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Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 1 Practice Set 6.3 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.3 Chapter 6 Statistics Questions With Answers Maharashtra Board

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.

Solution:

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

∴ 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.

Solution:

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

∴ 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.

Solution:

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

∴ 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.

Solution:

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

∴ The mode of the ages of the patients is 12.31 years (approx.).

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 1 Practice Set 6.2 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.2 Chapter 6 Statistics Questions With Answers Maharashtra Board

Statistics Practice Set 6.2 Question 1.
The following table shows classification of number of workers and the number of hours they work in a software company. Find the median of the number of hours they work.

Solution:

Cumulative frequency which is just greater than (or equal) to 500 is 650.
∴ The median class is 10 – 12.
Now, L = 10, f = 500, cf = 150, h = 2

∴ The median of the number of hours the workers work is 11.4 hours.

10th Class Algebra Practice Set 6.2 Question 2.
The frequency distribution table shows the number of mango trees in a grove and their yield of mangoes. Find the median of data.

Solution:

Here, total frequency = ∑f_i = N = 250
∴ \(\frac { N }{ 2 } \) = \(\frac { 250 }{ 2 } \) = 125
Cumulative frequency which is just greater than (or equal) to 125 is 153.
∴ The median class is 150 – 200.
Now, L = 150, f = 90, cf = 63, h = 50

∴ The median of the given data is 184 mangoes (approx).

Statistics Class 10 Practice Set 6.2 Question 3.
The following table shows the classification of number of vehicles and their speeds on Mumbai-Pune express way. Find the median of the data.

Solution:

Here, total frequency = ∑f_i = N = 200
∴ \(\frac { N }{ 2 } \) = \(\frac { 200 }{ 2 } \) = 100
Cumulative frequency which is just greater than (or equal) to 100 is 184.
∴ The median class is 74.5 – 79.5.
Now, L = 74.5, f = 85, cf = 99, h = 5

∴ The median of the given data is 75 km/hr (approx.).

Practice Set 6.2 Geometry Class 10 Question 4.
The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.

Solution:

Cumulative frequency which is just greater than (or equal) to 52.5 is 67.
∴ The median class is 50 – 60.
Now, L = 50, f = 20, cf = 47, h = 10

∴ The median of the productions is 52750 bulbs (approx.).

Practice Set 6.2 Question 1.
If the number of scores is odd, then the (\(\frac { n+1 }{ 2 } \))^th score is the median of the data. That is, the number of scores below as well as above \({ K }_{ \frac { n+1 }{ 2 } }\) is \(\frac { n-1 }{ 2 } \) Verify the fact by taking n = 2m + I. (Textbk pg. no. 139)
Solution:

The sequence of the terms of scores is 1,2, 3, …….., m, m + 1, m + 2, …, 2m + 1
Thus, we have to prove that m + 1 is the middle term if the number of scores is 2m + 1
i.e. to prove
number of terms from 1 to m = number of terms from m + 2 to 2m + 1 …(i)
Consider the L.H.S. of equation (i)
The sequence is an A.P. with a = 1,d = 1, t_n1 = m
t_n1 = a + (n₁ – 1) d
∴ m = 1 + (n₁ – 1)1
∴ m = 1 + n₁ – 1
∴ m = n₁
Consider the R.H.S. of equation (ii)
The sequence is an A.P. with a = m + 2, d = 1, t_n2 = 2m + 1
t_n2 = a + (n₂ – 1)d
∴ 2m + 1 = m + 2 + (n₂ – 1)1
∴ 2m + 1 = m + n₂ + 1
∴ m = n₂
∴ number of terms from 1 to m = number of terms from m + 2 to 2m + 1 = m = \(\frac { n-1 }{ 2 } \)
∴ m + 1 is the middle term if the number of scores is 2m + 1.

Question 2.
If the number of the scores is even, then the mean of the middle two terms is the median. This is because the number of terms below \({ K }_{ \frac { n }{ 2 } }\) and above \({ K }_{ \frac { n+2 }{ 2 } }\) is equal, which is \(\frac { n-2 }{ 2 } \). Verify this by taking n = 2m. (Textbook pg. no. 139)
Solution:

