which are doublets of odd numbers which are (1, 1), (3, 3) and (5, 5)
∴ P(D) = \(\frac { 3 }{ 36 }\) = \(\frac { 1 }{ 12 }\)
(v) Let E be the occurrence of favourable events whose sum is greater than 8 i.e, (3,6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6) which are 6 in numbers
∴ P(E) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)
(vi) Let F be the occurrence of favourable events in which is an even number is on first i.e (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1),(4,2), (4,3) (4,4), (4,5), (4,6), (6,1), (6,2), (6, 3), (6,4 ), (6, 5), (6,6) which are 18 in numbers.
∴ P(F) = \(\frac { 18 }{ 36 }\) = \(\frac { 1 }{ 2 }\)
(vii) Let G be the occurrence of favourable events in which an even number on the one and a multiple of 3 on the other which are (2,3), (2,6), (4, 3), (4, 6), (6, 3), (6, 6), (3, 2), (3, 4), (3, 6), (6,2), (6,4) = which are 11th number
∴ P(G) = \(\frac { 11 }{ 36 }\)
(viii) Let H be the occurrence of favourable events in which neither 9 or 11 as the sum of the numbers on the faces which are (1,1), (1,2), (1,3) , (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2,4) , (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3,5) , (4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (5, 1), (5,2), (5,3), (5,5), (6,1), (6,2), (6,4), (6,6) which are 30
∴ P(H) = \(\frac { 30 }{ 36 }\) = \(\frac { 5 }{ 6 }\)
(ix) Let I be the occurrence of favourable events, such that a sum less than 6, which are (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4,1) which are 10
∴ P(I) = \(\frac { 10 }{ 36 }\) = \(\frac { 5 }{ 18 }\)
(x) Let J be the occurrence of favourable events such that a sum is less than 7, which are
(1.1) , (1,2), (1,3), (1,4), (1,5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5.1) which are 15
∴ P(J) = \(\frac { 15 }{ 36 }\) = \(\frac { 5 }{ 6 }\)
(xi) Let K be the occurrence of favourable events such that the sum is more than 7, which are (2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) which 15
∴ P(K) = \(\frac { 15 }{ 36 }\) = \(\frac { 5 }{ 12 }\)
(xii) Let L be the occurrence of favourable events such that at least P (L) one is black card
∴ P(L) = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)
(xiii) Let M is the occurrence of favourable events such that a number other than 5 on any dice which can be (1,1), (1,2), (1,3), (1,4), (1,6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 6), (3,1), (3,2), (3, 3), (3,4), (3,6), (4,1), (4, 2), (4, 3), (4,4), (4,6), (6, 1), (6,2), (6, 3), (6,4), (6, 6) which are 25
∴ P(M) = \(\frac { 25 }{ 36 }\)

Question 4.
Three coins are tossed together. Find the probability of getting:
(i) exactly two heads
(ii) at least two heads
(iii) at least one head and one tail
(iv) no tails
Solution:
Total number of events tossed by 3 coins each having one head and one tail = 2x2x2 = 8
(i) Let A be the occurrence of favourable events which is exactly two heads, which can be 3 in number which are HTH, HHT, THH.
∴ P(A) = \(\frac { 3 }{ 8 }\)
(ii) Let B be the occurrence of favourable events which is at least two heads, which will be 4 which are HHT, HTH, THH, and HHH.
∴ P(B) = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)
(iii) Let C be the occurrence 6f favourable events which is at least one head and one tail which are 6 which can be HHT, HTH, THH, TTH, THT, HTT
∴ P(C) = \(\frac { 6 }{ 8 }\) = \(\frac { 3 }{ 4 }\)
(iv) Let D be the occurrence of favourable events in which there is no tail which is only 1 (HHH)
∴ P(D) = \(\frac { 1 }{ 8 }\)

