## 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

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

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:
(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
(vii) neither an ace nor a king
(viii) neither a red card nor a queen
(ix) other than an ace
(x) a ten
(xii) a black card
(xiii) the seven of clubs
(xiv) jack
(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

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 …………

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

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 27.1

Other Exercises

Question 1.
Plot the points (5,0 ), (5,1), (5, 8). Do they lie on a line ? What is your observation ?
Solution:
Draw XOX’ and YOY’ the co-ordinates axis on the graph.
Take 1 cm = 1 unit
Point A (5, 0), B (5, 1) and C (5, 8) have been plotted on the graph. By joining A, B and C, we see that these points lie on the same line which is 5 units from y-axis.

Question 2.
Plot the points (2, 8), (7, 8) and (12, 8). Join these points in pairs. Do they lie on a line ? What do you observe ?
Solution:
Draw XOX’ and YOY’, the co-ordinate axis on the graph.
Take 0.5 cm = 1 unit.
Now points A (2, 8), B (7, 8), and C (12, 8) have been plotted on the graph. By joining them, we see that these points lie on the same line which is at a distance of 8 unit from x-axis.

Question 3.
Locate the points :
(i) (1,1), (1,2), (1,3), (1,4)
(ii) (2,1), (2, 2), (2,3), (2,4)
(iii) (1,3), (2,3), (3,3), (4,3)
(iv) (1,4), (2,4), (3,4), (4,4).
Solution:
The points given in (i) and (ii) are locates in first graph.

(i) (1,1), (1,2), (1,3), (1,4)
(ii) (2,1), (2, 2), (2, 3), (2, 4)
Points of (iii) and (iv) are located in the adjoining graph.
(iii) (1, 3), (2, 3), (3, 3), (4, 3)
(iv) (1,4), (2, 4), (3, 4), (4,4)

Question 4.
Find the coordinates of points A, B, C, D in the figure

Solution:
Draw perpendicular from A, B, C and D on x-axis and also on y-axis.
A is 1 unit from y-axis and 1 unit from x-axis.
∴ Co-ordinates of A are (1,1)
B is 1 unit fr onr y-axis and 4 units from x-axis
∴ Co-ordinates of B are (1,4)
C is 4 units from y-axis and 6 units from x-axis
∴ Co-ordinates of C are (4, 6)
D is 5 units fromy-axis and 3 units from x-axis
∴ Co-ordinates of D are (5,3)

Question 5.
Find the coordinates of points P, Q, R and S in Fig.

Solution:
Through P, Q, R and S, draw perpendiculars on x-axis and also on y-axis.
(i) P is 10 units form >’-axis and 70 units from y-axis
∴ Co-ordinates of P are (10, 70)
(ii) Q is 12 unit from x-axis and 80 units from y-axis
∴ Co-ordinates of Q are (12, 80)
(iii) R is 16 units from x-axis and 100 units from y-axis
∴ Co-ordinates of R are (16, 100)
(iv) S is 20 units from x-axis and 120 units from y-axis
∴ Co-ordinates of S are (20, 120)

Question 6.
Write the coordinates of each of the vertices of each polygon in the figure.

Solution:
(i) In figure OXYZ
Co-ordinates of O are (0, 0) ∵It is the origin
Co-ordinates of X are (0, 2) ∵ It lies on y – axis
Co-ordinates of Y are (2, 2)
Co-ordinates of Z are (2, 0) ∵It lies on x-axis
(ii) In figure ABCD, draw perpendicular from A, B, C, D on x-axis and y-axis.
A is 4 unit from y-axis and 5 units from x- axis
∴ Co-ordinates of A are (4, 5)
B is 7 units from y-axis and 5 units from x – axis
∴ Co-ordinates of B are (7, 5)
C is 6 units from y-axis and 3 units from x- axis
∴ Co-ordinates of C are (6, 3)
D is 3 units from y-axis as well x-axis
∴ Co-ordinates of D are (3, 3)
(iii) In figure PQR, perpendiculars for P, Q, R are drawn on x-axis and also on y-axis.
∴ Co-ordinates of P are (7, 4), of Q are (9, 5) and of R are (9, 3).