The sequence of the terms of scores is 1, 2, 3 … m – 1, m, m + 1, m + 2,…., 2m
Thus, we have to prove that m and m + 1 are the middlemost terms if the number of scores is 2m.
i.e. to prove
number of terms from 1 to m – 1 = number of terms from m + 2 to 2m …(i)
Consider the L.H.S. of equation (i)
The sequence is an A.P. with a = 1, d = 1, t_n1 = m – 1
t_n1 = a + (n₁ – 1)d
∴ m – 1 = 1 + (n₁ – 1)1
∴m – 1 = 1 + n₁ – 1
∴ n₁ = m – 1
Consider the R.H.S. of equation (i)
The sequence is an A.P. with a = m + 2, d = 1, t_n2= 2m
t_n2= a + (n₂ – 1) d
∴ 2m = m + 2 + (n₂ – 1)1
∴ 2m = m + 2 + n₂ – 1
∴ n₂ = m – 1
∴ number of terms from 1 to m – 1 = number of terms from m + 2 to 2m = m – 1 = \(\frac { n-2 }{ 2 } \)
∴ m and m + 1 are the middlemost terms if the number of scores is 2m.

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 5 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

10th Standard Maths 1 Problem Set 5 Chapter 5 Probability Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Problem Set 5 Chapter 5 Probability Questions With Answers Maharashtra Board

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.

ii. Let B be the event that the captain selected will be a defender.
There are 6 defenders in the hockey team.

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

∴ 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.

∴ The probability that Pranali gets a red balloon is \(\frac { 2 }{ 9 } \)

ii. Let B be the event that Pranali gets a blue balloon.

∴ The probability that Pranali gets a blue balloon is \(\frac { 1 }{ 3 } \).

iii. Let C be the event that Pranali gets a green balloon.

∴ 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}

∴ The probability that Rutuja picks a blue pen is \(\frac { 1 }{ 2 } \).

Question 7.
Six faces of a die are as shown below.

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

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

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

ii. Let B be the event that the ticket drawn bears a complete square number.
∴ B = {1,4,9,16,25}
∴ n(B) = 5

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.

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

∴ 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.

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.

ii. Let B be the event that the spinning arrow comes to rest at an odd number.

iii. Let C be the event that the spinning arrow comes to rest at a number greater than 2.

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}

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

ii. Let B be the event that the card drawn shows a number less than 1.
∴ B = {0}
∴ n(B) = 1

iii. Let C be the event that the card drawn shows a whole number.
∴ C = {0,1, 2, 3, 4, 5}
∴ n(C) = 6

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

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

ii. Let B be the event that the ball drawn is not red.
B = {W₁,W₂,W₃,G₁,G₂,G₃}
∴ n(B) = 6

iii. Let C be the event that the ball drawn is red or white.
∴ C = {R₁, R₂, R₃, W₁, W₂, W₃}
∴ n(C) = 6

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

∴ 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

∴ 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

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

iii. Let C be the event that the number so formed is a multiple of 11.
∴ C = {11,22,33,44}
∴ n(C) = 4

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

∴ The probability that the product of the digits on the upper face is zero is \(\frac { 11 }{ 36 } \).

Class 10 Maths Digest

• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Practice Set 5.4 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 6.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

10th Standard Maths 1 Practice Set 6.1 Chapter 6 Statistics Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 6.1 Chapter 6 Statistics Questions With Answers Maharashtra Board

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.

Solution:

∴ 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.

Solution:
Let us take the assumed mean (A) = 550

∴ 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.

Solution:

∴ 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.

Solution:
Let us take the assumed mean (A) = 37.5

∴ 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.

Solution:
Here, we take A = 1250 and g = 500

∴ 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.

Solution:
Here, we take A = 2500 and g = 1000.

∴ 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)

Solution:

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)

Solution:

∴ The mean of investments in health insurance is ₹ 2140.
Assumed mean method:

∴ 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.

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

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:

∴ The mean of the funds is ₹ 1390.

ii. Assumed mean method:
Here, A = 1250

∴ The mean calculated by assumed mean method is the same.

iii. Step deviation method:
Here, we take A = 1750 and g = 250

∴ The mean of the funds is ₹ 1390.

Class 10 Maths Digest

• Practice Set 6.1 Class 10 Answers
• Practice Set 6.2 Class 10 Answers
• Practice Set 6.3 Class 10 Answers
• Practice Set 6.4 Class 10 Answers
• Practice Set 6.5 Class 10 Answers
• Practice Set 6.6 Class 10 Answers
• Problem Set 6 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 5 Probability.

10th Standard Maths 1 Practice Set 5.4 Chapter 5 Probability Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 5.4 Chapter 5 Probability Questions With Answers Maharashtra Board

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) = \(\frac { 1 }{ 4 } \)

Class 10 Maths Digest

• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Practice Set 5.4 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 1.1 8th Std Maths Answers Solutions Chapter 1 Rational and Irrational Numbers.

