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Permutations and Combinations Class 11 Commerce Maths 2 Chapter 6 Exercise 6.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.1 Questions and Answers.

Std 11 Maths 2 Exercise 6.1 Solutions Commerce Maths

Question 1.
A teacher wants to select the class monitor in a class of 30 boys and 20 girls. In how many ways can he select a student if the monitor can be a boy or a girl?
Solution:
There are 30 boys and 20 girls in a class.
The teacher wants to select a class monitor from these boys and girls.
A boy can be selected in 30 ways and a girl can be selected in 20 ways.
∴ By using the fundamental principle of addition,
in a number of ways either a boy or a girl is selected as a class monitor = 30 + 20 = 50.

Question 2.
In question 1, in how many ways can the monitor be selected if the monitor must be a boy? What is the answer if the monitor must be a girl?
Solution:
(i) Since there are 30 boys in the class
∴ A boy monitor can be selected in 30 ways.
(ii) Since there are 20 girls in the class
∴ A girl monitor can be selected in 20 ways.

Question 3.
A Signal is generated from 2 flags by putting one flag above the other. If 4 flags of different colours are available, how many different signals can be generated?
Solution:
A signal is generated from 2 flags and there are 4 flags of different colours available.
∴ 1st flag can be any one of the available 4 flags.
∴ It can be selected in 4 ways.
Now, 2nd flag is to be selected for which 3 flags are available for a different signal.
∴ 2nd flag can be anyone from these 3 flags.
∴ It can be selected in 3 ways.
∴ By using the fundamental principle of multiplication,
Total number of ways in which a signal can be generated = 4 × 3 = 12
∴ 12 different signals can be generated.

Question 4.
How many two-letter words can be formed using letters from the word SPACE when repetition of letters
(i) is allowed
(ii) is not allowed
Solution:
A two-letter word is to be formed out of the letters of the word SPACE.
(i) When repetition of the letters is allowed
1st letter can be selected in 5 ways
2nd letter can be selected in 5 ways
∴ By using the fundamental principle of multiplication,
total number of 2-letter words = 5 × 5 = 25

(ii) When repetition of the letters is not allowed
1st letter can be selected in 5 ways
2nd letter can be selected in 4 ways
∴ By using the fundamental principle of multiplication,
total number of 2-letter words = 5 × 4 = 20

Question 5.
How many three-digit numbers can be formed from the digits 0, 1, 3, 5, 6 if repetitions of digits
(i) are allowed
(ii) are not allowed
Solution:
The three-digit number is to be formed from the digits 0, 1, 3, 5, 6
(i) When repetition of digits is allowed:
100’s place digit should be a non-zero number.
Hence, it can be anyone from digits 1, 3, 5, 6
∴ 100’s place digit can be selected in 4 ways.
0 can appear in 10’s and unit’s place and digits can be repeated.
∴ 10’s place digit can be selected in 5 ways and the unit’s place digit can be selected in 5 ways.
∴ By using the fundamental principle of multiplication,
the total number of three-digit numbers = 4 × 5 × 5 = 100

(ii) When repetition of digits is not allowed:
100’s place digit should be a non-zero number.
Hence, it can be anyone from digits 1, 3, 5, 6
∴ 100’s place digit can be selected in 4 ways
0 can appear in 10’s and unit’s place and digits can’t be repeated.
∴ 10’s place digit can be selected in 4 ways and the unit’s place digit can be selected in 3 ways
∴ By using the fundamental principle of multiplication,
total number of three-digit numbers = 4 × 4 × 3 = 48

Question 6.
How many three-digit numbers can be formed using the digits 2, 3, 4, 5, 6 if digits can be repeated?
Solution:
A 3-digit number is to be formed from the digits 2, 3, 4, 5, 6 where digits can be repeated.
∴ The unit’s place digit can be selected in 5 ways.
10’s place digit can be selected in 5 ways.
100’s place digit can be selected in 5 ways.
∴ By using fundamental principle of multiplication,
the total number of 3-digit numbers = 5 × 5 × 5 = 125

Question 7.
A letter lock has 3 rings and each ring has 5 letters. Determine the maximum number of trials that may be required to open the lock.
Solution:
A letter lock has 3 rings, each ring containing 5 different letters.
∴ A letter from each ring can be selected in 5 ways.
∴ By using fundamental principle of multiplication,
the total number of trials that can be made = 5 × 5 × 5 = 125
Out of these 124 wrong attempts are made and in the 125th attempt,
the lock gets opened, for a maximum number of trials.
∴ A maximum number of trials required to open the lock is 125.

Question 8.
In a test that has 5 true/false questions, no student has got all correct answers and no sequence of answers is repeated. What is the maximum number of students for this to be possible?
Solution:
For a set of 5 true/false questions, each question can be answered in 2 ways.
∴ By using fundamental principle of multiplication,
the total number of possible sequences of answers = 2 × 2 × 2 × 2 × 2 = 32
Since no student has written all the correct answers.
∴ Total number of sequences of answers given by the students in the class = 32 – 1 = 31
Also, no student has given the same sequence of answers.
∴ Maximum number of students in the class = Number of sequences of answers given by the students = 31

Question 9.
How many numbers between 100 and 1000 have 4 in the unit’s place?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 where the unit place digit is 4.
Since Unit’s place digit is 4.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 10 ways.
For 3-digit number 100’s place digit should be a non-zero number.
∴ 100’s place digit can be selected in 9 ways.
∴ By using fundamental principle of multiplication,
total number of numbers between 100 and 1000 which have 4 in the units place = 1 × 10 × 9 = 90

Question 10.
How many numbers between 100 and 1000 have the digit 7 exactly once?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where exactly one of the digits is 7.
When 7 is in the unit’s place
The unit’s place digit is 7.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 9 ways.
100’s place digit can be selected in 8 ways.
∴ total number of numbers which have 7 in the unit’s place = 1 × 9 × 8 = 72
When 7 is in 10’s place
The unit’s place digit can be selected in 9 ways.
10’s place digit is 7
∴ it can be selected in 1 way only.
100’s place digit can be selected in 8 ways.
∴ total number of numbers which have 7 in 10’s place = 9 × 1 × 8 = 72
When 7 is in 100’s place
The unit’s place digit can be selected in 9 ways.
10’s place digit can be selected in 9 ways.
100’s place digit is 7
∴ it can be selected in 1 way.
∴ total numbers which have 7 in 100’s place = 9 × 9 × 1 = 81
∴ total number of numbers between 100 and 1000 having digit 7 exactly once = 72 + 72 + 81 = 225.

Question 11.
How many four-digit numbers will not exceed 7432 if they are formed using the digits 2, 3, 4, 7 without repetition?
Solution:
Among many set’s of digits, the greatest number is possible when digits are arranged in descending order.
∴ 7432 is the greatest number, formed from the digits 2, 3, 4, 7.
∴ Since a 4-digit number is to be formed from the digits 2, 3, 4, 7, where repetition of the digit is not allowed.
∴ 1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
∴ Total number of numbers not exceeding 7432 that can be formed from the digits 2, 3, 4, 7
= Total number of four-digit numbers formed from the digits 2, 3, 4, 7
= 4 × 3 × 2 × 1
= 24

Question 12.
If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
Solution:
Case I: Three-digit numbers with 4 occurring in hundred’s place:
100’s place digit can be selected in 1 way.
Ten’s place can be filled by any one of the numbers 2, 3, 5, 6.
∴ 10’s place digit can be selected in 4 ways.
The unit’s place digit can be selected in 3 ways.
∴ total number of numbers which have 4 in 100’s place = 1 × 4 × 3 = 12

Case II: Three-digit numbers more than 500
100’s place digit can be selected in 2 ways.
10’s place digit can be selected in 4 ways.
Unit’s place digit can be selected in 3 ways.
∴ total number of three digit numbers more than 500 = 2 × 4 × 3 = 24

Case III: Number of four digit numbers formed from 2, 3, 4, 5, 6
Since, repetition of digits is not allowed
∴ total four digit numbers formed = 5 × 4 × 3 × 2 = 120

Case IV: Number of five digit numbers formed from 2, 3, 4, 5, 6
Since, repetition of digits is not allowed
∴ total five digit numbers formed = 5 × 4 × 3 × 2 × 1 = 120
∴ total number of numbers that exceed 400 = 12 + 24 + 120 + 120 = 276

Question 13.
How many numbers formed with the digits 0, 1, 2, 5, 7, 8 will fall between 13 and 1000 if digits can be repeated?
Solution:
Case I: 2-digit numbers more than 13, less than 20, formed from the digits 0, 1, 2, 5, 7, 8
Number of such numbers = 3

Case II: 2-digit numbers more than 20 formed from 0, 1, 2, 5, 7, 8
Ten’s place digit is selected from 2, 5, 7, 8.
∴ Ten’s place digit can be selected in 4 ways.
Unit’s place digit is anyone from 0, 1, 2, 5, 7, 8
∴ The unit’s place digit can be selected in 6 ways.
Using the multiplication principle,
the number of such numbers (repetition allowed) = 4 × 6 = 24

Case III: 3-digit numbers formed from 0, 1, 2, 5, 7, 8
100’s place digit is anyone from 1, 2, 5, 7, 8.
∴ 100’s place digit can be selected in 5 ways.
As digits can be repeated, the 10’s place and unit’s place digits are selected from 0, 1, 2, 5, 7, 8
∴ 10’s place and unit’s place digits can be selected in 6 ways each.
Using multiplication principle,
the number of such numbers (repetition allowed) = 5 × 6 × 6 = 180
All cases are mutually exclusive and exhaustive.
∴ Required number = 3 + 24 + 180 = 207

Question 14.
A school has three gates and four staircases from the first floor to the second floor. How many ways does a student have to go from outside the school to his classroom on the second floor?
Solution:
A student can go inside the school from outside in 3 ways and from the first floor to the second floor in 4 ways.
∴ Number of ways to choose gates = 3
Number of ways to choose staircase = 4
∴ By using fundamental principle of multiplication,
number of ways in which a student has to go from outside the school to his classroom = 4 × 3 = 12

Question 15.
How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 3 if digits are not repeated?
Solution:
For a number to be divisible by 3.
The sum of digits must be divisible by 3.
Given 6 digits are 0, 1,2, 3, 4, 5.
Sum of 1, 2, 3, 4, 5 = 15, which is divisible by 3.
∴ There are two cases of 5 digit numbers formed from 0, 1, 2, 3, 4, 5 and divisible by 3.
Either 3 is selected in 5 digits (and 0 not selected) or 3 is not selected in 5 digits (and 0 is selected)
Case I:
3 is not selected (and 0 is selected) i.e., the digits are 0, 1, 2, 4, 5.
10000’s place digit can be selected in 4 ways (as 0 cannot appear).
As digits are not repeated, 1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
∴ Using multiplication theorem,
Number of 5-digit number formed from 0, 1, 2, 4, 5 (with no repetition of digits) = 4 × 4 × 3 × 2 × 1 = 96

Case II:
3 is selected (and 0 is not selected) i.e., 1, 2, 3, 4, 5
10000’s place digit can be selected in 5 ways.
1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
Using multiplication theorem,
Number of 5-digit numbers formed from 1, 2, 3, 4, 5 = 5 × 4 × 3 × 2 × 1 = 120
Both the cases are mutually exclusive and exhaustive.
∴ Required number = 96 + 120 = 216

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 6.1 Solutions
• 11th Commerce Maths Exercise 6.2 Solutions
• 11th Commerce Maths Exercise 6.3 Solutions
• 11th Commerce Maths Exercise 6.4 Solutions
• 11th Commerce Maths Exercise 6.5 Solutions
• 11th Commerce Maths Exercise 6.6 Solutions
• 11th Commerce Maths Exercise 6.7 Solutions
• 11th Commerce Maths  Miscellaneous Exercise 6 Solutions

Permutations and Combinations Class 11 Commerce Maths 2 Chapter 6 Exercise 6.6 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.6 Questions and Answers.

