Yearly Archives: 2022

Exercise 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 (A¯ ) = 0.85
∴ P (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) = 36 = 12
(ii) Let B be the favourable occurrence which are 2 or 4
∴ P(B) = 26 = 13
(iii) Let C be the favourable occurrence which are multiple of 2 or 3 i.e., 2, 3, 4, 6.
∴ P(C) = 46 = 23

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) = 536
(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) = 636 = 16
(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) = 336 = 112
(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) = 336 = 112
(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) = 636 = 16
(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) = 1836 = 12
(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) = 1136
(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) = 3036 = 56
(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) = 1036 = 518
(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) = 1536 = 56
(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) = 1536 = 512
(xii) Let L be the occurrence of favourable events such that at least P (L) one is black card
∴ P(L) = 2652 = 12
(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) = 2536

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) = 38
(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) = 48 = 12
(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) = 68 = 34
(iv) Let D be the occurrence of favourable events in which there is no tail which is only 1 (HHH)
∴ P(D) = 18

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) = 252 = 126
(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) = 2852 = 713
(iii) Let C be the occurrence of favourable events such that it is black and a king which can be 2.
∴ P(C) = 252 = 126
(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) = 1252 = 313
(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) = 3652 = 913
(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) = 1652 = 413
(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) = 4452 = 1113
(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) = 2452 = 613
(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) = 4852 = 1213
(x) Let J be the occurrence of favourable event such that it is ten
∴ Number of events = 4
∴ P(J) = 452 = 113
(xi) Let K be the occurrence of favourable event such that it is a spade.
∴ Number of events =13
∴ P(K) = 1352 = 14
(xii) Let L be the occurrence of favourable event such that it is a black card.
∴ Number of events = 26
∴ P(L) = 2652 = 12
(xiii) Let M be the occurrence of favourable event such that it is the seven of clubs.
∴ Number of events = 1
∴ P(M) = 152
(xiv) Let N be the occurrence of favourable event such that it is a jack.
∴ Number of events = 1
∴ P(N) = 452 = 113
(xv) Let O be the occurrence of favourable event such that it is an ace of spades.
∴ Number of events = 1
∴ P(O) = 152
(xvi) Let Q be the occurrence of favourable event such that it is a queen.
∴ VNumber of events = 4
∴ P(P) = 452 = 113
(xvii) Let R be the occurrence of favourable event such that it is a heart card.
∴ Number of events =13
∴ P(P) = 1352 = 14
(xviii) Let S be the occurrence of favourable event such that it is a red card
∴ Number of events = 26
∴ P(P) = 2652 = 12

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) = 818 = 49

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) = 412 = 13
(ii) Let B be the favourable event such that it is a red ball.
∴ P(B) = 312 = 14
(iii) Let C be the favourable event such that it is a black ball.
∴ P(C) = 512
(iv) Let D be the favourable event such that it is not red.
∴ Number of favourable events = 5 + 4 = 9
∴ P(D) = 912 = 34

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) = 315 = 15

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) = 1018 = 59

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) = 512

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) = 615 = 25
(ii) Let B be the favourable event such that it is a red
∴ P(B) = 415
(iii) Let C be the favourable event such that it is not black.
∴ Number of favourable events = 4 + 6=10
∴ P(C) = 1015 = 23

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) = 38
(ii) Let B be the favourable event such that it is black.
∴ P(B) = 58

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) = 57
(ii) Let B be the favourable event such that it is white
Then P (B) = 87
(iii) Let C be the favourable event such that it is not green.
∴ Number of favourable events = 5 + 8 = 13
∴ P(C) = 1317

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) = 2126
(iii) Let B be the favourable event such that it is a vowel.
∴ P(B) = 526

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) = 1520 = 34

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) = 69 = 23
(ii) Let B be the favourable outcome which is a black pair.
∴ P(B) = 39 = 13

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) = 35
∴ 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) = 1836 = 12
(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) = 636 = 16

Exercise 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
10+152 = 252 =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 20+252= 452 = 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

Exercise 19.1

Question 1.
What is the least number of planes that can enclose a solid ? What is the name of the solid ?
Solution:
The least number of planes that can enclose a solid is called a Tetrahedron.

Question 2.
Can a polyhedron have for its faces :
(i) three triangles ?
(ii) four triangles ?
(iii) a square and four triangles ?
Solution:
(i) No, polyhedron has three faces.
(ii) Yes, tetrahedron has four triangles as its faces.
(iii) Yes, a square pyramid has a square as its base and four triangles as its faces.

Question 3.
Is it possible to have a polyhedron with any given number of faces ?
Solution:
Yes, it is possible if the number of faces is 4 or more.

Question 4.
Is a square prism same as a cube ?
Solution:
Yes, a square prism is a cube.

Question 5.
Can a polyhedron have 10 faces, 20 edges and 15 vertices ?
Solution:
No, it is not possible as By Euler’s formula
F + V = E + 2
⇒ 10 + 15 = 20 + 2
⇒ 25 = 22
Which is not possible

Question 6.
Verify Euler’s formula for each of the following polyhedrons :

Solution:
(i) In this polyhedron,
Number of faces (F) = 7
Number of edges (E) = 15
Number of vertices (V) = 10
According to Euler’s formula,
F + V = E + 2
⇒ 7 + 10 = 15 + 2
⇒ 17 = 17
Which is true.
(ii) In this polyhedron,
Number of faces (F) = 9
Number of edges (E) = 16
Number of vertices (V) = 9
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 9 = 16 + 2
⇒ 18 = 18
Which is true.
(iii) In this polyhedron,
Number of faces (F) = 9
Number of edges (E) =18
Number of vertices (V) = 11
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 11 = 18 + 2
⇒ 20 = 20
Which is true.
(iv) In this polyhedron,
Number of faces (F) = 5
Number of edges (E) = 8
Number of vertices (V) = 5
According to Euler’s formula,
F + V = E + 2
⇒ 5 + 5 = 8 + 2
⇒ 10 = 10
Which is true.
(v) In the given polyhedron,
Number of faces (F) = 9
Number of edges (E) = 16
Number of vertices (V) = 9
According to Euler’s formula,
F + V = E + 2
⇒ 9 + 9 = 16 + 2
⇒ 18 = 18
Which is true.

Question 7.
Using Euler’s formula, find the unknown:

Solution:
We know that Euler’s formula is
F + V = E + 2
(i) F + 6 = 12 + 2
⇒ F + 6 = 14
⇒ F = 14 – 6 = 8
Faces = 8
(ii) F + V = E + 2
⇒ 5 + V = 9 + 2
⇒ 5 + V = 11
⇒ V = 11 – 5 = 6
Vertices = 6
(iii) F + V = E + 2
⇒ 20 + 12 = E + 2
⇒ 32 = E + 2
⇒ E = 32 – 2 = 30
Edges = 30

Exercise 19.2

Question 1.
Which among the following are nets for a cube ?

Solution:
Nets for a cube are (ii), (iv) and (vi)

Question 2.
Name the polyhedron that can be made by folding each net:

Solution:
(i) This net is for a square

(ii) This net is for triangular prism.

(iii) This net is for triangular prism.

(iv) This net is for hexagonal prism.

(v) This net is for hexagon pyramid.

(vi) This net is for cuboid.

Question 3.
Dice are cubes where the numbers on the opposite faces must total 7. Which of the following are dice ?


Solution:
Figure (i) shows the net of cube or dice.

Question 4.
Draw nets for each of the following polyhedrons:


Solution:
(i) Net for cube is given below :

(ii) Net of a triangular prism is as under :

(iii) Net of hexagonal prism is as under :

(iv) The net for pentagonal pyramid is as under:

Question 5.
Match the following figures:


Solution:
(a) (iv)
(b) (i)
(c) (ii)
(d) (iii)

Exercise 16.1

Question 1.
Define the following terms:
(i) Quadrilateral
(ii) Convex Quadrilateral.
Solution:
(i) Quadrilateral: A closed figure made of four line segments is called a quadrilateral such that:

(a) no three points of them are collinear
(b) the line segments do not intersect except at their ends points.
(ii) Convex quadrilateral: A quadrilateral is called a convex quadrilateral of the line containing any side of the quadrilateral has the remaining vertices on the same side of it. In the figure, quadrilateral ABCD is a convex quadrilateral.