Rational and Irrational Numbers Class 8 Maths Chapter 1 Practice Set 1.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 1.1 Chapter 1 Solutions Answers

Question 1.
Show the following numbers on a number line. Draw a separate number line for each example.
i. \(\frac{3}{2}, \frac{5}{2},-\frac{3}{2}\)
ii. \(\frac{7}{5}, \frac{-2}{5}, \frac{-4}{5}\)
iii. \(\frac{-5}{8}, \frac{11}{8}\)
iv. \(\frac{13}{10}, \frac{-17}{10}\)
Solution:
i. \(\frac{3}{2}, \frac{5}{2},-\frac{3}{2}\)

Here, the denominator of each fraction is 2.
∴ Each unit will be divided into 2 equal parts.

ii. \(\frac{7}{5}, \frac{-2}{5}, \frac{-4}{5}\)

Here, the denominator of each fraction is 5.
∴ Each unit will be divided into 5 equal parts.

iii. \(\frac{-5}{8}, \frac{11}{8}\)

Here, the denominator of each fraction is 8.
∴ Each unit will be divided into 8 equal parts.

iv. \(\frac{13}{10}, \frac{-17}{10}\)

Here, the denominator of each fraction is 10.
∴ Each unit will be divided into 10 equal parts.

Question 2.
Observe the number line and answer the questions.

i. Which number is indicated by point B?
ii. Which point indicates the number \(1\frac { 3 }{ 4 }\) ?
iii. State whether the statement, ‘the point D denotes the number \(\frac { 5 }{ 2 }\) is true or false.
Solution:
Here, each emit is divided into 4 equal parts.
i. Point B is marked on the 10th equal part on the left side of O.
∴ The number indicated by point B is \(\frac { -10 }{ 4 }\).

ii.

Point C is marked on the 7th equal part on the right side of O.
∴ The number \(1\frac { 3 }{ 4 }\) is indicated by point C.

iii. True
Point D is marked on the 10th equal part on the right side of O.
∴ D denotes the number \(\frac{10}{4}=\frac{5 \times 2}{2 \times 2}=\frac{5}{2}\)

Std 8 Maths Digest

• Practice Set 1.1 Class 8 Answers
• Practice Set 1.2 Class 8 Answers
• Practice Set 1.3 Class 8 Answers
• Practice Set 1.4 Class 8 Answers
• Practice Set 2.1 Class 8 Answers
• Practice Set 2.2Class 8 Answers
• Practice Set 2.3 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 11.2 8th Std Maths Answers Solutions Chapter 11 Statistics.

Statistics Class 8 Maths Chapter 11 Practice Set 11.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 11.2 Chapter 11 Solutions Answers

practice set 11.2 8th class Question 1.
Observe the following graph and answer the questions.

i. State the type of the graph.
ii. How much is the savings of Vaishali in the month of April?
iii. How much is the total of savings of Saroj in the months March and April?
iv. How much more is the total savings of Savita than the total savings of Megha?
v. Whose savings in the month of April is the least?
Solution:
i. The given graph is a subdivided bar graph.
ii. Vaishali’s savings in the month of April is Rs 600.
iii. Total savings of Saroj in the months of March and April is Rs 800.
iv. Savita’s total saving = Rs 1000, Megha’s total saving = Rs 500
∴ difference in their savings = 1000 – 500 = Rs 500.
Savita’s saving is Rs 500 more than Megha.
v. Megha’s savings in the month of April is the least.

practice set 11.2 Question 2.
The number of boys and girls, in std 5 to std 8 in a Z.P. School is given in the table. Draw a subdivided bar graph to show the data. (Scale : On Y axis, 1cm = 10 students)

Standard	5th	6th	7th	8th
Boys	34	26	21	25
Girls	17	14	14	20

Solution:

Standard	5th	6th	7th	8th
Boys	34	26	21	25
Girls	17	14	14	20
Total	51	40	35	45

Statistics class 8 practice set 11.1 Question 3.
In the following table number of trees planted in the year 2016 and 2017 in four towns is given. Show the data with the help of subdivided bar graph.