Question 7.
Decide which of the following statements is true and which is false. Give reasons for your answer.
(i) A point whose x-coordinate is zero, will lie on they-axis.
(ii) A point whose y-coordinate is zero, will lie on x-axis.
(iii) The coordinates of the origin are (0, 0).
(iv) Points whose x and y coordinates are equal, lie on a line passing through the origin.
Solution:
(i) Correct: ∵ every point on y-axis, its x = 0
(ii) Correct: ∵ every point on x-axis, its y = 0
(iii) Correct
(iv) Correct

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## RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.4

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.4

Other Exercises

Question 1.
Find the cube roots of each of the following integers :
(i) -125
(ii) -5832
(iii) -2744000
(iv) -753571
Solution:

Question 2.
Show that :

Solution:

Question 3.
Find the cube root of each of the following numbers :
(i) 8 x 125
(ii) -1728 x 216
(iii) -27 x 2744
(iv) -729 x -15625
Solution:

Question 4.
Evaluate :

Solution:

Question 5.
Find the cube root of each of the following rational numbers.

Solution:

Question 6.
Find the cube root of each of the following rational numbers :
(i) 0.001728
(ii) 0.003375
(iii) 0.001
(iv) 1.331
Solution:

Question 7.
Evaluate each of the following :

Solution:

Question 8.
Show that :

Solution:

Question 9.
Fill in the Blanks :

Solution:

Question 10.
The volume of a cubical box is 474.552 cubic metres. Find the length of each side of the box.
Solution:

Question 11.
Three numbers are to one another 2:3: 4. The sum of their cubes is 0.334125. Find the numbers.
Solution:

Question 12.
Find side of a cube whose volume is

Solution:

Question 13.
Evaluate :

Solution:

Question 14.
Find the cube root of the numbers : 2460375,20346417,210644875,57066625 using the fact that
(i) 2460375 = 3375 x 729
(i) 20346417 = 9261 x 2197
(iii) 210644875 = 42875 x 4913
(iv) 57066625 = 166375 x 343
Solution:

Question 15.
Find the units digit of the cube root of the following numbers ?
(i) 226981
(ii) 13824
(iii) 571787
(iv) 175616
Solution:
(i) 226981
In it unit digit is 1
∴The units digit of its cube root will be = 1
(∵ 1 x 1 x 1 = 1)
∴Tens digit of the cube root will be = 6
(ii) 13824
∵ The units digit of 13824 = 4
(∵ 4 X 4 X 4 = 64)
∴Units digit of the cube root of it = 4
(iii) 571787
∵ The units digit of 571787 is 7
∴The units digit of its cube root = 3
(∵ 3 x 3 x 3 = 27)
(iv) 175616
∵ The units digit of 175616 is 6
∴The units digit of its cube root = 6
(∵ 6 x 6 x 6=216)

Question 16.
Find the tens digit of the cube root of each of the numbers in Question No. 15.
Solution:
(i) In 226981
∵ Units digit is 1
∴Units digit of its cube root = 1
We have 226
(Leaving three digits number 981)
63 = 216 and 73 = 343
∴63 ∠226 ∠ T
∴The ten’s digit of cube root will be 6
(ii) In 13824
Leaving three digits number 824, we have 13
∵ (2)3 = 8, (3)3 = 27
∴23 ∠13 ∠3′
∴Tens digit of cube root will be 2
(iii) In 571787
Leaving three digits number 787, we have 571
83 = 512, 93 = 729
∴ 83 ∠571 ∠93
Tens digit of the cube root will be = 8
(iv) In 175616
Leaving three digit number 616, we have 175
∵ 53 = 125, 63 = 216
∴53 ∠175 ∠63
∴Tens digit of the cube root will be = 5

Hope given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.4 are helpful to complete your math homework.

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## RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.3

Other Exercises

Question 1.
Find the cube roots of the following numbers by successive subtraction of numbers : 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397,
(i) 64
(ii) 512
(iii) 1728
Solution:
(i) 64
64 – 1 = 63
63 – 7 = 56
56 – 19 = 37
37 – 37 = 0
∴ 64 = (4)3
∴ Cube root of 64 = 4

(ii) 512
512 -1 =511
511- 7 = 504
504 – 19 = 485
485 – 37 = 448
448 – 61 = 387
387 – 91 =296
296 – 127 = 169
169 – 169 = 0
∴ 512 = (8)3
∴ Cube root of 512 = 8

(iii) 1728
1728 – 1= 1727
1727 -7 = 1720
1720 -19 = 1701
1701 -37= 1664
1664 – 61 = 1603
1603 – 91 = 1512
1512 -127= 1385 .
1385 – 169= 1216
1216 – 217 = 999
999 – 271 =728
728 – 331 = 397
397 – 397=0
∴ 1728 = (12)3
∴ Cube root of 1728 = 12

Question 2.
Using the method of successive subtraction, examine whether or not the following numbers are perfect cubes :
(i) 130
(ii) 345
(iii) 792
(iv) 1331
Solution:
(i) 130
130 – 1 = 129
129 -7 = 122
122 -19 = 103
103 -37 = 66
66 – 61 = 5
We see that 5 is left
∴ 130 is not a perfect cube.