Question 2.
In a quadrilateral, define each of the following:
(i) Sides
(ii) Vertices
(iii) Angles
(iv) Diagonals
(v) Adjacent angles
(vi) Adjacent sides
(vii) Opposite sides
(viii) Opposite angles
(ix) Interior
(x) Exterior
Solution:
(i) Sides: In a quadrilateral ABCD, form line segments AB, BC, CD and DA are called sides of the quadrilateral.
(ii) Vertices : The ends points are called the vertices of the quadrilateral. Here in the figure, A, B, C and D are its vertices.
(iii) Angles: A quadrilateral has four angles which are at their vertices. In the figure, ∠A, ∠B, ∠C and ∠D are its angles.
(iv) Diagonals: The line segment joining the opposite vertices is called diagonal. A quadrilateral has two diagonals.
(v) Adjacent Angles : The angles having a common arm (side) are called adjacent angles.
(vi) Adjacent sides : If two sides of a quadrilateral have a common end-point, these are called adjacent sides.
(vii) Opposite sides: If two sides do not have a common end-point of a quadrilateral, they are called opposite sides.
(viii) Opposite angles : The angles which are not adjacent are called opposite angles.
(ix) Interior: The region which is surrounded by the sides of the quadrilateral is called its interior.
(x) Exterior : The part of the plane made up by all points as the not enclosed by the quadrilateral, is called its exterior.

Question 3.
Complete each of the following, so as to make a true statement:
(i) A quadrilateral has ………… sides.
(ii) A quadrilateral has ………… angles.
(iii) A quadrilateral has ……….. vertices, no three of which are …………
(iv) A quadrilateral has …………. diagonals.
(v) The number of pairs of adjacent angles of a quadrilateral is ………….
(vi) The number of pairs of opposite angles of a quadrilateral is ……………
(vii) The sum of the angles of a quadrilateral is …………
(viii) A diagonal of a quadrilateral is a line segment that joins two ………. vertices of the quadrilateral.
(ix) The sum of the angles of a quadrilateral is …………. right angles.
(x) The measure of each angle of a convex quadrilateral is …………. 180°.
(xi) In a quadrilateral the point of intersection of the diagonals lies in ………….. of the quadrilateral.
(xii) A point is in the interior of a convex quadrilateral, if it is in the ……….. of its two opposite angles.
(xiii) A quadrilateral is convex if for each side, the remaining …………. lie on the same side of the line containing the side.
Solution:
(i) A quadrilateral has four sides.
(a) A quadrilateral has four angles.
(iii) A quadrilateral has four vertices, no three of which are collinear .
(iv) A quadrilateral has two diagonals.
(v) The number of pairs of adjacent angles of a quadrilateral is four .
(vi) The number of pairs of opposite angles ot a quadrilateral is two.
(vii) The sum of the angles of a quadrilateral is 360°.
(viii) A diagonal of a quadrilateral is a line segment that join two opposite vertices of the quadrilateral.
(ix) The sum of the angles of a quadrilateral is 4 right angles.
(x) The measure of each angle of a convex quadrilateral is less than 180°.
(xi) In a quadrilateral the point of intersection of the diagonals lies in interior of the quadrilateral.
(xii) A point is in the interior of a convex quadrilateral, if it is in the interior of its two opposite angles.
(xiii) A quadrilateral is convex if for each side, the remaining vertices lie on the same side of the line containing the side.

Question 4.
In the figure, ABCD is a quadrilateral.
(i) Name a pair of adjacent sides.
(ii) Name a pair of opposite sides.
(iii) How many pairs of adjacent sides are there?
(iv) How many pairs of Opposite sides are there ?

(v) Name a pair of adjacent angles.
(vi) Name a pair of opposite angles.
(vii) How many pairs of adjacent angles are there ?
(viii) How many pairs of opposite angles are there ?
Solution:
In the figure, ABCD is a quadrilateral
(i) Pairs of adjacent sides are AB, BC, BC, CD, CD, DA, DA, AB.
(ii) Pairs of opposite sides are AB and CD; BC and AD.
(iii) There are four pairs of adjacent sides.
(iv) There are two pairs of opposite sides.
(v) Pairs of adjacent angles are ∠A, ∠B; ∠B, ∠C; ∠C, ∠D; ∠D, ∠A.
(vi) Pairs of opposite angles are ∠A and ∠C; ∠B and ∠D.
(vii) There are four pairs of adjacent angles.
(viii) There are two pairs of opposite angles.

Question 5.
The angles of a quadrilateral are 110°, 72°, 55° and x°. Find the value of x.
Solution:
Sum of four angles of quadrilateral is 360°
110° + 12° + 55° + x° = 360°
⇒ 237° + x° = 360°
⇒ x° = 360° – 237° = 123°
x = 123°

Question 6.
The three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.
Solution:
The sum of four angles of a quadrilateral = 360°
Three angles are 110°, 50° and 40°
Let fourth angle = x
Then 110° + 50° + 40° + x° = 360°
⇒ 200° + x° = 360°
⇒ x = 360° – 200° = 160°
x = 160°

Question 7.
A quadrilateral has three acute angles each measures 80°. What is the measure of fourth angle ?
Solution:
Sum of four angles of a quadrilateral = 360°
Sum of three angles having each angle equal to 80° = 80° x 3 = 240°
Let fourth angle = x
Then 240° + x = 360°
⇒ x° = 360° – 240°
⇒ x° = 120°
Fourth angle = 120°

Question 8.
A quadrilateral has all its four angles of the same measure. What is the measure of each ?
Solution:
Let each equal angle of a quadrilateral = x
4x° = 360°
⇒ x° = 3604 = 90°
Each angle will be = 90°

Question 9.
Two angles of a quadrilateral are of measure 65° and the other two angles are equal. What is the measure of each of these two angles ?
Solution:
Measures of two angles each = 65°
Sum of these two angles = 2 x 65°= 130°
But sum of four angles of a quadrilateral = 360°
Sum of the remaining two angles = 360° – 130° = 230°
But these are equal to each other
Measure of each angle = 2302 = 115°

Question 10.
Three angles of a quadrilateral are equal. Fourth angle is of measure 150°. What is the measure of equal angles ?
Solution:
Sum of four angles of a quadrilateral = 360°
One angle = 150°
Sum of remaining three angles = 360° – 150° = 210°
But these three angles are equal
Measure of each angle = 2103 = 70°

Question 11.
The four angles of a quadrilateral are as 3 : 5 : 7 : 9. Find the angles.
Solution:
Sum of four angles of a quadrilateral = 360°
and ratio in angles = 3 : 5 : 7 : 9
Let first angles = 2x
Then second angle = 5x
third angle = 7x
and fourth angle = 9x
3x + 5x + 7x + 9x = 360°
⇒ 24x = 369°
⇒ x = 36024 = 15°
First angle = 3x = 3 x 15° = 45°
second angle = 5x = 5 x 15° = 75°
third angle = 7x = 7 x 15° = 105°
and fourth angle = 9x = 9 x 15° = 135°

Question 12.
If the sum of the two angles of a quadrilateral is 180°, what is the sum of the remaining two angles ?
Solution:
Sum of four angles of a quadrilateral = 360°
and sum of two angle out of these = 180°
Sum of other two angles will be = 360° – 180° = 180°

Question 13.
In the figure, find the measure of ∠MPN.