Std 11 Maths 2 Exercise 6.6 Solutions Commerce Maths

Question 1.
Find the value of
(i) ¹⁵C₄
Solution:

(ii) ⁸⁰C₂
Solution:

(iii) ¹⁵C₄ + ¹⁵C₅
Solution:

(iv) ²⁰C₁₆ – ¹⁹C₁₆
Solution:

Question 2.
Find n if
(i) ⁶P₂ = n ⁶C₂
Solution:

(ii) ²ⁿC₃ : ⁿC₂ = 52 : 3
Solution:

(iii) ⁿC_n-3 = 84
Solution:
ⁿC_n-3 = 84
∴ \(\frac{n !}{(n-3) ![n-(n-3)] !}\) = 84
∴ \(\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) !}{(\mathrm{n}-3) ! \times 3 !}\) = 84
∴ n(n – 1) (n – 2) = 84 × 6
∴ n(n – 1) (n – 2) = 9 × 8 × 7
Comparing on both sides, we get
∴ n = 9

Question 3.
Find r if ¹⁴C_2r : ¹⁰C_2r-4 = 143 : 10
Solution:

∴ \(\frac{14 \times 13 \times 12 \times 11}{2 \mathrm{r}(2 \mathrm{r}-1) \times(2 \mathrm{r}-2)(2 \mathrm{r}-3)}=\frac{143}{10}\)
∴ 2r(2r – 1)(2r – 2)(2r – 3) = 14 × 12 × 10
∴ 2r(2r – 1)(2r – 2)(2r – 3) = 8 × 7 × 6 × 5
Comparing on both sides, we get
∴ r = 4

Question 4.
Find n and r if.
(i) ⁿP_r = 720 and ⁿC_n-r = 120
Solution:

(ii) ⁿC_r-1 : ⁿC_r : ⁿC_r+1 = 20 : 35 : 42
Solution:

Question 5.
If ⁿP_r = 1814400 and ⁿC_r = 45, find r.
Solution:

∴ r! = 40320
∴ r! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
∴ r! = 8!
∴ r = 8

Question 6.
If ⁿC_r-1 = 6435, ⁿC_r = 5005, ⁿC_r+1 = 3003, find ^rC₅.
Solution:

Question 7.
Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 5 green balls and 7 blue balls so that 3 balls of every colour are drawn.
Solution:
9 balls are to be selected from 6 red, 5 green, 7 blue balls such that the selection consists of 3 balls of each colour.
∴ 3 red balls can be selected from 6 red balls in ⁶C₃ ways.
3 green balls can be selected from 5 green balls in ⁵C₃ ways.
3 blue balls can be selected from 7 blue balls in ⁷C₃ ways.
∴ Number of ways selection can be done if the selection consists of 3 balls of each colour

Question 8.
Find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.
Solution:
There are 6 boys and 4 girls.
A team of 3 boys and 2 girls is to be selected.
∴ 3 boys can be selected from 6 boys in ⁶C₃ ways.
2 girls can be selected from 4 girls in ⁴C₂ ways.
∴ Number of ways the team can be selected = ⁶C₃ × ⁴C₂
= \(\frac{6 !}{3 ! 3 !} \times \frac{4 !}{2 ! 2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !} \times \frac{4 \times 3 \times 2 !}{2 \times 2 !}\)
= 20 × 6
= 120
∴ The team of 3 boys and 2 girls can be selected in 120 ways.

Question 9.
After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting.
Solution:
Let there be n participants present in the meeting.
A handshake occurs between 2 persons.
∴ Number of handshakes = ⁿC₂
Given 66 handshakes were exchanged.
∴ 66 = ⁿC₂
∴ 66 = \(\frac{n !}{2 !(n-2) !}\)
∴ 66 × 2 = \(\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{(\mathrm{n}-2) !}\)
∴ 132 = n(n – 1)
∴ n(n – 1) = 12 × 11
Comparing on both sides, we get n = 12
∴ 12 participants were present at the meeting.

Question 10.
If 20 points are marked on a circle, how many chords can be drawn?
Solution:
To draw a chord we need to join two points on the circle.
There are 20 points on a circle.
∴ Total number of chords possible from these points = ²⁰C₂
= \(\frac{20 !}{2 ! 18 !}\)
= \(\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}\)
= 190

Question 11.
Find the number of diagonals of an n-sided polygon. In particular, find the number of diagonals when
(i) n = 10
(ii) n = 15
(iii) n = 12
Solution:
In n-sided polygon, there are ‘n’ points and ‘n’ sides. .
∴ Through ‘n’ points we can draw ⁿC₂ lines including sides.
∴ Number of diagonals in n sided polygon = ⁿC₂ – n (∴ n = number of sides)

Question 12.
There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.
Solution:
There are 20 lines such that no two of them are parallel and no three of them are concurrent.
Since no two lines are parallel
∴ they intersect at a point
∴ Number of points of intersection if no two lines are parallel and no three lines are concurrent = ²⁰C₂
= \(\frac{20 !}{2 ! 18 !}\)
= \(\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}\)
= 190

Question 13.
Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if
(i) no three points are collinear
(ii) four points are collinear
Solution:
There are 10 points on a plane.
(i) No three of them are collinear:
Since a line is obtained by joining 2 points,
number of lines passing through these points if no three points are collinear = ¹⁰C₂
= \(\frac{10 !}{2 ! 8 !}\)
= \(\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}\)
= 5 × 9
= 45

(ii) When 4 of them arc collinear:
∴ Number of lines passing through these points if 4 points are collinear
= ¹⁰C₂ – ⁴C₂ + 1
= 45 – \(\frac{4 !}{2 ! 2 !}\) + 1
= 45 – \(\frac{4 \times 3 \times 2 !}{2 \times 2 !}\) + 1
= 45 – 6 + 1
= 40

Question 14.
Find the number of triangles formed by joining 12 points if
(i) no three points are collinear
(ii) four points are collinear
Solution:
There are 12 points on the plane
(i) When no three of them are collinear:
Since a triangle can be drawn by joining any three non-collinear points.
∴ Number of triangles that can be obtained from these points = ¹²C₃
= \(\frac{12 !}{3 ! 9 !}\)
= \(\frac{12 \times 11 \times 10 \times 9 !}{3 \times 2 \times 1 \times 9 !}\)
= 220

(ii) When 4 of these points are collinear:
∴ Number of triangles that can be obtained from these points = ¹²C₃ – ⁴C₃
= 220 – \(\frac{4 !}{3 ! \times 1 !}\)
= 220 – \(\frac{4 \times 3 !}{3 !}\)
= 220 – 4
= 216

Question 15.
A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen?
Solution:
Out of 8 consonants, 4 can be selected in ⁸C₄
= \(\frac{8 !}{4 ! 4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 \times 3 \times 2 \times 1 \times 4 !}\)
= 70 ways
From 3 vowels, 2 can be selected in ³C₂
= \(\frac{3 !}{2 ! 1 !}\)
= \(\frac{3 \times 2 !}{2 !}\)
= 3 ways
Now, to form a word, these 6 letters (i.e., 4 consonants and 2 vowels) can be arranged in ⁶P₆ i.e., 6! ways.
∴ Total number of words that can be formed = 70 × 3 × 6!
= 70 × 3 × 720
= 151200
∴ 151200 words of 4 consonants and 2 vowels can be formed.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 6.1 Solutions
• 11th Commerce Maths Exercise 6.2 Solutions
• 11th Commerce Maths Exercise 6.3 Solutions
• 11th Commerce Maths Exercise 6.4 Solutions
• 11th Commerce Maths Exercise 6.5 Solutions
• 11th Commerce Maths Exercise 6.6 Solutions
• 11th Commerce Maths Exercise 6.7 Solutions
• 11th Commerce Maths  Miscellaneous Exercise 6 Solutions

Partition Values Class 11 Commerce Maths 2 Chapter 1 Miscellaneous Exercise 1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Partition Values Miscellaneous Exercise 1 Questions and Answers.

Std 11 Maths 2 Miscellaneous Exercise 1 Solutions Commerce Maths

Question 1.
The data gives the number of accidents per day on a railway track. Compute Q₂, P₁₇, and D₇.
4, 2, 3, 5, 6, 3, 4, 1, 2, 3, 2, 3, 4, 3, 2
Solution:
The given data can be arranged in ascending order as follows:
1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 6
Here, n = 15
Q₂ = value of 2\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{15+1}{4}\right)^{\text {th }}\) observation
= value of (2 × 4)th observation
= value of 8th observation
∴ Q₂ = 3
P₁₇ = value of 17\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 17\(\left(\frac{15+1}{100}\right)^{\text {th }}\) observation
= value of (17 × 0.16)th observation
= value of (2.72)th observation
= value of 2nd observation + 0.72 (value of 3rd observation – value of 2nd observation)
= 2 + 0.72 (2 – 2)
∴ P₁₇ = 2
D₇ = value of 7\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 7\(\left(\frac{15+1}{10}\right)^{\text {th }}\) observation
= value of (7 × 1.6)th observation
= value of (11.2)th observation
= value of 11th observation + 0.2(value of 12th observation – value of 11th observation)
= 4 + 0. 2(4 – 4)
∴ D₇ = 4

Question 2.
The distribution of daily sales of shoes (size-wise) for 100 days from a certain shop is as follows:

Compute Q₁, D₂, and P₉₅.
Solution:
By arranging the given data in ascending order, we construct the less than cumulative frequency table as given below:

Here, n = 100
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{100+1}{4}\right)^{\text {th }}\) observation
= value of (25.25)th observation
Cumulative frequency which is just greater than (or equal) to 25.25 is 27.
∴ Q₁ = 3
D₂ = value of 2\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{100+1}{10}\right)^{\text {th }}\) observation
= value of (2 × 10.1)th observation
= value of (20.2)th observation
Cumulative frequency which is just greater than (or equal) to 20.2 is 27.
∴ D₂ = 3
P₉₅ = value of 95\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 95\(\left(\frac{100+1}{100}\right)^{\text {th }}\) observation
= value of (95 × 1.01)th observation
= value of (95.95)th observation
The cumulative frequency which is just greater than (or equal) to 95.95 is 100.
∴ P₉₅ = 8

Question 3.
Ten students appeared for a test in Mathematics and Statistics and they obtained the marks as follows:

If the median will be the criteria, in which subject, the level of knowledge of the students is higher?
Solution:
Marks in Mathematics can be arranged in ascending order as follows:
23, 23, 25, 25, 32, 35, 36, 37, 38, 42
Here, n = 10
∴ Median = value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
Median = value of \(\left(\frac{10+1}{2}\right)^{\text {th }}\) observation
= value of (5.5)th observation
= value of 5th observation + 0.5(value of 6th observation – value of 5th observation)
= 32 + 0.5 (35 – 32)
= 32 + 0.5(3)
= 32 + 1.5
= 33.5
Marks in Statistics can be arranged in ascending order as follows:
22, 23, 26, 28, 29, 32, 34, 36, 45, 50
Here, n = 10
∴ Median = value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
= value of \(\left(\frac{10+1}{2}\right)^{\text {th }}\) observation
= value of (5.5)th observation
= value of 5th observation + 0.5(value of 6th observation – value of 5th observation)
= 29 + 0.5(32 – 29)
= 29 + 0.5(3)
= 29 + 1.5
= 30.5
∴ Median marks for Mathematics = 33.5 and
Median marks for Statistics = 30.5
∴ The level of knowledge in Mathematics is higher than that of Statistics.

Question 4.
In the frequency distribution of families given below, the number of families corresponding to expenditure group 2000 – 4000 is missing from the table. However, the value of the 25th percentile is 2880. Find the missing frequency.

Solution:
Let x be the missing frequency of expenditure group 2000 – 4000.
We construct the less than cumulative frequency table as given below:

Here, N = 75 + x
Given, P₂₅ = 2880
∴ P₂₅ lies in the class 2000 – 4000.
∴ L = 2000, h = 2000, f = x, c.f. = 14
∴ P₂₅ = L + \(\frac{h}{f}\left(\frac{25 \mathrm{~N}}{100}-\text { c.f. }\right)\)
∴ 2880 = 2000 + \(\frac{2000}{x}\left(\frac{75+x}{4}-14\right)\)
∴ 2880 – 2000 = \(\frac{2000}{x}\left(\frac{75+x-56}{4}\right)\)
∴ 880x = 500(x + 19)
∴ 880x = 500x + 9500
∴ 880x – 500x = 9500
∴ 380x = 9500
∴ x = 25
∴ 25 is the missing frequency of the expenditure group 2000 – 4000.