Year\Town	karjat	Wadgaon	Shivapur	Khandala
2016	150	250	200	100
2017	200	300	250	150

Solution:

Year\Town	karjat	Wadgaon	Shivapur	Khandala
2016	150	250	200	100
2017	200	300	250	150
Total	350	550	450	250

Statistics class 8 Question 4.
In the following table, data of the transport means used by students in 8th standard for commutation between home and school is given. Draw a subdivided bar diagram to show the data.
(Scale: On Y axis: 1 cm = 500 students)

Means of commutation\Town	Paithan	Yeola	Shahapur
Cycle	3250	1500	1250
Bus and auto	750	500	500
On foot	1000	1000	500

Solution:

Means of commutation\Town	Paithan	Yeola	Shahapur
Cycle	3250	1500	1250
Bus and auto	750	500	500
On foot	1000	1000	500
Total	5000	3000	2250

Std 8 Maths Digest

• Practice Set 11.1 Class 8 Answers
• Practice Set 11.2 Class 8 Answers
• Practice Set 11.3 Class 8 Answers
• Practice Set 12.1 Class 8 Answers
• Practice Set 12.2 Class 8 Answers
• Practice Set 13.1 Class 8 Answers
• Practice Set 13.2 Class 8 Answers
• Practice Set 14.1 Class 8 Answers
• Practice Set 14.2 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 10.1 8th Std Maths Answers Solutions Chapter 10 Division of Polynomials.

Division of Polynomials Class 8 Maths Chapter 10 Practice Set 10.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 10.1 Chapter 10 Solutions Answers

Question 1.
Divide and write the quotient and the remainder.
i. 21m² ÷ 7m
ii. 40a³ ÷ (-10a)
iii. (- 48p⁴) ÷ (- 9p²)
iv. 40m⁵ ÷ 30m³
v. (5x³ – 3x²) ÷ x²
vi. (8p³ – 4p²) ÷ 2p²
vii. (2y³ + 4y² + 3 ) ÷ 2y²
viii. (21x⁴ – 14x² + 7x) ÷ 7x³
ix. (6x⁵ – 4x⁴ + 8x³ + 2x²) ÷ 2x²
x. (25m⁴ – 15m³ + 10m + 8) ÷ 5m³
Solution:
i. 21m² ÷ 7m

∴ Quotient = 3m
Remainder = 0

ii. 40a³ ÷ (-10a)

∴ Quotient = -4a²
Remainder = 0

iii. (- 48p⁴) ÷ (- 9p²)

∴ Quotient = \(\frac { 16 }{ 3 }\) p²
Remainder = 0

iv. 40m⁵ ÷ 30m³

∴ Quotient = \(\frac { 4 }{ 3 }\) m²
Remainder = 0

v. (5x³ – 3x²) ÷ x²

∴ Quotient = 5x – 3
Remainder = 0

vi. (8p³ – 4p²) ÷ 2p²

∴ Quotient = 4p – 2
Remainder = 0

vii. (2y³ + 4y² + 3 ) ÷ 2y²

∴ Quotient = y + 2
Remainder = 3

viii. (21x⁴ – 14x² + 7x) ÷ 7x³

∴ Quotient = 3x
Remainder = -14x² + 7x

ix. (6x⁵ – 4x⁴ + 8x³ + 2x²) ÷ 2x²

∴ Quotient = 3x³ – 2x² + 4x + 1
Remainder = 0

x. (25m⁴ – 15m³ + 10m + 8) ÷ 5m³

∴ Quotient = 5m – 3
Remainder = 10m + 8

Maharashtra Board Class 8 Maths Chapter 10 Division of Polynomials Practice Set 10.1 Intext Questions and Activities

Question 1.
Fill in the blanks in the following examples. (Textbook pg. no. 61)

1. 2a + 3a = __
2. 7b – 4b = __
3. 3p × p² = __
4. 5m² × 3m² = __
5. (2x + 5y) × \(\frac { 3 }{ x }\) = __
6. (3x² + 4y) × (2x + 3y) = __

Solution:

1. 2a + 3a = 5a
2. 7b – 4b = 3b
3. 3p × p² = 3p³
4. 5m² × 3m² = 15m⁴
5. (2x + 5y) × \(\frac { 3 }{ x }\) = \(6+\frac { 15y }{ x }\)
6. (3x² + 4y) × (2x + 3y) = 6x³ + 9x²y + 8xy + 12y²

Std 8 Maths Digest

• Practice Set 8.1 Class 8 Answers
• Practice Set 8.2 Class 8 Answers
• Practice Set 8.3 Class 8 Answers
• Practice Set 9.1 Class 8 Answers
• Practice Set 9.2 Class 8 Answers
• Practice Set 10.1 Class 8 Answers
• Practice Set 10.2 Class 8 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 7 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

