Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5

Maharashtra State Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5

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:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 1
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:
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 2
∴ 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
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 3
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 4
∴ 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 5th 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
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 5
= 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)
∴ 18th 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 18th 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
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 6
= 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.
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 7
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.
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 8
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.
Maharashtra Board Class 9 Maths Solutions Chapter 7 Statistics Practice Set 7.5 9
= 27.31 marks (approximately)
∴ The mean of the mark is 27.31.

Maharashtra Board Class 9 Maths Solutions

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Maharashtra Board Class 9 Maths Solutions Chapter 6 Financial Planning Problem Set 6

Maharashtra State Board Class 9 Maths Solutions Chapter 6 Financial Planning Problem Set 6

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

i. For different types of investments what is the maximum permissible amount under section 80C of income tax ?
(A) ₹ 1,50,000
(B) ₹ 2,50,000
(C) ₹ 1,00,000
(D) ₹ 2,00,000
Answer:
(A) ₹ 1,50,000

ii. A person has earned his income during the financial year 2017-18. Then his assessment year is….
(A) 2016 – 17
(B) 2018 – 19
(C) 2017 – 18
(D) 2015 – 16
Answer:
(B) 2018 – 19

Question 2.
Mr. Shekhar spends 60% of his income. From the balance he donates ₹ 300 to an orphanage. He is then left with ₹ 3,200. What is his income ?
Solution:
Let the income of Shekhar be ₹ x.
Shekhar spends 60% of his income.
∴ Shekhar’s expenditure = 60% of x
∴ Amount remaining with Shekhar = (100 – 60)% of x
= 40% of x
= \(\frac { 1 }{ 2 }\) × x
= 0.4x
From the balance left, he donates ₹ 300 to an orphanage.
∴ Amount left with Shekhar = 0.4x – 300
Now, the amount left with him is ₹ 3200.
∴ 3200 = 0.4x- 300
∴ 0.4x = 3500
Maharashtra Board Class 9 Maths Solutions Chapter 6 Financial Planning Problem Set 6 1
∴ The income of Mr. Shekhar is ₹ 8750.

Question 3.
Mr. Hiralal invested ₹ 2,15,000 in a Mutual Fund. He got ₹ 3,05,000 after 2 years. Mr. Ramniklal invested ₹ 1,40,000 at 8% compound interest for 2 years in a bank. Find out the percent gain of each of them. Whose investment was more profitable ?
Solution:
Mr. Hiralal:
Amount invested by Mr. Hiralal in mutual fund = ₹ 2,15,000
Amount received by Mr. Hiralal = ₹ 3,05,000
∴ Mr. Hiralal’s profit = Amount received – Amount invested
= 305000 – 215000 = ₹ 90000
Mr. Hirala’s percentage of profit
= \(\frac { 90000 }{ 215000 }\) × 100
= 41.86%

Mr. Ramniklal:
P = ₹ 140000, R = 8%, n = 2 years
∴ Compound interest (I)
= A – P
Maharashtra Board Class 9 Maths Solutions Chapter 6 Financial Planning Problem Set 6 2
= 140000 [(1 + 0.08)2 – 1]
= 140000 [ (1.08)2 – 1]
= 140000(1.1664 – 1)
= 140000 x 0.1664
= ₹ 23296
∴ Mr. Ramniklal’s percentage of profit
= \(\frac { 23296 }{ 140000 }\) × 100
= 16.64%
∴ The percentage gains of Mr. Hiralal and Mr. Ramniklal are 41.86% and 16.64% respectively, and hence, Mr. Hiralal’s investment was more profitable.

Question 4.
At the start of a year there were ₹ 24,000 in a savings account. After adding ₹ 56,000 to this the entire amount was invested in the bank at 7.5% compound interest. What will be the total amount after 3 years ?
Solution:
Here, P = 24000 + 56000
= ₹ 80000
R = 7.5%, n = 3 years
Total amount after 3 years
Maharashtra Board Class 9 Maths Solutions Chapter 6 Financial Planning Problem Set 6 3
= 80000 (1 + 0.075)3
= 80000 (1.075)3
= 80000 x 1.242297
= 99383.76
∴ The total amount after 3 years is ₹ 99383.76.

