Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.4

Maharashtra State Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.4

Question 1.
Draw a histogram of the following data.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.4 1
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.4 2

Question 2.
The table below shows the yield of jowar per acre. Show the data by histogram.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.4 3
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.4 4 Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.4 5

Question 3.
In the following table, the investment made by 210 families is shown. Present it in the form of a histogram.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.4 6
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.4 7

Question 4.
Time allotted for the preparation of an examination by some students is shown in the table. Draw a histogram to show the information.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.4 8
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.4 9

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3

Maharashtra State Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3

Question 1.
The following table shows the information regarding the milk collected from farmers on a milk collection centre and the content of fat in the milk, measured by a lactometer. Find the mode of fat content.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 1
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 2
Here, the maximum frequency is 80.
∴ The modal class is 4 – 5.
L = lower class limit of the modal class = 4
h = class interval of the modal class = 1
f1 = frequency of the modal class = 80
f0 = frequency of the class preceding the modal class = 70
f2 = frequency of the class succeeding the modal class = 60
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 3
∴ The mode of the fat content is 4.33%.

Question 2.
Electricity used by some families is shown in the following table. Find the mode of use of electricity.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 20
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 5
Here, the maximum frequency is 100.
∴ The modal class is 60 – 80.
L = lower class limit of the modal class = 60
h = class interval of the modal class = 20
f1 = frequency of the modal class = 100
f0 = frequency of the class preceding the modal class = 70
f2 = frequency of the class succeeding the modal class = 80
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 6
∴ The mode of use of electricity is 72 units.

Question 3.
Grouped frequency distribution of supply of milk to hotels and the number of hotels is given in the following table. Find the mode of the supply of milk.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 7
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 8
Here, the maximum frequency is 35.
∴ The modal class is 9 – 11.
L = lower class limit of the modal class = 9
h = class interval of the modal class = 2
f1 = frequency of the modal class = 35
f0 = frequency of the class preceding the modal class = 20
f2 = frequency of the class succeeding the modal class = 18
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 9
∴ The mode of the supply of milk is 9.94 litres (approx.).

Question 4.
The following frequency distribution table gives the ages of 200 patients treated in a hospital in a week. Find the mode of ages of the patients.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 10
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 11
Here, the maximum frequency is 50.
The modal class is 9.5 – 14.5.
L = lower class limit of the modal class = 9.5
h = class interval of the modal class = 5
f1 = frequency of the modal class = 50
f0 = frequency of the class preceding the modal class = 32
f2 = frequency of the class succeeding the modal class = 36
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.3 12
∴ The mode of the ages of the patients is 12.31 years (approx.).

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1

Maharashtra State Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1

Question 1.
The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by students for their studies by direct method.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 1
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 2
∴ The mean of the time spent by the students for their studies is 4.36 hours.

Question 2.
In the following table, the toll paid by drivers and the number of vehicles is shown. Find the mean of the toll by ‘assumed mean’ method.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 3
Solution:
Let us take the assumed mean (A) = 550
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 4
∴ The mean of the toll paid by the drivers is ₹ 521.43.

Question 3.
A milk centre sold milk to 50 customers. The table below gives the number of customers and the milk they purchased. Find the mean of the milk sold by direct method.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 5
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 6
∴ The mean of the milk sold is 2.82 litres.

Question 4.
A frequency distribution table for the production of oranges of some farm owners is given below. Find the mean production of oranges by ‘assumed mean’ method.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 7
Solution:
Let us take the assumed mean (A) = 37.5
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 8
∴ The mean of the production of oranges is ₹ 35310.

Question 5.
A frequency distribution of funds collected by 120 workers in a company for the drought affected people are given in the following table. Find the mean of he funds by ‘step deviation’ method.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 9
Solution:
Here, we take A = 1250 and g = 500
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 10
∴ The mean of the funds collected is ₹ 987.5.

Question 6.
The following table gives the information of frequency distribution of weekly wages of 150 workers of a company. Find the mean of the weekly wages by ‘step deviation’ method.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 11
Solution:
Here, we take A = 2500 and g = 1000.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 12
∴ The mean of the weekly wages is ₹ 3070.

Question 1.
The daily sale of 100 vegetable vendors is given in the following table. Find the mean of the sale by direct method. (Textbook pg. no. 133 and 134)
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 13
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 14
The mean of the sale is 2150.

Question 2.
The amount invested in health insurance by 100 families is given in the following frequency table. Find the mean of investments using direct method and assumed mean method. Check whether the mean found by the two methods is the same as calculated by step deviation method (Ans: ₹ 2140). (Textbook pg. no. 135 and 136)
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 15
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 16
∴ The mean of investments in health insurance is ₹ 2140.
Assumed mean method:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 17
∴ The mean of investments in health insurance is ₹ 2140.
∴ Mean found by direct method and assumed mean method is the same as calculated by step deviation method.

Question 3.
The following table shows the funds collected by 50 students for flood affected people. Find the mean of the funds.
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 18
If the number of scores in two consecutive classes is very low, it is convenient to club them. So, in the above example, we club the classes 0 – 500, 500 – 1000 and 2000 – 2500, 2500 – 3000. Now the new table is as follows
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 19
i. Solve by direct method.
ii. Verily that the mean calculated by assumed mean method is the same.
iii. Find the mean in the above example by taking A = 1750. (Textbook pg. no. 137)
Solution:
i. Direct method:
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 20
∴ The mean of the funds is ₹ 1390.

ii. Assumed mean method:
Here, A = 1250
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 21
∴ The mean calculated by assumed mean method is the same.

iii. Step deviation method:
Here, we take A = 1750 and g = 250
Maharashtra Board Class 10 Maths Solutions Chapter 6 Statistics Practice Set 6.1 22
∴ The mean of the funds is ₹ 1390.