(ii) 345
345 – 1 = 344
344 – 7 = 337
337 – 19 = 318
318 – 37 = 281
81 – 61 =220
220 – 91 = 129
129 – 127 = 2
We see that 2 is left
∴ 345 is not a perfect cube.

(iii) 792
792 – 1 = 791
791 – 7 = 784
784 – 19 = 765
765 – 37 = 728
728 – 61 = 667
667 – 91 = 576
576 – 127 = 449
449 – 169 = 280
∴ We see 280 is left as 280 <217
∴ 792 is not a perfect cube.

(iv) 1331
1331 – 1 = 1330
1330 -7 = 1323
1323 – 19 = 1304
1304 – 37 = 1267
1267 – 61 = 1206
1206 – 91 = 1115
1115 – 127 = 988
988 – 169 = 819
819 – 217 = 602
602 – 271 = 331
331 – 331 =0
∴ 1331 is a perfect cube

Question 3.
Find the smallest number that must be subtracted from those of the numbers in question 2, which are not perfect cubes, to make them perfect cubes. What are the corresponding cube roots ?
Solution:
We have examined in Question 2, the numbers 130, 345 and 792 are not perfect cubes. Therefore
(i) 130
130 – 1 = 129
129 -7= 122
122 -19 = 103
103 – 37 = 66
66 – 61 = 5
Here 5 is left
∴ 5 < 91 5 is to be subtracted to get a perfect cube.
Cube root of 130 – 5 = 125 is 5

(ii) 345
345 – 1 = 344
344 -7 = 337
337 – 19 = 318
318 – 37 = 281
281 – 61 =220
220 – 91 = 129
129 – 127 = 2
Here 2 is left ∵ 2 < 169
∴ Cube root of 345 – 2 = 343 is 7
∴ 2 is to be subtracted to get a perfect cube.

(iii) 792
792 – 1 = 791
791 – 7 = 784
784 – 19 = 765
765 – 37 = 728
728 – 61 =667
667 – 91 = 576 5
76 – 127 = 449
449 – 169 = 280
280 – 217 = 63
∴ 63 <217
∴ 63 is to be subtracted
∴ Cube root of 792 – 63 = 729 is 9

Question 4.
Find the cube root of each of the following natural numbers :
(i) 343
(ii) 2744
(iii) 4913
(iv) 1728
(v) 35937
(vi) 17576
(vii) 134217728
(viii) 48228544
(ix) 74088000
(x) 157464
(xi) 1157625
(xii) 33698267
Solution:

Question 5.
Find the smallest number which when multiplied with 3600 will make the product a perfect cube. Further, find the cube root of the product.
Solution:

Grouping the factors in triplets of equal factors, we see that 2, 3 x 3 and 5 x 5 are left
∴ In order to complete the triplets, we have to multiply it by 2, 3 and 5.
∴ The smallest number to be multiplied = 2×2 x 3 x 5 = 60
Now product = 3600 x 60 = 216000 and cube root of 216000

Question 6.
Multiply 210125 by the smallest number so that the product is a perfect cube. Also, find out the cube root of the product.
Solution:

Grouping the factors in triplets of equal factors, we see that 41 x 41 is left
∴ In order to complete the triplet, we have to multiply it by 41
∴ Smallest number to be multiplied = 41
∴ Product = 210125 x 41 = 8615125
∴ Cube root of 8615125

Question 7.
What is the smallest number by which 8192 must be divided so that quotient is a perfect cube ? Also, find the cube root of the quotient so obtained.
Solution:

Grouping the factors in triplets of equal factors, we see that 2 is left
∴ Dividing by 2, we get the quotient a perfect cube
∴ Perfect cube = 8192 + 2 = 4096

Question 8.
Three numbers are in the ratio 1:2:3. The sum of their cubes is 98784. Find the numbers.
Solution:
Ratio in numbers =1:2:3
Let first number = x
Then second number = 2x
and third number = 3x
∴ Sum of cubes of there numbers = (x)3 + (2x)3+(3x)3

Question 9.
The volume of a cube is 9261000 m3. Find the side of the cube.
Solution:

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## RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2

Other Exercises

Question 1.
Find the cubes of:
(i) -11
(ii) -12
(iii) – 21
Solution:
(i) (-11)3=(-11)3=(11 x 11 x 11) =-1331
(ii) (-12)3=(-12)3=(12 x 12 x 12) =  -1728
(iii) (-21)3=(-21)3=(21 x 21 x 21) = -9261

Question 2.
Which of the following numbers are cubes of negative integers.
(i) -64
(ii) -1056
(iii) -2197
(iv) -2744
(v)  -42875
Solution:

∴ All factors of 64 can be grouped in triplets of the equal factors completely.
∴ -64 is a perfect cube of negative integer.