Question 5.
Mr. Manohar gave 20% of his income to his elder son and 30% to his younger son. He gave 10% of the balance income as donation to a school. He still had ₹ 1,80,000 for himself. What was Mr. Manohar’s income ?
Solution:
Let the income of Mr. Manohar be ₹ x.
Amount given to elder son = 20% of x
Amount given to younger son = 30% of x
Total amount given to both sons = (20 + 30)% of x = 50% of x
∴ Amount remaining with Mr. Manohar = (100 – 50)% of x
= 50% of x 50
= \(\frac { 50 }{ 100 }\) × 100
= 0.5 x
He gave 10% of the balance income as donation to a school.
Amount donated to school = 10% of 0.5x
= \(\frac { 10 }{ 100 }\) × 0.5x
= 0.05x
∴ Amount remaining with Mr. Manohar after donating to school = 0.5x – 0.05x
= 0.45x
Mr. Manohar still had 1,80,000 for himself after donating to school.
∴ 180000 = 0.45x
∴ \(x=\frac{180000}{0.45}=\frac{180000 \times 100}{0.45 \times 100}=\frac{18000000}{45}=400000\)
∴ The income of Mr. Manoliar is ₹4,00,000.

Question 6.
Kailash used to spend 85% of his income. When his income increased by 36% his expenses also increased by 40% of his earlier expenses. How much percentage of his earning he saves now ?
Solution:
Let the income of Kailash be ₹ x.
Kailash spends 85% of his income.
∴ Kailash’s expenditure = 85% of x
= \(\frac { 85 }{ 100 }\) × x = 0.85 x
Kailash’s income increased by 36%.
∴ Kailash’s new income = x + 36% of x
= x + \(\frac { 36 }{ 100 }\) × x
= x + 0.36x
= 1.36x
Kailash’s expenses increased by 40%.
∴ Kailash’s new expenditure = 0.85x + 40% of 0.85x
= 0.85x + \(\frac { 40 }{ 100 }\) × 0.85 × 100
= 0.85x + 0.4 × 0.85x
= 0.85x (1 + 0.4)
= 0.85x × 1.4
= 1.19x
∴ Kailash’s new saving = Kailash’s new income – Kailash’s new expenditure
= 1.36x – 1.19x
= 0.17x
Percentage of Kailash’s new saving
= \(\frac { 0.17x }{ 1.36x }\) × 100
= 12.5%
∴ Kailash saves 12.5% of his new earning.

Question 7.
Total income of Ramesh, Suresh and Preeti is ₹ 8,07,000. The percentages of their expenses are 75%, 80% and 90% respectively. If the ratio of their savings is 16 : 17 : 12, then find the annual saving of each of them.
Solution:
Let the annual income of Ramesh, Suresh and Preeti be ₹ x, t y and ₹ z respectively.
Total income of Ramesh, Suresh and Preeti = ₹ 8,07,000
∴ x + y + z = 807000 …(i)
Maharashtra Board Class 9 Maths Solutions Chapter 6 Financial Planning Problem Set 6 4
∴ Savings of Ramesh = 25% of x
= ₹ \(\frac { 25x }{ 100 }\) ..(ii)
Savings of Suresh = 20% of y
= ₹\(\frac { 20y }{ 100 }\) …(iii)
Savings of Preeti = 10% of z
= ₹\(\frac { 10z }{ 100 }\) …..(iv)

Ratio of their savings = 16 : 17 : 12
Let the common multiple be k.
Savings of Ramesh = ₹ 16 k … (v)
Savings of Suresh = ₹ 17 k … (vi)
Savings of Preeti = ₹ 12 k .. .(vii)
Maharashtra Board Class 9 Maths Solutions Chapter 6 Financial Planning Problem Set 6 5
∴ z = 120k …(x)
From (i), (viii), (ix) and (x), we get
64k + 85k + 120k = 807000
269k = 807000
k = \(\frac { 807000 }{ 269 }\)
k = 3000
∴ Annual saving of Ramesh = 16k
= 16 x 3000
= ₹ 48,000
Annual saving of Suresh = 17k
= 17 x 3000
= ₹ 51,000
Annual saving of Preeti = 12k
= 12 x 3000
= ₹ 36,000
The annual savings of Ramesh, Suresh and Preeti are ₹ 48,000, ₹ 51,000 and ₹ 36,000 respectively.