9th Standard Maths 1 Problem Set 7 Chapter 7 Statistics Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Problem Set 7 Chapter 7 Statistics Questions With Answers Maharashtra Board

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

i. Which of the following data is not primary ?
(A) By visiting a certain class, gathering information about attendance of students.
(B) By actual visit to homes, to find number of family members.
(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.
(D) Review the cleanliness status of canals by actually visiting them.
Answer:
(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.

ii. What is the upper class limit for the class 25 – 35?
(A) 25
(B) 35
(C) 60
(D) 30
Answer:
(B) 35

iii. What is the class-mark of class 25 – 35?
(A) 25
(B) 35
(C) 60
(D) 30
Answer:
(D) 30

iv. If the classes are 0 – 10, 10 – 20, 20 – 30, …, then in which class should the observation 10 be included?
(A) 0 – 10
(B) 10 – 20
(C) 0 – 10 and 10-20 in these 2 classes
(D) 20 – 30
Answer:
(B) 10 – 20

v. If \(\overline { x }\) is the mean of x₁, x₂, ……. , x_n and \(\overline { y }\) is the mean of y₁, y₂, ….. y_n and \(\overline { z }\) is the mean of x₁,x₂, …… , x_n , y₁, y₂, …. y_n , then z = ?

Answer:
x₁, x₂, x₃, ……. , x_n
∴ \(\overline{x}=\frac{\sum x}{\mathrm{n}}\)
∴ n\(\overline{x}\) = ∑x
Similarly, n\(\overline{y}\) = ∑y
Now,

\(\text { (A) } \frac{\overline{x}+\overline{y}}{2}\)

vi. The mean of five numbers is 50, out of which mean of 4 numbers is 46, find the 5th number.
(A) 4
(B) 20
(C) 434
(D) 66
Answer:
5^th number = Sum of five numbers – Sum of four numbers
= (5 x 50) – (4 x 46)
= 250 – 184
= 66
(D) 66

vii. Mean of 100 observations is 40. The 9th observation is 30. If this is replaced by 70 keeping all other observations same, find the new mean.
(A) 40.6
(B) 40.4
(C) 40.3
(D) 40.7
Answer:
New mean = \(\frac { 4000-30+70 }{ 100 }\)
= 40.4
(B) 40.4

viii. What is the mode of 19, 19, 15, 20, 25, 15, 20, 15?
(A) 15
(B) 20
(C) 19
(D) 25
Answer:
(A) 15

ix. What is the median of 7, 10, 7, 5, 9, 10 ?
(A) 7
(B) 9
(C) 8
(D) 10
Answer:
(C) 8

x. From following table, what is the cumulative frequency of less than type for the class 30 – 40?

(A) 13
(B) 15
(C) 35
(D) 22
Answer:
Cumulative frequency of less than type for the class 30 – 40 = 7 + 3 + 12 + 13 = 35
(C) 35

Question 2.
The mean salary of 20 workers is ₹10,250. If the salary of office superintendent is added, the mean will increase by ₹ 750. Find the salary of the office superintendent.
Solution:
\( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations = Mean x Total number of observations
The mean salary of 20 workers is ₹ 10,250.
∴ Sum of the salaries of 20 workers
= 20 x 10,250
= ₹ 2,05,000 …(i)
If the superintendent’s salary is added, then mean increases by 750
new mean = 10, 250 + 750 = 11,000
Total number of people after adding superintendent = 20 + 1 = 21
∴ Sum of the salaries including the superintendent’s salary = 21 x 11,000 = ₹ 2,31,000 …(ii)
∴ Superintendent salary = sum of the salaries including superintendent’s salary – sum of salaries of 20 workers
= 2, 31,00 – 2,05,000 …[From (i) and (ii)]
= 26,000
∴ The salary of the office superintendent is ₹ 26,000.

Question 3.
The mean of nine numbers is 77. If one more number is added to it, then the mean increases by 5. Find the number added in the data.
Solution:
∴ \( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations = Mean x Total number of observations mean of nine numbers is 77
∴ sum of 9 numbers = 11 x 9 = 693 …(i)
If one more number is added, then mean increases by 5
mean of 10 numbers = 77 + 5 = 82
∴ sum of the 10 numbers = 82 x 10 = 820 …(ii)
∴ Number added = sum of the 10 numbers – sum of the 9 numbers = 820 – 693 … [From (i) and (ii)]
= 127
∴ The number added in the data is 127.