Question 5.
Calculate Q₁, D₆, and P₁₅ for the following data:

Solution:
Since the difference between any two consecutive mid values is 50, the width of each class interval is 50.
∴ the class intervals will be 0 – 50, 50 – 100, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 500
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{500}{4}\) = 125
Cumulative frequency which is just greater than (or equal) to 125 is 160.
Q₁ lies in the class 100 – 150.
∴ L = 100, h = 50, f = 80, c.f. = 80
∴ Q₁ = L + \(\left(\frac{\mathrm{N}}{4}-\text { c.f. }\right)\)
= 100 + \(\frac{50}{80}\)(125 – 80)
= 100 + \(\frac{5}{8}\)(45)
= 100 + 28.125
= 128.125
D₆ class = class containing \(\left(\frac{6 \mathrm{~N}}{10}\right)^{\text {th }}\) observation
∴ \(\frac{6 \mathrm{~N}}{10}=\frac{6 \times 500}{10}\) = 300
Cumulative frequency which is just greater than (or equal) to 300 is 410.
∴ D₆ lies in the class 200 – 250.
∴ L = 200, h = 50, f = 150, c.f. = 260
∴ D₆ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{6 \mathrm{~N}}{10}-\text { c.f. }\right)\)
= 200 + \(\frac{50}{150}\)(300 – 260)
= 200 + \(\frac{1}{3}\)(40)
= 200 + 13.33
= 213.33
P₁₅ class = class containing \(\left(\frac{15 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{15 \mathrm{~N}}{100}=\frac{15 \times 500}{100}\) = 75
Cumulative frequency which is just greater than (or equal) to 75 is 80.
∴ P15 lies in the class 50 – 100.
∴ L = 50, h = 50, f = 70, c.f. = 10
∴ P₁₅ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{15 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 50 + \(\frac{50}{70}\) (75 – 10)
= 50 + \(\frac{5}{7}\) (65)
= 50 + \(\frac{325}{7}\)
= 50 + 46.4286
= 96.4286
∴ Q₁ = 128.125, D₆ = 213.33, P₁₅ = 96.4286

Question 6.
Daily income for a group of 100 workers are given below:

P₃₀ for this group is ₹ 110. Calculate the missing frequencies.
Solution:
Let a and b be the missing frequencies of class 50 – 100 and class 200 – 250 respectively.
We construct the less than cumulative frequency table as given below:

Here, N = 62 + a + b
Since, N = 100
∴ 62 + a + b = 100
∴ a + b = 38 ……(i)
Given, P30 = 110
∴ P30 lies in the class 100 – 150.
∴ L = 100, h = 50, f = 25, c.f. = 7 + a
\(\frac{30 \mathrm{~N}}{100}=\frac{30 \times 100}{100}\) = 30
∴ P₃₀ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{30 \mathrm{~N}}{100}-\text { c.f. }\right)\)
∴ 110 = 100 + \(\frac{50}{25}\) [30 – (7 + a)]
∴ 110 – 100 = 2(30 – 7 – a)
∴ 10 = 2(23 – a)
∴ 5 = 23 – a
∴ a = 23 – 5
∴ a = 18
Substituting the value of a in equation (i), we get
18 + b = 38
∴ b = 38 – 18
∴ b = 20
∴ 18 and 20 are the missing frequencies of the class 50 – 100 and class 200 – 250 respectively.

Question 7.
The distribution of a sample of students appearing for a C.A. examination is:

Help C.A. institute to decide cut-off marks for qualifying for an examination when 3% of students pass the examination.
Solution:
To decide cut-off marks for qualifying for an examination when 3% of students pass, we have to find P97.
We construct the less than cumulative frequency table as given below:

Here, N = 1100
P97 class = class containing \(\left(\frac{97 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{97 \mathrm{~N}}{100}=\frac{97 \times 1100}{100}\) = 1067
Cumulative frequency which is just greater than (or equal) to 1067 is 1100.
∴ P₉₇ lies in the class 500 – 600.
∴ L = 500, h = 100, f = 130, c.f. = 970
∴ P₉₇ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{97 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 500 + \(\frac{100}{130}\)(1067 – 970)
= 500 + \(\frac{10}{13}\) (97)
= 500 + 74.62
= 574.62 ~ 575
∴ the cut off marks for qualifying an examination is 575.

Question 8.
Determine graphically the value of median, D₃, and P₃₅ for the data given below:

Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (15, 8), (20, 22), (25, 30), (30, 55), (35, 70), (40, 84), (45, 90).

N = 90
For median, consider \(\frac{\mathrm{N}}{2}=\frac{90}{2}\) = 45
For D3, consider \(\frac{3 \mathrm{~N}}{10}=\frac{3 \times 90}{10}\) = 27
For P35, consider \(\frac{35 \mathrm{~N}}{100}=\frac{35 \times 90}{100}\) = 31.5
∴ We take the values 45, 27 and 31.5 on the Y-axis and draw lines from these points parallel to X-axis.
From the points where they intersect the less than ogive, we draw perpendicular on the X-axis.
Foot of the perpendicular represent the values of median, D3 and P35 respectively.
∴ Median ~ 29, D₃ ~ 23.5, P₃₅ ~ 26

Question 9.
The I.Q. test of 500 students of a college is as follows:

Find graphically the number of students whose I.Q. is more than 55 graphically.
Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (30, 41), (40, 93), (50, 157), (60, 337), (70, 404), (80, 449), (90, 489), (100, 500)

To find the number of students whose I.Q. is more than 55, we consider the value 55 on the X-axis.
From this point, we draw a line that is parallel to Y-axis.
From the point this line intersects the less than ogive, we draw a perpendicular on the Y-axis.
The foot of perpendicular gives the number of students whose I.Q. is less than 55.
∴ The foot of perpendicular ~ 244
∴ No. of students whose I.Q. is less than 55 ~ 244
∴ No. of Students whose I.Q. is more than 55 = 500 – 244 = 256

Question 10.
Draw an ogive for the following distribution. Determine the median graphically and verify your result by a mathematical formula.

Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (150, 2), (155, 7), (160, 16), (165, 31), (170, 47), (175, 54), (180, 59) and (185, 60).

N = 60
∴ \(\frac{\mathrm{N}}{2}=\frac{60}{2}\) = 30
∴ We take the value 30 on the Y-axis and from this point, we draw a line parallel to X-axis.
From the point where this line intersects the less than ogive, we draw a perpendicular on X-axis.
The foot perpendicular gives the value of the median.
∴ Median ~ 164.67
Now, let us calculate the median from the mathematical formula.
∴ \(\frac{\mathrm{N}}{2}\) = 30
The median lies in the class interval 160 – 165.
∴ L = 160, h = 5, f = 15, c.f. = 16
Median = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{2}-\mathrm{c} . \mathrm{f} .\right)\)
= 160 + \(\frac{5}{15}\) (30 – 16)
= 160+ \(\frac{1}{3}\) × 14
= 160 + 4.67
= 164.67

Question 11.
In a group of 25 students, 7 students failed and 6 students got distinction and the marks of the remaining 12 students are 61, 36, 44, 59, 52, 56, 41, 37, 39, 38, 41, 64. Find the median marks of the whole group.
Solution:
n = 25
Median = \(\frac{\mathrm{n}+1}{2}=\frac{25+1}{2}\) = 13th observation
We have been stated that 7 students failed (assuming passing marks on 35) and 6 students got distinction (assuming distinction as 70+), and the marks of the remaining 12 students (who will be situated between the two groups mentioned above, if arranged in ascending order), we have,
F, F, F, F, F, F, F, 36, 37, 38, 39, 41, 41, 44, 52, 56, 59, 61, 64, D, D, D, D, D, D
∴ median = 13th observation = 41.

Question 12.
The median weight of a group of 79 students is found to be 55 kg. 6 more students are added to this group whose weights are 50, 51, 52, 59.5, 60, 61 kg. What will be the value of the median of the combined group if the lowest and the highest weights were 53 kg and 59 kg respectively?
Solution:
n = 79
Median = 55kg
Lowest observation = 53 kg
Flighest observation = 59 kg
6 new students are added to the group having weights in Kg as follows:
50, 51, 52, 59.5, 60, 61
From the above, we see that of the 6 new students, 3 have weights which are below the lowest weight of the earlier group and 3 have weights which are above the highest weight of the earlier group.
∴ the median remains the same
∴ median = 55 kg.

Question 13.
The median of the following incomplete table is 92. Find the missing frequencies:

Solution:
Let a and b be the missing frequencies of class 50 – 70 and class 110 – 130 respectively.
We construct the less than cumulative frequency table as given below:

Here, N = 54 + a + b
Since, N = 80
∴ 54 + a + b = 80
∴ a + b = 26 …..(i)
Given, Median = Q₂ = 92
∴ Q₂ lies in the class 90 – 110.
∴ L = 90, h = 20, f = 20, c.f. = 24 + a
\(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 80}{4}\) = 40
∴ Q₂ = L + \(\frac{h}{f}\left(\frac{2 N}{4}-\text { c.f. }\right)\)
∴ 92 = 90 + \(\frac{20}{20}\) [40 – (24 + a)
∴ 92 – 90 = 40 – 24 – a
∴ 2 = 16 – a
∴ a = 14
Substituting the value of a in equation (i), we get
14 + b = 26
∴ b = 26 – 14 = 12
∴ 14 and 12 are the missing frequencies of the class 50 – 70 and class 110 – 130 respectively.

Question 14.
A company produces tables which are packed in batches of 100. An analysis of the defective tubes in different batches has received the following information:

estimate the number of defective tubes in the central batch.
Solution:
To find the number of defective tubes in the central batch, we have to find Q₂.
Since the given data is not continuous, we have to convert it into a continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
∴ the class intervals will be Less than 4.5, 4.5 – 9.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 251
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 251}{4}\) = 125.5
Cumulative frequency which is just greater than (or equal to) 125.5 is 180.
∴ Q₂ lies in the class 9.5 – 14.5.
∴ L = 9.5, h = 5, f = 84, c.f. = 96
∴ Q₂ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)
= 9.5 + \(\frac{5}{84}\) (125.5 – 96)
= 9.5 + \(\frac{5}{84}\) × 29.5
= 9.5 + \(\frac{147.5}{84}\)
= 9.5 + 1.76
= 11.26

Question 15.
In a college, there are 500 students in junior college, 5% score less than 25 marks, 68 scores from 26 to 30 marks, 30% score from 31 to 35 marks, 70 scores from 36 to 40 marks, 20% score from 41 to 45 marks and the rest score 46 and above marks. What are the median marks?
Solution:
Given data can be written in tabulated form as follows:

Since the given data is not continuous, we have to convert it into the continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
∴ the class intervals will be Less than 25.5, 25.5 – 30.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 500
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 500}{4}\) = 250
Cumulative frequency which is just greater than (or equal to) 250 is 313.
∴ Q₂ lies in the class 35.5 – 40.5.
∴ L = 35.5, h = 5, f = 70, c.f. = 243
∴ Median = Q₂ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)
= 35.5 + \(\frac{5}{70}\) (250 – 243)
= 35.5 + \(\frac{1}{14}\) (7)
= 35.5 + 0.5
= 36

Question 16.
Draw a cumulative frequency curve more than typical for the following data and hence locate Q₁ and Q₃. Also, find the number of workers with daily wages
(i) Between ₹ 170 and ₹ 260
(ii) less than ₹ 260

Solution:
For more than ogive points to be plotted are (100, 200), (150, 188), (200, 160), (250, 124), (300, 74), (350, 49), (400, 31), (450, 15), (500, 5)

Here, N = 200
For Q₁, \(\frac{\mathrm{N}}{4}=\frac{200}{4}\) = 4
For Q₃, \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 200}{4}\) = 150
We take the points having Y co-ordinates 50 and 150 on Y-axis.
From these points, we draw lines which are parallel to X-axis.
From the points of intersection of these lines with the curve, we draw perpendicular on X-axis.
X-Co-ordinates of these points gives the values of Q₁ and Q₃.
Since X-axis has daily wages more than and not less than the given amounts.
∴ Q₁ = Q₃ and Q₃ = Q₁
∴ Q₂ ~ 215 , Q₃ ~ 348

(i) To find the number of workers with daily wages between ₹ 170 and ₹ 260,
Take the values 170 and 260 on X-axis. From these points, we draw lines parallel to Y-axis.
From the point where they intersect the more than ogive, we draw perpendiculars on Y-axis.
The points where they intersect the Y-axis gives the values 178 and 114.
∴ Number of workers having daily wages between ₹ 170 and ₹ 260 = 178 – 114 = 64

(ii) To find the number of workers having daily wages less than ₹ 260, we consider the value 260 on the X-axis.
From this point, we draw a line that is parallel to Y-axis.
From the point where the line intersects the more than ogive, we draw a perpendicular on the Y-axis.
The foot of perpendicular gives the number of workers having daily wages of more than 260.
The foot of perpendicular ~ 114
∴ No. of workers whose daily wages are more than ₹ 260 ~ 114
∴ No. of workers whose daily wages are less than ₹ 260 = 200 – 114 = 86

Question 17.
Draw ogive of both the types for the following frequency distribution and hence find the median.