Question 8.
Compute the income tax payable by following individuals.
i. Mr. Kadam who is 35 years old and has a taxable income of ₹13,35,000.
ii. Mr. Khan is 65 years of age and his taxable income is ₹4,50,000.
iii. Miss Varsha (Age 26 years) has a taxable income of ₹2,30,000.
Solution:
i. Mr. Kadam is 35 years old and his taxable income is ₹13,35,000.
Mr. Kadam’s income is more than ₹ 10,00,000.
∴ Income tax = ₹1,12,500 + 30% of (taxable income -10,00,000)
= ₹ 1,12,500 + 30% of (13,35,000 – 10,00,000)
= 112500+ \(\frac { 30 }{ 100 }\) x 335000 100
= 112500+ 100500
= ₹ 213000
Education cess = 2% of income tax
= \(\frac { 2 }{ 100 }\) x 213000
= ₹ 4260.
Secondary and Higher Education cess
= 1% of income tax
= \(\frac { 1 }{ 100 }\) x 213000 100
= 2130
Total income tax = Income tax + Education cess + Secondary and higher education cess
= 213000 + 4260 + 2130 = ₹ 2,19,390
∴ Mr. Kadam will have to pay income tax of ₹ 2,19,390.

ii. Mr. Khan is 65 years old and his taxable income is ₹ 4,50,000.
Mr. Khan’s income falls in the slab ₹ 3,00,001 to ₹ 5,00,000.
∴ Income tax
= 5% of (taxable income – 300000)
= 5% of (450000 – 300000)
= \(\frac { 5 }{ 100 }\) x 150000 100
= ₹ 7500
Education cess = 2% of income tax
= \(\frac { 2 }{ 100 }\) x 7500
= ₹ 150
Secondary and Higher Education cess = 1 % of income tax
= \(\frac { 1 }{ 100 }\) x 7500
= 75
Total income tax = Income tax + Education cess + Secondary and higher education cess
= 7500+ 150 + 75
= ₹ 7725
Mr. Khan will have to pay income tax of ₹7725.

iii. Taxable income = ₹2,30,000
age = 26 years
The yearly income of Miss Varsha is less than ₹ 2,50,000.
Hence, Miss Varsha will not have to pay income tax.

Maharashtra Board Class 9 Maths Chapter 6 Financial Planning Problem Set 6 Intext Questions and Activities

Question 1.
With your parent’s help write down the income and expenditure of your family for one week. Make 7 columns for the seven days of the week. Write all expenditure under such heads as provisions, education, medical expenses, travel, clothes and miscellaneous. On the credit side write the amount received for daily expenses, previous balance and any other new income. (Textbook pg. no. 98)

Question 2.
In the holidays, write the accounts for the whole month. (Textbook pg. no. 98)

Question 3.
What is a tax? Which are different types of taxes? Find out more information on following websites
www.incometaxindia.gov.in,
www.mahavat.gov.in
www.gst.gov.in (Textbook pg. no. 99)

Question 4.
Obtain more information about different types of taxes from employees and professionals who pay taxes. (Textbook pg. no. 99)

Question 5.
Obtain information about sections 80C, 80G, 80D of the Income Tax Act. (Textbook pg. no. 103)

Question 6.
Study a PAN card and make a note of all the information it contains. (Textbook pg.no. 103)

Question 7.
Obtain information about all the devices and means used for carrying out cash minus transactions. (Textbook pg, no, 103)

Question 8.
Visit www.incometaxindia.gov.in which is a website of the Government of India. Click on the ‘incometax calculator’ menu. Fill in the form that gets downloaded using an imaginary income and imaginary deductible amounts and try to compute the income tax payable for this income. (Textbook pg.no. 107)
[Students should attempt the above activities on their own.]