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5

Maharashtra State Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5

Question 1.
Choose the correct alternative answer for each of the following questions.

i. Which number cannot represent a probability?
(A) \(\frac { 2 }{ 3 } \)
(B) 1.5
(C) 15%
(D) 0.7
Answer:
The probability of any 0 to 1 or 0% to 100%. event is from
(B)

ii. A die is rolled. What is the probability that the number appearing on upper face is less than 3?
(A) \(\frac { 1 }{ 6 } \)
(B) \(\frac { 1 }{ 3 } \)
(C) \(\frac { 1 }{ 2 } \)
(D) 0
Answer:
(B)

iii. What is the probability of the event that a number chosen from 1 to 100 is a prime number?
(A) \(\frac { 1 }{ 5 } \)
(B) \(\frac { 6 }{ 25 } \)
(C) \(\frac { 1 }{ 4 } \)
(D) \(\frac { 13 }{ 50 } \)
Answer:
n(S) = 100
Let A be the event that the number chosen is a prime number.
∴ A = {2, 3, 5. , 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
∴ n(A) = 25
∴ P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac { 25 }{ 100 } \) = \(\frac { 1 }{ 4 } \)
(C)

iv. There are 40 cards in a bag. Each bears a number from 1 to 40. One card is drawn at random. What is the probability that the card bears a number which is a multiple of 5?
(A) \(\frac { 1 }{ 5 } \)
(B) \(\frac { 3 }{ 5 } \)
(C) \(\frac { 4 }{ 5 } \)
(D) \(\frac { 1 }{ 3 } \)
Answer:
(A)

v. If n(A) = 2, P(A) = \(\frac { 1 }{ 5 } \), then n(S) = ?
(A) 10
(B) \(\frac { 5 }{ 2 } \)
(C) \(\frac { 2 }{ 5 } \)
(D) \(\frac { 1 }{ 3 } \)
Answer:
(A)

Question 2.
Basketball players John, Vasim, Akash were practising the ball drop in the basket. The probabilities of success for John, Vasim and Akash are \(\frac { 4 }{ 5 } \), 0.83 and 58% respectively. Who had the greatest probability of success ?
Solution:
The probability that the ball is dropped in the basket by John = \(\frac { 4 }{ 5 } \) = 0.80
The probability that the ball is dropped in the basket by Vasim = 0.83
The probability that the ball is dropped in the basket by Akash = 58% = \(\frac { 58 }{ 100 } \) = 0.58
0.83 > 0.80 > 0.58
∴ Vasim has the greatest probability of success.

Question 3.
In a hockey team there are 6 defenders , 4 offenders and 1 goalie. Out of these, one player is to be selected randomly as a captain. Find the probability of the selection that:
i. The goalie will be selected.
ii. A defender will be selected.
Solution:
Total number of players in the hockey team
= 6 + 4 + 1 = 11
∴ n(S) = 11

i. Let A be the event that the captain selected will be a goalie.
There is only one goalie in the hockey team.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 1

ii. Let B be the event that the captain selected will be a defender.
There are 6 defenders in the hockey team.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 2

Question 4.
Joseph kept 26 cards in a cap, bearing one English alphabet on each card. One card is drawn at random. What is the probability that the card drawn is a vowel card ?
Solution:
Each card bears an English alphabet.
∴ n(S) = 26
Let A be the event that the card drawn is a vowel card.
There are 5 vowels in English alphabets.
∴ A = {a, e, i, o, u}
∴ n(A) = 5
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 3
∴ The probability that the card drawn is a vowel card is \(\frac { 5 }{ 26 } \).

Question 5.
A balloon vendor has 2 red, 3 blue and 4 green balloons. He wants to choose one of them at random to give it to Pranali. What is the probability of the event that Pranali gets,
i. a red balloon.
ii. a blue balloon,
iii. a green balloon.
Solution:
Let the 2 red balloon be R1, R2,
3 blue balloons be B1, B2, B3, and
4 green balloons be G1, G2, G3, G4.
∴ Sample space
S = {R1, R2, B1, B2, B3, G1, G2, G3, G4}
∴ n(S) = 9

i. Let A be the event that Pranali gets a red balloon.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 4
∴ The probability that Pranali gets a red balloon is \(\frac { 2 }{ 9 } \)

ii. Let B be the event that Pranali gets a blue balloon.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 5
∴ The probability that Pranali gets a blue balloon is \(\frac { 1 }{ 3 } \).

iii. Let C be the event that Pranali gets a green balloon.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 6
∴ The probability that Pranali gets a green balloon is \(\frac { 4 }{ 9 } \).

Question 6.
A box contains 5 red, 8 blue and 3 green pens. Rutuja wants to pick a pen at random. What is the probability that the pen is blue?
Solution:
Let 5 red pens be R1, R2, R3, R4, R5.
8 blue pens be B1, B2, B3, B4, B5, B6, B7, B8. and
3 green pens be G1, G2, G3.
∴ Sample space
S = {R1, R2, R3, R4, R5, B1, B2, B3, B4, B5, B6, B7, B8, G1, G2, G3}
∴ n(S) = 16
Let A be the event that Rutuja picks a blue pen.
∴ A = {B1, B2, B3, B4, B5, B6, B7, B8}
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 7
∴ The probability that Rutuja picks a blue pen is \(\frac { 1 }{ 2 } \).