All the factors of 1056 can be grouped in triplets of equal factors grouped completely
∴ 1058 is not a perfect cube of negative integer.

All the factors of -2197 can be grouped in triplets of equal factors completely
∴ 2197 is a perfect cube of negative integer,

All the factors of -2744 can be grouped in triplets of equal factors completely
∴ 2744 is a perfect cube of negative integer

All the factors of -42875 can be grouped in triplets of equal factors completely
∴ 42875 is a perfect cube of negative integer.

Question 3.
Show that the following integers are cubes of negative integers. Also, find the integer whose cube is the given integer :
(i) -5832 (ii) -2744000
Solution:

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ -5832 is a perfect cube
Now taking one factor from each triplet we find that
-5832 is a cube of – (2 x 3 x 3) = -18
∴ Cube root of-5832 = -18

Grouping the factors in tuplets of equal factors, we see that no factor is left. Therefore it is a perfect cube.
Now taking one factor from each triplet, we find that.
-2744000 is a cube of – (2 x 2 x 5 x 7) ie. -140
∴ Cube root of -2744000 = -140

Question 4.
Find the cube of :

Solution:

Question 5.
Which of the following numbers are cubes of rational numbers :

Solution:

Hope given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2 are helpful to complete your math homework.

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## RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1

Other Exercises

Question 1.
Find the cubes of the following numbers:
(i) 7
(ii) 12
(iii) 16
(iv) 21
(v) 40
(vi) 55
(vii) 100
(viii) 302
(ix) 301
Solution:
(i) (7)3 = 7 x 7 x 7 = 343
(ii) (12)3 = 12 x-12 x 12 = 1728
(iii) (16)3 = 16 x 16 x 16 = 4096
(iv) (21)3 = 21 x 21 x 21 = 441 x 21 =9261
(v) (40)3 = 40 x 40 x 40 = 64000
(vi) (55)3 = 55 x 55 x 55 = 3025 x 55 = 166375
(vii) (100)3 = 100 x 100 x 100 =1000000
(viii)(302)3 = 302 x 302 x 302 = 91204 x 302 = 27543608
(ix) (301)3 = 301 x 301 x 301 = 90601 x 301 =27270901

Question 2.
Write the cubes of all natural numbers between 1 and 10 and verify the following statements :
(i) Cubes of all odd natural numbers are odd.
(ii) Cubes of all even natural numbers are even.
Solution:
Cubes of first 10 natural numbers :
(1)3 = 1 x 1 x 1 = 1
(2)3 = 2 x 2 x 2 = 8
(3)3 = 3 x 3 x 3 = 27
(4)3= 4 x 4 x 4 = 64
(5)3 = 5 x 5 x 5 = 125
(6)3 = 6 x 6 x 6 = 216
(7)3 = 7 x 7 x 7 = 343
(8)3 = 8 x 8 x 8 = 512
(9)3 = 9 x 9 x 9= 729
(10)3 = 10 x 10 x 10= 1000
We see that the cubes of odd numbers is also odd and cubes of even numbers is also even.

Question 3.
Observe the following pattern :

Write the next three rows and calculate the value of 13 + 23 + 33 +…. + 93 + 103 by the above pattern.
Solution:
We see the pattern

Question 4.
Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings :
The cube of a natural number which is a multiple of 3 is a multiple of 27′
Solution:
5 natural numbers which are multiples of 3
3,6,9,12,15.
(3)3 = 3 x 3 x 3 = 27
Which is multiple of 27
(6)3 = 6 x 6 x 6 = 216 ÷ 27 = 8
Which is multiple of 27
(9)3 = 9 x 9 x 9 = 729 + 27 = 27
Which is multiple of 27
(12)3= 12 x 12 x 12 = 1728 ÷ 27 = 64
Which is multiple of 27
(15)3 = 15 x 15 x 15 = 3375 ÷ 27 = 125
Which is multiple of 27
Hence, cube of multiple of 3 is a multiple of 27