Maharashtra Board Class 9 Maths Solutions

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Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3

Maharashtra State Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3

Question 1.
State the order of the surds given below.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 1
Answer:
i. 3, ii. 2, iii. 4, iv. 2, v. 3

Question 2.
State which of the following are surds Justify. [2 Marks each]
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 2
Answer:
i. \(\sqrt [ 3 ]{ 51 }\) is a surd because 51 is a positive rational number, 3 is a positive integer greater than 1 and \(\sqrt [ 3 ]{ 51 }\) is irrational.

ii. \(\sqrt [ 4 ]{ 16 }\) is not a surd because
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 3
= 2, which is not an irrational number.

iii. \(\sqrt [ 5 ]{ 81 }\) is a surd because 81 is a positive rational number, 5 is a positive integer greater than 1 and \(\sqrt [ 5 ]{ 81 }\) is irrational.

iv. \(\sqrt { 256 }\) is not a surd because
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 4
= 16, which is not an irrational number.

v. \(\sqrt [ 3 ]{ 64 }\) is not a surd because
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 5
= 4, which is not an irrational number.

vi. \(\sqrt { \frac { 22 }{ 7 } }\) is a surd because \(\frac { 22 }{ 7 }\) is a positive rational number, 2 is a positive integer greater than 1 and \(\sqrt { \frac { 22 }{ 7 } }\) is irrational.

Question 3.
Classify the given pair of surds into like surds and unlike surds. [2 Marks each]
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 6
Solution:
If the order of the surds and the radicands are same, then the surds are like surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 7
Here, the order of 2\(\sqrt { 13 }\) and 5\(\sqrt { 13 }\) is same and their radicands are also same.
∴ \(\sqrt { 52 }\) and 5\(\sqrt { 13 }\) are like surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 8
Here, the order of 2\(\sqrt { 17 }\) and 5\(\sqrt { 3 }\) is same but their radicands are not.
∴ \(\sqrt { 68 }\) and 5\(\sqrt { 3 }\) are unlike surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 9
Here, the order of 12\(\sqrt { 2 }\) and 7\(\sqrt { 2 }\) is same and their radicands are also same.
∴ 4\(\sqrt { 18 }\) and 7\(\sqrt { 2 }\) are like surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 10
Here, the order of 38\(\sqrt { 3 }\) and 6\(\sqrt { 3 }\) is same and their radicands are also same.
∴ 19\(\sqrt { 12 }\) and 6\(\sqrt { 3 }\) are like surds.

v. 5\(\sqrt { 22 }\), 7\(\sqrt { 33 }\)
Here, the order of 5\(\sqrt { 22 }\) and 7\(\sqrt { 33 }\) is same but their radicands are not.
∴ 5\(\sqrt { 22 }\) and 7\(\sqrt { 33 }\) are unlike surds.

Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 11
Here, the order of 5√5 and 5√3 is same but their radicands are not.
∴ 5√5 and √75 are unlike surds.

Question 4.
Simplify the following surds.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 12
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 13

Question 5.
Compare the following pair of surds.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 14
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 15
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 16
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 17

Question 6.
Simplify.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 18
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 19

Question 7.
Multiply and write the answer in the simplest form.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 20
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 21

Question 8.
Divide and write form.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 22
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 23

Question 9.
Rationalize the denominator.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 24
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 25
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 26

Question 1.
\(\sqrt { 9+16 }\) ? + \(\sqrt { 9 }\) + \(\sqrt { 16 }\) (Texbookpg. no. 28)
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 27

Question 2.
\(\sqrt { 100+36 }\) ? \(\sqrt { 100 }\) + \(\sqrt { 36 }\) (Textbook pg. no. 28)
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 28

Question 3.
Follow the arrows and complete the chart by doing the operations given. (Textbook pg. no. 34)
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 29
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 30

Question 4.
There are some real numbers written on a card sheet. Use these numbers and construct two examples each of addition, subtraction, multiplication and division. Solve these examples. (Textbook pg. no. 34)
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 31
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.3 32

Maharashtra Board Class 9 Maths Solutions

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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

New Profit Sharing and Gaining Ratio: Calculations with Solved Examples

The compilation of these Retirement or Death of a Partner Notes makes students exam preparation simpler and organised.