Solution:

For less than given points to be plotted are (10, 5), (20, 10), (30, 18), (40, 30), (50, 46), (60, 61), (70, 71), (80, 79), (90, 84), (100, 86)
For more than given points to be plotted are (0, 86), (10, 81), (20, 76), (30, 68), (40, 56), (50, 40), (60, 25), (70, 15), (80, 7), (90, 2)

From the point of intersection of two ogives. We draw a perpendicular on X-axis.
The point where it meets the X-axis gives the value of the median.

Question 18.
Find Q1, D6 and P78 for the following data:

Solution:
Since the given data is not in the form of a continuous frequency distribution, we have to convert it into that form by subtracting 0.025 from the lower limit and adding 0.025 to the upper limit of each class interval.
∴ the class intervals will be 7.975 – 8.975, 8.975 – 9.975, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 50
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{N}{4}=\frac{50}{4}\) = 12.5
Cumulative frequency which is just greater than (or equal) to 12.5 is 15.
∴ Q₁ lies in the class 8.975 – 9.975.
∴ L = 8.975, h = 1, f = 10, c.f. = 5
Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\mathrm{c} . \mathrm{f} .\right)\)
= 8.975 + \(\frac{1}{10}\) (12.5 – 5)
= 8.975 + 0.1(7.5)
= 8.975 + 0.75
= 9.725
D₆ class = class containing \(\left(\frac{6 \mathrm{~N}}{10}\right)^{\text {th }}\) observation
∴ \(\frac{6 \mathrm{~N}}{10}=\frac{6 \times 50}{10}\) = 30
Cumulative frequency which is just greater than (or equal) to 30 is 35.
∴ D6 lies in the class 9.975 – 10.975.
∴ L = 9.975, h = 1, f = 20, c.f. = 15
D₆ = L + \(\frac{h}{f}\left(\frac{6 N}{10}-\text { c.f. }\right)\)
= 9.975 + \(\frac{1}{20}\) (30 – 15)
= 9.975 + 0.05(15)
= 9.975 + 0.75
= 10.725
P₇₈ class = class containing \(\left(\frac{78 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
\(\frac{78 \mathrm{~N}}{100}=\frac{78 \times 50}{100}\) = 39
Cumulative frequency which is just greater than (or equal) to 39 is 45.
∴ P₇₈ lies in the class 10.975 – 11.975.
∴ L = 10.975, h = 1, f = 10, c.f. = 35
∴ P₇₈ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{78 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 10.975 + \(\frac{1}{10}\) (39 – 35)
= 10.975 + 0.1(4)
= 10.975 + 0.4
= 11.375

Question 19.

For the above data, find all quartiles and number of persons weighing between 57 kg and 72 kg.
Solution:
We construct the less than cumulative frequency table as given below:

Here, N = 111
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{N}{4}=\frac{111}{4}\) = 27.75
Cumulative frequency which is just greater than (or equal) to 27.75 is 39.
∴ Q₁ lies in the class 50 – 55.
∴ Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\text { c.f. }\right)\)
= 50 + \(\frac{5}{20}\) (27.75 – 19)
= 50 + \(\frac{1}{4}\) × 8.75
= 50 + 2.1875
= 52.1875
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{2 N}{4}=\frac{2 \times 111}{4}\) = 55.5
Cumulative frequency which is just greater than (or equal) to 55.5 is 69.
∴ Q₂ lies in the class 55 – 60.
∴ L = 55, h = 5, f = 30, c.f. = 39
∴ Q₂ = L + \(\frac{h}{f}\left(\frac{2 N}{4}-\text { c.f. }\right)\)
= 55 + \(\frac{5}{30}\) (55.5 – 39)
= 55 + \(\frac{1}{6}\) × 16.5
= 55 + 2.75
= 57.75
Q₃ class = class containing \(\left(\frac{3 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 111}{4}\) = 83.25
Cumulative frequency which is just greater than (or equal) to 83.25 is 89.
∴ Q₃ lies in the class 60 – 65.
∴ L = 60, h = 5, f = 20, c.f. = 69
∴ Q₃ = L + \(\frac{h}{f}\left(\frac{3 N}{4}-c . f .\right)\)
= 60 + \(\frac{5}{20}\) (83.25 – 69)
= 60 + \(\frac{1}{4}\) × 14.25
= 60 + 3.5625
= 63.5625
In order to find the number of persons between 57 kg and 72 kg,
We need to find x in P_x, where P_x = 57 kg and y in P_y, where P_y = 72 kg
Then (y – x) would be the % of persons weighing between 57 kg and 72 kg
P_x = 57
∴ L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{x \times \mathrm{N}}{100}-\mathrm{c} . \mathrm{f} .\right)\) = 57
∴ 55 + \(\frac{5}{30}\) (1.11x – 39) = 57
∴ \(\frac{1}{6}\) (1.11x – 39) = 2
∴ 1.11x – 39 = 12
∴ 1.11x = 51
∴ x = 45.95
∴ P_y = 72
∴ L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{y \times \mathrm{N}}{100}-\mathrm{c} . \mathrm{f} .\right)\) = 72
∴ 70 + \(\frac{5}{8}\) (1.11y – 99) = 72
∴ 0.625(1.11y – 99) = 2
∴ 1.11y – 99 = 3.2
∴ 1.11y = 102.2
∴ y = 92.07
∴ % of people weighing between 57 kg and 72 kg = 92.07 – 45.95 = 46.12 %
∴ No. of people weighing between 57 kg and 72 kg = 111 × 46.12% = 51.1932 ~ 51

Question 20.
For the following data showing weights of 100 employees, find the maximum weight of the lightest 25% of employees.

Solution:
We construct the less than cumulative frequency table as given below:

Here, N = 100
Q₁ class = class containing \(\left(\frac{N}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{100}{4}\) = 25
Cumulative frequency which is just greater than (or equal) to 25 is 29.
∴ Q₁ lies in the class 55 – 60.
∴ L = 55, h = 5, f = 15, c.f. = 14
∴ Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\mathrm{c} . \mathrm{f} .\right)\)
= 55 + \(\frac{5}{15}\) (25 – 14)
= 55 + \(\frac{1}{3}\) × 11
= 55 + 3.67
= 58.67
∴ Maximum weight of the lightest 25% of employees is 58.67 kg.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 1.1 Solutions
• 11th Commerce Maths Exercise 1.2 Solutions
• 11th Commerce Maths Exercise  1.3 Solutions
• 11th Commerce Maths Miscellaneous Exercise 1 Solutions

Commercial Mathematics Class 11 Commerce Maths 2 Chapter 9 Exercise 9.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.1 Questions and Answers.

Std 11 Maths 2 Exercise 9.1 Solutions Commerce Maths

Question 1.
Find 77% of 580 + 34% of 390.
Solution:

Question 2.
240 candidates appeared for an examination, of which 204 passed. What is the pass percentage?
Solution:
We find the pass percentage using the unitary method

∴ The pass percentage for the examination is 85%.

Question 3.
What percent of 8.4 kg are 168 grams?
Solution:
Let 168 gms be x% of 8.4 kg
i.e., let 168 gms be \(\frac{x}{100}\) of 8400 gms
∴ 168 = \(\frac{x}{100}\) × 8400
∴ x = \(\frac{168}{84}\) = 2
∴ 168 gms is 2% of 8.4 kg.

Question 4.
If the length of a rectangle is decreased by 20%, what should be the increase in the breadth of the rectangle so that the area remains the same?
Solution:
Let x and y represent the length and breadth of the rectangle respectively.
∴ The original area of the rectangle = xy
There is a 20% decrease in length.

Let k % be the required increase in breadth

Given that the new and old areas should be equal.

∴ 100 + k = 125
∴ k = 125 – 100 = 25
∴ Breadth should be increased by 25% so that the area remains same.

Question 5.
The price of rice increased by 20%, as a result, a person can have 5kg rice for ₹ 600. What was the initial price of rice per kg?
Solution:
A person can buy 5 kg of rice for ₹ 600 after the increase in price
∴ New price of rice = \(\frac{600}{5}\) = ₹ 120/kg …..(i)
Let ‘x’ be the initial price per kg of rice.
There is a 20% increase in the price of rice.
Thus the new price of the rice will be given as

∴ The initial price of rice is ₹ 100 per kg

Question 6.
What percent is 3% of 5%?
Solution:
Let 3% be x % of 5%.
Then \(\frac{3}{100}=\frac{x}{100} \times \frac{5}{100}\)
∴ x = \(\frac{3 \times 100}{5}\) = 60
∴ 3% is 60% of 5%.

Question 7.
After availing of two successive discounts of 20% each, Madhavi paid ₹ 64 for a book. If she would have got only one discount of 20%, how much additional amount would she have paid?
Solution:
Let the price of the book be ₹ x.
After the first 20% discount, the price of the book becomes

After another 20% discount, the price of the book becomes

This price = ₹ 64 …..[Given]
∴ \(\frac{16}{25}\)x = 64
∴ x = 4 × 25 = 100
Thus, Amount of the book after one discount = \(\frac{4}{5}\)(100) = 80 …..[from (i)]
∴ The additional amount that Madhavi would have paid = 80 – 64 = ₹ 16

Question 8.
The price of the table is 40% more than the price of a chair. By what percent price of a chair is less than the price of a table?
Solution:
Let ₹ x and ₹ y be the price of a table and chair respectively.
The price of the table is 40% more than the price of a chair
∴ \(\frac{x-y}{y}\) × 100 = 40

We need to find by how much percent is the price of a chair less than that of a table.

∴ The price of a chair is 28.57% less than the price of a table.

Question 9.
A batsman scored 92 runs which includes 4 boundaries 5 sixes. He scored other runs by running between the wickets. What percent of his total score did he make by running between the wickets?
Solution:
Batsman scores 4 fours (boundaries) and 5 sixes in 92 runs.
Number of runs scored by fours and sixes = 4 × 4 + 5 × 6 = 46
∴ 92 – 46 = 46
Let 46 be x% of 92.
Then 46 = \(\frac{x}{100}\) × 92
∴ x = \(\frac{46 \times 100}{92}=\frac{100}{2}\) = 50
∴ 50% of the total runs were scored by running between the wickets.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 9.1 Solutions
• 11th Commerce Maths Exercise 9.2 Solutions
• 11th Commerce Maths Exercise 9.3 Solutions
• 11th Commerce Maths Exercise 9.4 Solutions
• 11th Commerce Maths Exercise 9.5 Solutions
• 11th Commerce Maths Exercise 9.6 Solutions
• 11th Commerce Maths Exercise 9.7 Solutions
• 11th Commerce Maths Miscellaneous Exercise 9 Solutions

Commercial Mathematics Class 11 Commerce Maths 2 Chapter 9 Exercise 9.6 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.6 Questions and Answers.

Std 11 Maths 2 Exercise 9.6 Solutions Commerce Maths

Question 1.
M/s Janaseva sweet mart sold sweets of ₹ 3,86,000. What CGST and SGST he will pay if the rate of GST is 5%?
Solution:
Given that M/s Janaseva sweet mart sold sweets of ₹ 3,86,000
∴ Bill amount = ₹ 3,86,000
GST payable at the rate 5%
∴ CGST and SGST applicable is 2.5% each
∴ CGST on the bill = \(\frac{2.5}{100}\) × 3,86,000 = ₹ 9650
and SGST on the bill = \(\frac{2.5}{100}\) × 3,86,000 = ₹ 9650

Question 2.
Janhavi Gas Agency purchased some gas cylinders for ₹ 5,00,000 and sold them to the customers for ₹ 5,90,000. Find the amount of GST payable and the amount of ITC. 5% GST is applicable.
Solution:
Given that, Janhavi Gas Agency purchased some gas cylinders for ₹ 5,00,000 and GST applicable is 5%.
∴ Input tax (ITC) = 5% of 5,00,000
= \(\frac{5}{100}\) × 5,00,000
= ₹ 25,000
Janhavi Gas Agency sold the gas cylinders for ₹ 5,90,000
∴ Output tax for Janhavi Gas Agency = 5% of 5,90,000
= \(\frac{5}{100}\) × 5,90,000
= ₹ 29,500
GST payable = Output tax – Input tax (ITC)
= 29,500 – 25,000
= ₹ 4500
∴ GST payable for Janhavi Gas Agency is ₹ 4,500 and ITC is ₹ 25,000.

Question 3.
A company dealing in mobile phones purchased mobile phones worth ₹ 5,00,000 and sold the same to customers at ₹ 6,00,000. Find the amount of ITC and amount of GST if the rate of GST is 12%.
Solution:
Given that the rate of GST applicable is 12%.
The company purchased mobile phones worth ₹ 5,00,000.
∴ Input tax (ITC) = 12% of 5,00,000
= \(\frac{12}{100}\) × 5,00,000
= ₹ 60,000
The company dealing in mobile phones sold the same to customers at ₹ 6,00,000.
∴ Output tax of the company = 12% of 6,00,000
= \(\frac{12}{100}\) × 6,00,000
= ₹ 72,000
GST payable for the company = Output tax – Input tax (ITC)
= 72,000 – 60,000
= ₹ 12,000
∴ The ITC for the company is ₹ 60,000 and GST payable is ₹ 12,000.