Question 4.
The monthly maximum temperature of a city is given in degree Celsius in the following data. By taking suitable classes, prepare the grouped frequency distribution table
29.2, 29.0, 28.1, 28.5, 32.9, 29.2, 34.2, 36.8, 32.0, 31.0, 30.5, 30.0, 33, 32.5, 35.5, 34.0, 32.9, 31.5, 30.3, 31.4, 30.3, 34.7, 35.0, 32.5, 33.5.29.0. 29.5.29.9.33.2.30.2
From the table, answer the following questions.
i. For how many days the maximum temperature was less than 34°C?
ii. For how many days the maximum temperature was 34°C or more than 34°C?
Solution:

i. Number of days for which the maximum temperature was less than 34°C
= 8 + 8 + 8 = 24
ii. Number of days for which the maximum temperature was 34°C or more than 34°C
= 5 + 1 = 6

Question 5.
If the mean of the following data is 20.2, then find the value of p.

Solution:

∴ 20.2 (30 + p) = 610 + 20p
∴ 606 + 20.2p = 610 + 20p
∴ 20.2p – 20p = 610 – 606
∴ 0.2p = 4
∴ p = \(\frac { 4 }{ 0.2 }\) = \(\frac { 40 }{ 2 }\) = 20
∴ p = 20

Question 6.
There are 68 students of 9th standard from Model Highschool, Nandpur. They have scored following marks out of 80, in written exam of mathematics.
70, 50, 60, 66, 45, 46, 38, 30, 40, 47, 56, 68,
80, 79, 39, 43, 57, 61, 51, 32, 42, 43, 75, 43,
36, 37, 61, 71, 32, 40, 45, 32, 36, 42, 43, 55,
56, 62, 66, 72, 73, 78, 36, 46, 47, 52, 68, 78,
80, 49, 59, 69, 65, 35, 46, 56, 57, 60, 36, 37,
45, 42, 70, 37,45, 66, 56, 47
By taking classes 30 – 40, 40 – 50, …. prepare the less than type cumulative frequency table. Using the table, answer the following questions:

i. How many students have scored marks less than 80?
ii. How many students have scored marks less than 40?
iii. How many students have scored marks less than 60?
Solution:
Class

i. 66 students have scored marks less than 80.
ii. 14 students have scored marks less than 40.
iii. 45 students have scored marks less than 60.

Question 7.
By using data in example (6), and taking classes 30 – 40, 40 – 50,… prepare equal to or more than type cumulative frequency table and answer the following questions based on it.
i. How many students have scored marks 70 or more than 70?
ii. How many students have scored marks 30 or more than 30?
Solution:

i. 11 students have scored marks 70 or more than 70.
ii. 68 students have scored marks 30 or more than 30.

Question 8.
There are 10 observations arranged in ascending order as given below.
45, 47, 50, 52, JC, JC + 2, 60, 62, 63, 74. The median of these observations is 53.
Find the value of JC. Also find the mean and the mode of the data.
Solution:
i. Given data in ascending order:
45,47, 50, 52, x, JC+2, 60, 62, 63, 74.
∴ Number of observations (n) = 10 (i.e., even)
∴ Median is the average of middle two observations
Here, the 5th and 6th numbers are in the middle position.
∴
∴ 106 = 2x + 2
∴ 106 – 2 = 2x
∴ 104 = 2x
∴ x = 52
∴ The given data becomes:
45, 47, 50, 52, 52, 54, 60, 62, 63, 74.

∴ The mean of the given data is 55.9.

iii. Given data in ascending order:
45, 47, 50, 52, 52, 54, 60, 62, 63, 74.
∴ The observation repeated maximum number of times = 52
∴ The mode of the given data is 52.

Maharashtra Board Class 9 Maths Chapter 7 Statistics Problem Set 7 Intext Questions and Activities

Question 1.
To show following information diagrammatically, which type of bar- diagram is suitable?
i. Literacy percentage of four villages.
ii. The expenses of a family on various items.
iii. The numbers of girls and boys in each of five divisions.
iv. The number of people visiting a science exhibition on each of three days.
v. The maximum and minimum temperature of your town during the months from January to June.
vi. While driving a two-wheeler, number of people wearing helmets and not wearing helmet in 100 families.
(Textbook pg. no. 112)
Solution:
i. Percentage bar diagram
ii. Sub-divided bar diagram
iii. Sub-divided bar diagram
iv. Sub-divided bar diagram
v. Sub-divided bar diagram
vi. Sub-divided bar diagram

Question 2.
You gather information for several reasons. Take a few examples and discuss whether the data is primary or secondary.
(Textbook pg. no, 113)
[Students should attempt the above activity on their own.]