New Profit Sharing and Gaining Ratio

On the death or retirement of a partner, the partnership firm will be reconstituted. One major change will be the change in the profit-sharing ratios of the remaining partners. Let us see how the gaining ratio and sacrificing ratio will be calculated.

New Profit Sharing Ratio

There are different cases when a partnership can have a new profit sharing ratio:

  • Sometimes the partners may decide to change their existing profit sharing ratio, without any admission or retirement of a partner.
  • At the time of admission of the new partner.
  • At the time of retirement or death of an old partner.

This may result in again to a few partners and loss to others. The partners who are in profit due to this change in the profit-sharing ratio should compensate the sacrificing partner/partners.

New profit sharing ratio: Ratio in which the partners decide to share profits/losses in the future.

Gaining ratio: Ratio in which the partners have agreed to gain their share of profit from other partners.

Sacrificing ratio: Ratio in which the partners have agreed to sacrifice their share of profit in favour of other partners.
Sacrificing ratio = Old Ratio – New Ratio

New Profit Sharing and Gaining Ratio

Gaining Ratio

The gaining ratio is calculated at the time of retirement or the death of a partner. It is the ratio in which the remaining partners acquire the outgoing partner’s share of profit.

When the partner retires, the profit-sharing ratio of the continuing partners gets changed. Continuing partners distribute the share of retiring partners among them.

Gaining ratio = New Ratio – Old Ratio (if positive)

Examples:

Question:
Various cases of new ratio and gaining ratio are explained as follows:

Case 1: When the share of retiring partner is acquired by old partners in an old ratio
Amit, Sumit, and Punit share profit and losses in the ratio of 3 : 2 : 1, respectively. Amit retires and the remaining partners decide to share to take Amit’s share in the existing ratio i.e. 2 : 1. Calculate the new ratio and gaining ratio.
Solution:
The existing ratio between Sumit and Punit = 2/6 and 1/6
Amit’s ratio (retiring partner) = 3/6
Amit’s share taken by Sumit and Punit in the ratio of 2 : 1
Sumit gets = 3/6 × 2/3 = 6/18
Punit gets = 3/6 × 1/3 = 3/18
New ratio between Sumit and Punit is = 6 : 3 = 2 : 1
Gaining ratio = New Ratio – Old Ratio
Sumit’s gain = 2/3 – 2/6 = 2/6
Punit’s gain = 1/3 – 1/6 = 1/6
Gaining ratio = 2 : 1
New Ratio = 2 : 1

Case 2: When the share of retiring partner is acquired by old partners in old specified proportions
Amit, Sumit, and Punit share profit and losses in the ratio of 2 : 3 : 1, respectively. Amit retires and the remaining partners decide to share to take Amit’s share equally. Calculate the new ratio and gaining ratio.
Solution:
The existing ratio between Sumit and Punit = 3/6 and 1/6
Amit’s ratio (retiring partner) = 2/6
Amit’s share taken by Sumit and Punit in the ratio of 1 : 1
Sumit gets = 2/6 × 1/2 = 1/6
Sumit’s new share = 3/6 + 1/6 = 4/6
Punit gets = 2/6 × 1/2 = 1/6
Punit’s new share = 1/6 + 1/6 = 2/6
New ratio between Sumit and Punit is = 4 : 2 = 2 : 1
Gaining ratio is given the question i.e. 1 : 1
Gaining ratio = 1 : 1
New Ratio = 2 : 1

Case 3: When the share of the retiring partner is acquired fully by one of the continuing partners
Amit, Sumit, and Punit share profit and losses in the ratio of 4 : 5 : 2, respectively. Amit retires and Punit acquires Amit’s share. Calculate the new ratio and gaining ratio.
Solution:
Punit’s new share = 2/11 + 4/11 = 6/11
Sumit share remains unchanged = 5/11
The new ratio between Sumit and Punit is = 5 : 6
The gaining ratio in this case between Sumit and Punit will be
Sumit’s gain = 5/11 – 5/11 = Nil
Punit’s gain = 6/11 – 2/11 = 4/11
This shows that the entire gain is taken by Punit.

Read More:

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