Question 4.
Prepare business to customers (B2C) tax invoice using given information. Write the name of supplier, address, state, Date, Invoice Number, GSTIN etc. as per your choice
Supplier: ___________
Address: ___________
State: ___________
Date: ___________
Invoice No: ___________
GSTIN: ___________
Particular: Rate of Sarees – ₹ 2750
Rate of GST 5% HSN 5407 – 2 pcs
Rate of Kurta – ₹ 750
Rate of GST 12% HSN 5408
Solution:
Supplier: M/s Swaglife Fashions
Address: 143, Shivaji Rasta, Mumbai 400001
Mobile No. 9263692111
Email: abc@gmail.com
State: Maharashtra
Date: 31/08/19
Invoice No: GST/110
GSTIN: 27ABCDE1234HIZS

∴ Rate of 1 saree = ₹ 2750
∴ Rate of 2 sarees = 2 x 2750 = ₹ 5500
∴ GST on sarees = 12% of 5500
= \(\frac{12}{100}\) × 5500
= ₹ 660
∴ CGST = SGST = ₹ 330
∴ Rate of 1 Kurta = ₹ 750
∴ GST on Kurta = 12% of 750
= \(\frac{12}{100}\) × 750
= ₹ 90
∴ CGST = SGST = ₹ 45

Question 5.
Heena Enterprise sold cosmetics worth ₹ 25,000 to Leena traders, a retailer. Leena Traders sold it further to Meena Beauty Products for ₹ 30,000. Meena Beauty Product sold it further to the customers for ₹ 40,000. The rate of GST is 18%. Find
(i) GST Payable by each party
(ii) CGST and SGST
Solution:
The trading chain,

∴ Output tax for Heena Enterprises = 18% of 25,000
= \(\frac{18}{100}\) × 25,000
= ₹ 4,500
∴ GST payable by Heena Enterprises
Now output tax for Leena traders = 18% of 30,000
= \(\frac{18}{100}\) × 30,000
= ₹ 5,400
∴ GST payable by Leena traders = Output tax – Input tax
= 5,400 – 4,500
= ₹ 900
∴ Output tax for Meena beauty products = 18% of 40,000
= \(\frac{18}{100}\) × 40,000
= ₹ 7,200
∴ GST payable by Meena beauty products = Output tax – Input tax
= 7,200 – 5,400
= ₹ 1,800

(ii) Now, CGST = SGST = \(\frac{\text { GST }}{2}\) = 9%
∴ Statement of GST payable at each stage can be tabulated as:

Question 6.
‘Chitra furnishings’ purchased tapestry (curtain cloth) for ₹ 28,00,000 and sold for ₹ 44,80,000. Rate of GST is 5%. Find
(i) Input Tax
(ii) Output Tax
(iii) ITC
(iv) CGST and SGST
Solution:
Given, that ‘Chitra furnishings’ purchased tapestry (curtain cloth) for ₹ 28,00,000 and rate of GST is 5%
(i) Input tax = 5% of 28,00,000
= \(\frac{5}{100}\) × 28,00,000
= ₹ 1,40,000
The tapestry was sold at ₹ 44,80,000

(ii) Output tax = 5% of 44,80,000
= \(\frac{5}{100}\) × 44,80,000
= ₹ 2,24,000

(iii) Now ITC = Input tax = ₹ 1,40,000
GST payable = Output tax – ITC
= 2,24,000 – 1,40,000
= ₹ 84,000

(iv) CGST = SGST = \(\frac{\text { GST Payable }}{2}\)
= \(\frac{84,000}{2}\)
= ₹ 42,000
∴ CGST = SGST = ₹ 42,000

Question 7.
Two friends ‘Aditi’ and ‘Vaishali’ went to a restaurant. They ordered 2 Masala Dosa costing ₹ 90 each 2 coffee costing ₹ 60 each and 1 sandwich costing ₹ 80. GST is charged at 5%. Find the Total amount of the bill including GST.
Solution:
Aditi and Vaishali ordered for 2 Masala Dosas, 2 Coffees and 1 Sandwich
∴ Total price of their order = 2 × 90 + 2 × 60 + 80 = ₹ 380
GST is charged at 5%
∴ GST on the total order = 5% × 380
= \(\frac{5}{100}\) × 380
= ₹ 19
∴ Total bill amount including GST = 380 + 19 = ₹ 399

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 9.1 Solutions
• 11th Commerce Maths Exercise 9.2 Solutions
• 11th Commerce Maths Exercise 9.3 Solutions
• 11th Commerce Maths Exercise 9.4 Solutions
• 11th Commerce Maths Exercise 9.5 Solutions
• 11th Commerce Maths Exercise 9.6 Solutions
• 11th Commerce Maths Exercise 9.7 Solutions
• 11th Commerce Maths Miscellaneous Exercise 9 Solutions

Probability Class 11 Commerce Maths 2 Chapter 7 Exercise 7.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.2 Questions and Answers.

Std 11 Maths 2 Exercise 7.2 Solutions Commerce Maths

Question 1.
A fair die is thrown two times. Find the chance that
(i) product of the numbers on the upper face is 12.
(ii) sum of the numbers on the upper face is 10.
(iii) sum of the numbers on the upper face is at least 10.
(iv) sum of the numbers on the upper face is 4.
(v) the first throw gives an odd number and the second throw gives a multiple of 3.
(vi) both the times die to show the same number (doublet).
Solution:
If a fair die is thrown twice, the sample space is
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 product of the numbers on uppermost face is 12.
∴ A = {(2, 6), (3, 4), (4, 3), (6, 2)}
∴ n(A) = 4
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{4}{36}=\frac{1}{9}\)

(ii) Let B be the event that sum of the numbers on uppermost face is 10.
∴ B = {(4, 6), (5, 5), (6, 4)}
∴ n(B) = 3
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{3}{36}=\frac{1}{12}\)

(iii) Let C be the event that sum of the numbers on uppermost face is at least 10 (i.e., 10 or more than 10 which are 10 or 11 or 12)
∴ C = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), ( 6, 6)}
∴ n(C) = 6
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}\)

(iv) Let D be the event that sum of the numbers on uppermost face is 4.
∴ D = {(1, 3), (2, 2), (3, 1)}
∴ n(D) = 3
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{3}{36}=\frac{1}{12}\)

(v) Let E be the event that 1st throw gives an odd number and 2nd throw gives multiple of 3.
∴ E = {(1, 3), (1, 6), (3, 3), (3, 6), (5, 3), (5, 6)}
∴ n(E) = 6
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}\)

(vi) Let F be the event that both times die shows same number.
∴ F = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
∴ n(F) = 6
∴ P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}\)

Question 2.
Two cards are drawn from a pack of 52 cards. Find the probability that
(i) both are black.
(ii) both are diamonds.
(iii) both are ace cards.
(iv) both are face cards.
(v) one is a spade and the other is a non-spade.
(vi) both are from the same suit.
(vii) both are from the same denomination.
Solution:
Two cards can be drawn from a pack of 52 cards in ⁵²C₂ ways.
∴ n(S) = ⁵²C₂
(i) Let A be the event that both the cards drawn are black.
The pack of 52 cards contains 26 black cards.
∴ 2 cards can be drawn from them in ²⁶C₂ ways
∴ n(A) = ²⁶C₂
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(ii) Let B be the event that both the cards drawn are diamond.
There are 13 diamond cards in a pack of 52 cards.
∴ 2 diamond cards can be drawn from 13 diamond cards in ¹³C₂ ways
∴ n(B) = ¹³C₂
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(iii) Let C be the event that both the cards drawn are aces.
In a pack of 52 cards, there are 4 ace cards.
∴ 2 ace cards can be drawn from 4 ace cards in ⁴C₂ ways
∴ n(C) = ⁴C₂
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(iv) Let D be the event that both the cards drawn are face cards.
There are 12 face cards in a pack of 52 cards.
∴ 2 face cards can be drawn from 12 face cards in ¹²C₂ ways.
∴ n(D) = ¹²C₂
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(v) Let E be the event that out of the two cards drawn one is a spade and other is non-spade.
There are 13 spade cards and 39 cards are non-spade cards in a pack of 52 cards.
∴ One spade card can be drawn from 13 spade cards in ¹³C₁ ways and one non-spade card can be drawn from 39 non-spade cards in 39C1 ways.
∴ n(E) = ¹³C₁ . ³⁹C₁
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1} \cdot{ }^{39} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{2}}\)

(vi) Let F be the event that both the cards drawn are of the same suit.
A pack of 52 cards consists of 4 suits each containing 13 cards.
2 cards can be drawn from a suit in ¹³C₂ ways.
A suit can be selected in 4 ways.
∴ n(F) = ¹³C₂ × 4
∴ P(F) = \(\frac{\mathrm{n}(\mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{4 \times{ }^{13} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

(vii) Let G be the event that both the cards drawn are of same denominations.
A pack of cards has 13 denominations and 4 different cards for each denomination
∴ n(G) = 13 × ⁴C₂
∴ P(G) = \(\frac{\mathrm{n}(\mathrm{G})}{\mathrm{n}(\mathrm{S})}=\frac{13 \times{ }^{4} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}\)

Question 3.
Four cards are drawn from a pack of 52 cards. Find the probability that
(i) 3 are Kings and 1 is Jack.
(ii) all the cards are from different suits.
(iii) at least one heart.
(iv) all cards are club and one of them is a jack.
Solution:
4 cards can be drawn out of 52 cards in ⁵²C₄ ways.
∴ n(S) = ⁵²C₄
(i) Let A be the event that out of the four cards drawn, 3 are kings and 1 is a jack.
There are 4 kings and 4 jacks in a pack of 52 cards.
∴ 3 kings can be drawn from 4 kings in ⁴C₃ ways.
Similarly, 1 jack can be drawn out of 4 jacks in ⁴C₁ ways.
∴ Total number of ways in which 3 kings and 1 jack can be drawn is ⁴C₃ × ⁴C₁
∴ n(A) = ⁴C₃ × ⁴C₁
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{3} \times{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{4}}\)

(ii) Let B be the event that all the cards drawn are of different suits.
A pack of 52 cards consists of 4 suits each containing 13 cards.
∴ A card can be drawn from each suit in ¹³C₁ ways.
∴ 4 cards can be drawn from 4 different suits in ¹³C₁ × ¹³C₁ × ¹³C₁ × ¹³C₁ ways.
∴ n(B) = ¹³C₁ × ¹³C₁ × ¹³C₁ × ¹³C₁
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1} \times{ }^{13} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{4}}\)

(iii) Let C be the event that out of the four cards drawn at least one is a heart.
∴ C’ is the event that all 4 cards drawn are non-heart cards.
In a pack of 52 cards, there are 39 non-heart cards.
∴ 4 non-heart cards can be drawn in ³⁹C₄ ways.
∴ n(C’) = ³⁹C₄
∴ P(C’) = \(\frac{\mathrm{n}\left(\mathrm{C}^{\eta}\right)}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{39} \mathrm{C}_{4}}{{ }^{52} \mathrm{C}_{4}}\)
∴ P(C) = 1 – P(C’) = 1 – \(\frac{{ }^{39} \mathrm{C}_{4}}{{ }^{52} \mathrm{C}_{4}}\)

(iv) Let D be the event that all the 4 cards drawn are clubs and one of them is a jack.
In a pack of 52 cards, there are 13 club cards having 1 jack card.
∴ 1 jack can be drawn in ¹C₁ way and the other 3 cards can be drawn from remaining 12 club cards in ¹²C₃ ways.
∴ n(D) = ¹²C₃ × ¹C₁
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{12} \mathrm{C}_{3} \times{ }^{1} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{4}}\)

Question 4.
A bag contains 15 balls of three different colours: Green, Black, and Yellow. A ball is drawn at random from the bag. The probability of a green ball is 1/3. The probability of yellow is 1/5.
(i) What is the probability of blackball?
(ii) How many balls are green, black, and yellow?
Solution:
(i) The bag contains 15 balls of three different colours i.e., green (G), black (B) and yellow (Y)
∴ P(G) = \(\frac{1}{3}\) and P(Y) = \(\frac{1}{5}\)
If a ball is drawn from the bag, then it can be any one of the green, black and yellow.
∴ P(G) + P(B) + P(Y) = 1
∴ \(\frac{1}{3}\) + P(B) + \(\frac{1}{5}\) = 1
∴ P(B) + \(\frac{8}{15}\) = 1
∴ P(B) = 1 – \(\frac{8}{15}\) = \(\frac{7}{15}\)
∴ Probability of black ball is \(\frac{7}{15}\)

(ii) Total number of balls = 15 and
P(G) = \(\frac{1}{3}\), P(B) = \(\frac{7}{15}\), P(Y) = \(\frac{1}{5}\)
∴ number of green balls = \(\frac{1}{3}\) × 15 = 5
number of black balls = \(\frac{7}{15}\) × 15 = 7
and number of yellow balls = \(\frac{1}{5}\) × 15 = 3.