Question 7.
Six faces of a die are as shown below.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 8
If the die is rolled once, find the probability of
i. ‘A’ appears on upper face.
ii. ‘D’ appears on upper face.
Solution:
Sample space
S = {A, B, C, D, E, A}
∴ n (S) = 6
i. Let R be the event that ‘A’ appears on the upper face.
∴ R = {A, A}
∴ n(R) = 2
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 9

ii. Let Q be the event that ‘D’ appears on the upper face.
Total number of faces having ‘D’ on it = 1
Q = {D}
∴ n(Q) = 1
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 10
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 11

Question 8.
A box contains 30 tickets, bearing only one number from 1 to 30 on each. If one ticket is drawn at random, find the probability of an event that the ticket drawn bears
i. an odd number.
ii. a complete square number.
Solution:
Sample space,
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}
∴ n(S) = 30

i. Let A be the event that the ticket drawn bears an odd number.
∴ A = {1,3,5,7,9,11,13,15,17,19,21, 23,25,27,29}
∴ n(A) =15
E:\Prasanna\Learncram\Class 10 Maths\ch 5\Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 14.png

ii. Let B be the event that the ticket drawn bears a complete square number.
∴ B = {1,4,9,16,25}
∴ n(B) = 5
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 13

Question 9.
Length and breadth of a rectangular garden are 77 m and 50 m. There is a circular lake in the garden having diameter 14 m. Due to wind, a towel from a terrace on a nearby building fell into the garden. Then find the probability of the event that it fell in the lake.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 14
Solution:
Area of the rectangular garden
= length × breadth
= 77 × 50
∴ Area of the rectangular garden = 3850 sq.m
Radius of the lake = \(\frac { 14 }{ 2 } \) = 7 m
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 15
∴ The probability of the event that the towel tell in the lake is \(\frac { 1 }{ 25 } \).

Question 10.
In a game of chance, a spinning arrow comes to rest at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8. All these are equally likely outcomes. Find the probability that it will rest at
i. 8.
ii. an odd number.
iii. a number greater than 2.
iv. a number less than 9.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 16
Solution:
Sample space (S) = {1,2, 3, 4, 5, 6, 7, 8}
∴ n(S) = 8
i. Let A be the event that the spinning arrow comes to rest at 8.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 17
ii. Let B be the event that the spinning arrow comes to rest at an odd number.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 18
iii. Let C be the event that the spinning arrow comes to rest at a number greater than 2.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 19
iv. Let D be the event that the spinning arrow comes to rest at a number less than 9.
∴ D = {1,2, 3, 4, 5, 6, 7, 8}
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 20

Question 11.
There are six cards in a box, each bearing a number from 0 to 5. Find the probability of each of the following events, that a card drawn shows,
i. a natural number.
ii. a number less than 1.
iii. a whole number.
iv. a number greater than 5.
Solution:
Sample space (S) = {0, 1, 2, 3, 4, 5}
∴ n(S) = 6

i. Let A be the event that the card drawn shows a natural number.
∴ A = {1,2,3,4,5}
∴ n(A) = 5
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 21

ii. Let B be the event that the card drawn shows a number less than 1.
∴ B = {0}
∴ n(B) = 1
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 22

iii. Let C be the event that the card drawn shows a whole number.
∴ C = {0,1, 2, 3, 4, 5}
∴ n(C) = 6
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 23

iv. Let D be the event that the card drawn shows a number greater than 5.
Here, the greatest number is 5.
∴ Event D is an impossible event.
∴ D = { }
∴ n(D) = 0
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 24

Question 12.
A bag contains 3 red, 3 white and 3 green balls. One ball is taken out of the bag at random. What is the probability that the ball drawn is:
i. red.
ii. not red.
iii. either red or white.
Solution:
Let the three red balls be R1, R2, R3, three white balls be W1, W2, W3 and three green balls be G1, G2, G3.
∴ Sample space,
S = {R1, R2, R3, W1, W2, W3, G1, G2, G3}
∴ n(S) = 9

i. Let A be the event that the ball drawn is red.
∴ A = {R1, R2, R3}
∴ n(A) = 3
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 25

ii. Let B be the event that the ball drawn is not red.
B = {W1,W2,W3,G1,G2,G3}
∴ n(B) = 6
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 26

iii. Let C be the event that the ball drawn is red or white.
∴ C = {R1, R2, R3, W1, W2, W3}
∴ n(C) = 6
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 27

Question 13.
Each card bears one letter from the word ‘mathematics’. The cards are placed on a table upside down. Find the probability that a card drawn bears the letter ‘m’.
Solution:
Sample space
= {m, a, t, h, e, m, a, t, i, c, s}
∴ n(S) = 11
Let A be the event that the card drawn bears the letter ‘m’
∴ A = {m, m}
∴ n(A) = 2
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 28
∴ The probability that a card drawn bears letter ‘m’ is \(\frac { 2 }{ 11 } \).

Question 14.
Out of 200 students from a school, 135 like Kabaddi and the remaining students do not like the game. If one student is selected at random from all the students, find the probability that the student selected dosen’t like Kabaddi.
Solution:
Total number of students in the school = 200
∴ n(S) = 200
Number of students who like Kabaddi = 135
∴ Number of students who do not like Kabaddi
= 200 – 135 = 65
Let A be the event that the student selected does not like Kabaddi.
∴ n(A) = 65
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 29
∴ The probability that the student selected doesn’t like kabaddi is \(\frac { 13 }{ 40 } \).

Question 15.
A two digit number is to be formed from the digits 0, 1, 2, 3, 4. Repetition of the digits is allowed. Find the probability that the number so formed is a:
i. prime number.
ii. multiple of 4.
iii multiple of 11.
Solution:
Sample space
(S) = {10, 11, 12, 13, 14,
20, 21, 22, 23, 24,
30, 31, 32, 33, 34,
40, 41, 42, 43, 44}
∴ n(S) = 20

i. Let A be the event that the number so formed is a prime number.
∴ A = {11,13,23,31,41,43}
∴ n(A) = 6
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 30

ii. Let B be the event that the number so formed is a multiple of 4.
∴ B = {12,20,24,32,40,44}
∴ n(B) = 6
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 31

iii. Let C be the event that the number so formed is a multiple of 11.
∴ C = {11,22,33,44}
∴ n(C) = 4
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 32

Question 16.
The faces of a die bear numbers 0,1, 2, 3,4, 5. If the die is rolled twice, then find the probability that the product of digits on the upper face is zero.
Solution:
Sample space,
S = {(0, 0), (0,1), (0,2),
(1,0), (1,1), (1,2),
(2,0), (2,1), (2,2),
(3.0), (3,1), (3,2),
(4.0), (4,1), (4,2),
(5.0), (5,1), (5,2),
∴ n(S) = 36
Let A be the event that the product of digits on the upper face is zero.
∴ A = {(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (1,0), (2, 0), (3,0), (4, 0), (5,0)}
∴ n(A) = 11
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Problem Set 5 33
∴ The probability that the product of the digits on the upper face is zero is \(\frac { 11 }{ 36 } \).