Solution:
In the figure, OMPN is a quadrilateral in which
∠O = 45°, ∠M = ∠N = 90° (PM ⊥ OA and PN ⊥ OB)
Let ∠MPN = x°
∠O + ∠M + ∠N + ∠MPN = 360° (Sum of angles of a quadrilateral)
⇒ 45° + 90° + 90° + x° = 360°
⇒ 225° + x° = 360°
⇒ x° = 360° – 225°
⇒x = 135°
∠MPN = 135°

Question 14.
The sides of a quadrilateral are produced in order. What is the sum of the four exterior angles ?
Solution:
The sides of a quadrilateral ABCD are produced in order, forming exterior angles ∠1, ∠2, ∠3 and ∠4.

Now ∠DAB + ∠1 = 180° (Linear pair) ……(i)
Similarly,
∠ABC + ∠2 = 180°
∠BCD + ∠3 = 180°
and ∠CDA + ∠4 = 180°
Adding, we get
∠DAB + ∠1 + ∠ABC + ∠2 + ∠BCD + ∠3 + ∠CDA + ∠4 = 180° + 180° + 180° + 180° = 720°
⇒ ∠DAB + ∠ABC + ∠CDA + ∠ADC + ∠1 + ∠2 + ∠3 + ∠4 = 720°
But ∠DAB + ∠ABC + ∠CDA + ∠ADB = 360° (Sum of angles of a quadrilateral)
360° + ∠1 + ∠2 + ∠3 + ∠4 = 720°
⇒ ∠l + ∠2 + ∠3 + ∠4 = 720° – 360° = 360°
Sum of exterior angles = 360°

Question 15.
In the figure, the bisectors of ∠A and ∠B meet at a point P. If ∠C = 100° and ∠D = 50°, find the measure of ∠APB.
Solution:
In quadrilateral ABCD,

∠D = 50°, ∠C = 100°
PA and PB are the bisectors of ∠A and ∠B.
In quadrilateral ABCD,
∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)
⇒ ∠A + ∠B + 100° + 50° = 360°
⇒ ∠A + ∠B + 150° = 360°’
⇒ ∠A + ∠B = 360° – 150° = 210°
and 12 ∠A + 12 ∠B = 2102 = 105°
(PA and PB are bisector of ∠A and ∠B respectively)
∠PAB + ∠PBA = 105°
⇒ ∠PAB + ∠PBA + ∠APB = 180° (Sum of angles of a triangle)
⇒ 105° + ∠APB = 180°
⇒ ∠APB = 180° – 105° = 75°
∠APB = 75°

Question 16.
In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measure of each angle of the quadrilateral.
Solution:
Sum of angles A, B, C and D of a quadrilateral = 360°
i.e. ∠A + ∠B + ∠C + ∠D = 360°
But ∠A = ∠B = ∠C = ∠D = 1 : 2 : 4 : 5

Let ∠A = x,
Then ∠B = 2x
∠C = 4x
∠D = 5x
x + 2x + 4x + 5x = 360°
⇒ 12x = 360°
⇒ x = 36012 = 30°
∠A = x = 30°
∠B = 2x = 2 x 30° = 60°
∠C = 4x = 4 x 30° = 120°
∠D = 5A = 5 x 30° = 150°

Question 17.
In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that ∠COD = 12 (∠A + ∠B).
Solution:
In quadrilateral ABCD,

Question 18.
Find the number of sides of a regular polygon when each of its angles has a measures of
(i) 160°
(ii) 135°
(iii) 175°
(iv) 162°
(v) 150°.
Solution:
In a n-sided regular polygon, each angle

Question 19.
Find the number of degrees in each exterior angle of a regular pentagon.
Solution:
In a pentagon or a polygon, sum of exterior angles formed by producing the sides in order, is four right angles or 360°
Each exterior angle = 3605 = 72°

Question 20.
The measure of angles of a hexagon are x°, (x – 5)° (x – 5)°, (2x – 5)°, (2x – 5)°, (2x + 20)°. Find the value of x.
Solution:
We know that the sum of interior angels of a hexagon = 720° (180° x 4)
⇒ x + x – 5 + x – 5 + 2x – 5 + 2x – 5 + 2x + 20 = 720°
⇒ 9x – 20 + 20 = 720
⇒ 9x = 720
⇒ x = 7209 = 80°
x = 80°

Question 21.
In a convex hexagon, prove that the sum of all interior angles is equal to twice the sum of its exterior angles formed by producing the sides in the same order.
Solution:
In a convex hexagon ABCDEF, its sides AB, BG, CD, DE, EF and FA are produced in order forming exterior angles a, b, c, d, e, f

∠a + ∠b + ∠c + ∠d + ∠e + ∠f = 4 right angles (By definition)
By joining AC, AD, and AE, 4 triangles ABC, ACD, ADE and AEF are formed
In ∆ABC,
∠1 + ∠2 + ∠3 = 180° = 2 right angle (Sum of angles of a triangle) …… (i)
Similarly,
In ∆ACD,
∠4 +∠5 + ∠6 = 180° = 2 right angles
In ∆ADE,
∠1 + ∠8 + ∠9 = 2 right angles …(iii)
In ∆AEF,
∠10 + ∠11 + ∠12 = 2 right angles …(iv)
Joining (i), (ii), (iii) and (iv)
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 + ∠9 + ∠10 + ∠11 + ∠12 = 8 right angles
⇒ ∠2 + ∠3 + ∠5 + ∠6 + ∠8 + ∠9 + ∠11 + ∠12 + ∠1 + ∠4 + ∠7 + ∠10 = 8 right angles
⇒ ∠B + ∠C + ∠D + ∠E +∠F + ∠A = 8 right angles
⇒ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 2 (∠a + ∠b + ∠c + ∠d + ∠e + ∠f)
Sum of all interior angles = 2(the sum of exterior angles)
Hence proved.

Question 22.
The sum of the interior angles of a polygon is three times the sum of its exterior angles. Determine the number of sides of the polygon.
Solution:
Let number of sides of a regular polygon = n
Each interior angle = 2n–4n right angles
Sum of all interior angles = 2n–4n x n
right angles = (2n – 4) right angles
But sum of exterior angles = 4 right angles
According to the condition,
(2n – 4) = 3 x 4 (in right angles)
⇒ 2n – 4 = 12
⇒ 2n = 12 + 4 = 16
⇒ n = 8
Number of sides of the polygon = 8

Question 23.
Determine the number of sides of a polygon whose exterior and interior angles are in the ratio 1 : 5.
Solution:
Ratio in exterior angle and interior angles of a regular polygon = 1 : 5
But sum of interior and exterior angles = 180° (Linear pair)

By cross multiplication:
6n – 12 = 5n
⇒ 6n – 5n = 12
⇒ n = 12
Number of sides of polygon is 12

Question 24.
PQRSTU is a regular hexagon. Determine each angle of ∆PQT.
Solution:
In regular hexagon, PQRSTU, diagonals PT and QT are joined.

In ∆PUT, PU = UT
∠UPT = ∠UTP
But ∠UPT + ∠UTP = 180° – ∠U = 180° – 120° = 60°
∠UPT = ∠UTP = 30°
∠TPQ = 120° – 30° = 90° (QT is diagonal which bisect ∠Q and ∠T)
∠PQT = 1202 = 60°
Now in ∆PQT,
∠TPQ + ∠PQT + ∠PTQ = 180° (Sum of angles of a triangle)
⇒ 90° + 60° + ∠PTQ = 180°
⇒ 150° + ∠PTQ = 180°
⇒ ∠PTQ = 180° – 150° = 30°
Hence in ∆PQT,
∠P = 90°, ∠Q = 60° and ∠T = 30°

Exercise 15.1

Question 1.
Draw rough diagrams to illustrate the following:
(i) Open curve
(ii) Closed curve
Solution:

Question 2.
Classify the following curves as open or closed.

Solution:
Open curves : (i), (iv) and (v) are open curves.
(ii) , (iii), and (vi) are closed curves.

Question 3.
Draw a polygon and shade its interior. Also draw its diagonals, if any.
Solution:
In the given polygon, the shaded portion is its interior region AC and BD are the diagonals of polygon ABCD.