Question 5.
A box contains 75 tickets numbered 1 to 75. A ticket is drawn at random from the box. What is the probability that the
(i) number on the ticket is divisible by 6?
(ii) number on the ticket is a perfect square?
(iii) number on the ticket is prime?
(iv) number on the ticket is divisible by 3 and 5?
Solution:
The box contains 75 tickets numbered 1 to 75.
∴ 1 ticket can be drawn from the box in ⁷⁵C₁ = 75 ways.
∴ n(S) = 75
(i) Let A be the event that number on ticket is divisible by 6.
∴ A = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72}
∴ n(A) = 12
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{12}{75}=\frac{4}{25}\)

(ii) Let B be the event that number on ticket is a perfect square.
∴ B = {1, 4, 9, 16, 25, 36, 49, 64}
∴ n(B) = 8
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{8}{75}\)

(iii) Let C be the event that the number on the ticket is a prime number.
∴ C = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73}
∴ n(C) = 21
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{s})}=\frac{21}{75}=\frac{7}{25}\)

(iv) Let D be the event that number on ticket is divisible by 3 and 5 i.e., divisible by L.C.M. of 3 and 5 i.e., 15
∴ D = {15, 30, 45, 60, 75}
∴ n(D) = 5
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{5}{75}=\frac{1}{15}\)

Question 6.
From a group of 8 boys and 5 girls, a committee of five is to be formed. Find the probability that the committee contains
(i) 3 boys and 2 girls
(ii) at least 3 boys.
Solution:
The group consists of 8 boys and 5 girls i.e., 8 + 5 = 13 persons.
A committee of 5 is to be formed from this group.
∴ 5 persons from 13 persons can be selected in ¹³C₅ ways
∴ n(S) = ¹³C₅
(i) Let A be the event that the committee contains 3 boys and 2 girls.
3 boys from 8 boys can be selected in ⁸C₃ ways and 2 girls from 5 girls can be selected in ⁵C₂ ways
∴ n(A) = ⁸C₃ . ⁵C₂
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{8} \mathrm{C}_{3} \cdot{ }^{5} \mathrm{C}_{2}}{{ }^{13} \mathrm{C}_{5}}\)

(ii) Let B be the event that the committee contains at least 3 boys (i.e., 3 boys and 2 girls or 4 boys and 1 girl or 5 boys and no girl)
∴ n(B) = ⁸C₃ . ⁵C₂ + ⁸C₄ . ⁵C₁ + ⁸C₅ . ⁵C₀
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{8} \mathrm{C}_{3} \cdot{ }^{5} \mathrm{C}_{2}+{ }^{8} \mathrm{C}_{4} \cdot{ }^{5} \mathrm{C}_{1}+{ }^{8} \mathrm{C}_{5} \cdot{ }^{5} \mathrm{C}_{0}}{{ }^{13} \mathrm{C}_{5}}\)

Question 7.
A room has three sockets for lamps. From a collection of 10 light bulbs of which 6 are defective, a person selects 3 bulbs at random and puts them in a socket. What is the probability that the room is lit?
Solution:
Total number of bulbs = 10
Number of defective bulbs = 6
∴ Number of non-defective bulbs = 4
3 bulbs can be selected out of 10 light bulbs in ¹⁰C₃ ways.
∴ n(S) = ¹⁰C₃
Let A be the event that room is lit.
∴ A’ is the event that the room is not lit.
For A’ the bulbs should be selected from the 6 defective bulbs.
This can be done in ⁶C₃ ways.
∴ n(A’) = ⁶C₃
∴ P(A’) = \(\frac{\mathrm{n}\left(\mathrm{A}^{\prime}\right)}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{6} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}\)
∴ P(Room is lit) = 1 – P(Room is not lit)

Question 8.
The letters of the word LOGARITHM are arranged at random. Find the probability that
(i) Vowels are always together.
(ii) Vowels are never together.
(iii) Exactly 4 letters between G and H
(iv) begins with O and ends with T
(v) Start with a vowel and ends with a consonant.
Solution:
There are 9 letters in the word LOGARITHM.
These letters can be arranged among themselves in ⁹P₉ = 9! ways.
∴ n(S) = 9!
(i) Let A be the event that vowels are always together.
The word LOGARITHM consists of 3 vowels (O, A, I) and 6 consonants (L, G, R, T, H, M).
3 vowels can be arranged among themselves in = ³P₃ = 3! ways.
Considering 3 vowels as one group, 6 consonants and this group (i.e., altogether 7) can be arranged in ⁷P₇ = 7! ways.
∴ n(A) = 3! × 7!
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{3 ! \times 7 !}{9 !}\)

(ii) Let B be the event that vowels are never together.
Consider the following arrangement
_C_C_C_C_C_C_
6 consonants create 7 gaps.
∴ 3 vowels can be arranged in 7 gaps in ⁷P₃ ways.
Also 6 consonants can be arranged among themselves in ⁶P₆ = 6! ways.
∴ n(B) = 6! × ⁷P₃
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{6 ! \times{ }^{7} \mathrm{P}_{3}}{9 !}\)

(iii) Let C be the event that exactly 4 letters are arranged between G and H.
Consider the following arrangement
1 2 3 4 5 6 7 8 9
∴ Out of 9 places, G and H can occupy any one of following 4 positions in 4 ways.
1st and 6th, 2nd and 7th, 3rd and 8th, 4th and 9th
Now, G and H can be arranged among themselves in ²P₂ = 2! =2 ways.
Also, the remaining 7 letters can be arranged in remaining 7 places in ⁷P₇ = 7! ways.
∴ n(C) = 4 × 2 × 7! = 8 × 7! = 8!
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{S})}=\frac{8 !}{9 !}=\frac{8 !}{9 \times 8 !}=\frac{1}{9}\)

(iv) Let D be the event that word begins with O and ends with T.
Thus first and last letter can be arranged in one way each and the remaining 7 letters can be arranged in remaining 7 places in ⁷P₇ = 7! ways
∴ n(D) = 7! × 1 × 1 = 7!
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{S})}=\frac{7 !}{9 !}\)

(v) Let E be the event that word starts with vowel and ends with consonant.
There are 3 vowels and 6 consonants in the word LOGARITHM.
∴ The first place can be filled in 3 different ways and the last place can be filled in 6 ways.
Now, remaining 7 letters can be arranged in 7 places in ⁷P₇ = 7! ways
∴ n(E) = 3 × 6 × 7!
∴ P(E) = \(\frac{n(E)}{n(S)}=\frac{3 \times 6 \times 7 !}{9 !}\)

Question 9.
The letters of the word SAVITA are arranged at random. Find the probability that vowels are always together.
Solution:
The word SAVITA contains 6 letters. Out of 6 letters, 3 are vowels (A, A, I) and 3 are consonants (S, V, T).
6 letters in which A repeats twice can be arranged among themselves in \(\frac{6 !}{2 !}\) ways.
∴ n(S) = \(\frac{6 !}{2 !}\)
Let A be the event that vowels are always together.
3 vowels (A, A, I) can be arranged among themselves in \(\frac{3 !}{2 !}\) ways.
Considering 3 vowels as one group, 3 consonants and this group (i.e. altogether 4) can be arranged in ⁴P₄ = 4! ways.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 7.1 Solutions
• 11th Commerce Maths Exercise 7.2 Solutions
• 11th Commerce Maths Exercise 7.3 Solutions
• 11th Commerce Maths Exercise7.4 Solutions
• 11th Commerce Maths Miscellaneous Exercise 7 Solutions

Commercial Mathematics Class 11 Commerce Maths 2 Chapter 9 Miscellaneous Exercise 9 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Miscellaneous Exercise 9 Questions and Answers.

Std 11 Maths 2 Miscellaneous Exercise 9 Solutions Commerce Maths

Question 1.
A man buys a house for ₹ 10 lakh and rents it. He puts 10% of the annual rent aside for repairs, pays ₹ 1,000 as annual taxes, and realizes 8% on his investment thereafter. Find the annual rent of the house.
Solution:
Let ₹ ‘x’ be the annual rent of the house.
The man keeps 10% of the annual rent aside for repairs.
i.e., \(\frac{10}{100}\) × x or ₹ \(\frac{x}{10}\) aside tor repairs.
In addition, he pays ₹ 1000 as annual taxes.
After incurring these expenses he is left with an amount which is 8% of his investment for the house.

∴ The annual rent of the house is ₹ 90,000.

Question 2.
Rose got 30% of the maximum marks in an examination and failed by 10 marks. However, Lily who appeared for the same examination got 40% of the total marks and got 15 marks more than the passing marks. What were the passing marks in the examination?
Solution:
Let maximum marks be x
Rose scored 30% of maximum marks
i.e. Rose scored \(\frac{30}{100}\)x
Rose failed by 10 marks
∴ passing marks = \(\frac{30}{100}\)x + 10 …..(i)
Lily scored 40% of maximum marks
i.e. Lily scored \(\frac{40}{100}\)x
Lily scored 15 marks more than passing marks
∴ passing marks = \(\frac{40}{100}\)x – 15 ……(ii)
equating (i) and (ii),
\(\frac{30x}{100}\) + 10 = \(\frac{40x}{100}\) – 15
∴ 10 + 15 = \(\frac{40 x-30 x}{100}\)
∴ 10x = (25)(100)
∴ x = 250
From (i), passing marks = \(\frac{30}{100}\)(250) + 10
= 75 + 10
= 85
∴ Passing marks for the examination were 85.

Question 3.
Ankita’s Salary was reduced by 50%. Again the reduced salary was increased by 50%. Find loss in terms of percentage.
Solution:
Let Ankita’s initial salary be ₹ ‘x’.
Her salary was reduced by 50%.
∴ Ankita’s salary after reduction = x(1 – \(\frac{50}{100}\))
= x(1 – \(\frac{1}{2}\))
= \(\frac{x}{2}\)
Ankita’s reduced salary was then increased by 50%
∴ Ankita’s final salary after the increase

∴ Loss in Ankita’s salary after the decrease and increase = x – \(\frac{3 x}{4}\) = \(\frac{x}{4}\)

∴ Ankita lost 25% of her salary.

Question 4.
By selling 300 lunch boxes, a shopkeeper gains the selling price of 100 lunch boxes. Find his gain percent.
Solution:
Let ₹ x be the selling price (S.P.) of one lunch box.
∴ S.P. of 300 lunch boxes = 300x
and S.P. of 100 lunch boxes = 100x
Gain = 100x ……[given]
C.P. of 300 lunch boxes = S.P. – Gain
= 300x – 100x
= 200x

∴ The shopkeeper’s gain percentage is 50%.

Question 5.
A salesman sold an article at a loss of 10%. If the selling price has been increased by ₹ 80, there would have been a gain of 10%. What was the cost of the article?
Solution:
Let ₹ x be the cost price of the article.
S.P. of the article = x – \(\frac{10}{100}\)x = \(\frac{9x}{100}\) …….(i)
Given that, S.P. increased by ₹ 80 would have given 10% gain

∴ The cost price of the article is ₹ 400

Question 6.
Find the single discount equivalent to a series discount of 10%, 20%, and 15%.
Solution:
Let the marked price be ₹ 100
After 1st discount the price = 100(1 – \(\frac{10}{100}\)) = 90
After 2nd discount the price = 90(1 – \(\frac{20}{100}\)) = 72
After 3rd discount the price = 72(1 – \(\frac{15}{100}\)) = 61.2
∴ The selling price after 3 discounts is ₹ 61.2.
∴ Single equivalent discount = marked price – selling price
= 100 – 61.2
= ₹ 38.8
∴ The single equivalent discount is ₹ 38.8 on ₹ 100.
i.e. The single equivalent discount is 38.8%.

Question 7.
Reshma put an amount at simple interest at a certain rate for 3 years. Had it been put at 2% higher rate, she would have received ₹ 360 more. Find the sum.
Solution:
Let P and R represent the principal amount and rate of interest p.a. respectively.
Given duration = T = 3 years
Simple interest = \(\frac{\mathrm{PRT}}{100}=\frac{3 \mathrm{PR}}{100}\)
Given that, had the amount been kept at 2% more, then the gain would have been ₹ 360 more.