Question 5.
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is:
(i) a black king
(ii) either a black card or a king
(iii) black and a king
(iv) a jack, queen or a king
(v) neither a heart nor a king
(vi) spade or an ace
(vii) neither an ace nor a king
(viii) neither a red card nor a queen
(ix) other than an ace
(x) a ten
(xi) a spade
(xii) a black card
(xiii) the seven of clubs
(xiv) jack
(xv) the ace of spades
(xvi) a queen
(xvii) a heart
(xviii) a red card
Solution:
A pack of cards have 52 cards, 26 black and 26 red and four kinds each of 13 cards from 2 to 10, one ace, one jack, one queen and one king.
∴ Total number of possible events = 52
(i) Let A be the occurrence of favourable events which is a black king which are 2.
∴ P(A) = \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26 }\)
(ii) Let B be the occurrence of favourable events such that it is either a black card or a king.
Total = number of black cards = 26 + 2 red kings = 28
∴ P(B) = \(\frac { 28 }{ 52 }\) = \(\frac { 7 }{ 13 }\)
(iii) Let C be the occurrence of favourable events such that it is black and a king which can be 2.
∴ P(C) = \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26 }\)
(iv) Let D be the occurrence of favourable events such that it is a jack, queen or a king which will be4 + 4 + 4 = 12
∴ P(D) = \(\frac { 12 }{ 52 }\) = \(\frac { 3 }{ 13 }\)
(v) Let E be the occurrence of favourable events such that it is neither a heart nor a king.
∴ Number of favourable event will be 13 x 3 -3 = 39 – 3 = 36
∴ P(E) = \(\frac { 36 }{ 52 }\) = \(\frac { 9 }{ 13 }\)
(vi) Let F be the occurrence of favourable events such that it is a spade or an ace.
∴ Number of events = 13 + 3 = 16
∴ P(F) = \(\frac { 16 }{ 52 }\) = \(\frac { 4 }{ 13 }\)
(vii) Let G be the occurrence of favourable events such that it neither an ace nor a king.
∴Number of events = 52 – 4 – 4 = 44
∴ P(G) = \(\frac { 44 }{ 52 }\) = \(\frac { 11 }{ 13 }\)
(viii) Let H be the occurrence of favourable events such that it is neither a red card nor a queen.
∴Number of events = 26 – 2 = 24,
∴ P(H) = \(\frac { 24 }{ 52 }\) = \(\frac { 6 }{ 13 }\)
(ix) Let 1 be the occurrence of favourable events such that it is other than an ace.
∴ Number of events = 52 – 4 = 48
∴ P(I) = \(\frac { 48 }{ 52 }\) = \(\frac { 12 }{ 13 }\)
(x) Let J be the occurrence of favourable event such that it is ten
∴ Number of events = 4
∴ P(J) = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)
(xi) Let K be the occurrence of favourable event such that it is a spade.
∴ Number of events =13
∴ P(K) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
(xii) Let L be the occurrence of favourable event such that it is a black card.
∴ Number of events = 26
∴ P(L) = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)
(xiii) Let M be the occurrence of favourable event such that it is the seven of clubs.
∴ Number of events = 1
∴ P(M) = \(\frac { 1 }{ 52 }\)
(xiv) Let N be the occurrence of favourable event such that it is a jack.
∴ Number of events = 1
∴ P(N) = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)
(xv) Let O be the occurrence of favourable event such that it is an ace of spades.
∴ Number of events = 1
∴ P(O) = \(\frac { 1 }{ 52 }\)
(xvi) Let Q be the occurrence of favourable event such that it is a queen.
∴ VNumber of events = 4
∴ P(P) = \(\frac { 4 }{ 52 }\) = \(\frac { 1 }{ 13 }\)
(xvii) Let R be the occurrence of favourable event such that it is a heart card.
∴ Number of events =13
∴ P(P) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
(xviii) Let S be the occurrence of favourable event such that it is a red card
∴ Number of events = 26
∴ P(P) = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)

Question 6.
An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.
Solution:
Number of possible events = 10 + 8 = 18
Let A be the occurrence of favourable event such that it is a white ball.
∴ Number of events = 8
∴ P(A) = \(\frac { 8 }{ 18 }\) = \(\frac { 4 }{ 9 }\)

Question 7.
A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) white ? (ii) red ? (iii) black ? (iv) not red ?
Solution:
Number of possible events = 3 + 5 + 4 = 12 balls
(i) Let A be the favourable event such that it is a white ball.
∴ P(A) = \(\frac { 4 }{ 12 }\) = \(\frac { 1 }{ 3 }\)
(ii) Let B be the favourable event such that it is a red ball.
∴ P(B) = \(\frac { 3 }{ 12 }\) = \(\frac { 1 }{ 4 }\)
(iii) Let C be the favourable event such that it is a black ball.
∴ P(C) = \(\frac { 5 }{ 12 }\)
(iv) Let D be the favourable event such that it is not red.
∴ Number of favourable events = 5 + 4 = 9
∴ P(D) = \(\frac { 9 }{ 12 }\) = \(\frac { 3 }{ 4 }\)

Question 8.
What is the probability that a number selected from the numbers 1,2,3,…………, 15 is a multiple of 4 ?
Solution:
Number of possible events =15
Let A be the favourable event such that it is a multiple of 4 which are 4, 8, 12
∴ P(A) = \(\frac { 3 }{ 15 }\) = \(\frac { 1 }{ 5 }\)

Question 9.
A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that ball drawn is not black ?
Solution:
Number of possible events = 6 + 8 + 4 = 18 balls
Let A be the favourable event such that it is not a black
∴ Number of favourable events = 6 + 4=10
∴ P(A) = \(\frac { 10 }{ 18 }\) = \(\frac { 5 }{ 9 }\)