Question 4.
Illustrate, if possible, each one of the following with a rough diagram:
(i) A closed curve that is not a polygon.
(ii) An open curve made up entirely of line segments.
(iii) A polygon with two sides.
Solution:
(i) Close curve but not a polygon.

(ii) An open curve made up entirely of line segments.

(iii) A polygon with two sides. It is not possible. At least three sides are necessary

Question 5.
Following are some figures : Classify each of these figures on the basis of the following:


(i) Simple curve
(ii) Simple closed curve
(iii) Polygon
(iv) Convex polygon
(v) Concave polygon
(vi) Not a curve
Solution:
(i) It is a simple closed curve and a concave polygon.
(ii) It is a simple closed curve and convex polygon.
(iii) It is neither a curve nor polygon.
(iv) it is neither a curve not a polygon.
(v) It is a simple closed curve but not a polygon.
(vi) It is a simple closed curve but not a polygon.
(vii) It is a simple closed curve but not a polygon.
(viii) It is a simple closed curve but not a polygon.

Question 6.
How many diagonals does each of the following have ?
(i) A convex quadrilateral
(ii) A regular hexagon
(iii) A triangle.
Solution:
(i) A convex quadrilateral
Here n = 4

Question 7.
What is a regular polygon ? State the name of a regular polygon of:
(i) 3 sides
(ii) 4 sides
(iii) 6 sides.
Solution:
A regular polygon is a polygon which has all its sides equal and so all angles are equal,
(i) 3 sides : It is an equilateral triangle.
(ii) 4 sides : It is a square.
(iii) 6 sides : It is a hexagon.

Exercise 12.1

Question 1.
Write each of the following as percent: Solution—
(i) 725
(ii) 16625
(iii) 58
(iv) 0.8
(v) 0.005
(vi) 3 : 25
(vii) 11 : 80
(viii) 111 : 125
(ix) 13 : 75
(x) 15 : 16
(xi) 0.18
(xii) 7125
Solution:

Question 2.
Convert the following percentages to fractions and ratios :
(i) 25%
(ii) 2.5%
(iii) 0.25%
(iv) 0.3%
(v) 125%
Solution:

Question 3.
Express the following as decimal fractions :
(i) 27%
(ii) 6.3%
(iii) 32%
(iv) 0.25%
(v) 7.5%
(vi) 18 %
Solution:

Exercise 12.2

Question 1.
Find :
(i) 22% of 120
(ii) 25% of Rs. 1000
(iii) 25% of 10 kg
(iv) 16.5% of 5000 metres
(v) 135% of 80 cm
(vi) 2.5% of 10000 ml
Solution:

Question 2.
Find the number ‘a’, if
(i) 8.4% of a is 42
(ii) 0.5% of a is 3
(iii) 12 % of a is 50
(iv) 100% of a is 100
Solution:

Question 3.
x is 5% of y, y is 24% of z. If x = 480, find the values of y and z.
Solution:
x = 5% of y, y = 24% of z.
x = 480

Question 4.
A coolie deposits Rs. 150 per month in his post office Saving Bank account. If this is 15% of his monthly income, find his monthly income.
Solution:
Let his monthly income = Rs. x
15% of x = Rs. 150

Question 5.
Asha got 86.875% marks in the annual examination. If she got 695 marks, find the number of marks of the Examination.
Solution:
Let total marks of the examination = x
86.875% of x = 695
=> 86.875 x 1100 x x = 695

Question 6.
Deepti went to school for 216 days in a full year. If her attendance is 90%, find the number of days on which the school was opened ?
Solution:
Let the school opened for = x days = 90% of x = 216

Question 7.
A garden has 2000 trees. 12% of these are mango trees, 18% lemon and the rest are orange trees. Find the number of orange trees.
Solution:
Number of total trees = 2000

Rest trees = 2000 – (240 + 360) = 2000 – 600 = 1400
Number of orange trees = 1400

Question 8.
Balanced diet should contain 12% of protein, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in this food daily, find in calories the amount of each of these in his daily food in take.
Solution:
Balance diet contains
Protein = 12%
Fats = 25%
Carbohydrates = 63%
Number of total calories = 2600
Number of calories of proteins = 12% of 2600 = 12100 x 2600 = 312
Number of calories of fats = 25% of 2600 = 25100 x 2600 = 650
Number of calories of carbohydrates = 63% of 2600 = 63100 x 2600 = 1638

Question 9.
A cricketer scored a total of 62 runs in 96 balls. He hits 3 sixes, 8 fours, 2 twos and 8 singles. What percentage of the total runs came in :
(i) Sixes
(ii) Fours
(iii) Twos
(iv) Singles
Solution:
Total score of a cricketer = 62 runs
(z) Number of sixes = 3
Run from 3 sixes = 3 x 6 = 18
Percentage = 1862 x 100 = 29.03%
(ii) Number of fours = 8
Total run from 8 fours = 4 x 8 = 32
Percentage = 3262 x 100 = 51.61%
(iii) Number of twos = 2
Total score from 2 twos = 2 x 2 = 4
Percentage = 462 x 100 = 40062 = 6.45%
(iv) Number of single run = 8
Percentage = 862 x 100 = 80062 = 12.9%

Question 10.
A cricketer hits 120 runs in 150 balls during a test match. 20% of the runs came in 6’s, 30% in 4’s, 25% in 2’s and the rest in 1’s. How many runs did he score in :
(i) 6’s
(ii) 4’s
(iii) 2’s
(iv) singles
What % of his shots were scoring ones ?
Solution:
Total runs scored by a cricketer =120
(i) Number of runs from sixes (6’s) = 20% of 120

Question 11.
Radha earns 22% of her investment. If she earns Rs. 187, then how much did she invest ?
Solution:
Total earning from investment = Rs. 187
Percent earning = 22%
Let his investment = x
Then 22% of x = Rs. 187

Question 12.
Rohit deposits 12% his income in a bank. He deposited Rs. 1440 in the bank during 1997. What was his total income for the year 1997 ?
Solution:
Deposit in the bank = Rs. 1440
Percentage = 12% of his total income
Let his total income = Rs. x

Question 13.
Gunpowder contains 75% nitre and 10% sulphur. Find the amount of the gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.3 kg sulphur ?
Solution:
(i) In gunpowder,
Nitre = 75%
Sulphur = 10%
Let total amount of gunpowder = x kg
Nitre = 9 kg

Question 14.
An alloy of tin and copper consists of 15 parts of tin and 105 parts of copper. Find the percentage of copper in the alloy ?
Solution:
In an alloy,
Number of parts of tin = 15
and number of parts of copper = 105
Total parts = 15 + 105 = 120
Percentage of copper in the alloy = 105120 x 100 = 87.5%

Question 15.
An alloy contains 32% copper, 40% nickel and rest zinc. Find the mass of the zinc in 1 kg of the alloy.
Solution:
In an alloy,
Copper = 32%
Nickel = 40%
Rest is zinc = 100 – (32 + 40) = 100 – 72 = 28%
Mass of zinc in 1 kg = 28% of 1 kg = 28100 x 100 gm = 280 gm.