∴ The sum of money is ₹ 6,000.

Question 8.
The compound interest on ₹ 30000 at 7% p.a. is ₹ 4347. What is the period in years?
Solution:
Given that,
Principal (P) = ₹ 30,000
Rate of interest (R) = 7% p.a.
Compound interest = ₹ 4,347
Amount after compound interest

= \(\left(\frac{107}{100}\right)^{2}\)
= (1.07)²
∴ T = 2
∴ Amount is invested for 2 years.

Question 9.
The value of the machine depreciates at the rate of 15% p.a. It was purchased 2 years ago. Its present value is ₹ 7,225. What was the purchase price of the machine?
Solution:
Given,
Rate of depreciation = r = 15%
Number of years = n = 2 years
Present value of machine = P.V. = ₹ 7,225
The purchase price (V) of the machine can be found using

∴ The purchase price of the machine was ₹ 10,000/-.

Question 10.
A tree increases annually by \(\frac{1}{8}\) of its height. By how much will it increase after 2\(\frac{1}{2}\) years. If its length today is 8 m?
Solution:
The height of the tree today is 8m.
The height of the tree increases by \(\frac{1}{8}\)th of its height every year.
At the end of 1st year, height of the tree will be = 8 + \(\frac{1}{8}\) × 8 = 9 m
And, at the end of the 2nd year, height of the tree will be = 9 + \(\frac{1}{8}\) × 9
= 9(1 + \(\frac{1}{8}\))
= 9 × \(\frac{9}{8}\)
= \(\frac{81}{8}\)
After six more months, the height of the tree will be

∴ Increase in the height of the tree after 2\(\frac{1}{2}\) years = 10.75 – 8 = 2.75 m.

Question 11.
A building worth ₹ 1,21,000 is constructed on land worth ₹ 81,000. After how many years will the value of both be the same if land appreciates at 10% p.a and buildings depreciate at 10% p.a.
Solution:
Given,
Value of the building = V.B. = ₹ 1,21,000
Value of land = V.L. = ₹ 81,000/-
Rate of appreciation of land = rate of depreciation of building = r = 10%.
For the value of building and land to be the same.

∴ n = 2 years.
∴ After two years value of the building and land will be the same.

Question 12.
Varun invested 25%, 30%, and 20% of his savings in buying shares of three different companies, ‘A’, ‘B’, and ‘C’ which declared dividends, 10%, 12%, and 15% respectively. If his total income on account of dividends is ₹ 6,370/-, find the amount he invested in buying shares of company ‘B’.
Solution:
Let ‘T’ be Varan’s total savings.
∴ Investment of Varan in:
Company A = 25% of T = \(\frac{25}{100}\) × T = \(\frac{T}{4}\),
Company B = 30% of T = \(\frac{30}{100}\) × T = \(\frac{3T}{10}\),
Company C = 20% of T = \(\frac{20}{100}\) × T = \(\frac{T}{5}\)
Company A, B and C declared dividends 10%, 12% and 15% respectively.
∴ Dividend from company A = 10% of \(\frac{T}{4}\)

∴ Varan invested ₹ 21,000 in company B.

Question 13.
Find the annual dividend received from ₹ 25,000, 8% stock at ₹ 108.
Solution:
Amount invested = ₹ 25,000
Dividend = 8%
Assuming face value F.V. as ₹ 100
Annual income per share = \(\frac{\text { Dividend }}{100} \times \text { Face value }\)
= \(\frac{8}{100}\) × 100
= ₹ 8
Market value of the share M.V. = ₹ 108

Annual dividend on amount invested = Rate of return × amount invested
= \(\frac{7.4}{100}\) × 25,000
= ₹ 1850
∴ Annual dividend of ₹ 1,850 is received from 8% stock at ₹ 108.
Alternate approach
Assuming ₹ 25,000 as the total face value of all the shares.
Since the dividend is 8%,
Annual dividend = \(\frac{8}{100}\) × 25,000 = ₹ 2,000

Question 14.
A, B, and C enter into a partnership. A invests 3 times as much as B invests and B invests two-thirds of what ‘C’ invests. At the end of the year, the profit earned is ₹ 8,800. What is the share of ‘B’?
Solution:
Let ‘a’, ‘b’ and ‘c’ be the amounts invested by A, B and C respectively.
Given that, A invests 3 times as much as B and B invests two third of what ‘C’ invests.
∴ a = 3b and b = \(\frac{2}{3}\)c
∴ \(\frac{a}{b}=\frac{3}{1}\) and \(\frac{b}{c}=\frac{2}{3}\)
or \(\frac{a}{b}=\frac{6}{2}\) and \(\frac{b}{c}=\frac{2}{3}\)
∴ a : b = 6 : 2 and b : c = 2 : 3
∴ a : b : c = 6 : 2 : 3
Given that profit earned = ₹ 8800
∴ Share of ‘B’ in profit = \(\frac{2}{11}\) × 8800 = ₹ 1600
∴ B’share in profit is ₹ 1600.

Question 15.
The ratio of investment of two partners Santa and Banta is 11 : 12 and the ratio of their profits is 2 : 3. If Santa invested the money for 8 months, then for how much time did Banta his money?
Solution:
Let ‘x’ be the time in months for which Banta invested his money
Santa and Banta invested their money in the ratio 11 : 12.
Santa invested his money for 8 months and the ratio of their profits is 2 : 3.
∴ 11 × 8 : 12 × x = 2 : 3
∴ \(\frac{88}{12 x}=\frac{2}{3}\)
∴ x = \(\frac{88 \times 3}{2 \times 12}\)
∴ x = 11
∴ Banta invested his money for 11 months.

Question 16.
Akash, Sameer, and Sid took a house on rent for one year for ₹ 16,236. They stayed together for 4 months and then Sid left the house. After 5 more months, Sameer also left the house. How much rent should each pay?
Solution:
Let ‘R’ be the rent per month to be paid to the landlord.
Given that, Sid left the house after 4 months
∴ Rent paid by Sid = \(\frac{R}{3}\) × 4 = \(\frac{4R}{3}\)
Sameer left the house after another 5 months,
∴ Rent paid by Sameer = \(\frac{R}{2}\) × 5 + \(\frac{R}{3}\) × 4
= R(\(\frac{5}{2}+\frac{4}{3}\))
= \(\frac{23R}{6}\)
Akash stayed in the house for the entire year.
∴ Rent paid by Akash = 3R + \(\frac{R}{2}\) × 5 + \(\frac{R}{3}\) × 4
= R(3 + \(\frac{5}{2}+\frac{4}{3}\))
= \(\frac{41R}{6}\)
∴ The rent paid by the three of them, over that period of one year must be in the proportion.
\(\frac{41 \mathrm{R}}{6}: \frac{23 \mathrm{R}}{6}: \frac{4 \mathrm{R}}{3}\)
i.e. in the proportion
41 : 23 : 8 …..(multiplying throughout by \(\frac{6}{R}\))
Let x be the constant of proportionality.
Rent to be paid by Akash = ₹ 41x
Rent to be paid by Sameer = ₹ 23x
and rent to be paid by Sid = ₹ 8x
The total rent for the house was ₹ 16236.
∴ 41x + 23x + 8x = ₹ 16236
∴ 72x = 16236
∴ x = 225.5
∴ Akash should pay 41x = 41 × 225.5 = ₹ 9245.5
Sameer should pay 23x = 23 × 225.5 = ₹ 5186.5
and Sid should pay 8x = 8 × 225.5 = ₹ 1804

Question 17.
Ashwin Auto Automobiles sold 10 motorcycles. Total sales amount was ₹ 6,80,000. 18% GST is applicable. Calculate how much CGST and SGST the firm has to pay.
Solution:
Given, total sales amount for Ashwin Automobiles was ₹ 6,80,000.
18% GST is applicable.
∴ GST payable = 18% of 6,80,000
= \(\frac{18}{100}\) × 6,80,000
= ₹ 1,22,400
Now CGST = SGST = 9%
= \(\frac{\text { GST payable }}{2}\)
= \(\frac{1,22,400}{2}\)
= ₹ 61,200
∴ CGST = SGST = ₹ 61,200

Question 18.
‘Sweet 16’ A ready made garments shop for Women’s garments, purchased stock for ₹ 4,00,000 and sold that stock for ₹ 5,50,000 (12% GST is applicable) Find,
(i) Input Tax Credit
(ii) CGST and SGST paid by the firm.
Solution:
Given that, stock purchased by ‘Sweet 16’ was worth ₹ 4,00,000
GST applicable is 12%.
∴ Input tax = 12% of 4,00,000
= \(\frac{12}{100}\) × 4,00,000
= ₹ 48,000
∴ Input tax Credit (ITC) = ₹ 48,000
The garment stock was sold for ₹ 5,50,000
Output tax = 12% of 5,50,000
= \(\frac{12}{100}\) × 5,50,000
= ₹ 66,000
∴ GST payable = output tax – ITC
= 66,000 – 48,000
= ₹ 18,000
∴ CGST = SGST = \(\frac{\text { GST payable }}{2}\) = ₹ 9,000

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 9.1 Solutions
• 11th Commerce Maths Exercise 9.2 Solutions
• 11th Commerce Maths Exercise 9.3 Solutions
• 11th Commerce Maths Exercise 9.4 Solutions
• 11th Commerce Maths Exercise 9.5 Solutions
• 11th Commerce Maths Exercise 9.6 Solutions
• 11th Commerce Maths Exercise 9.7 Solutions
• 11th Commerce Maths Miscellaneous Exercise 9 Solutions

Bivariate Frequency Distribution and Chi Square Statistic Class 11 Commerce Maths 2 Chapter 4 Exercise 4.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Ex 4.2 Questions and Answers.

Std 11 Maths 2 Exercise 4.2 Solutions Commerce Maths

Question 1.
The following table shows the classification of applications for secretarial and for sales positions according to gender. Calculate the value of ϰ² statistic.

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{225 \times 100}{300}\) = 75
E₁₂ = \(\frac{225 \times 200}{300}\) = 150
E₂₁ = \(\frac{75 \times 100}{300}\) = 25
E₂₂ = \(\frac{75 \times 200}{300}\) = 50
Table of expected frequencies.

Question 2.
200 teenagers were asked which takeaway food do they prefer – French fries, burgers, or pizza. The results were-

Compute ϰ² statistic.
Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{50 \times 24}{200}\) = 6
E₁₂ = \(\frac{50 \times 60}{200}\) = 15
E₁₃ = \(\frac{50 \times 116}{200}\) = 29
E₂₁ = \(\frac{150 \times 24}{200}\) = 18
E₂₂ = \(\frac{150 \times 60}{200}\) = 45
E₂₃ = \(\frac{150 \times 116}{200}\) = 87
Table of expected frequencies.

Question 3.
A sample of men and women who had passed their driving test either in 1st attempt or in 2nd attempt
were surveyed. Compute ϰ² statistic.

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{60 \times 40}{80}\) = 30
E₁₂ = \(\frac{60 \times 40}{80}\) = 30
E₂₁ = \(\frac{20 \times 40}{80}\) = 10
E₂₂ = \(\frac{20 \times 40}{80}\) = 10
Table of expected frequencies.

Question 4.
800 people were asked whether they wear glasses for reading with the following results.

Compute the ϰ² square statistic.
Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{400 \times 600}{800}\) = 300
E₁₂ = \(\frac{400 \times 200}{800}\) = 100
E₂₁ = \(\frac{400 \times 600}{800}\) = 300
E₂₂ = \(\frac{400 \times 200}{800}\) = 100
Table of expected frequencies.

Question 5.
Out of a sample of 120 persons in a village, 80 were administered a new drug for preventing influenza, and out of the 18 were attacked by influenza. Out of those who are not administered the new drug, 10 persons were not attacked by influenza:
(i) Prepare a two-way table showing frequencies.
(ii) Compute the ϰ² square statistic.
Solution:
(i) The given data can be arranged in the following table.

The observed frequency table can be prepared as follows:

(ii) Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{48 \times 80}{120}\) = 32
E₁₂ = \(\frac{48 \times 40}{120}\) = 16
E₂₁ = \(\frac{72 \times 80}{120}\) = 48
E₂₂ = \(\frac{72 \times 40}{120}\) = 24
Table of expected frequencies.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 4.1 Solutions
• 11th Commerce Maths Exercise 4.2 Solutions
• 11th Commerce Maths Miscellaneous Exercise 4 Solutions

Linear Inequations Class 11 Commerce Maths 2 Chapter 8 Miscellaneous Exercise 8 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Linear Inequations Miscellaneous Exercise 8 Questions and Answers.