Question 10.
A bag contains 5 white and 7 red balls. One ball is drawn at random, what is the probability that ball drawn is white ?
Solution:
Number of possible events = 5 + 7 = 12
Let A be the favourable event such that it is a while which are 5.
∴ P(A) = \(\frac { 5 }{ 12 }\)

Question 11.
A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is (i) white (if) red (iii) not black (iv) red or white.
Solution:
Number of possible events = 4 + 5 + 6 = 15
(i) Let A be the favourable events such that it is a white.
∴ P(A) = \(\frac { 6 }{ 15 }\) = \(\frac { 2 }{ 5 }\)
(ii) Let B be the favourable event such that it is a red
∴ P(B) = \(\frac { 4 }{ 15 }\)
(iii) Let C be the favourable event such that it is not black.
∴ Number of favourable events = 4 + 6=10
∴ P(C) = \(\frac { 10 }{ 15 }\) = \(\frac { 2 }{ 3 }\)

Question 12.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is: (i) red (ii) black.
Solution:
Number of possible events = 3 + 5 = 8
(i) Let A be the favourable events such that it is red.
∴ P(A) = \(\frac { 3 }{ 8 }\)
(ii) Let B be the favourable event such that it is black.
∴ P(B) = \(\frac { 5 }{ 8 }\)

Question 13.
A bag contains 5 red marbles, 8 white marbles, 4 green marbles. What is the probability that if one marble is taken out of the bag at random, it will be
(i) red
(ii) white
(iii) not green.
Solution:
Total number of possible events = 5 + 8+4=17
(i) Let A be the favourable event such that it is red.
∴ P(A) = \(\frac { 5 }{ 7 }\)
(ii) Let B be the favourable event such that it is white
Then P (B) = \(\frac { 8 }{ 7 }\)
(iii) Let C be the favourable event such that it is not green.
∴ Number of favourable events = 5 + 8 = 13
∴ P(C) = \(\frac { 13 }{ 17 }\)

Question 14.
If you put 21 consonants and 5 vowels in a bag. What would carry greater probability ? Getting a consonant or a vowel ? Find each probability.
Solution:
Total number of possible events = 21 + 5 = 26
(i) Probability of getting a consonant is greater as to number is greater than the other.
(ii) Let A be the favourable event such that it is a consonant.
∴ P(A) = \(\frac { 21 }{ 26 }\)
(iii) Let B be the favourable event such that it is a vowel.
∴ P(B) = \(\frac { 5 }{ 26 }\)

Question 15.
If we have 15 boys and 5 girls in a class which carries a higher probability ? Getting a copy belonging to a boy or a girl ? Can you give it a value ?
Solution:
Number of possible outcome (events) = 15 + 5 = 20
∵ The number of boys is greater than the girls
∴ The possibility of getting a copy belonging to a boy is greater.
Let A be the favourable outcome (event) then
P(A) = \(\frac { 15 }{ 20 }\) = \(\frac { 3 }{ 4 }\)

Question 16.
If you have a collection of 6 pairs of white socks and 3 pairs of black socks. What is the probability that a pair you pick without looking is (i) white ? (if) black ?
Solution:
Total number of possible outcomes =6+3=9
(i) Let A be the favourable outcome which is white pair.
∴ P(A) = \(\frac { 6 }{ 9 }\) = \(\frac { 2 }{ 3 }\)
(ii) Let B be the favourable outcome which is a black pair.
∴ P(B) = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)

Question 17.
If you have a spinning wheel with 3 green sectors, 1-blue sector and 1-red sector, what is the probability of getting a green sector ? Is it the maximum ?
Solution:
Total number of possible outcomes = 3 + 1 + 1 =5
Let A be the favourable outcome which is green sector
∴ P(A) = \(\frac { 3 }{ 5 }\)
∴ Number of green sectors is greater.
∴ It’s probability is greater.

Question 18.
When two dice are rolled :
(i) List the outcomes for the event that the total is odd.
(ii) Find probability of getting an odd total.
(iii) List the outcomes for the event that total is less than 5.
(iv) Find the probability for getting a total less than 5.
Solution:
∵ Every dice has 6 number from 1 to 6.
∴ Total outcomes = 6 x 6 = 36.
(i) List of outcomes for event then the total is odd will be (1,2), (1,4), (1,6), (2,1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6,1), (6, 3), (6, 5)
(ii) Probability of getting an odd total
Let A be the favourable outcomes which are
P(A) = \(\frac { 18 }{ 36 }\) = \(\frac { 1 }{ 2 }\)
(iii) List of outcomes for the event that total is less than 5 are (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2,1), which are 6.
(iv) Probability of getting a total less than 5 Let B be the favourable outcome,
Then P (B) = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)

Hope given RD Sharma Class 8 Solutions Chapter 26 Data Handling IV (probability) Ex 26.1 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.