Question 16.
A motorist travelled 122 kilometres before his first stop. If he had 10% of his journey to complete at this point, how long was the total ride ?
Solution:
Distance travelled before first stop = 122 km
Let total journey = x km
10% of x = 122

Question 17.
A certain school has 300 students, 142 of whom are boys. It has 30 teachers, 12 of whom are men. What percent of the total number of students and teachers in the school is female ?
Solution:
Total numbers of teachers = 30
Number of male teachers = 12
Number of female teacher = 30 – 12 = 18
Percentage of female teachers = 18×10030 = 60%

Question 18.
Aman’s income is 20% less than that of Anil. How much percent is Anil’s income more than Aman’s income ?
Solution:
Let Anil’s income = Rs. 100
Then Aman’s income = Rs, 100 – 20 = Rs. 80
Now, difference of both’s incomes = 100 – 80 = Rs. 20
Anil income is Rs. 20 more than that of Aman’s
Percentage = 20×10080 = 25%

Question 19.
The value of a machine depreciates every year by 5%. If the present value of the machine be Rs. 100000, what will be its value after 2 years ?
Solution:
Present value of machine = Rs. 100000
Rate of depreciation per year = 5%
Period = 2 years
Value of machine after 2 years

Question 20.
The population of a town increases by 10% annually. If the present population is 60000, what will be its population after 2 years ?
Solution:
Present population of the town = 60000
Increase annually = 10%
Period = 2 years
Population after 2 years will be

Question 21.
The population of a town increases 10% annually. If the present population is 22000, find its population a year ago.
Solution:
Let the population of the town a year ago was = x
Increase in population = 10%

Question 22.
Ankit was given an increment of 10% on his salary. His new salary is Rs. 3575. What was his salary before increment ?
Solution:
Let the salary of Ankit before increment = x
Increment given = 10% of the salary
Salary after increment will be

Question 23.
In the new budget, the price of petrol rose by 10%. By how much percent must one reduce the consumption so that the expenditure does not increase ?
Solution:
Let price of petrol before budged = Rs. 100
Increase = 10%
Price after budget = Rs. 100 + 10 = Rs. 110
Let the consumption of petrol before budget = 100 l
Price pf 100 l = Rs. 110
Now of new price is Rs. 110, consumption = 100 l
are of new price will be 100, then

Question 24.
Mohan’s income is Rs. 15500 per month. He saves 11% of his income. If his income increases by 10% then he reduces his saving by 1%, how much does he save now ?
Solution:
Mohan’s income = Rs. 15500

We see that the savings is same
There is no change in savings.

Question 25.
Shikha’s income is 60% more than that of Shalu. What percent is Shalu’s income less than Shikha’s ?
Solution:
Let Shalu’s income = Rs. 100
Then Shikha’s income will be = Rs. 100 + 60 = Rs. 160
Now difference in their incomes = Rs. 160 – 100 = Rs. 60
Shalu’s income is less than Shikha’s income by Rs. 60
Percentage less = 60×100160 = 752 % = 37.5%

Question 26.
Rs. 3500 is to be shared among three people so that the first person gets 50% of the second who in turn gets 50% of the third. How much will each of them get ?
Solution:
Let the third person gets = Rs. x
Then second person will get

Question 27.
After a 20% hike, the cost of Chinese Vase is Rs. 2000. What was the original price of the object ?
Solution:
Let the original price of the vase = Rs. x
Hike in price = 20%

Original price of the vase = Rs. 1666.66

Exercise 11.1

Question 1.
Rakesh can do a piece of work in 20 days. How much work can he do in 4 days ?
Solution:
Rakesh can do it in 20 days = 1
his 1 day’s work = 120
and his 4 days work = 120 x 4 = 15 th work

Question 2.
Rohan can paint 13 of a painting in 6 days. How many days will he take to complete the painting ?
Solution:
Rohan can paint 13 of painting in = 6 days
he will complete the painting in = 6×31 = 18 days

Question 3.
Anil can do a piece of work in 5 days and Ankur in 4 days. How long will they take to do the same work, if they work together ?
Solution:
Anil’s 1 day’s work = 15

Question 4.
Mohan takes 9 hours to mow a large lawn. He and Sohan together can mow it in 4 hours. How long will Sohan take to mow the lawn if he works alone ?
Solution:

Question 5.
Sita can finish typing a 100 page document in 9 hours, Mita in 6 hours and Rita in 12 hours. How long will they take to type a 100 page document if they work together?
Solution:
Sita can do a work in 1 hour = 19

Question 6.
A, B and C working together can do a piece of work in 8 hours. A alone can do it in 20 hours and B alone can do it in 24 hours. In how many hours will C alone do the same work ?
Solution:

Question 7.
A and B can do a piece of work in 18 days; B and C in 24 days and A and C in 36 days. In what time can they do it, all working together ?
Solution:

Question 8.
A and B can do a piece of work in 12 days; B and C in 15 days; C and A in 20 days. How much time will A alone take to finish the work ?
Solution:

Question 9.
A, B and C can reap a field in 1534 days; B, C and D in 14 days; C, D and A in 18 days; D, A and B in 21 days. In what time can A, B, C and D together reap it ?
Solution:

Question 10.
A and B can polish the floors of a building in 10 days A alone can do 14 th of it in 12 days. In how many days can B alone polish the floor ?
Solution:

Question 11.
A and B can finish a work in 20 days. A alone can do 15 th of the work in 12 days. In how many days can B alone do it ?
Solution:

Question 12.
A and B can do a piece of work in 20 days and B in 15 days. They work together for 2 days and then A goes away. In how many days will B finish the remaining work ?
Solution:

Question 13.
A can do a piece of work in 40 days and B in 45 days. They work together for 10 days and then B goes away. In how many days will A finish the remaining work ?
Solution:

Question 14.
Aasheesh can paint his doll in 20 minutes and his sister Chinki can do so in 25 minutes. They paint the doll together for five minutes. At this juncture they have a quarrel and Chinki withdraws from painting. In how many minutes will Aasheesh finish the painting of the remaining doll ?
Solution:

Question 15.
A and B can do a piece of work in 6 days and 4 days respectively. A started the work; worked at it for 2 days and then was joined by B. Find the total time taken to complete the work.
Solution:

Question 16.
6 men can complete the electric fitting in a building in 7 days. How many days will it take if 21 men do the job ?
Solution:
6 men can complete the work in = 7 days
1 man will complete the same work in = 7 x 6 days (Less men, more days)
21 men will finish the work in = 7×621 days (More men, less days) = 2 days

Question 17.
8 men can do a piece of work in 9 days. In how many days will 6 men do it ?
Solution:
8 men can do a work in = 9 days
1 men will do the work in = 9 x 8 days (Less men, more days)
6 men will do the work in = 9×86 days (More men, less days)
= 726 = 12 days

Question 18.
Reema weaves 35 baskets in 25 days. In how many days will she weave 55 baskets?
Solution:
Reema can weave 35 baskets in = 25 days

Question 19.
Neha types 75 pages in 14 hours. How many pages will she type in 20 hours ?
Solution:
Neha types pages in 14 hours = 75 pages

Question 20.
If 12 boys earn Rs. 840 in 7 days, what will 15 boys earn in 6 days ?
Solution:
12 boys in 7 days earn an amount of = Rs. 840

Question 21.
If 25 men earn Rs. 1000 in 10 days, how much will 15 men earn in 15 days ?
Solution:
25 men can earn in 10 days = Rs. 1000

Question 22.
Working 8 hours a day, Ashu can copy a book in 18 days. How many hours a day should he work so as to finish the work in 12 days ?
Solution:
Ashu can copy a book in 18 days working in a day = 8 hours
He will copy the book in 1 day working = 8 x 18 hours a day (Less days, more hours a day)
He will copy the book in 12 days working in a day = 8×1812 hours
(More days, less hours a day)
= 14412 = 12 hours a day

Question 23.
If 9 girls can prepare 135 garlands in 3 hours, how many girls are needed to prepare 270 garlands in 1 hour.
Solution:
135 garlands in 3 hours are prepared by = 9 girls
1 garland in 3 hours will be prepared by

Question 24.
A cistern can be filled by one tap in 8 hours, and by another in 4 hours. How long will it take to fill the cistern if both taps are opened together ?
Solution:
First tap’s 1 hour work to fill the cistern = 18

Question 25.
Two taps A and B can fill an overhead tank in 10 hours and 15 hours respectively. Both the taps are opened for 4 hours and then B is turned off. How much time will A take to fill the remaining tank ?
Solution:

Question 26.
A pipe can fill a cistern in 10 hours. Due to a leak in the bottom, it is filled in 12 hours. When the cistern is full, in how much time will it be emptied by the leak?
Solution:

Question 27.
A cistern has two inlets A and B which can fill it in 12 hours and 15 hours respectively. An outlet can empty the full cistern in 10 hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely ?
Solution:

Question 28.
A cistern can be filled by a tap in 4 hours and emptied by an outlet pipe in 6 hours. How long will it take to fill the cistern of both the tap and the pipe are opened together ?
Solution:

Exercise 10.1

Question 1.
Explain the concept of direct variation.
Solution:
If two quantifies a and b vary with each other in such a way that the ratio ab remains constant and is positive, then we say that a and b vary directly with each other or a and b are in direct variation.