Std 11 Maths 2 Miscellaneous Exercise 8 Solutions Commerce Maths

Solve the following system of inequalities graphically.

Question 1.
x ≥ 3, y ≥ 2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 2.
3x + 2y ≤ 12, x ≥ 1, y ≥ 2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 3.
2x + y ≥ 6, 3x + 4y < 12
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 4.
x + y ≥ 4, 2x – y ≤ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 5.
2x – y ≥1, x – 2y ≤ -1
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 6.
x + y ≤ 6, x + y ≥ 4
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 7.
2x + y ≥ 8, x + 2y ≥ 10
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 8.
x + y ≤ 9, y > x, x ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 9.
5x + 4y ≤ 20, x ≥ 1, y ≥ 2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 10.
3x + 4y ≤ 60, x +3y ≤ 30, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 11.
2x + y ≥ 4, x + y ≤ 3, 2x – 3y ≤ 6
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 12.
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 13.
4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 14.
3x + 2y ≤ 150, x + 4y ≥ 80, x ≤ 15, y ≥ 0, x ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 15.
x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 8.1 Solutions
• 11th Commerce Maths Exercise 8.2 Solutions
• 11th Commerce Maths Exercise 8.3 Solutions
• 11th Commerce Maths Miscellaneous Exercise 8 Solutions

Commercial Mathematics Class 11 Commerce Maths 2 Chapter 9 Exercise 9.5 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.5 Questions and Answers.

Std 11 Maths 2 Exercise 9.5 Solutions Commerce Maths

Question 1.
Three partners shared the profit in a business in the ratio 5 : 6 : 7. They had partnered for 12 months, 10 months, and 8 months respectively. What was the ratio of their investments?
Solution:
Let the ratio of investments of the three partners be p : q : r.
They partnered for 12 months, 10 months, and 8 months respectively.
∴ The profit shared by the partners will be in proportion to the product of capital invested and their respective time periods.
∴ 12 × p : 10 × q : 8 × r = 5 : 6 : 7

From (i) & (ii), we have
p : q : r = 50 : 72 : 105
∴ The ratio of their investments was 50 : 72 : 105.

Question 2.
Kamala, Vimala and Pramila enter into a partnership. They invest ₹ 40,000, ₹ 80,000 and ₹ 1,20,000 respectively. At the end of the first year, Vimala withdraws ₹ 40,000, while at the end of the second year, Pramila withdraws ₹ 80,000. In what ratio will the profit be shared at the end of 3 years?
Solution:
Given that, Kamala, Vimala, and Pramila invest ₹ 40,000, ₹ 80,000, and ₹ 1,20,000 respectively.
The ratio of profits is to be calculated at the end of 3 years.
Vimala withdraws ₹ 40,000 at the end of the first year.
∴ Vimala invested ₹ 80,000 for one year and 40,000 for 2 years.
Pramila withdraws ₹ 80,000 at the end of the second year.
∴ Pramila invested ₹ 1,20,000 for two years and 40,000 for one year.
Kamala invested ₹ 40,000 for all the 3 years.
∴ The ratio of profits to be shared at the end of 3 years will be
= 40,000 × 3 : 80,000 × 1 + 40,000 × 2 : 1,20,000 × 2 + 40,000 × 1
= 1,20,000 : 1,60,000 : 2,80,000
= 12 : 16 : 28
= 3 : 4 : 7

Alternate Method:
Given that, Kamala, Vimala and Pramila invest ₹ 40,000, ₹ 80,000 & ₹ 1,20,000 respectively.
Given, information can be tabulated as:

∴ The profits to be shared at the end of 3 years will be
= 1,20,000 : 1,60,000 : 2,80,000
= 12 : 16 : 28
= 3 : 4 : 7

Question 3.
Sanjeev started a business investing ₹ 25,000 in 1999. In 2000, he invested an additional amount of ₹ 10,000 and Rajeev joined him with an amount of ₹ 35,000. In 2001, Sanjeev invested another additional amount of ₹ 10,000 and Pawan joined them with an amount of ₹ 35,000. What will be Rajeev’s share in the profit of ₹ 1,50,000 earned at the end of 3rd year from the start of the business in 1999?
Solution:
The given information can be tabulated as:

∴ The ratio of profits to be shared at the end of 3 years will be 1,05,000 : 70,000 : 35,000
i.e. in the proportion 3 : 2 : 1
Given, profit earned ₹ 1,50,000/-
∴ Rajeev’s share in the profit = \(\frac{2}{6}\) × 1,50,000 = ₹ 50,000/-

Question 4.
Teena, Leena, and Meena invest in a partnership in the ratio: 7/2, 4/3, 6/5. After 4 months, Teena increases her share by 50%. If the total profit at the end of one year is ₹ 21,600, then what is Leena’s share in the profit?
Solution:
Investment of Teena, Leena and Meena are in the ratio \(\frac{7}{2}: \frac{4}{3}: \frac{6}{5}\)
After 4 months, Teena’s share increases by 50%.
i.e. \(\frac{7}{2}+\left(\frac{7}{2} \times \frac{50}{100}\right)=\frac{7}{2}+\frac{7}{4}\)
i.e. \(\frac{21}{4}\)
The profit will be shared in the proportion of product of capitals and respective time periods in months.
i.e. \(\frac{7}{2} \times 4+\frac{21}{4} \times 8: \frac{4}{3} \times 12: \frac{6}{5} \times 12\)
i.e. 56 : 16 : \(\frac{72}{5}\)
i.e. 7 : 2 : \(\frac{9}{5}\)
i.e. in the proportion 35 : 10 : 9 …..[Multiplying throughout by 5]
Given that profit at the end of one year = ₹ 21,600/-
∴ Leena’s share in the profit = \(\frac{10}{54}\) × 21,600
= 5 × 800
= 4000
∴ Leena’s share in the profit is ₹ 4000/-.

Question 5.
Dilip and Pradeep invested amounts in the ratio 2 : 1, whereas the ratio between amounts invested by Dilip and Sudip was 3 : 2. If ₹ 1,49,500 was their profit, how much amount did Sudip receive?
Solution:
Let the amounts invested by Dilip, Pradeep and Sudip be ₹ ‘d’, ₹ ‘p’ and ₹ ‘s’ respectively.
Given that, d : p = 2 : 1
∴ d : p = 6 : 3 …..(i)
and d : s = 3 : 2
∴ d : s = 6 : 4 …..(ii)
From (i) and (ii),
d : p : s = 6 : 3 : 4
∴ The ratio of profits to be shared among Dilip, Pradeep and Sudip will be 6 : 3 : 4.
Given, profit earned = ₹ 1,49,500/-
∴ Sudip’s share in the profit = \(\frac{4}{13}\) × 1,49,500
= 4 × 11,500
= ₹ 46,000/-

Question 6.
The ratio of investments of two partners Jatin and Lalit is 11 : 12 and the ratio of their profits is 2 : 3. If Jatin invested the money for 8 months, find for how much time Lalit invested his money.
Solution:
Let ‘x’ be the time in months for which Lalit invested his money
Jatin and Lalit invested their money in the ratio 11 : 12.
Jatin invested his money for 8 months and the ratio of their profits is 2 : 3.
∴ 11 × 8 : 12 × x = 2 : 3
∴ \(\frac{88}{12 x}=\frac{2}{3}\)
∴ x = \(\frac{88 \times 3}{2 \times 12}\)
∴ x = 11
∴ Lalit invested his money for 11 months.

Question 7.
Three friends had dinner at a restaurant. When the bill was received, Alpana paid \(\frac{2}{3}\) as much as Beena paid and Beena paid \(\frac{1}{2}\) as much as Catherin paid. What fraction of the bill did Beena pay?
Solution:
Let ‘T’ be the total bill amount at the restaurant and ‘a’, ‘b’, and ‘c’ be the share of Alpana, Beena, and Catherin respectively.
Given, that Alpana paid \(\frac{2}{3}\) as much as Beena paid
∴ a = \(\frac{2}{3}\) b …..(i)
Also, Beena paid \(\frac{1}{2}\) as much as Catherin paid.
∴ b = \(\frac{1}{2}\) c
∴ c = 2b …….(ii)
∴ Three friends paid the total bill amount.
∴ a + b + c = T …..(iii)
Using (i) and (ii) in (iii), we get

Thus, Beena paid \(\left(\frac{3}{11}\right)^{\text {th }}\) fraction of the total bill amount.

Question 8.
Roy starts a business with ₹ 10,000, Shikha joins him after 2 months with 20% more investment than Roy, after 2 months Tariq joins him with 40% less than Shikha. If the profit earned by them at the end of the year is equal to twice the difference between the investment of Roy and ten times the investment of Tariq. Find the profit of Roy?
Solution:
Given that, Roy starts the business with ₹ 10,000.
Shikha joins him after 2 months with 20% more investment than Roy.
∴ Shikha’s investment = 10,000 + (10,000 × \(\frac{20}{100}\)) = ₹ 12,000
Tariq joins after two more months with an investment 40% less than Shikha.
∴ Tariq’s investment = 12,000 – (12,000 × \(\frac{40}{100}\)) = ₹ 7,200
Now, the profit will be shared in the proportion of product of capitals and respective periods in months.
i.e. 10,000 × 12 : 12,000 × 10 : 7,200 × 8
i.e. in the proportion, 25 : 25 : 12 …..(i) [Dividing throughout by 4,800]
Given that, profit at the end of the year = twice of the difference between investment of Roy and ten times the investment of Tariq.
∴ Profit = 2 [(10 × 7,200) – 10,000]
= 2[72,000 – 10,000]
= 2 × 62,000
= ₹ 1,24,000
∴ Roy’s share of profit = \(\frac{25}{62}\) × 1,24,000 …..[From (i)]
= ₹ 50,000/-

Question 9.
If 4(P’s Capital) = 6(Q’s Capital) = 10 (R’s Capital), then out of the total profit of ₹ 5,580, what is R’s share?
Solution:
Let ‘p’, ‘q’ and ‘r’ be P, Q and R’s Capital for the business respectively.
∴ 4p = 6q = 10r
L.C.M of 4, 6, 10 = 60
∴ We take 4p = 6q = 10r = 60x
∴ p = 15x, q = 10x, r = 6x
∴ p : q : r = 15 : 10 : 6
Given that total profit = ₹ 5580
R’s share in the profit = \(\frac{6}{31}\) × 5580 = ₹ 1080/-

Question 10.
A and B start a business, with A investing the total capital of ₹ 50,000, on the condition that B pays interest at the rate of 10% per annum on his half of the capital. A is a working partner and receives ₹ 1,500 per month from the total profit and any profit remaining is equally shared by both of them. At the end of the year, it was found that the income of A is twice that of B. Find the total profit for the year?
Solution:
Let ‘x’ and ‘y’ be the profits earned by A and B respectively and let ‘z’ be the total profit for the year.
A is the working partner and receives ₹ 1500 per month from the total profit.
i.e. 12 × 1500 = ₹ 18,000 at the end of the year.
The remaining profit is shared between A and B equally.
∴ y = \(\frac{z-18000}{2}\) …..(i)
Thus, profit earned by A at the end of that year is given by
x = 18000 + \(\left(\frac{z-18000}{2}\right)\)
∴ x = \(\frac{z+18000}{2}\) ……(ii)
A invests the entire capital on the condition that B pays A interest at the rate of 10% per annum on his half of the capital.
∴ At the end of the first year,
A will receive \(\frac{10}{100}\) × 25,000 i.e. ₹ 2500/- over and above his share of profit.
∴ A’s income = Profit of A + 2500 = x + 2500
Given that,
income of A = twice the income of B
∴ x + 2500 = 2y …..(iii)
Using (i) and (ii) in (iii), we get
\(\frac{z+18000}{2}\) + 2500 = 2\(\left(\frac{z-18000}{2}\right)\)
z + 18000 + 5000 = 2(z – 18000)
z + 23000 = 2z – 36000
∴ z = 59,000
∴ The total profit for the year = ₹ 59,000/-

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 9.1 Solutions
• 11th Commerce Maths Exercise 9.2 Solutions
• 11th Commerce Maths Exercise 9.3 Solutions
• 11th Commerce Maths Exercise 9.4 Solutions
• 11th Commerce Maths Exercise 9.5 Solutions
• 11th Commerce Maths Exercise 9.6 Solutions
• 11th Commerce Maths Exercise 9.7 Solutions
• 11th Commerce Maths Miscellaneous Exercise 9 Solutions