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.4

Maharashtra State Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.4

Question 1.
If two coins are tossed, find the probability of the following events.
i. Getting at least one head.
ii. Getting no head.
Solution:
Sample space,
S = {HH, HT, TH, TT}
∴ n(S) = 4

i. Let A be the event of getting at least one head.
∴ A = {HT, TH, HH}
∴ n(A) = 3
∴ P(A) = \(\frac { n(A) }{ n(S) } \)
∴ P(A) = \(\frac { 3 }{ 4 } \)

ii. Let B be the event of getting no head.
∴ B = {TT}
∴ n(B) = 1
∴ P(B) = \(\frac { n(B) }{ n(S) } \)
∴ P(B) = \(\frac { 1 }{ 4 } \)
∴ P(A) = \(\frac { 3 }{ 4 } \); P(B) = \(\frac { 1 }{ 4 } \)

Question 2.
If two dice are rolled simultaneously, find the probability of the following events.
i. The sum of the digits on the upper faces is at least 10.
ii. The sum of the digits on the upper faces is 33.
iii. The digit on the first die is greater than the digit on second die.
Solution:
Sample space,
s = {(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), (3, 6),
(4, 1), (4, 2), (4,3), (4,4), (4, 5), (4,6),
(5, 1), (5, 2), (5,3), (5,4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6,4), (6, 5), (6,6)}
∴ n(S) = 36

i. Let A be the event that the sum of the digits on the upper faces is at least 10.
∴ A = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
∴ n(A) = 6
∴ P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac { 6 }{ 36 } \)
∴ P(A) = \(\frac { 1 }{ 6 } \)

ii. Let B be the event that the sum of the digits on the upper faces is 33.
The sum of the digits on the upper faces can be maximum 12.
∴ Event B is an impossible event.
∴ B = { }
∴ n(B) = 0
∴ P(B) = \(\frac { n(B) }{ n(S) } \) = \(\frac { 0 }{ 36 } \)
∴ P(B) = 0

iii. Let C be the event that the digit on the first die is greater than the digit on the second die.
C = {(2, 1), (3, 1), (3,2), (4,1), (4,2), (4, 3), (5, 1), (5,2), (5,3), (5,4), (6,1), (6,2), (6, 3), (6, 4), (6, 5),
∴ n(C) = 15
∴ P(C) = \(\frac { n(c) }{ n(S) } \) = \(\frac { 15 }{ 36 } \)
∴ P(C) = \(\frac { 5 }{ 12 } \)
∴ P(A) = \(\frac { 1 }{ 6 } \) ; P(B) = 0; P(C) = \(\frac { 5 }{ 12 } \)

Question 3.
There are 15 tickets in a box, each bearing one of the numbers from 1 to 15. One ticket is drawn at random from the box. Find the probability of event that the ticket drawn:
i. shows an even number.
ii. shows a number which is a multiple of 5.
Solution:
Sample space,
S = {1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,14, 15}
∴ n(S) = 15

i. Let A be the event that the ticket drawn shows an even number.
∴ A = {2, 4, 6, 8, 10, 12, 14}
∴ n(A) = 7
∴ P(A) = \(\frac { n(A) }{ n(S) } \)
∴ P(A) = \(\frac { 7 }{ 15 } \)

ii. Let B be the event that the ticket drawn shows a number which is a multiple of 5.
∴ B = {5, 10, 15}
∴ n(B) = 3
∴ P(B) = \(\frac { n(B) }{ n(S) } \) = \(\frac { 3 }{ 15 } \)
∴ P(B) = \(\frac { 1 }{ 5 } \)
∴ P(A) = \(\frac { 7 }{ 15 } \) ; P(B) = \(\frac { 1 }{ 5 } \)

Question 4.
A two digit number is formed with digits 2, 3, 5, 7, 9 without repetition. What is the probability that the number formed is
i. an odd number?
ii. a multiple of 5?
Solution:
Sample space
(S) = {23, 25, 27, 29,
32, 35, 37, 39,
52, 53, 57, 59,
72, 73, 75, 79,
92, 93, 95, 97}
∴ n(S) = 20
i. Let A be the event that the number formed is an odd number.
∴ A = {23, 25, 27, 29, 35, 37, 39, 53, 57, 59, 73, 75,79,93,95,97}
∴ n(A) = 16
∴ P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac { 16 }{ 20 } \)
∴ P(A) = \(\frac { 4 }{ 5 } \)

ii. Let B be the event that the number formed is a multiple of 5.
∴ B = {25,35,75,95}
∴ n(B) = 4
∴ P(B) = \(\frac { n(B) }{ n(S) } \) = \(\frac { 4 }{ 20 } \)
∴ P(B) = \(\frac { 1 }{ 5 } \)
∴ P(A) = \(\frac { 4 }{ 5 } \) ; P(B) = \(\frac { 1 }{ 5 } \)

Question 5.
A card is drawn at random from a pack of well shuffled 52 playing cards. Find the probability that the card drawn is
i. an ace.
ii. a spade.
Solution:
There are 52 playing cards.
∴ n(S) = 52
i. Let A be the event that the card drawn is an ace.
∴ n(A) = 4
∴ P(A) = \(\frac { n(A) }{ n(S) } \) = \(\frac { 4 }{ 52 } \)
∴ P(A) = \(\frac { 1 }{ 13 } \)

ii. Let B be the event that the card drawn is a spade.
∴ n(B) = 13
∴ P(B) = \(\frac { n(B) }{ n(S) } \) = \(\frac { 13 }{ 52 } \)
∴ P(B) = \(\frac { 1 }{ 4 } \)
∴ P(A) = \(\frac { 1 }{ 13 } \) ; P(B) = \(\frac { 1 }{ 4 } \)