Question 2.
Which of the following quantities vary directly with each other ?
(i) Number of articles (x) and their price (y).
(ii) Weight of articles (x) and their cost (y).
(iii) Distance x and time y, speed remaining the same.
(iv) Wages (y) and number of hours (x) of work.
(v) Speed (x) and time (y) (distance covered remaining the same).
(vi) Area of a land (x) and its cost (y).
Solution:
(i) It is direct variation because more articles more price and less articles, less price.
(ii) It is direct variation because, more weight more price, less weight, less price.
(iii) It is not direct variation. The distance and time vqry indirectly or inversely.
(iv) It is direct variation as more hours, more wages, less hours, less wages.
(v) It is not direct variation, as more speed, less time, less speed, more time.
(vi) It is direct variation, as more area more cost, less area, less cost.
Hence (i), (ii), (iv) and (vi) are in direct variation.

Question 3.
In which of the following tables x and y vary directly ?

Solution:

All are different.
It is not in direct variation.
Hence (i) and (ii) are in direct variation.

Question 4.
Fill in the blanks in each of the following so as to make the statement true :
(i) Two quantities are said to vary ……….. with each other if they increase (decrease) together in such a way that the ratio of the corresponding values remains same.
(ii) x and y are said to vary directly with each other if for some positive number k = k.
(iii) If u = 3v, then u and v vary ……….. with each other.
Solution:
(i) Two quantities are said to vary directly with each other if they increase (decrease) together in such a way that the ratio of the corresponding values remains same.
(ii) x and y are said to vary directly with each other if for some positive number k, xy = k.
(iii) If u = 3v, then u and v vary directly with each other.

Question 5.
Complete the following tables given that x varies directly as y.

Solution:

Question 6.
Find the constant of variation from the table given below :

Solution:

Set up a table and solve the following problems. Use unitary method to verify the answer.
Question 7.
Rohit bought 12 registers for Rs. 156, find the cost of 7 such registers.
Solution:
Price of 12 registers = Rs. 156
Let cost of 7 registers = Rs. x. Therefore

Question 8.
Anupama takes 125 minutes in walking a distance of 100 metre. What distance would she cover in 315 minutes.
Solution:
For walking 100 m, time is taken = 125 minutes
Let in 315 minutes, distance covered = m
Therefore,

Question 9.
If the cost of 93 m of a certain kind of plastic sheet is Rs. 1395, then what would it cost to buy 105 m of such plastic sheet.
Solution:
Cost of 93 m of plastic sheet = Rs. 1395
Let cost of 105 m of such sheet = Rs. x
Therefore,

Question 10.
Suneeta types 1080 words in one hour. What is GWAM (gross words a minute rate) ?
Solution:
1080 words were typed in = 1 hour = 60 minutes
Let x words will be typed in 1 minute
Therefore,

Question 11.
A car is travelling at the average speed of 50 km/hr. How much distance would it travel in 12 minutes.
Solution:
Speed of car = 50 km/hr = 50 km in 60 minutes
Let it travel x km in 12 minutes. Therefore

Question 12.
68 boxes of a certain commodity require a shelf length of 13.6 m. How many boxes of the same commodity would occupy a shelf of 20.4 m ?
Solution:
For 68 boxes of certain commodity is required a shelf length of 13.6 m
Let x boxes are require for 20.4 m shelf Then

Question 13.
In a library 136 copies of a certain book require a shelf length of 3.4 metre. How many copies of the same book would occupy a shelf-length of 5.1 metres ?
Solution:
For 136 copies of books require a shelf of length = 3.4 m
For 5.1 m shelf, let books be required = x Therefore :

Question 14.
The second class railway fare for 240 km of journey is Rs. 15.00. What would be the fare for a journey of 139.2 km ?
Solution:
Fare of second class for 240 km = Rs. 15.00
Let fare for 139.2 km journey = Rs. x
Therefore :

Question 15.
If the thickness of a pile of 12 cardboards is 35 mm, find the thickness of a pile of 294 cardboards.
Solution:
Thickness of a pile of 12 cardboards = 35 mm.
Let the thickness of a pile of 294 cardboards = x mm
Therefore :

Question 16.
The cost of 97 metre of cloth is Rs. 242.50. What length of this can be purchased for Rs. 302.50 ?
Solution:
Cost of 97 m of cloth = Rs. 242.50
Let x m can be purchase for Rs. 302.50
Therefore :

Question 17.
men can dig 634 metre long trench in one day. How many men should be employed for digging 27 metre long trench of the same type in one day ?
Solution:
11 men can dig a trench = 634 m long
Let x men will dig a trench 27 m long.
Therefore,

Question 18.
A worker is paid Rs. 210 for 6 days work. If his total income of the month is Rs. 875, for how many days did he work ?
Solution:
Payment for 6 day’s work = Rs. 210
Let payment for x day’s work = Rs. 875
Therefore :

Question 19.
A worker is paid Rs. 200 for 8 days work. If he works for 20 days, how much will he get ?
Solution:
Labour for 8 days work = Rs. 200
Let x be the labour for 20 days work, then

Question 20.
The amount of extension in an elastic string varies directly as the weight hung on it. If a weight of 150 gm produces an extension of 2.9 cm, then what weight would produce an extension of 17.4 cm ?
Solution:
150 gm of weight produces an extension = 2.9 cm
Let x gm of weight will produce an extension of 17.4 cm
Therefore :

Question 21.
The amount of extension in an elastic spring varies directly with the weight hung on it. If a weight of 250 gm produces an extension of 3.5 cm, find the extension produced by the weight of 700 gm.
Solution:
A weight of 250 gm produces an extension of 3.5 cm.
Let a weight of 700 gm will produce an extension of x cm. Therefore :

Question 22.
In 10 days, the earth picks up 2.6 x 108 pounds of dust from the atmosphere. How much dust will it pick up in 45 days.
Solution:
In 10 days dust is picked up = 2.6 x 108 pounds
Let x pounds of dust is picked up in = 45 days
Therefore,

Question 23.
In 15 days, the earth picks up 1.2 x 108 kg of dust from the atmosphere. In how many days it will pick up 4.8 x 10s kg of dust ?
Solution:
Dust of 1.2 x 108 kg is picked up in = 15 days
Let the dust of 4.8 x 108 will be picked up in x days
Therefore,

Exercise 10.2

Question 1.
In which of the following tables x and y vary inversely :



Solution:


We see that it in 15 x 4 and 3 x 25 are not equal to 36 others are 72
In it x and y do not vary.

Question 2.
It x and y vary inversely, fill in the following blanks :

Solution:

Question 3.
Which of the following quantities vary inversely as each other ?
(i) The number of x men hired to construct a wall and the time y taken to finish the job.
(ii) The length x of a journey by bus and price y of the ticket.
(iii) Journey (x km) undertaken by a car and the petrol (y litres) consumed by it.
Solution:
(i) Here x and’y var inversely
More men less time and more time less men.
(ii) More journey more price, less journey less price
x and y do not vary inversely.
(iii) More journey more petrol, less journey, less petrol
x and y do not vary inversely.
In (i) x and y, vary inversely.

Question 4.
It is known that for a given mass of gas, the volume v varies inversely as the pressure p. Fill in the missing entries in the following table :

Solution:

Question 5.
If 36 men can do a piece of work in 25 days, in how many days will 15 men do it ?
Solution:
Here less men, more days.
Let in x days, 15 men can finish the work
Therefore.