Class 9 Maths Digest

• Practice Set 6.1 Class 9 Answers
• Practice Set 6.2 Class 9 Answers
• Problem Set 6 Class 9 Answers
• Practice Set 7.1 Class 9 Answers
• Practice Set 7.2 Class 9 Answers
• Practice Set 7.3 Class 9 Answers
• Practice Set 7.4 Class 9 Answers
• Practice Set 7.5 Class 9 Answers
• Problem Set 7 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

9th Standard Maths 1 Practice Set 7.5 Chapter 7 Statistics Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 7.5 Chapter 7 Statistics Questions With Answers Maharashtra Board

Question 1.
Yield of soyabean per acre in quintal in Mukund’s field for 7 years was 10, 7, 5,3, 9, 6, 9. Find the mean of yield per acre.
Solution:

Mean = 7
The mean of yield per acre is 7 quintals.

Question 2.
Find the median of the observations, 59, 75, 68, 70, 74, 75, 80.
Solution:
Given data in ascending order:
59, 68, 70, 74, 75, 75, 80
∴ Number of observations(n) = 7 (i.e., odd)
∴ Median is the middle most observation
Here, 4th number is at the middle position, which is = 74
∴ The median of the given data is 74.

Question 3.
The marks (out of 100) obtained by 7 students in Mathematics examination are given below. Find the mode for these marks.
99, 100, 95, 100, 100, 60, 90
Solution:
Given data in ascending order:
60, 90, 95, 99, 100, 100, 100
Here, the observation repeated maximum number of times = 100
∴ The mode of the given data is 100.

Question 4.
The monthly salaries in rupees of 30 workers in a factory are given below.
5000, 7000, 3000, 4000, 4000, 3000, 3000,
3000, 8000, 4000, 4000, 9000, 3000, 5000,
5000, 4000, 4000, 3000, 5000, 5000, 6000,
8000, 3000, 3000, 6000, 7000, 7000, 6000,
6000, 4000
From the above data find the mean of monthly salary.
Solution:

∴ The mean of monthly salary is ₹ 4900.

Question 5.
In a basket there are 10 tomatoes. The weight of each of these tomatoes in grams is as follows:
60, 70, 90, 95, 50, 65, 70, 80, 85, 95.
Find the median of the weights of tomatoes.
Solution:
Given data in ascending order:
50, 60, 65, 70, 70, 80 85, 90, 95, 95
∴ Number of observations (n) = 10 (i.e., even)
∴ Median is the average of middle two observations
Here, 5th and 6th numbers are in the middle position
∴ Median = \(\frac { 70+80 }{ 2 }\)
∴ Median = \(\frac { 150 }{ 2 }\)
∴ The median of the weights of tomatoes is 75 grams.

Question 6.
A hockey player has scored following number of goals in 9 matches: 5, 4, 0, 2, 2, 4, 4, 3,3.
Find the mean, median and mode of the data.
Solution:
i. Given data: 5, 4, 0, 2, 2, 4, 4, 3, 3.
Total number of observations = 9


∴ The mean of the given data is 3.

ii. Given data in ascending order:
0,2, 2, 3, 3, 4, 4, 4,5
∴ Number of observations(n) = 9 (i.e., odd)
∴ Median is the middle most observation
Here, the 5^th number is at the middle position, which is 3.
∴ The median of the given data is 3.

iii. Given data in ascending order:
0,2, 2, 3, 3, 4, 4, 4,5
Here, the observation repeated maximum number of times = 4
∴ The mode of the given data is 4.

Question 7.
The calculated mean of 50 observations was 80. It was later discovered that observation 19 was recorded by mistake as 91. What Was the correct mean?
Solution:
Here, mean = 80, number of observations = 50
\( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations = Mean x Total number of observations
∴ The sum of 50 observations = 80 x 50
= 4000
One of the observation was 19. However, by mistake it was recorded as 91.
Sum of observations after correction = sum of 50 observation + correct observation – incorrect observation
= 4000 + 19 – 91
= 3928
∴ Corrected mean

= 78.56
∴ The corrected mean is 78.56.