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.3

Maharashtra State Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.3

Question 1.
Write sample space ‘S’ and number of sample points n(S) for each of the following experiments. Also write events A, B, C in the set form and write n(A), n(B), n(C).

i. One die is rolled,
Event A: Even number on the upper face.
Event B: Odd number on the upper face.
Event C: Prime number on the upper face.

ii. Two dice are rolled simultaneously,
Event A: The sum of the digits on upper faces is a multiple of 6.
Event B: The sum of the digits on the upper faces is minimum 10.
Event C: The same digit on both the upper faces.

iii. Three coins are tossed simultaneously.
Condition for event A: To get at least two heads.
Condition for event B: To get no head.
Condition for event C: To get head on the second coin.

iv. Two digit numbers are formed using digits 0, 1, 2, 3, 4, 5 without repetition of the digits.
Condition for event A: The number formed is even.
Condition for event B: The number is divisible by 3.
Condition for event C: The number formed is greater than 50.

v. From three men and two women, environment committee of two persons is to be formed.
Condition for event A: There must be at least one woman member.
Condition for event B: One man, one woman committee to be formed.
Condition for event C: There should not be a woman member.

vi. One coin and one die are thrown simultaneously.
Condition for event A: To get head and an odd number.
Condition for event B: To get a head or tail and an even number.
Condition for event C: Number on the upper face is greater than 7 and tail on the coin.
Solution:
i. Sample space (S) = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
Condition for event A: Even number on the upper face.
∴ A = {2,4,6}
∴ n(A) = 3
Condition for event B: Odd number on the upper face.
∴ B = {1, 3, 5}
∴ n(B) = 3
Condition for event C: Prime number on the upper face.
∴ C = {2, 3, 5}
∴ n(C) = 3

ii. Sample space,
S = {(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), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
∴ n(S) = 36
Condition for event A: The sum of the digits on the upper faces is a multiple of 6.
A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)}
∴ n(A) = 6

Condition for event B: The sum of the digits on the upper faces is minimum 10.
B = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
∴ n(B) = 6

Condition for event C: The same digit on both the upper faces.
C = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
∴ n(C) = 6

iii. Sample space,
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
∴ n(S) = 8

Condition for event A: To get at least two heads.
∴ A = {HHT, HTH, THH, HHH}
∴ n(A) = 4

Condition for event B: To get no head.
∴ B = {TTT}
∴ n(B) = 1

Condition for event C: To get head on the second coin.
∴ C = {HHH, HHT, THH, THT}
∴ n(C) = 4

iv. Sample space (S) = {10, 12, 13, 14, 15,
20, 21, 23, 24, 25,
30, 31, 32, 34, 35,
40, 41, 42, 43,
45, 50, 51, 52, 53, 54}
∴ n(S) = 25
Condition for event A: The number formed is even
∴ A = {10, 12, 14, 20, 24, 30, 32, 34, 40, 42, 50, 52, 54)
∴ n(A) = 13
Condition for event B: The number formed is divisible by 3.
∴ B = {12, 15, 21, 24, 30, 42, 45, 51, 54}
∴ n(B) = 9
Condition for event C: The number formed is greater than 50.
∴ C = {51,52, 53,54}
∴ n(C) = 4

v. Let the three men be M1, M2, M3 and the two women be W1, W2.
Out of these men and women, a environment committee of two persons is to be formed.
∴ Sample space,
S = {M1M2, M1M3, M1W1, M1W2, M2M3, M2W1, M2W2, M3W1, M3W2, W1W2}
∴ n(S) = 10
Condition for event A: There must be at least one woman member.
∴ A = {M1W1, M1W2, M2W1, M2W2, M3W1, M3W2, W1W2}
∴ n(A) = 7
Condition for event B: One man, one woman committee to be formed.
∴ B = {M1W1, M1W2, M2W1, M2W2, M3W2, M3W2}
∴ n(B) = 6
Condition for event C: There should not be a woman member.
∴ C = {M1M2, M1M3, M2M3}
∴ n(C) = 3

vi. Sample space,
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
∴ n(S) = 12
Condition for event A: To get head and an odd number.
∴ A = {(H, 1), (H, 3), (H, 5)}
∴ n(A) = 3
Condition for event B: To get a head or tail and an even number.
∴ B = {(H, 2), (H, 4), (H, 6), (T, 2), (T, 4), (T, 6)}
∴ n(B) = 6
Condition for event C: Number on the upper face is greater than 7 and tail on the coin.
The greatest number on the upper face of a die is 6.
∴ Event C is an impossible event.
∴ C = { }
∴ n(C) = 0

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.2

Maharashtra State Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.2

Question 1.
Write the correct number in the given boxes from the following A.P.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.2 1
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.2 2

Question 2.
Decide whether following sequence is an A.P., if so find the 20th term of the progression.
-12, -5, 2, 9,16, 23,30,…
Solution:
i. The given sequence is
-12, -5,2, 9, 16, 23,30,…
Here, t1 = -12, t2 = -5, t3 = 2, t4 = 9
∴ t2 – t1 – 5 – (-12) – 5 + 12 = 7
t3 – t2 = 2 – (-5) = 2 + 5 = 7
∴ t4 – t3 – 9 – 2 = 7
∴ t2 – t1 = t3 – t2 = … = 7 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P.

ii. tn = a + (n – 1)d
∴ t20 = -12 + (20 – 1)7 …[∵a = -12, d = 7]
= -12 + 19 × 7
= -12 + 133
∴ t20 = 121
∴ 20th term of the given A.P. is 121.