Question 6.
A work force of 50 men with a contractor can finish a piece of work in 5 months. In how many months the same work can be completed by 125 men.
Solution:
Let in x months, the work will be completed by 125 men

Question 7.
A work-force of 420 men with contractor can finish a certain piece of work in 9 months. How many extra men must he employ to complete the job in 7 months?
Solution:
Let total x men can finish the work in 7 months.
Therefore,


Total men = 540
Number of men already employed = 420
Extra men required = 540 – 420 = 120

Question 8.
1200 men can finish a stock of food in 35 days. How many more men should join them so that the same stock may last for 25 days ?
Solution:
Let x men can finish the stock, then

Total men required = 1680
Already men working = 1200
More men required = 1680 – 1200 = 480

Question 9.
In a hostel of 50 girls, there are food provisions for 40 days. If 30 more girls join the hostel. How long will these provisions last ?
Solution:
Number of girls in the beginning = 50
More girls joined = 30
Total number of girls = 50 + 30 = 80
Let the provisions last for x days.

Question 10.
A car can finish a certain journey in 10 hours at the speed of 48 km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance ?
Solution:
Let x km/hr be the speed. Then

Speed required = 60 km/hr.
Already speed = 48 km/hr
Speed to be increase = 60 – 48 = 12 km/hr

Question 11.
1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldiers were transferred to another fort and thus the food lasted now for 32 more days. How many soldiers left the fort ?
Solution:
Period = 28 days
After 4 day, the remaining period = 28 – 4 = 24 days
In the beginning number of soldiers in the fort = 1200
Period for which the food lasted = 32 days
Let for x soldier, the food was sufficient, then

Question 12.
Three spraying machines working together can finish painting a house in 60 minutes. How long will it take 5 machines of the same capacity to do the same job ?
Solution:
Let in x minutes, 5 machines can do the work
Now

Question 13.
A group of 3 friends staying together, consume 54 kg of wheat every month. Some more friends join this group and they find that the same amount of wheat lasts for 18 days. How new many numbers are there in this group now ?
Solution:
Let x members can finish the wheat in 18 day.
Therefore :


5 member can consume the wheat
Number of members already = 3
5 – 3 = 2 more member joined them.

Question 14.
55 cows can graze a field in 16 days. How many cows will graze the same field in 10 days ?
Solution:
Let number of cows required = x
Therefore :

Question 15.
18 men can reap a field in 35 days. For reaping the same field in 15 days, how many men are required ?
Solution:
Let x men are required,
Therefore,

Question 16.
A person has money to buy 25 cycles worth Rs. 500 each. How many cycles he will be able to buy if each cycle is costing Rs. 125 more ?
Solution:
Price of one cycle = Rs. 500
Number of cycle purchased = 25
New price of the cycle = Rs. 500 + Rs. 125 = Rs. 625
Let number of cycle will be purchase = x

Question 17.
Raghu has enough money to buy 75 machines worth Rs. 200 each. How many machines can he buy if he gets a discount of Rs. 50 on each machine ?
Solution:
Price of each machine = Rs. 200
Price after given discount of Rs. 50 = Rs. 200 – 50 = Rs. 150
Let machine can be purchase = x

Number of machines can be purchased = 100

Question 18.
If x and y vary inversely as each other and
(i) x = 3 when y = 8, find y when x = 4
(ii) x = 5 when y = 15, find x when y = 12
(iii) x = 30, find y when constant of variation = 900.
(iv) y = 35, find x when constant of variation = 7.
Solution:
x and y vary inversely
x x y is constant of variation
(i) x = 3, y = 8
Constant = xy = 3 x 8 = 24

Exercise 5.1

Question 1.
Without performing actual addition and division, write the quotient when the sum of 69 and 96. is divided by
(i) 11
(ii) 15
Solution:
Two numbers are 69 and 96 whose digits are reversed Here a = 6,= 9
(i) Sum if 69 + 96 is divisible by 11, then quotient = a + 6 = 6 + 9 = 15
(ii) If it is divided by a + b i.e., 6 + 9 = 15, then quotient = 11

Question 2.
Without performing actual computations, find the quotient when 94 – 49 is divided by
(i) 9
(ii) 5
Solution:
Two given numbers are 94 and 49. Whose digits are reversed.
(i) If 94 – 49 is divided by 9, then the quotient = a-b = 9-4 = 5
(ii) and when it is divided by a – b i.e. 9-4 = 5, then quotient will be = 9

Question 3.
If sum of the number 985 and two other numbers obtained by arranging the digits of 985 in cyclic order is divided by 111, 22 and 37 respectively. Find the quotient in each case.
Solution:
The given number is 985
The other two numbers by arranging its digits
in cyclic order, will be 859, 598 of the form
abc¯¯¯¯¯¯¯,bca¯¯¯¯¯¯¯,cba¯¯¯¯¯¯¯
Therefore,
If 985 + 859 + 598 is divided by 111, then quotient will bea + 6 + c = 9 + 8 + 5 = 22
If this sum is divided by 22, then the quotient = 111
and if it is divided by 37, then quotient = 3 (a + b + c) = 3 (22) = 66

Question 4.
Find the quotient when difference of 985 and 958 is divided by 9.
Solution:
The numbers of three digits are
985 and 958 in which tens and ones digits are reversed, then
abc¯¯¯¯¯¯¯−acb¯¯¯¯¯¯¯ = 9 (b – c)
985 – 958 = 9 (8 – 5) = 9 x 3
i. e., it is divisible by 9, then quotient = b-c =8-5=3

Exercise 5.2

Question 1.
Given that the number  35a64¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 3, where a is a digit, what are the possible volues of a ?
Solution:
The number 35a64¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 3
∵The sum of its digits will also be divisible by 3
∴ 3 + 5 + a + b + 4 is divisible by 3
⇒ 18 + a is divisible by 3
⇒ a is divisible by 3 (∵ 18 is divisible by 3)
∴ Values of a can be, 0, 3, 6, 9

Question 2.
If x is a digit such that the number 18×71¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 3,’ find possible values of x.
Solution:
∵ The number 18×71¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
is divisible by 3
∴ The sum of its digits will also be divisible by 3
⇒ l + 8+ x + 7 + 1 is divisible by 3
⇒ 17 + x is divisible by 3
The sum greater than 17, can be 18, 21, 24, 27…………
∴ x can be 1, 4, 7 which are divisible by 3.

Question 3.
If is a digit of the number 66784x¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ such that it is divisible by 9, find the possible values of x.
Solution:
∵ The number 66784 x is divisible by 9
∴ The sum of its digits will also be divisible by 9
⇒ 6+6+7+8+4+x is divisible by 9
⇒ 31 + x is divisible by 9
Sum greater than 31, are 36, 45, 54………
which are divisible by 9
∴ Values of x can be 5 on 9
∴ x = 5

Question 4.
Given that the number 67y19¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 9, where y is a digit, what are the possible values of y ?
Solution:
∵ The number 67y19¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 9
∴The sum of its digits will also be divisible by 9
⇒ 6 + 7+ y+ 1+ 9 is divisible by 9
⇒ 23 + y is divisible by 9
∴ The numbers greater than 23 are 27, 36, 45,……..
Which are divisible by 9
∴y = A

Question 5.
If 3×2¯¯¯¯¯¯¯¯ is a multiple of 11, where .v is a digit, what is the value of * ?
Solution:
∵ The number 3×2¯¯¯¯¯¯¯¯ is multiple of 11
∴ It is divisible by 11
∴ Difference of the sum of its alternate digits is zero or multiple of 11
∴ Difference of (2 + 3) and * is zero or multiple of 11
⇒ If x – (2 + 3) = 0 ⇒ x-5 = 0
Then x = 5

Question 6.
If 98125×2¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is a number with x as its tens digits such that it is divisible by 4. Find all the possible values of x.
Solution:
∵ The number 98125×2¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 4
∴ The number formed by tens digit and units digit will also be divisible by 4
∴ x2¯¯¯¯¯ is divisible by 4
∴ Possible number can be 12, 32, 52, 72, 92
∴ Value of x will be 1,3, 5, 7, 9