RD Sharma Class 8 Solutions Chapter 27 Introduction to Graphs Ex 27.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 27 Introduction to Graphs Ex 27.2

Other Exercises

• RD Sharma Class 8 Solutions Chapter 27 Introduction to Graphs Ex 27.1
• RD Sharma Class 8 Solutions Chapter 27 Introduction to Graphs Ex 27.2

Question 1.
The following table shows the number of patients discharged from a hospital with HIV diagnosis in different years :

Represent this information by a graph.
Solution:
Represent years along x-axis and number of patients along y-axis. Now, plot the points (2002, 150), (2003, 170), (2004, 195), (2005, 225) and (2006, 230) on the graph and join them in order. We get the required graph as shown on the graph.

Question 2.
The following table shows the amount of rice grown by a farmer in different years :

Plot a graph to illustrate this information.
Solution:
We represent years along x-axis and rice (in quintals) along y-axis. Now we plot the points (2000,200), (2001, 180), (2002,240), (2003,260), (2004,250), (2005,200) and (2006,270) on the graph and join them in order as shown on the graph.

Question 3.
The following table gives the information regarding the number of persons employed to a piece of work and time taken to complete the work:

Plot a graph of this information.
Solution:
We represent number of person along x-axis and time taken in day along y-axis. Now plot the points (2,12), (4,6), (6,4) and (8, 3) on the graph and join them in order as shown on the graph.

Question 4.
The following table gives the information regarding length of a side of a square and its area :

Draw a graph to illustrate this information.
Solution:
We represent length of a side (in cm) along x-axis and Area of square (in cm2) along the y-axis. Now plot the points (1,1), (2, 4), (3, 9), (4, 16) and (5, 25) on the graph and join them in order. We get the required graph as shown on the graph.

Question 5.
The following table shows the sales of a commodity during the years 2000 to 2006.

Draw a graph of this information.
Solution:
We represent years on x-axis and sales (in lakh rupees) along y-axis. Now plot the points
(2000, 1.5), (2001, 1.8), (2002, 2.4), (2003, 3.2), (2004, 5.4), (2005, 7.8) and (2006, 8.6) on the graph and join them in order we get the required graph as shown.

Question 6.
Draw the temperature-time graph in each of the following cases :

Solution:
(i) We represent time along x-axis and temperature (in °F) alongy-axis. Now plot the points (7:00, 100), (9:00, 101), (11:00, 104), (13:00, 102), (15:00, 100), (17:00, 99), (19:00, 100) and (21:00, 98) on the graph and join them in order to get the required graph as shown on the graph.


(ii) We represent time along x-axis and temperature (in °F) along j-axis. Now plot the points (8:00, 100), (i0:00, 101), (12:00, 104), (14:00, 103), (16:00, 99), (18:00,98), (20:00,100) on the graph and join them in order to get the required graph as shown on the graph.

Question 7.
Draw the velocity-time graph from the following data :

Solution:
We represent time (in hours) along x- axis and speed (in km/hr) along y-axis. Now plot the points (7:00,30), (8:00,45), (9:00,60), (10:00, 50), (11:00, 70), (12:00, 50), (13:00, 40) and (14:00,45) on the graph and join them in order to get the required graph as shown.

Question 8.
The runs scored by a cricket team in first 15 overs are given below :

Draw the graph representing the above data in two different ways as a graph and as a bar chart.
Solution:
We represent overs along.t-axis and runs along v-axis. Now plot the points (1,2), (II, 1), (III, 4), (IV, 2), (V, 6), (VI, 8), (VII, 10), (VIII, 21), (IX, 5), (X, 8), (XI, 3), (XII, 2), (XIII, 6), (XIV, 8), (XV, 12) on the graph and join them to get the graph, as shown bar graph of the given data is given below:

Question 9.
The runs scored by two teams A and B in first 10 overs are given below :

Draw a graph depicting the data, making the graphs on the same axes in each case in two different ways as a graph and as a bar chart.
Solution:
We represent overs along x-axis and runs scored by team A and team B with different types of lines along y-axis.
Plot the points for team A : (I, 2), (II, 1), (III,8), (IV, 9), (V, 4), (VI, 5), (VII, 6), (VIII, 10), (IX, 6) and (X, 2)
and for team B, the points will be : (I, 5), (II, 6), (III, 2), (IV, 10), (V, 5), (VI, 6), (VII, 3), (VIII, 4), (IX, 8), (X, 10).
Then join them in order to get the required graph for team A and team B as shown.
Note : For team A ………
for team B …………

Hope given RD Sharma Class 8 Solutions Chapter 27 Introduction to Graphs Ex 27.2 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.