Question 3.
Given Arithmetic Progression is 12, 16, 20, 24, … Find the 24th term of this progression.
Solution:
The given A.P. is 12, 16, 20, 24,…
Here, a = 12, d = 16 – 12 = 4 Since,
tn = a + (n – 1)d
∴ t24 = 12 + (24 – 1)4
= 12 + 23 × 4
= 12 + 92
∴ t24 = 104
∴ 24th term of the given A.P. is 104.

Question 4.
Find the 19th term of the following A.P. 7,13,19,25…..
Solution:
The given A.P. is 7, 13, 19, 25,…
Here, a = 7, d = 13 – 7 = 6
Since, tn = a + (n – 1)d
∴ t19 = 7 + (19 – 1)6
= 7 + 18 × 6
= 7 + 108
∴ t19 = 115
∴ 19th term of the given A.P. is 115.

Question 5.
Find the 27th term of the following A.P. 9,4,-1,-6,-11,…
Solution:
The given A.P. is 9, 4, -1, -6, -11,…
Here, a = 9, d = 4- 9 = -5
Since, tn = a + (n – 1)d
∴ t27 = 9 + (27 – 1)(-5)
= 9 + 26 × (-5)
= 9 – 130
∴ t27 = -121
∴ 27th term of the given A.P. is -121.

Question 6.
Find how many three digit natural numbers are divisible by 5.
Solution:
The three digit natural numbers divisible by
5 are 100, 105, 110, …,995
The above sequence is an A.P.
∴ a = 100, d = 105 – 100 = 5
Let the number of terms in the A.P. be n.
Then, tn = 995
Since, tn = a + (n – 1)d
∴ 995 = 100 +(n – 1)5
∴ 995 – 100 = (n – 1)5
∴ 895 = (n – 1)5
∴ n – 1 = \(\frac { 895 }{ 5 } \)
∴ n – 1 = 179
∴ n = 179 + 1 = 180
∴ There are 180 three digit natural numbers which are divisible by 5.

Question 7.
The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41st term of that A.P.
Solution:
Bor an A.P., let a be the first term and d be the common difference,
t11 = 16, t21 = 29 …[Given]
tn = a + (n – 1)d
∴ t11, = a + (11 – 1)d
∴ 16 = a + 10d
i.e. a + 10d = 16 …(i)
Also, t21 = a + (21 – 1)d
∴ 29 = a + 20d
i.e. a + 20d = 29 …(ii)
Subtracting equation (i) from (ii), we get a
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.2 3

Question 8.
8. 11, 8, 5, 2, … In this A.P. which term is number-151?
Solution:
The given A.P. is 11, 8, 5, 2,…
Here, a = 11, d = 8 – 11 = -3
Let the nth term of the given A.P. be -151.
Then, tn = – 151
Since, tn = a + (n – 1)d
∴ -151= 11 + (n – 1)(-3)
∴ -151 – 11 =(n – 1)(-3)
∴ -162 = (n – 1)(-3)
∴ n – 1 = \(\frac { -162 }{ -3 } \)
∴ n – 1 = 54
∴ n = 54 + 1 = 55
∴ 55th term of the given A.P. is -151.

Question 9.
In the natural numbers from 10 to 250, how many are divisible by 4?
Solution:
The natural numbers from 10 to 250 divisible
by 4 are 12, 16, 20, …,248
The above sequence is an A.P.
∴ a = 12, d = 16 – 12 = 4
Let the number of terms in the A.P. be n.
Then, tn = 248
Since, tn = a + (n – 1)d
∴ 248 = 12 + (n – 1)4
∴ 248 – 12 = (n – 1)4
∴ 236 = (n – 1)4
∴ n – 1 = \(\frac { 236 }{ 4 } \)
∴ n – 1 = 59
∴ n = 59 + 1 = 60
∴ There are 60 natural numbers from 10 to 250 which are divisible by 4.

Question 10.
In an A.P. 17th term is 7 more than its 10th term. Find the common difference.
Solution:
For an A.P., let a be the first term and d be the common difference.
According to the given condition,
t17 = t10 + 7
∴ a + (17 – 1)d = a + (10 – 1)d + 7 …[∵ tn = a + (n – 1)d]
∴ a + 16d = a + 9d + 7
∴ a + 16d – a – 9d = 7
∴ 7d = 7
∴ d = \(\frac { 7 }{ 7 } \) = 1
∴ The common difference is 1.

Question 1.
Kabir’s mother keeps a record of his height on each birthday. When he was one year old, his height was 70 cm, at 2 years he was 80 cm tall and 3 years he was 90 cm tall. His aunt Meera was studying in the 10th class. She said, “it seems like Kabir’s height grows in Arithmetic Progression”. Assuming this, she calculated how tall Kabir will be at the age of 15 years when he is in 10th! She was shocked to find it. You too assume that Kabir grows in A.P. and find out his height at the age of 15 years. (Textbook pg. no. 63)
Solution:
Height of Kabir when he was 1 year old = 70 cm Height of Kabir when he was 2 years old = 80 cm
Height of Kabir when he was 3 years old = 90 cm The heights of Kabir form an A.P.
Here, a = 70, d = 80 – 70 = 10
We have to find height of Kabir at the age of 15years i.e. t15.
Now, tn = a + (n – 1)d
∴ t15 = 70 + (15 – 1)10
= 70 + 14 × 10 = 70 + 140
∴ t15 = 210
∴ The height of Kabir at the age of 15 years will be 210 cm.