Question 5.
Write the cubes of 5 natural numbers which are of the form 3n+ 1 (e.g.,4, 7, 10, …………) and verify the following :
‘The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1’.
Solution:
3n + 1
Let n = 1, 2, 3, 4, 5, then
If n = 1, then 3n +1= 3 x 1+1= 3+1= 4
If n = 2, then 3n +1=3 x 2+1=6+1=7
If n = 3, then 3n + 1= 3 x 3 + 1= 9 + 1 = 10
If n = 4, then 3n + 1= 3 x 4+1 = 12 + 1= 13
If n = 5, then 3n +1=3 x 5 + 1 = 15 +1 = 16
Now
(4)3 = 4 x 4 x 4 = 64
Which is $$\frac {64 }{ 3 }$$=21, Remainder = 1
(7)3 = 7 x 7 x 7 = 343
Which is $$\frac {343 }{ 3 }$$ =114, Remainder = 1
(10)3 = 10 x 10 x 10 = 1000 ÷ 3 = 333, Remainder = 1
(13)3 = 13 x 13 x 13 = 2197 ÷ 3 = 732, Remainder = 1
(16)3 = 16 x 16 x 16 = 4096 ÷ 3 = 1365, Remainder = 1
Hence cube of natural number of the form, 3n + 1, is a natural of the form 3n + 1

Question 6.
Write the cubes of 5 natural numbers of the form 3n + 2 (i.e. 5, 8, 11,……… ) and verify the following :
‘The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2’.
Solution:
Natural numbers of the form 3n + 2, when n
is a natural number i.e. 1, 2, 3, 4, 5,………….
If n = 1, then 3n + 2 = 3 x 1+2 = 3+ 2 = 5
If n = 2, then 3n + 2 = 3 x 2 + 2 = 6 + 2 = 8
If n = 3, then 3n + 2 = 3 x 3 + 2 = 9 + 2 = 11
If n = 4, then 3n + 2 = 3 x 4 + 2 = 12 + 2 = 14
and if n = 5, then 3n + 2 = 3 x 5 + 2 = 15 + 2= 17
Now (5)3 = 5 x 5 x 5 = 125
125 + 3 = 41, Remainder = 2
(8)2 = 8 x 8 x 8 = 512 512 -s- 3 = 170, Remainder = 2
(11)3 = 11 x 11 x 11 = 1331
1331 + 3 = 443, Remainder = 2
(14)3 = 14 x 14 x 14 = 2744
2744 + 3 = 914, Remainder = 2
(17)3 = 17 x 17 x 17 = 4913
4913 = 3 = 1637, Remainder = 2
We see the cube of the natural number of the
form 3n + 2 is also a natural number of the
form 3n + 2.

Question 7.
Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following :
‘The cube of a multiple of 7 is a multiple of 73′.
Solution:
5 natural numbers which are multiple of 7,are 7, 14, 21, 28, 35
(7)3 = (7)3 which is multiple of 73
(14)3 = (2 x 7)3 = 23 x 73, which is multiple of 73
(21)3 = (3 x 7)3 = 33 x 73, which is multiple of 73
(28)3 = (4 x 7)3 = 43 x 73, which is multiple of 73 (35)3 = (5 x 7)3 = 53 x 73 which is multiple of 73
Hence proved.

Question 8.
Which of the following are perfect cubes?
(i) 64
(ii) 216
(iii) 243
(iv) 1000
(v) 1728
(vi) 3087
(vii) 4608
(viii) 106480
(ix) 166375
(x) 456533
Solution:
(i) 64 = 2 x 2 x 2 x 2 x 2 x 2

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 64 is a perfect cube
(ii) 216 = 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets of equal factors, we see that no factor is left
216 is a perfect cube.
(iii) 243 = 3 x 3 x 3 x 3 x 3

Grouping the factors in triplets, we see that two factors 3 x 3 are left
∴ 243 is not a perfect cube.
(iv) 1000 = 2 x 2 x 2 x 5 x 5 x 5

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 1000 is a perfect cube.
(v) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets of the equal factors, we see that no factor is left
∴ 1728 is a perfect cube,
(vi) 3087 = 3 x 3 x 7 x 7 x 7

Grouping the factors in triplets of the equal factors, we see that two factor 3×3 are left
∴ 3087 is not a perfect cube.
(vii) 4608 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

Grouping the factors in triplets of equal factors, we see that two factors 3, 3 are left
∴ 4609 is not a perfect cube.
(viii) 106480 = 2 x 2 x 2 x 2 x 5 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, we see that factors 2, 5 are left
∴ 106480 is not a perfect cube.
(ix) 166375 = 5 x 5 x 5 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 166375 is a perfect cube.
(x) 456533 = 7 x 7 x 7 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 456533 is a perfect cube.

Question 9.
Which of the following are cubes of even natural numbers ?
216, 512, 729,1000, 3375, 13824
Solution:
We know that the cube of an even natural number is also an even natural number
∴ 216, 512, 1000, 13824 are even natural numbers.
∴ These can be the cubes of even natural number.