Question 8.
Following 10 observations are arranged in ascending order as follows. 2, 3 , 5 , 9, x + 1, x + 3, 14, 16, 19, 20. If the median of the data is 11, find the value of x.
Solution:
Given data in ascending order :
2, 3, 5, 9, x + 1, x + 3, 14, 16, 19, 20.
∴ Number if observations (n) = 10 (i.e., even)
∴ Median is the average of middle two observations
Here, the 5th and 6th numbers are in the middle position.
∴ \( \text { Median }=\frac{(x+1)+(x+3)}{2}\)
∴ 11 = \(\frac { 2x+4 }{ 2 }\)
∴ 22 = 2x + 4
∴ 22 – 4 = 2x
∴ 18 = 2x
∴ x = 9

Question 9.
The mean of 35 observations is 20, out of which mean of first 18 observations is 15 and mean of last 18 observations is 25. Find the 18th observation.
Solution:
\( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations
= Mean x Total number of observations
The mean of 35 observations is 20
∴ Sum of 35 observations = 20 x 35 = 700 ,..(i)
The mean of first 18 observations is 15
Sum of first 18 observations =15 x 18
= 270 …(ii)
The mean of last 18 observations is 25 Sum of last 18 observations = 25 x 18
= 450 …(iii)
∴ 18^th observation = (Sum of first 18 observations + Sum of last 18 observations) – (Sum of 35 observations)
= (270 + 450) – (700) … [From (i), (ii) and (iii)]
= 720 – 700 = 20
The 18^th observation is 20.

Question 10.
The mean of 5 observations is 50. One of the observations was removed from the data, hence the mean became 45. Find the observation which was removed.
Solution:
\( \text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}\)
∴ The sum of all observations = Mean x Total number of observations
The mean of 5 observations is 50
Sum of 5 observations = 50 x 5 = 250 …(i)
One observation was removed and mean of remaining data is 45.
Total number of observations after removing one observation = 5 – 1 = 4
Now, mean of 4 observations is 45.
∴ Sum of 4 observations = 45 x 4 = 180 …(ii)
∴ Observation which was removed
= Sum of 5 observations – Sum of 4 observations = 250 – 180 … [From (i) and (ii)]
= 70
∴ The observation which was removed is 70.

Question 11.
There are 40 students in a class, out of them 15 are boys. The mean of marks obtained by boys is 33 and that for girls is 35. Find out the mean of all students in the class.
Solution:
Total number of students = 40
Number of boys =15
∴ Number of girls = 40 – 15 = 25
The mean of marks obtained by 15 boys is 33
Here, sum of the marks obtained by boys
= 33 x 15
= 495 …(i)
The mean of marks obtained by 25 girls is 35 Sum of the marks obtained by girls = 35 x 25
= 875 …(ii)
Sum of the marks obtained by boys and girls = 495 + 875 … [From (i) and (ii)]
= 1370
∴ Mean of all the students

= 34.25
∴ The mean of all the students in the class is 34.25.

Question 12.
The weights of 10 students (in kg) are given below:
40, 35, 42, 43, 37, 35, 37, 37, 42, 37. Find the mode of the data.
Solution:
Given data in ascending order:
35, 35, 37, 37, 37, 37, 40, 42, 42, 43
∴ The observation repeated maximum number of times = 37
∴ Mode of the given data is 37 kg

Question 13.
In the following table, the information is given about the number of families and the siblings in the families less than 14 years of age. Find the mode of the data.

Solution:
Here, the maximum frequency is 25.
Since, Mode = observations having maximum frequency
∴ The mode of the given data is 2.

Question 14.
Find the mode of the following data.

Solution:
Here, the maximum frequency is 9.
Since, Mode = observations having maximum frequency
But, this is the frequency of two observations.
∴ Mode = 35 and 37

Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.5 Intext Questions and Activities

Question 1.
The first unit test of 40 marks was conducted for a class of 35 students. The marks obtained by the students were as follows. Find the mean of the marks.
40, 35, 30, 25, 23, 20, 14, 15, 16, 20, 17, 37, 37, 20, 36, 16, 30, 25, 25, 36, 37, 39, 39, 40, 15, 16, 17, 30, 16, 39, 40, 35, 37, 23, 16.
(Textbook pg, no. 123)
Solution:
Here, we can add all observations, but it will be a tedious job. It is easy to make frequency distribution table to calculate mean.

= 27.31 marks (approximately)
∴ The mean of the mark is 27.31.

Class 9 Maths Digest

• Practice Set 6.1 Class 9 Answers
• Practice Set 6.2 Class 9 Answers
• Problem Set 6 Class 9 Answers
• Practice Set 7.1 Class 9 Answers
• Practice Set 7.2 Class 9 Answers
• Practice Set 7.3 Class 9 Answers
• Practice Set 7.4 Class 9 Answers
• Practice Set 7.5 Class 9 Answers
• Problem Set 7 Class 9 Answers