Question 2.
Is 5, 8, 11, 14, …. an A.P.? If so then what will be the 100th term? Check whether 92 is in this A.P.? Is number 61 in this A.P.? (Textbook pg. no, 62)
Solution:
i. The given sequence is
5, 8,11,14,…
Here, t1 = 5, t2 = 8, t3 = 11, t4 = 14
∴ t2 – t1 = 8 – 5 = 3
t3 – t2 = 11 – 8 = 3
t4 – t3 = 14 – 11 = 3
∴ t2 – t1 = t3 – t2 = t4 – t3 = 3 = d = constant
The difference between two consecutive terms is constant
∴ The given sequence is an A.P.

ii. tn = a + (n – 1)d
∴ t100 = 5 + (100 – 1)3 …[∵ a = 5, d = 3]
= 5 + 99 × 3
= 5 + 297
∴ t100 = 302
∴ 100th term of the given A.P. is 302.

iii. To check whether 92 is in given A.P., let tn = 92
∴ tn = a + (n – 1)d
∴ 92 = 5 + (n – 1)3
∴ 92 = 5 + 3n – 3
∴ 92 = 2 + 3n
∴ 90 = 3n
∴ n = \(\frac { 90 }{ 3 } \) = 30
∴ 92 is the 30th term of given A.P.

iv. To check whether 61 is in given A.P., let tn = 61
61 = 5 + (n – 1)3
∴ 61 = 5 + 3n – 3
∴ 61 = 2 + 3n
∴ 61 – 2 = 3n
∴ 59 = 3n
∴ n = \(\frac { 59 }{ 3 } \)
But, n is natural number 59
∴ n ≠ \(\frac { 59 }{ 3 } \)
∴ 61 is not in given A.P.

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2

Maharashtra State Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2

Question 1.
For each of the following experiments write sample space ‘S’ and number of sample Point n(S)
i. One coin and one die are thrown simultaneously.
ii. Two digit numbers are formed using digits 2,3 and 5 without repeating a digit.
Solution:
i. Sample space,
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
∴ n(S) =12
ii. Sample space,
S = {23,25,32, 35, 52, 53}
∴ n(S) = 6

Question 2.
The arrow is rotated and it stops randomly on the disc. Find out on which colour it may stop.
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 1
Solution:
There are total six colours on the disc.
Sample space,
S = {Red, Orange, Yellow, Blue, Green, Purple}
∴ n(S) = 6
∴ Arrow may stop on any one of the six colours.

Question 3.
In the month of March 2019, find the days on which the date is a multiple of 5. (see the given page of the calendar).
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 2
Solution:
Dates which are multiple of 5:
5,10, 15,20,25,30
∴ S = {Tuesday, Sunday, Friday, Wednesday, Monday, Saturday}
∴ n(S) = 6
∴ The days on which the date will be a multiple of 5 are Tuesday, Sunday, Friday, Wednesday, Monday and Saturday.

Question 4.
Form a ‘Road safety committee’ of two, from 2 boys (B1 B2) and 2 girls (G1, G2). Complete the following activity to write the sample space.
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 3

Question 1.
Sample Space

  • The set of all possible outcomes of a random experiment is called sample space.
  • It is denoted by ‘S’ or ‘Ω’ (omega).
  • Each element of a sample space is called a sample point.
  • The number of elements in the set S is denoted by n(S).
  • If n(S) is finite, then the sample space is called a finite sample space.

Some examples of finite sample space. (Textbook pg. no, 117)
Solution:
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 4 Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.2 5

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.1

Maharashtra State Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.1

Question 1.
How many possibilities are there in each of the following?
i. Vanita knows the following sites in Maharashtra. She is planning to visit one of them in her summer vacation. Ajintha, Mahabaleshwar, Lonar Sarovar, Tadoba wild life sanctuary, Amboli, Raigad, Matheran, Anandavan.
ii. Any day of a week is to be selected randomly.
iii. Select one card from the pack of 52 cards.
iv. One number from 10 to 20 is written on each card. Select one card randomly.
Solution:
i. Here, 8 sites of Maharashtra are given.
∴ There are 8 possibilities in a random experiment of visiting a site out of 8 sites in Maharashtra.

ii. There are 7 days in a week.
∴ There are 7 possibilities in a random experiment of selecting a day of the week.

iii. There are 52 cards in a pack of cards.
∴ There are 52 possibilities in a random experiment of selecting one card from the pack of 52 cards.

iv. There are 11 cards numbered from 10 to 20.
∴ There are 11 possibilities in a random experiment of selecting one card from the given set of cards.

Question 1.
In which of the following experiments possibility of expected outcome is more? (Textbook pg, no. 116)
i. Getting 1 on the upper face when a die is thrown.
ii. Getting head by tossing a coin.
Solution:
i. On a die there are 6 numbers.
∴ There are 6 possibilities of getting any one number from 1 to 6 on the upper face i.e. \(\frac { 1 }{ 6 } \) is the possibility.

ii. There are two possibilities (H or T) on tossing a coin i.e. \(\frac { 1 }{ 2 } \) possibility.
∴ In the second experiment, the possibility of expected outcome is more.

Question 2.
Throw a die, once. What are the different possibilities of getting dots on the upper face? (Textbook pg. no. 114)
Answer:
There are six different possibilities of getting dots on the upper face. They are
Maharashtra Board Class 10 Maths Solutions Chapter 5 Probability Practice Set 5.1

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1

Maharashtra State Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1

Question 1.
Which of the following sequences are A.P.? If they are A.P. find the common difference.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 1
Solution:
i. The given sequence is 2, 4, 6, 8,…
Here, t1 = 2, t2 = 4, t3 = 6, t4 = 8
∴ t2 – t1 = 4 – 2 = 2
t3 – t2 = 6 – 4 = 2
t4 – t3 = 8 – 6 = 2
∴ t2 – t1 =  t3 – t2 = … = 2 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 2.

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 2
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 3
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = \(\frac { 1 }{ 2 } \).

iii. The given sequence is -10, -6, -2, 2,…
Here, t1 = -10, t2 = – 6, t3 = -2, t4 = 2
∴ t2 – t1 = -6 – (-10) = -6 + 10 = 4
t3 – t2 = -2 -(-6) = -2 + 6 = 4
t4 – t3 = 2 – (-2) = 2 + 2 = 4
∴ t2 – t1 = t3 – t2 = … = 4 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 4.

iv. The given sequence is 0.3, 0.33, 0.333,…
Here, t1 = 0.3, t2 = 0.33, t3 = 0.333
∴ t2 -t1 = 0.33 – 0.3 = 0.03
t3 – t2 = 0.333 – 0.33 = 0.003
∴ t2 – t1 ≠ t3 – t2
The difference between two consecutive terms is not constant.
∴ The given sequence is not an A.P.

v. The given sequence is 0, -4, -8, -12,…
Here, t1 = 0, t2 = -4, t3 = -8, t4 = -12
∴ t2 – t1 = -4 – 0 = -4
t3 – t2 = -8 – (-4) = -8 + 4 = -4
t4 – t3 = -12 – (-8) = -12 + 8 = -4
∴ t2 – t1 = t3 – t2 = … = —4 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = -4.

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 4
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 0.

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 5
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = √2.

viii. The given sequence is 127, 132, 137,…
Here, t1 = 127, t2 = 132, t3 = 137
∴ t2 – t1 = 132 – 127 = 5
t3 – t2 = 137 – 132 = 5
∴ t2 – t1 = t3 – t2 = … = 5 = d = constant
The difference between two consecutive terms is constant.
∴ The given sequence is an A.P. and common difference (d) = 5.

Question 2.
Write an A.P. whose first term is a and common difference is d in each of the following.
i. a = 10, d = 5
ii. a = -3, d = 0
iii. a = -7, d = \(\frac { 1 }{ 2 } \)
iv. a = -1.25, d = 3
v. a = 6, d = -3
vi. a = -19, d = -4
Solution:
i. a = 10, d = 5 …[Given]
∴ t1 = a = 10
t2 = t1 + d = 10 + 5 = 15
t3 = t2 + d = 15 + 5 = 20
t4 = t3 + d = 20 + 5 = 25
∴ The required A.P. is 10,15, 20, 25,…

ii. a = -3, d = 0 …[Given]
∴ t1 = a = -3
t2 = t1 + d = -3 + 0 = -3
t3 = t2 + d = -3 + 0 = -3
t4 = t3 + d = -3 + 0 = -3
∴ The required A.P. is -3, -3, -3, -3,…

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 6
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 7
∴ The required A.P. is -7, – 6.5, – 6, – 5.5,

iv. a = -1.25, d = 3 …[Given]
t1 = a = -1.25
t2 = t1 + d = – 1.25 + 3 = 1.75
t3 = t2 + d = 1.75 + 3 = .4.75
t4 = t3 + d = 4.75 + 3 = 7.75
∴ The required A.P. is -1.25, 1.75, 4.75, 7.75,…

v. a = 6, d = -3 …[Given]
∴ t1 = a = 6
t2 = t1 + d = 6 – 3 = 3
t3 = t2 + d = 3 – 3 = 0
t4 = t3 + d = 0- 3 = -3
∴ The required A.P. is 6, 3, 0, -3,…

vi. a = -19, d = -4 …[Given]
t1 = a = -19
t2 = t1 + d = -19 – 4 = -23
t3 = t2 + d = -23 – 4 = -27
t4 = t3 + d = -27 – 4 = -31
∴ The required A.P. is -19, -23, -27, -31,…

Question 3.
Find the first term and common difference for each of the A.P.
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 8
Solution:
i. The given A.P. is 5, 1,-3,-7,…
Here, t1 = 5, t2 = 1
∴ a = t1 = 5 and
d = t2 – t1 = 1 – 5 = -4
∴ first term (a) = 5,
common difference (d) = -4

ii. The given A.P. is 0.6, 0.9, 1.2, 1.5,…
Here, t1 = 0.6, t2 = 0.9
∴ a = t1 = 0.6 and
d = t2 – t1 = 0.9 – 0.6 = 0.3
∴ first term (a) = 0.6,
common difference (d) = 0.3

iii. The given A.P. is 127, 135, 143, 151,…
Here, t1 = 127, t2 = 135
∴ a = t1 = 127 and
d = t2 – t1 = 135 – 127 = 8
∴ first term (a) = 127,
common difference (d) = 8

Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 9

Question 1.
Complete the given pattern. Look at the pattern of the numbers. Try to find a rule to obtain the next number from its preceding number. Write the next numbers. (Textbook pg, no. 55 and 56)
Answer:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 10
Every pattern is formed by adding a circle in horizontal and vertical rows to the preceding pattern.
∴ The sequence for the above pattern is 1,3, 5, 7, 9,11,13,15,17,….
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 11
Every pattern is formed by adding 2 triangles horizontally and 1 triangle vertically to the preceding pattern.
∴ The sequence for the above pattern is 5,8,11,14,17,20,23,…

Question 2.
Some sequences are given below. Show the positions of the terms by t1, t2, t3,…
Answer:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 12

Question 3.
Some sequences are given below. Check whether there is any rule among the terms. Find the similarity between two sequences. To check the rule for the terms of the sequence look at the arrangements and fill the empty boxes suitably. (Textbook pg. no. 56 and 57)
i. .1,4,7,10,13,…
ii. 6,12,18,24,…
iii. 3,3,3,3,…
iv. 4, 16, 64,…
v. -1, -1.5, -2, -2.5,…
vi. 13, 23, 33, 43
Answer:
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 13
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 14
The similarity in the sequences i., ii., iii. and v. is that the next term is obtained by adding a particular fixed number to the previous term.

Note : A Geometric Progression is a sequence in which the ratio of any two consecutive terms is a constant,
Maharashtra Board Class 10 Maths Solutions Chapter 3 Arithmetic Progression Practice Set 3.1 15
Sequence iv. is a geometric progression.

Question 4.
Write one example of finite and infinite A.P. each. (Textbook pg. no. 59)
Answer:
Finite A.P.:
Even natural numbers from 4 to 50:
4, 6, 8, ………………. 50.
Infinite A. P.:
Positive multiples of 5:
5, 10, 15, ……………..

Maharashtra Board Class 10 Maths Solutions