Question 10.
Which of the following are cubes of odd natural numbers ?
125, 343, 1728, 4096, 32768, 6859
Solution:
We know that the cube of an odd natural number is also an odd natural number,
∴ 125, 343, 6859 are the odd natural numbers
∴ These can be the cubes of odd natural numbers.

Question 11.
What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes ?
(i) 675
(ii) 1323
(iii) 2560
(iv) 7803
(v) 107311
(vi) 35721
Solution:
(i) 675 = 3 x 3 x 3 x 5 X 5

Grouping the factors in triplet of equal factors, 5 x 5 are left without triplet
So, by multiplying by 5, the triplet will be completed.
∴ Least number to be multiplied = 5
(ii) 1323 = 3 x 3 x 3 x 7 x 7

Grouping the factors in triplet of equal factors. We find that 7 x 7 has been left
So, multiplying by 7, we get a triplet
∴ The least number to be multiplied = 7
(iii) 2560 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5

Grouping the factors in triplet of equal factors, 5 is left.
∴ To complete a triplet 5 x 5 is to multiplied
∴ Least number to be multiplied = 5 x 5 = 25
(iv) 7803 = 3 x 3 x 3 x 17 x 17

Grouping the factors in triplet of equal factors, we find the 17 x 17 are left
So, to complete the triplet, we have to multiply by 17
∴ Least number to be multiplied = 17
(v) 107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11

Grouping the factors in triplet of equal factors, factor 3 is left
So, to complete the triplet 3 x 3 is to be multiplied
∴ Least number to be multiplied = 3 x 3 = 9
(vi) 35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7

Grouping the factors in triplet of equal factors, we find that 7 x 7 is left
So, in order to complete the triplets, we have to multiplied by 7
∴ Least number to be multiplied = 7

Question 12.
By which smallest number must the following numbers be divided so that the quotient is a perfect cube ?
(i) 675
(ii) 8640
(iii) 1600
(iv) 8788
(v) 7803
(vi) 107811
(vii) 35721
(viii) 243000
Solution:
(i) 675 = 3 x 3 x 3 x 5 x 5

Grouping the factors in triplet of equal factors, 5 x 5 is left
5 x 5 is to be divided so that the quotient will be a perfect cube.
∴ The least number to be divided = 5 x 5 = 25
(ii) 8640 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets of equal factors, 5 is left
∴ In order to get a perfect cube, 5 is to divided
∴ Least number to be divided = 5
(iii) 1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5

Grouping the factors in triplets of equal factors, we find that 5 x 5 is left
∴ In order to get a perfect cube 5 x 5 = 25 is to be divided.
∴ Least number to be divide = 25
(iv) 8788 = 2 x 2 x 13 x 13 x 13

Grouping the factors in triplets of equal factors, we find that 2 x 2 has been left
∴ In order to get a perfect cube, 2 x 2 is to be divided
∴ Least number to be divided = 4
(v) 7803 = 3 x 3 x 3 x 17 x 17

Grouping the factors in triplets of equal factors, we see that 17 x 17 has been left.
So, in order to get a perfect cube, 17 x 17 is be divided
∴ Least number to be divided = 17 x 17 = 289
(vi) 107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, 3 is left
∴ In order to get a perfect cube, 3 is to be divided
∴ Least number to be divided = 3
(vii) 35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7

Grouping the factors in triplets of equal factors, we see that 7 x 7 is left
So, in order to get a perfect cube, 7 x 7 = 49 is to be divided
∴ Least number to be divided = 49
(viii) 243000 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 5 x 5 x 5

Grouping the factors in triplets of equal factors, 3 x 3 is left
∴ By dividing 3 x 3, we get a perfect cube
∴ Least number to be divided = 3 x 3=9

Question 13.
Prove that if a number is trebled then its cube is 27 times the cube of the given number.
Solution:
Let x be the number, then trebled number of x = 3x
Cubing, we get:
(3x)3 = (3)3 x3 = 27x3
27x3 is 27 times the cube of x i.e., of x3

Question 14.
What happenes to the cube of a number if the number is multiplied by
(i) 3 ?
(ii) 4 ?
(iii) 5 ?
Solution:
number (x)3 = x3
(i) If x is multiplied by 3, then the cube of
∴ (3x)3 = (3)3 x x3 = 27x3
∴ The cube of the resulting number is 27 times of cube of the given number
(ii) If x is multiplied by 4, then the cube of
(4x)3 = (4)3 x x3 = 64x3
∴ The cube of the resulting number is 64 times of the cube of the given number
(ii) If x is multiplied by 5, then the cube of
(5x)3 = (5)3 x x3 = 125x3
∴ The cube of the resulting number is 125 times of the cube of the given number

Question 15.
Find the volume of a cube, one face of which has an area of 64 m2.
Solution:
Area of one face of a cube = 64 m2
∴ Side (edge) of cube = √64
= √64 = 8 m
∴ Volume of the cube = (side)3 = (8 m)3 = 512 m3

Question 16.
Find the volume of a cube whose surface area is 384 m2.
Solution:
Surface area of a cube = 384 m2 Let side = a
Then 6a2 = 384 ⇒ a2 = $$\frac {384 }{ 6 }$$= 64 = (8)2
∴ a = 8 m
Now volume = a3 = (8)3 m3 = 512 m3

Question 17.
Evaluate the following :

Solution:

Question 18.
Write the units digit of the cube of each of the following numbers :
31,109,388,833,4276,5922,77774,44447, 125125125.
Solution:
We know that if unit digit of a number n is
= 1, then units digit of its cube = 1
= 2, then units digit of its cube = 8
= 3, then units digit of its cube = 7
= 4, then units digit of its cube = 4
= 5, then units digit of its cube = 5
= 6, then units digit of its cube = 6
= 7, then the units digit of its cube = 3
= 8, then units digit of its cube = 2
= 9, then units digit of its cube = 9
= 0, then units digit of its cube = 0
Now units digit of the cube of 31 = 1
Units digit of the cube of 109 = 9
Units digits of the cube of 388 = 2
Units digits of the cube of 833 = 7
Units digits of the cube of 4276 = 6
Units digit of the cube of 5922 = 8
Units digit of the cube of 77774 = 4
Units digit of tl. cube of 44447 = 3
Units digit of the cube of 125125125 = 5

Question 19.
Find the cubes of the following numbers by column method :
(i) 35
(ii) 56
(iii) 72
Solution:

Question 20.
Which of the following numbers are not perfect cubes ?
(i) 64
(ii) 216
(iii) 243
(iv) 1728
Solution:
(i) 64 = 2 x 2 x 2 X 2 x 2 x 2

Grouping the factors in triplets, of equal factors, we see that no factor is left
∴ 64 is a perfect cube.
(ii) 216 = 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets, of equal factors, we see that no factor is left
∴ 216 is a perfect cube.
(iii) 243 = 3 x 3 x 3 x 3 x 3

Grouping the factors in triplets, of equal factors, we see that 3 x 3 are left
∴ 243 is not a perfect cube.
(iv) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors m triplets, of equal factors, we see that no factor is left.
∴ 1728 is a perfect cube.

Question 21.
For each of the non-perfect cubes, in Q. 20, find the smallest number by which it must be
(a) multiplied so that the product is a perfect cube.
(b) divided so that the quotient is a perfect cube.
Solution:
In qustion 20, 243 is not a perfect cube and 243 = 3 x 3 x 3 x 3 x 3
Grouping the factors in triplets, of equal factors, we see that 3 x 3 is left.
(a) In order to make it a perfect cube, 3 is to be multiplied which makes a triplet.
(b) In order to make it a perfect cube, 3 x 3 or 9 is to be divided.

Question 22.
By taking three different values of n verify the truth of the following statements :
(i) If n is even, then n3 is also even.
(ii) If n is odd, then n3 is also odd.
(iii) If n leaves remainder 1 when divided by 3, then it3 also leaves 1 as remainder when divided by 3.
(iv) If a natural number n is of the form 3p + 2 then n3 also a number of the same type.
Solution:
(i) n is even number.
Let n = 2, 4, 6 then
(a) n3 = (2)3 = 2 x 2 x 2 = 8, which is an even number.
(b) (n)3= (4)3 = 4 x 4 x 4 = 64, which is an even number.
(c) (n)3 = (6)3 = 6 x 6 x 6 = 216, which is an even number.

(ii) n is odd number.
Letx = 3, 5, 7
(a) (n)3 = (3)3 = 3 x 3 x 3 = 27, which is an odd number.
(b) (n)3 = (5)3 = 5 x 5 x 5 = 125, which is an odd number.
(c) (n)3 = (7)3 = 7 x 7 x 7 = 343, which is an odd number.

(iii) If n leaves remainder 1 when divided by 3, then n3 is also leaves 1 as remainder,
Let n = 4, 7, 10 If n = 4,
then «3 = (4)3 = 4 x 4 x 4 = 64
= 64 + 3 = 21, remainder = 1
If n = 7, then
n3 = (7)3 = 7 x 7 x 7 = 343
343 + 3 = 114, remainder = 1
If n – 10, then
(n)3 = (10)3 = 10 x 10 x 10 = 1000
1000 + 3 = 333, remainder = 1

(iv) If the natural number is of the form 3p + 2, then n3 is also of the same type
Let p =’1, 2, 3, then
(a) If p = 1, then
n = 3p + 2 = 3 x 1+2=3+2=5
∴ n3 = (5)3 = 5 x 5 x 5 = 125
125 = 3 x 41 + 2 = 3p +2

(b) If p = 2, then
n = 3p + 2 = 3 x 2 + 2 = 6 + 2 = 8
∴ n3 = (8)3 = 8 x 8 x 8 = 512
∴ 512 = 3 x 170 + 2 = 3p + 2

(c) If p = 3, then
n = 3p + 2 = 3 x 3 + 2 = 9 + 2 = 11
∴ n3 = (11)3 = 11 x 11 x 11 = 1331
and 1331 =3 x 443 + 2 = 3p + 2
Hence proved.

Question 23.
Write true (T) or false (F) for the following statements :
(i) 0 392 is a perfect cube.
(ii) 8640 is not a perfect cube.
(iii) No cube can end with exactly two zeros.
(iv) There is no perfect cube which ends in 4.
(v) For an integer a, a3 is always greater than a> b2
(vi) If a and b are integers such that a> b2 , then a3 > b3.
(x) If a2 ends in an even number of zeros, then a ends in an odd number of zeros.
Solution:

∵ 5 is left.
(iii)True : A number ending three zeros can be a perfect cube.
(iv) False : v (4)3 = 4 x 4 x 4 = 64, which ends with 4.
(v) False : If n is a proper fraction, it is not possible.
(vi) False : It is not true if a and b are proper fraction.
(vii) True.
(viii) False : as a2 ends in 9, then a3 does not necessarily ends in 7. It ends in 3 also.
(ix) False : it is not necessarily that a3 ends in 25 it can end also in 75.
(x) False : If a2 ends with even zeros, then a3 will ends with odd zeros but of multiple of 3.

Hope given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1 are helpful to complete your math homework.

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## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.9

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.9

Other Exercises

Using square root table, find the square roots of the following :

Question 1.
7
Solution:

Question 2.
15
Solution:

Question 3.
74
Solution:

Question 4.
82
Solution:

Question 5.
198
Solution:

Question 6.
540
Solution:

Question 7.
8700
Solution:

Question 8.
3509
Solution:

Question 9.
6929
Solution:

Question 10.
25725
Solution:

Question 11.
1312
Solution:

Question 12.
4192
Solution:

Question 13.
4955
Solution:

Question 14.
$$\frac {99 }{ 144 }$$
Solution:

Question 15.
$$\frac {57 }{ 169 }$$
Solution:

Question 16.
$$\frac {101 }{ 169 }$$
Solution:

Question 17.
13.21
Solution:

Question 18.
21.97
Solution:

Question 19.
110
Solution:

Question 20.
1110
Solution:

Question 21.
11.11
Solution:

Question 22.
The area of a square field is 325 m2. Find the approximate length of one side of the field.
Solution:

Question 23.
Find the length of a side of a square, whose area is equal to the area bf the rectangle with sides 240 m and 70 m.
Solution:

Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.9 are helpful to complete your math homework.

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## RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.8

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.8

Other Exercises

Question 1.
Find the square root of each of the following correct to three places of decimal :
(i) 5
(ii) 7
(iii) 17
(iv) 20
(v) 66
(vi) 427
(vii) 1.7
(vii’) 23.1
(ix) 2.5
(x) 237.615
(xi) 15.3215
(xii) 0.9
(xiii) 0.1
(xiv) 0.016
(xv) 0.00064
(xvi) 0.019
(xvii) $$\frac {7 }{ 8 }$$
(xviii) $$\frac {5 }{ 12 }$$
(xix) 2 $$\frac {1 }{ 2 }$$
(xx) 287 $$\frac {8 }{ 8 }$$
Solution:

Question 2.
Find the square root of 12.0068 correct to four decimal places.
Solution:

Question 3.
Find the square root of 11 correct to five decimal places.
Solution:

Question 4.
Given that √2, = 1-414, √3 = 1.732, √5 = 2.236 and √7 = 2.646. Evaluate each of the following :

Solution:

Question 5.
Given that √2 = 1-414, √3 = 1-732, √5= 2.236 and √7= 2.646, find the square roots of the following :

Solution:

Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.8 are helpful to complete your math homework.

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