Question 7.
If x denotes the digit at hundreds place of the number 67×19¯¯¯¯¯¯¯¯¯¯¯¯¯ such that the
number is divisible by 11. Find all possible values of x.
Solution:
∵ The number 67×19¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 11
∴ The difference of the sums its alternate digits will be 0 or divisible by 11
∴ Difference of (9 + x + 6) and (1 + 7) is zero or divisible by 11
⇒ 15+x-8 = 0, or multiple of 11,
7 + x = 0 ⇒ x = -7, which is not possible
∴ 7 + x = 11, 7 + x = 22 etc.
⇒ x=11-7 = 4, x = 22 – 7
⇒ x = 15 which is not a digit
∴ x = 4

Question 8.
Find the remainder when 981547 is divided by 5. Do this without doing actual division.
Solution:
A number is divisible by 5 if its units digit is 0 or 5
But in number 981547, units digit is 7
∴ Dividing the number by 5,
Then remainder will be 7 – 5 = 2

Question 9.
Find the remainder when 51439786 is divided by 3. Do this without performing actual division.
Solution:
In the number 51439786, sum of digits is 5 + 1+ 4 + 3 + 9 + 7 + 8 + 6 = 43 and the given number is divided by 3.
∴ The sum of digits must by divisible by 3
∴ Dividing 43 by 3, the remainder will be = 1
Hence remainder = 1

Question 10.
Find the remainder, without performing actual division when 798 is divided by 11.
Solution:
Let n = 798 = a multiple of 11 + [7 + 8 – 9] 798 = a multiple of 11 + 6
∴ Remainder = 6

Question 11.
Without performing actual division, find the remainder when 928174653 is divided by 11.
Solution:
Let n = 928174653
= A multiple of 11+(9 + 8 + 7 + 6 + 3)-(2 + 1+4 + 5)
= A multiple of 11 + 33 – 12
= A multiple of 11 + 21
= A multiple of 11 + 11 + 10
= A multiple of 11 + 10
∴ Remainder =10

Question 12.
Given an example of a number which is divisible by :
(i) 2 but not by 4.
(ii) 3 but not by 6.
(iii) 4 but not by 8.
(iv) both 4 and 8 but not 32.
Solution:
(i) 2 but not by 4
A number is divisible by 2 if units do given is even but it is divisible by 4 if the number formed by tens digit and ones digit is divisible by 4.
∴ The number can be 222, 342 etc.
(ii) 3 but not by 6
A number is divisible by 3 if the sum of its digits is divisible by 3
But a number is divisible by 6, if it is divided by 2 and 3 both
∴ The numbers can be 333, 201 etc.
(iii) 4 but not by 8
A number is divisible by 4 if the number formed by the tens digit and ones digit is divisible by 4 but a number is divisible by 8, if the number formed by hundreds digit, tens digit and ones digit is divisible by 8.
∴ The number can be 244, 1356 etc.
(iv) Both 4 and 8 but not by 32
A number in which the number formed by the hundreds, tens and one’s digit, is divisible by 8 is divisible by 8. It will also divisible by 4 also.
But a number when is divisible by, 4 and 8 both is not necessarily divisible by 32 e.g., 328, 5400 etc.

Question 13.
Which of the following statements are true ?
(i) If a number is divisible by 3, it must be divisible by 9.
(ii) If a number is divisible by 9, it must be divisible by 3.
(iii) If a number is divisible by 4, it must be divisible by 8.
(iv) If a number is divisible by 8, it must be divisible by 4.
(v) A number is divisible by 18, if it is divisible by both 3 and 6.
(vi) If a number is divisible by both 9 and 10, it must be divisible by 90.
(vii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
(viii) If a number divides three numbers exactly, it must divide their sum exactly.
(ix) If two numbers are co-priirie, at least one of them must be a prime number.
(x) The sum of two consecutive odd numbers is always divisible by 4.
Solution:
(i) False, it is not necessarily that it must divide by 9.
(ii) Trae.
(iii) False, it is not necessarily that it must divide by 8.
(iv) True.
(v) False, it must be divisible by 9 and 2 both.
(vi) True.
(vii) False, it is not necessarily.
(viii)True.
(ix) False. It is not necessarily.
(x) True.

Exercise 5.3

Solve each of the following cryptarithms.
Question 1.

Solution:

Values of A and B be from 0 to 9 In ten’s digit 3 + A = 9
∴ A = 6 or less.
∴ 7 + B = A = 6 or less
∴ 7 + 9 or 8 = 16 or 15
∴ But it is two digit number
B = 8
Then A = 5

Question 2.

Solution:
Values of A and B can be between 0 and 9
In tens digit, A + 3 = 9
∴ A = 9 – 3 = 6 or less than 6
In ones unit B + 7 = A = 6or less
∴ 7 + 9 or 8 = 16 or 15
But it is two digit number
∴ B = 8 and
∴ A = 5

Question 3.

Solution:

Value of A and B can be between 0 and 9 In units place.
1+B = 0 ⇒1+B = 10
∴ B = 10 – 1 = 9
and in tens place
1 + A + 1 = B ⇒ A + 2 = 9
⇒ A = 9 – 2 = 7

Question 4.

Solution:
Values of A and.B can be between 0 and 9
In units place, B+1 = 8 ⇒ B = 8-1=7
In tens place A + B= 1 or A + B = 11
⇒ A + 7 = 11 ⇒ A =11-7 = 4

Question 5.

Solution:

Values of A and B can be between 0 and 9
In tens place, 2 + A = 0 or 2 + A=10
A = 10-2 = 8
In units place, A + B = 9
⇒ 8 + B = 9 ⇒ B = 9- 8 = 1

Question 6.

Solution:

Values of A and B can be between 0 and 9
In hundreds place,

Question 7.
Show that cryptarithm 4 x AB¯¯¯¯¯¯¯¯=CAB¯¯¯¯¯¯¯¯¯¯¯ does not have any solution.
Solution:

It means that 4 x B is a numebr whose units digit is B
Clearly, there is no such digit
Hence the given cryptarithm has no solution.

Exercise 4.1

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 643=21, Remainder = 1
(7)³ = 7 x 7 x 7 = 343
Which is 3433 =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 7^3′.
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 = 713&ifi=8&uci=a!8&btvi=7&fsb=1&xpc=6d2Qjo9kVP&p=https%3A//www.learninsta.com&dtd=6298

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 = 3846= 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.

Exercise 4.2

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:

Exercise 4.3

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:

Exercise 4.4

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

Exercise 4.5

Making use of the cube root table, find the cubes root of the following (correct to three decimal places) 
Question 1.
7
Solution:
7–√3 =1.913 (From the table)

Question 2.
70
Solution:
70−−√3 =4.121 (From the table)

Question 3.
700
Solution:
700−−−√3=7×100−−−−−−√3= 8.879 (from 10x−−−√3)

Question 4.
7000
Solution:
7000−−−−√3=70×100−−−−−−−√3 = 19.13 (from 100x−−−−√3)

Question 5.
1100
Solution:
1100−−−−√3=11×100−−−−−−−√3 = 10.32 (from 100x−−−−√3)

Question 6.
780
Solution:
780−−−√3=78×100−−−−−−−√3 = 9.205 (from 10x−−−√3)

Question 7.
7800
Solution:
7800−−−−√3=78×100−−−−−−−√3 = 19.83 (from 100x−−−−√3)

Question 8.
1346
Solution:

Question 9.
250
Solution:

Question 10.
5112
Solution:

Question 11.
9800
Solution:

Question 12.
732
Solution:

Question 13.
7342
Solution:

Question 14.
133100
Solution:

Question 15.
37800
Solution:

Question 16.
0.27
Solution:

Question 17.
8.6
Solution:

Question 18.
0.86
Solution:

Question 19.
8.65
Solution:

Question 20.
7532
Solution:

Question 21.
833
Solution:

Question 22.
34.2
Solution:

Question 23.
What is the length of the side of a cube whose volume is 275 cm³. Make use of the table for the cube root.
Solution: