Category: Class 9

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

Practice Set 7.5 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics

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

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

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

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

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

∴ The mean of monthly salary is ₹ 4900.

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

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


∴ The mean of the given data is 3.

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

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

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

= 78.56
∴ The corrected mean is 78.56.

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

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

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

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

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

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

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

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

Question 14.
Find the mode of the following data.

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

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

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

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

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

Practice Set 7.4 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics

Question 1.
Complete the following cumulative frequency table:

Solution:

Question 2.
Complete the following Cumulative Frequency Table:

Solution:

Question 3.
The data is given for 62 students in a certain class regarding their mathematics marks out of 100. Take the classes 0 – 10, 10 – 20,… and prepare frequency distribution
table and cumulative frequency table more than or equal to type.
55. 60, 81, 90, 45, 65, 45, 52, 30, 85, 20, 10,
75, 95, 09, 20, 25, 39, 45, 50, 78, 70, 46, 64,
42. 58. 31, 82, 27, 11, 78, 97, 07, 22, 27, 36,
35, 40, 75, 80, 47, 69, 48, 59, 32, 83, 23, 17,
77, 45, 05, 23, 37, 38, 35, 25, 46, 57, 68, 45.
47,49
From the prcparcd table, answer the following questions :
i. How many students obtained marks 40 or above 40?
ii. How many students obtained marks 90 or above 90?
iii. How many students obtained marks 60 or above 60?
iv. What is the cumulative frequency of equal to or more than type of the class 0 – 10?
Solution:

i. 38 students obtained marks 40 or above 40.
ii. 3 students obtained marks 90 or above 90.
iii. 19 students obtained marks 60 or above 60.
iv. Cumulative frequency of equal to or more than type of the class 0 – 10 is 62.

Question 4.
Using the data In example (3) above, prepare less than type cumulative frequency table and answer the following questions.
i. How many students obtained less than 40 marks?
ii. How many students obtained less than 10 marks?
iii. How many students obtained less than 60 marks?
iv. Find the cumulative frequency of the class 50 – 60.
Solution:

i. 24 students obtained less than 40 marks.
ii. 3 students obtained less than 10 marks.
iii. 43 students obtained less than 60 marks.
iv. Cumulative frequency of the class 50 – 60 is 43.

Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.4 Intext Questions and Activities

Question 1.
The following information is regarding marks in mathematics, obtained out of 40, scored by 50 students of 9th std. ¡n the first unit test. (Textbook pg. no. 120)

From the table, fill in the blanks in the following statements.
i. For class interval 10 – 20 the lower class limit is _____ and upper class limit is _____
ii. How many students obtained marks less than 10? 2
iii. How many students obtained marks less than 20? 2 + ____ = 14
iv. How many students obtained marks less than 30? ______ + _____ = 34
v. How many students obtained marks less than 40? ______ + ______ =50
Solution:
i. 10, 20
iii. 12
iv. 14 + 20
v. 34 + 16

Question 2.
A sports club has organised a table-tennis tournaments. The following table gives the distribution of players ages. Find the cumulative frequencies equal to or more than the lower class limit and complete the table (Textbook pg. no. 121)
Solution:
Equal to lower limit or more than lower limit type of cumulative table.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

Problem Set 3 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials

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

i. Which of the following is a polynomial?

Answer:
(D) √2x² + \(\frac { 1 }{ 2 }\)

ii. What is the degree of the polynomial √7 ?
(A) \(\frac { 1 }{ 2 }\)
(B) 5
(C) 2
(D) 0
Answer:
(D) 0

iii. What is the degree of the polynomial ?
(A) 0
(B) 1
(C) undefined
(D) any real number
Answer:
(C) undefined

iv. What is the degree of the polynomial 2x² + 5xsup>3 + 7?
(A) 3
(B) 2
(C) 5
(D) 7
Answer:
(A) 3

v. What is the coefficient form of x³ – 1 ?
(A) (1, -1)
(B) (3, -1)
(C) (1, 0, 0, -1)
(D) (1, 3, -1)
Answer:
(C) (1, 0, 0, -1)

vi. p(x) = x² – x + 3, then p (7√7) = ?
(A) 3
(B) 7√7
(C) 42√7+3
(D) 49√7
Answer:
(D) 49√7

vii. When x = – 1, what is the value of the polynomial 2x³ + 2x ?
(A) 4
(B) 2
(C) -2
(D) -4
Answer:
(A) 4

viii. If x – 1 is a factor of the polynomial 3x² + mx, then find the value of m.
(A) 2
(B) -2
(C) -3
(D) 3
Answer:
(C) -3

ix. Multiply (x² – 3) (2x – 7x³ + 4) and write the degree of the product.
(A) 5
(B) 3
(C) 2
(D) 0
Answer:
(A) 5

x. Which is the following is a linear polynomials?
(A)  x + 5
(B)  x² + 5
(C) x³ + 5
(D) x⁴ + 5
Answer:
(A)  x + 5

Hints:
v. x³ – 1 = x³ + 0x² + 0x – 1

vi. p(7√ 7) = (7√ 7)² (7√ 7) (7√ 7) + 3
= 3

vii. p(-1) = 2(-1)³ + 2(-1)
= -2 – 2 = -4

vii. p(1) = 0
∴ 3(1)² + m(1) = 0
∴ 3 + m =0
∴ m = -3

ix. Here, degree of first polynomial = 2 and
degree of second polynomial 3
∴ Degree of polynomial obtained by multiplication = 2 + 3 = 5

Question 2.
Write the degree of the polynomial for each of the following.
i. 5 + 3x⁴
ii. 7
iii. ax⁷ + bx⁹ (a, b are constants)
Answer:
i. 5 + 3x⁴
Here, the highest power of x is 4.
∴Degree of the polynomial = 4

ii. 7 = 7x°
∴ Degree of the polynomial = 0

iii. ax⁷ + bx⁹
Here, the highest power ofx is 9.
∴Degree of the polynomial = 9

Question 3.
Write the following polynomials in standard form. [1 Mark each]
i. 4x² + 7x⁴ – x³ – x + 9
ii. p + 2p³ + 10p² + 5p⁴ – 8
Answer:
i. 7x⁴ – x³ + 4x² – x + 9
ii. 5p⁴ + 2p³ + 10p² + p – 8

Question 4.
Write the following polynomial in coefficient form.
i. x⁴ + 16
ii. m⁵ + 2m² + 3m+15
Answer:
i. x⁴ + 16
Index form = x⁴ + 0x³ + 0x² + 0x + 16
∴ Coefficient form of the polynomial = (1,0,0,0,16)

ii. m⁵ + 2m² + 3m + 15
Index form = m⁵ + 0m⁴ + 0m³ + 2m² + 3m + 15
∴ Coefficient form of the polynomial = (1, 0, 0, 2, 3, 15)

Question 5.
Write the index form of the polynomial using variable x from its coefficient form.
i. (3, -2, 0, 7, 18)
ii. (6, 1, 0, 7)
iii. (4, 5, -3, 0)
Answer:
i. Number of coefficients = 5
∴ Degree = 5 – 1 = 4
∴Index form = 3x⁴ – 2x³ + 0x² + 7x + 18

ii. Number of coefficients = 4
∴Degree = 4 – 1 = 3
∴ Index form = 6x³ + x² + 0x + 7

iii. Number of coefficients = 4
∴ Degree = 4 – 1 = 3
∴ Index form = 4x³ + 5x² – 3x + 0

Question 6.
Add the following polynomials.
i. 7x⁴ – 2x³ + x + 10;
3x⁴ + 15x³ + 9x² – 8x + 2
ii. 3p³q + 2p²q + 7;
2p²q + 4pq – 2p³q
Solution:
i. (7x⁴ – 2x³ + x + 10) + (3x⁴ + 15x³ + 9x² – 8x + 2)
= 7x⁴ – 2x³ + x + 10 + 3x⁴ + 15x³ + 9x² – 8x + 2
= 7x⁴ + 3x⁴ – 2x³ + 1 5x³ + 9x² + x – 8x + 10 + 2
= 10x⁴ + 13x³ + 9x² – 7x + 12

ii. (3p³q + 2p²q + 7) + (2p²q + 4pq – 2p³q)
= 3p³q + 2p²q + 7 + 2p²q + 4pq – 2p³q
= 3p³q – 2p³q + 2p²q + 2p²q + 4pq + 7
= p³q + 4p²q + 4pq + 7

Question 7.
Subtract the second polynomial from the first.
i. 5x² – 2y + 9 ; 3x² + 5y – 7
ii. 2x² + 3x + 5 ; x² – 2x + 3
Solution:
i. (5x² – 2y + 9) – (3x² + 5y – 7)
= 5x² – 2y+ 9 – 3x² – 5y + 1
= 5x² – 3x² – 2y – 5y + 9 + 7
= 2x² – 1y + 16

ii. (2x²+ 3x + 5) – (x² – 2x + 3)
= 2x² + 3x + 5 – x² + 2x – 3
= 2x² – x² + 3x + 2x + 5 – 3
= x² + 5x + 2

Question 8.
Multiply the following polynomials.
i. (m³ – 2m + 3) (m⁴ – 2m² + 3m + 2)
ii. (5m³ – 2) (m² – m + 3)
Solution:
i. (m³ – 2m + 3) (m⁴ – 2m² + 3m + 2)
= m³(m⁴ – 2m² + 3m + 2) – 2m(m⁴ – 2m² + 3m + 2) + 3(m⁴ – 2m² + 3m + 2)
= m⁷ – 2m⁵ + 3m⁴ + 2m³ – 2m⁵ + 4m³ – 6m² – 4m + 3m⁴ – 6m² + 9m + 6
= m⁷ – 2m⁵ – 2m⁵ + 3m⁴ + 3m⁴ + 2m³ + 4m³ – 6m² – 6m² – 4m + 9m + 6
= m⁷ – 4m⁵ + 6m⁴ + 6m³ – 12m² + 5m + 6

ii. (5m³ – 2) (m² – m + 3)
= 5m³(m² – m + 3) – 2(m² – m + 3)
= 5m⁵ – 5m⁴ + 15m³ – 2m² + 2m – 6

Question 9.
Divide polynomial 3x³ – 8x² + x + 7 by x – 3 using synthetic method and write the quotient and remainder.
Solution:
Dividend = 3x³ – 8x² + x + 7
∴ Coefficient form of dividend = (3, – 8, 1,7)
Divisor = x – 3
∴ Opposite of – 3 is 3

Coefficient form of quotient = (3, 1,4)
∴ Quotient = 3x² + x + 4 and
Remainder =19

Question 10.
For which value of m, x + 3 is the factor of the polynomial x³ – 2mx + 21?
Solution:
Here, p(x) = x³ – 2mx + 21
(x + 3) is a factor of x³ – 2mx + 21.
∴ By factor theorem,
Remainder = 0
∴ P(- 3) = 0
p(x) = x³ – 2mx + 21
∴ p(-3) = (-3)³ – 2(m)(-3) + 21
∴ 0 = – 27 + 6m + 21
∴ 6 + 6m = 0
∴ 6m = 6
∴ m = 1
∴ x + 3 is the factor of x³ – 2mx + 21 for m = 1.

Question 11.
At the end of the year 2016, the population of villages Kovad, Varud, Chikhali is 5x² – 3y², 7y² + 2xy and 9x² + 4xy respectively. At the beginning of the year 2017, x² + xy – y², 5xy and 3x² + xy persons from each of the three villages respectively went to another village for education, then what is the remaining total population of these three villages ?
Solution:
Total population of villages at the end of 2016 = (5x² – 3y²) + (7y² + 2xy) + (9x² + 4xy)
= 5x² + 9x² – 3y² + 7y² + 2xy + 4xy
= 14x² + 4y² + 6xy …….(i)
Total number of persons who went to other village at the beginning of 2017 = (x² + xy – y²) + (5xy) + (3x² + xy)
= x² + 3x² – y² + xy + 5xy + xy
= 4x² – y² + 7xy … (ii)
Remaining total population of villages = Total population at the end of 2016 – total number of persons who went to other village at the beginning of 2017
= 14x² + 4y² + 6xy – (4x² – y² + 7xy) … [From (i) and (ii)]
= 14x² + 4y² + 6xy – 4x² + y² – 7xy
= 14x² – 4x² + 4y² + y² + 6xy – 7xy = 1
= 10x² + 5y² – xy
∴ The remaining total population of the three villages is 10x² + 5y² – xy.

Question 12.
Polynomials bx² + x + 5 and bx³ – 2x + 5 are divided by polynomial x – 3 and the remainders are m and n respectively. If m – n = 0, then find the value of b.
Solution:
When polynomial bx² + x + 5 is divided by (x – 3), the remainder is m.
∴ By remainder theorem,
Remainder = p(3) = m
p(x) = bx² + x + 5
∴ p(3) = b(3)² + 3 + 5
∴m = b(9) + 8
m = 9b + 8 …(i)
When polynomial bx³ – 2x + 5 is divided by x – 3 the remainder is n
∴ remainder = p(3) = n
p(x) = bx³ – 2x + 5
∴ P(3)= b(3)³ – 2(3) + 5
∴ n = b(27) – 6 + 5
∴ n = 27b – 1 …(ii)
Now, m – n = 0 …[Given]
∴ m = n
∴ 9b + 8 = 27b – 1 …[From (i) and (ii)]
∴ 8 + 1 = 27b – 9b
∴ 9 = 18b
∴ b = \(\frac { 1 }{ 2 }\)

Question 13.
Simplify.
(8m² + 3m – 6) – (9m – 7) + (3m² – 2m + 4)
Solution:
(8m² + 3m – 6) – (9m – 7) + (3m² – 2m + 4)
= 8m² + 3m – 6 – 9m + 7 + 3m² – 2m + 4
= 8m² + 3m² + 3m – 9m – 2m – 6 + 7 + 4
= 11m² – 8m + 5

Question 14.
Which polynomial is to be subtracted from x² + 13x + 7 to get the polynomial 3x² + 5x – 4?
Solution:
Let the required polynomial be A.
∴ (x² + 13x + 7) – A = 3x² + 5x – 4
∴ A = (x² + 13x + 7) – (3x² + 5x – 4)
= x² + 13x + 7 – 3x² – 5x + 4
= x² – 3x² + 13x – 5x + 7+4
= -2x² + 8x + 11
∴ – 2x² + 8x + 11 must be subtracted from x² + 13x + 7 to get 3x² + 5x – 4.

Question 15.
Which polynomial is to be added to 4m + 2n + 3 to get the polynomial 6m + 3n + 10?
Solution:
Let the required polynomial be A.
∴ (4m + 2n + 3) + A = 6m + 3n + 10
∴ A = 6m + 3n + 10 – (4m + 2n + 3)
= 6m + 3n + 10 – 4m – 2n – 3
= 6m – 4m + 3n – 2n + 10 – 3
= 2m + n + 7
∴ 2m + n + 7 must be added to 4m + 2n + 3 to get 6m + 3n + 10.

Question 1.
Read the following passage, write the appropriate amount in the boxes and discuss.
Govind, who is a dry land farmer from Shiralas has a 5 acre field. His family includes his wife, two children and his old mother. He borrowed one lakh twenty five thousand rupees from the bank for one year as agricultural loan at 10 p.c.p.a. He cultivated soyabean in x acres and cotton and tur in y acres. The expenditure he incurred was as follows :
He spent ₹10,000 on seeds. The expenses for fertilizers and pesticides for the soyabean crop was ₹ 2000x and ₹ 4000x² were spent on wages and cultivation of land. He spent ₹ 8000y on fertilizers and pesticides and ₹ 9000y2 for wages and cultivation of land for the cotton and tur crops.

Let us write the total expenditure on all the crops by using variables x and y.
₹ 10000 + 2000x + 4000×2 + 8000y + 9000y²
He harvested 5x² quintals soyabean and sold it at ₹ 2800 per quintal. The cotton crop yield was \(\frac { 5 }{ 3 }\) y² quintals which fetched ₹ 5000 per quintal.
The tur crop yield was 4y quintals and was sold at ₹ 4000 per quintal. Write the total income in rupees that was obtained by selling the entire farm produce, with the help of an expression using variables x and y. (Textbook pg. no. 44)
Answer:
Total income = income on soyabean crop + income on cotton crop + income on tur crop
= ₹ (5x² x 2800) + ₹(\(\frac { 5 }{ 3 }\) y² x 5000) + ₹ (4y x 4000)
= ₹ ( 14000x² + \(\frac { 25000 }{ 3 }\)y² + 16000y)

Question 2.
We have seen the example of expenditure and income (in terms of polynomials) of Govind who is a dry land farmer. He has borrowed rupees one lakh twenty-five thousand from the bank as an agriculture loan and repaid the said loan at 10 p.c.p.a. He had spent ₹ 10,000 on seeds. The expenses on soyabean crop was ₹ 2000x for fertilizers and pesticides and ₹ 4000x² was spent on wages and cultivation. He spent ₹ 8000y on fertilizers and pesticides and ₹9000y² on cultivation and wages for cotton and tur crop.
His total income was
₹ (14000x² + \(\frac { 25000 }{ 3 }\)y² + 16000y)
By taking x = 2, y = 3 write the income expenditure account of Govind’s farming. (Textbook pg. no. 52)
Solution:
–           Credit (Income)
₹ 1,25,000   Bank loan
₹ 56000      Income from soyabean
₹ 75000      Income from cotton
₹ 48000      Income from tur
₹ 304000     Total income

–                     Debit (Expenses)
₹ 1,37,000       loan paid with interest for seeds
₹ 10000          For seeds
₹ 4000            Fertilizers and pesticides for soyabean
₹ 16000         Wages and cultivation charges for soyabean
₹ 24000          Fertilizers and pesticides for cotton & tur
₹ 81000         Wages and cultivation charges for cotton & tur
₹ 272000       Total expenditure

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

Practice Set 2.2 Algebra 9th Std Maths Part 1 Answers Chapter 2 Real Numbers

Question 1.
Show that 4√2 is an irrational number.
Solution:
Let us assume that 4√2 is a rational number .
So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that
4√2 = \(\frac { a }{ b }\)
∴ √2 = \(\frac { a }{ 4b }\)
Since, a and b are integers, \(\frac { a }{ 4b }\) is a rational number and so √2 is a rational number.

Alternate Proof:
Let us assume that 4√2 is a rational number.
So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that

Since, 32 divides a², so 32 divides ‘a’ as well.
So, we write a = 32c, where c is an integer.
∴ a² = (32c)² … [Squaring both the sides]
∴ 32b² = 32 x 32c² …[From(i)]
∴ b² = 32c²
∴ c² = \(\frac { { b }^{ 2 } }{ 32 }\)
Since, 32 divides b², so 32 divides ‘b’.
∴ 32 divides both a and b.
a and b have at least 32 as a common factor.
But this contradicts the fact that a and b have no common factor other than 1.
∴ Our assumption that 4√2 is a rational number is wrong.
∴ 4√2 is an irrational number.

Question 2.
Prove that 3 + √5 is an irrational number.
Solution:
Let us assume that 3 + √5 is a rational number.
So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that

Since, a and b are integers, \(\frac { a }{ b }\) – 3 is a rational
number and so √5 is a rational number.
But this contradicts the fact that √5 is an irrational number.
∴ Our assumption that 3 – √5 is a rational number is wrong.
3 + √5 is an irrational number.

Question 3.
Represent the numbers √5 and √10 on a number line.
Solution:
i. Draw a number line and take point A at 2.
Draw AB perpendicular to the number line such that AB = 1 unit.
In ∆OAB, m∠OAB = 90°
∴ (OB)² = (OA)² + (AB)² … [Pythagoras theorem]
= (2)² + (1)²
∴ (OB)² = 5
∴ OB = √5 units. … [Taking square root of both sides]
With O as centre and radius equal to OB, draw an arc to intersect the number line at C.

The coordinate of the point C is √5 .

ii. Draw a number line and take point Pat 3.
Draw PR perpendicular to the number line such that PR = 1 unit.
In ∆OPR, m∠OPR = 90°
∴ (OR)² = (OP)² + (PR)² … [Pythagoras theorem]
= (3)² + (1)²
∴ (OR)² = 10
∴ OR= √10units. … [Taking square root of both sides]
With O as centre and radius equal to OR, draw an arc to intersect the number line at Q.

The coordinate of the point Q is √10 .

Question 4.
Write any three rational numbers between the two numbers given below.
i. 0.3 and – 0.5
ii. – 2.3 and – 2.33
iii. 5.2 and 5.3
iv. – 4.5 and – 4.6
Solution:
i. 0.3 = 0.30 and -0.5 = -0.50
We know that,
0. 30 >0.29 >….. >0.10>.. > – 0.10>…. > -0.30>…> -0.50
∴ the three rational numbers between 0.3 and -0.5 are -0.3, -0.1 and 0.1.

Alternate Method:
A rational number between two rational numbers a and b


∴ the three rational numbers between 0.3 and -0.5 are -0.3, -0.1 and 0.1.

ii. -2.3 = -2.300 and -2.33 = -2.330
We know that,
-2.300 > -2.301>… > -2.310>…> -2.320>…> -2.330
∴ the three rational numbers between -2.3 and -2.33 are -2.310, -2.320 and -2.325.

iii. 5.2 = 5.20 and 5.3 = 5.30
We know that,
5.20 < 5.21 < 5.22 < 5.23 < … < 5.30
∴ the three rational numbers between 5.2 and 5.3 are 5.21, 5.22 and 5.23.

iv. -4.5 = -4.50 and -4.6 = -4.60 We know that,
-4.50 > -4.51 > -4.52 >… > – 4.55 >…>- 4.60
∴ the three rational numbers between -4.5 and -4.6 are -4.51, -4.52 and -4.55.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.

Practice Set 7.1 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics

Question 1.
The following table shows the number of Buses and Trucks in nearest lakh units. Draw percentage bar diagram. (Approximate the percentages to the nearest integer)

Solution:


Question 2.
In the table given below, the information is given about roads. Using this draw sub-divided and percentage bar diagram (Approximate the percentages to the nearest integer)

Solution:
i. Sub-divided bar diagram:

ii. Percentage bar diagram:

Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.1 Intext Questions and Activities

Question 1.
A farmer has produced Wheat and Jowar in his field. The following joint bar diagram shows the production of Wheat and Jowar. From the gken diagram answer the following questions: (Textbook pg. no. 108)
i. Which crop production has increased consistently in 3 years?
ii. By how many quintals the production ofjowar has reduced in 2012 as compared to 2011?
iii. What is the difference between the production of wheat in 2010 and 2012 ?
iv. Complete the following table using this diagram.


Solution:
i. The crop production of wheat has increased consistently in 3 years.
ii. The production of jowar has reduced by 3 quintals in 2012 as compared to 2011.
iii. The difference between the production of wheat in 2010 and 2012 = 48 – 30 = 18 quintals
iv.

Question 2.
In the following table, the information of number of girls per 1000 boys is given for different states. Fill In the blanks and complete the table. (Textbook pg. no. 111)

Solution:
Draw percentage bar diagram from this information and discuss the findings from the diagram.

Question 3.
For the above given activity, the information of number of girls per 1000 boys is given for five states. The literacy percentage of these five states is given below. Assam (73%), Bihar (64%), Punjab (77%), Kerala (94%), Maharashtra (83%). Think of the number of girls and the literacy percentages in the respective states. Can you draw any conclusions from it? (Textbook pg. no. 112)
Solution:
By observing the number of girls per 1000 boys and literacy percentages in the given respective states, we can conclude that the literacy rate of girls is least in Bihar and is highest in Kerala.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 5 Linear Equations in Two Variables.

Problem Set 5 Algebra 9th Std Maths Part 1 Answers Chapter 5 Linear Equations in Two Variables

Question 1.
Choose the correct alternative answers for the following questions.

i. If 3x + 5y = 9 and 5x + 3y = 7, then what is the value of x + y ?
(A) 2
(B) 16
(C) 9
(D) 7
Answer:
(A) 2

ii. ‘When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes 26’. What is the mathematical form of the statement ?
(A) x – y = 8
(B) x + y = 8
(C) x + y = 23
(D) 2x + y = 21
Answer:
(C) x + y = 23

iii. Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay’s age?
(A) 20 years
(B) 15 years
(C) 10 years
(D) 5 years
Answer:
(C) 10 years

Hints:
i. Adding the given equations,
3x+ 5y = 9
5x + 3y = 7
8x + 8y = 16
∴ x + y = 2 .. [Dividing both sides by 8]

ii. Let the length of the rectangle be ‘x’ and that of breadth be ‘y’.
Perimeter of rectangle = 2[(x – 5) + (y – 5)]
∴ 26 = 2(x + y – 10)
∴ x + y – 10 = 13
∴ x + y = 23

iii. Let the age of Ajay bex years.
∴ x + (x + 5) = 25
∴ 2x = 20
∴ x = 10 years

Question 2.
Solve the following simultaneous equations.
i. 2x + y = 5 ; 3x – y = 5
ii. x – 2y = -1 ; 2x – y = 7
iii. x + y = 11 ; 2x – 3y = 7
iv. 2x + y = -2 ; 3x – y = 7
V. 2x – y = 5 ; 3x + 2y = 11
vi. x – 2y – 2 ; x + 2y = 10
Solution:
ii. 2x + y = 5 …(i)
3x – y = 5 …(ii)
Adding equations (i) and (ii),
2x + y = 5
+ 3x – y = 5
5x = 10
∴ x = 10/5
∴ x = 2
Substituting x = 2 in equation (i),
2(2) + y = 5
4 + y = 5
∴ y = 5 – 4 = 1
∴ (2, 1) is the solution of the given equations.

ii. x – 2y = -1
∴x = 2y – 1 … .(i)
∴ 2x – y = 7 ….(ii)
Substituting x = 2y – 1 in equation (ii),
2(2y – 1) – y = 7
∴ 4y – 2 – y = 7
∴ 3y = 7 + 2
∴ 3y = 9
∴ y = 9/3
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y – 1
∴ x = 2(3) – 1
∴ x = 6 – 1 = 5
∴ (5, 3) is the solution of the given equations.

iii. x + y = 11
∴ x = 11 – y …(i)
2x – 3y = 7 …….(ii)
Substituting x = 11 -y in equation (ii),
2(11 – y) – 3y = 7
∴ 22 – 2y – 3y = 1
∴ 22 – 5y = 7
∴ 22 – 7 = 5y
∴ 15 = 5y
∴ y = \(\frac { 15 }{ 5 }\)
∴ y = 3
Substituting y = 3 in equation (i),
x = 11 – y
∴ x = 11 – 3 = 8
∴ (8, 3) is the solution of the given equations.

iv. 2x + y = -2 …(i)
3x – y = 7 …(ii)
Adding equations (i) and (ii),
2x + y = -2
+ 3x – y = l
5x = 5
∴ x = \(\frac { 5 }{ 5 }\)
∴ x = 1
Substituting x = 1 in equation (i),
2x + y = -2
∴ 2(1) +y = -2
2 + y = -2
∴ y = – 2 – 2
∴ y = -4
∴ (1, -4) is the solution of the given equations.

v. 2x – y = 5
∴ -y = 5 – 2x
∴ y = 2x – 5 …(i)
3x + 2y = 11 ……(ii)
Substituting y = 2x – 5 in equation (ii),
3x + 2(2x – 5) = 11
∴ 3x + 4x- 10= 11
∴ 7x = 11 + 10
∴ 7x = 21
∴ x = \(\frac { 21 }{ 7 }\)
∴ x = 3
Substituting x = 3 in equation (i),
y = 2x – 5
∴ y = 2(3) – 5
∴ y = 6 – 5 = 1
∴(3,1) is the solution of the given equations.

vi. x – 2y = -2
∴ x = 2y – 2 …(i)
x + 2y = 10 …..(ii)
Substituting x = 2y – 2 in equation (ii),
2y – 2 + 2y = 10
∴ 4y = 10 + 2
∴ 4y= 12
∴ y = \(\frac { 12 }{ 7 }\)
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y – 2
∴ x = 2(3) – 2
∴ x = 6 – 2 = 4
∴ (4, 3) is the solution of the given equations.

Question 3.
By equating coefficients of variables, solve the following equations. [3 Marks each]
i. 3x – 4y = 7 ; 5x + 2y = 3
ii. 5x + ly= 17 ; 3x – 2y = 4
iii. x – 2y = -10 ; 3x – 3y = -12
iv. 4x+y = 34 ; x + 4y = 16
Solution:
i. 3x – 4y = 7 …(i)
5x + 2y = 3 ….(ii)
Multiplying equation (ii) by 2,
10x + 4y = 6 …(iii)
Adding equations (i) and (iii),

∴ x = 1
Substituting x = 1 in equation (i),
3x – 4y = 7
∴ 3(1) – 4y = 7
∴ 3 – 4y = 7
∴ 3 – 7 = 4y
∴ -4 = 4y
∴ y = \(\frac { -4 }{ 4 }\)
∴ y = -1
∴ (1, -1) is the solution of the given equations.

ii. 5x + 7y = 17 …(i)
3x – 2y = 4 ….(ii)
Multiplying equation (i) by 2,
10x + 14y = 34 …(iii)
Multiplying equation (ii) by 7,
21x – 14y = 28 …..(iv)
Adding equations (iii) and (iv),

∴ x = 2
Substituting x = 2 in equation (ii),
3x – 2y = 4
∴ 3(2) – 2y = 4
∴ 6 – 2y = 4
∴ 6 – 4 = 2y
∴ 2 = 2y
∴ y = \(\frac { 2 }{ 2 }\)
∴ y = 1
∴ (2,1) is the solution of the given equations.

iii. x – 2y = -10 ….(i)
3x – 5y = -12 …….(ii)
Multiplying equation (i) by 3,
3x – 6y = -30 …(iii)
Subtracting equation (ii) from (iii),

∴ y = 18
Substituting y = 18 in equation (i),
x – 2y = -10
∴ x – 2(18) = -10
∴ x – 36 = -10
∴ x = -10 + 36 = 26
∴ (26, 18) is the solution of the given equations.

iv. 4x + y = 34 …(i)
x + 4y = 16 …… (ii)
Multiplying equation (i) by 4,
16x + 4y = 136 …(iii)
Subtracting equation (ii) from (iii),

x = 8
Substituting x = 8 in equation (i),
4x + y = 34
∴ 4(8) + y = 34
∴ 32 + y = 34
∴ y = 34 – 32 = 2
∴ (8, 2) is the solution of the given equations.

Question 4.
Solve the following simultaneous equations.

Solution:
i. \(\frac{x}{3}+\frac{y}{4}=4\)
Multiplying both sides by 12,
4x + 3y = 48 …(i)
\(\frac{x}{2}-\frac{y}{4}=1\)
Multiplying both sides by 8,
4x – 2y = 8 …..(ii)
Subtracting equation (ii) from (i),

∴ y = 8
Substituting y = 8 in equation (ii),
4x – 2y = 8
∴ 4x – 2(8) = 8
∴ 4x – 16 = 8
∴ 4x = 8+ 16
∴ 4x = 24
∴ x = \(\frac { 24 }{ 4 }\)
∴ x = 6
∴ (6, 8) is the solution of the given equations.

ii. \(\frac { x }{ 3 }\) + 5y = 13
Multiplying both sides by 3,
x + 15y = 39 …(i)
2x + \(\frac { y }{ 2 }\) =19
Multiplying both sides by 2,
4x + y = 38 …….(ii)
Multiplying equation (i) by 4,
4x + 60y = 156 …(iii)
Subtracting equation (ii) from (iii),
4x + 60y =156 4x + y= 38

∴ y = 2
Substituting y = 2 in equation (i),
x + 15y = 39
∴ x+ 15(2) = 39
∴ x + 30 = 39
∴ x = 39 – 30 = 9
∴ (9,2) is the solution of the given equations.

iii. \(\frac { 2 }{ x }\) + \(\frac { 3 }{ y }\) = 13
Multiplying both sides by 5,

Question 5.
A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to this number, the sum is equal to the number obtained by interchanging the digits. Find the number.
Solution:
Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.

According to the first condition,
a two digit number is 3 more than 4 times the sum of its digits.
10y + x = 4(x + y) + 3
∴ 10y + x = 4x + 4y + 3
∴ x – 4x + 10y – 4y = 3
∴ – 3x + 6y = 3
Dividing both sides by -3,
x – 2y = -1 …(i)
According to the second condition,
if 18 is added to the number, the sum is equal to the number obtained by interchanging the digits.
10y + x + 18= 10x + y
∴ x – 10x + 10y – y = -18
∴ – 9x + 9y = -18
Dividing both sides by – 9,
x – y = 2 ……(ii)
Subtracting equation (ii) from (i),

∴ y = 3
Substituting y = 3 in equation (ii),
x – y = 2
∴ x – 3 = 2
∴ x = 2 + 3 = 5
∴ Original number = 10y + x
= 10(3) + 5
= 30 + 5
= 35
The required number is 35.

Question 6.
The total cost of 8 books and 5 pens is ₹ 420 and the total cost of 5 books and 8 pens is ₹321. Find the cost of 1 book and 2 pens.
Solution:
Let the cost of one book be ₹ x and the cost of one pen be ₹ y.
According to the first condition,
the total cost of 8 books and 5 pens is ₹ 420.
∴ 8x + 5y = 420 …(i)
According to the second condition, the total cost of 5 books and 8 pens is ₹ 321.
5x + 8y = 321 ….(ii)
Multiplying equation (i) by 5,
40x + 25y = 2100 …(iii)
Multiplying equation (ii) by 8,
40x + 64y = 2568 … (iv)
Subtracting equation (iii) from (iv),

∴ y = 12
Substituting y = 12 in equation (i),
8x + 5y = 420
∴ 8x + 5(12) = 420
∴ 8x + 60 = 420
∴ 8x = 420 – 60
∴ 8x = 360
∴ x = \(\frac { 360 }{ 8 }\)
∴ x = 45
Cost of 1 book and 2 pens = x + 2y
= 45 + 2(12)
= 45 + 24
= ₹69
∴ The cost of 1 book and 2 pens is ₹69.

Question 7.
The ratio of incomes of two persons is 9 : 7. The ratio of their expenses is 4 : 3. Every person saves ₹ 200, find the income of each.
Solution:
Let the income of first person be ₹ x and that of second person be ₹ y.
According to the first condition,
the ratio of their incomes is 9 : 7.
∴ \(\frac { x }{ y }\) = \(\frac { 9 }{ 7 }\)
∴ 7x = 9y
∴ 7x – 9y = 0 …….(i)
Each person saves ₹ 200.
Expenses of first person = Income – Saving = x – 200
Expenses of second person = y – 200
According to the second condition,
the ratio of their expenses is 4 : 3
∴ \(\frac { x – 200 }{ y – 200 }\) = \(\frac { 4 }{ 3 }\)
∴ 3(x – 200) = 4(y – 200)
∴ 3x – 600 = 4y – 800
∴ 3x – 4y = – 800 + 600
∴ 3x – 4y = -200 …(ii)
Multiplying equation (i) by 4,
28x-36y =0 …(iii)
Multiplying equation (ii) by 9,
27x-36y = -1800 …(iv)
Subtracting equation (iv) from (iii),

Substituting x = 1800 in equation (i),
7x – 9y = 0
∴ 7(1800) – 9y = 0
∴ 9y = 7 x 1800
∴ y = \(\frac { 7 \times 1800 }{ 9 }\)
y = 7 x 200
∴ y = 1400
∴ The income of first person is ₹ 1800 and that of second person is ₹ 1400.

Question 8.
If the length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 9 square units. If length is reduced by 3 units and breadth is increased by 2 units, then the area of rectangle will increase by 67 square units. Then find the length and breadth of the rectangle.
Solution:
Let the length of the rectangle be ‘x’ units and the breadth of the rectangle be ‘y’ units.
Area of the rectangle = xy sq. units
length of the rectangle is reduced by 5 units
∴ length = x – 5
breadth of the rectangle is increased by 3 units
∴ breadth = y + 3
area of the rectangle is reduced by 9 square units
∴ area of the rectangle = xy – 9
According to the first condition,
(x – 5)(y + 3) = xy – 9
∴ xy + 3x – 5y – 15 = xy – 9
∴ 3x – 5y = -9 + 15
∴ 3x – 5y = 6 …(i)
length of the rectangle is reduced by 3 units
∴ length = x – 3
breadth of the rectangle is increased by 2 units
∴ breadth = y + 2
area of the rectangle is increased by 67 square units
∴ area of the rectangle = xy + 61
According to the second condition,
(x – 3)(y + 2) = xy + 67
∴ xy + 2x – 3y – 6 = xy + 67
∴ 2x – 3y = 67 + 6
∴ 2x – 3y = 73 …(ii)
Multiplying equation (i) by 3,
9x – 15y = 18 . ..(iii)
Multiplying equation (ii) by 5,
10x – 15y = 365 …(iv)
Subtracting equation (iii) from (iv), 10x- 15y= 365 9x-15y= 18

Substituting x = 347 in equation (ii),
2x – 3y = 73
∴ 2(347) – 3y = 73
∴ 694 – 73 = 3y
∴ 621 = 3y
∴ y = \(\frac { 621 }{ 3 }\)
∴ y = 207
∴ The length and breadth of rectangle are 347 units and 207 units respectively.

Question 9.
The distance between two places A and B on a road is 70 kilometres. A car starts from A and the other from B. If they travel in the same direction, they will meet in 7 hours. If they travel towards each other they will meet in 1 hour, then find their speeds.
Solution:
Let the speed of the car starting from A (first car) be ‘x’ km/hr and that starting from B (second car) be ‘y’ km/hr. (x > y)
According to the first condition,
Distance covered by the first car in 7 hours = 7x km
Distance covered by the second car in 7 hours = 7y km

If the cars are travelling in the same direction, 7x – 7y = 70
Dividing both sides by 7,
x – y = 10 …(i)
According to the second condition,
Distance covered by the first car in
1 hour = x km
Distance covered by the second car in 1 hour = y km

If the cars are travelling in the opposite direction
x + y = 70 …(ii)
Adding equations (i) and (ii),

∴ x = 40
Substituting x = 40 in equation (ii), x + y = 70
∴ 40 +y = 70
∴ y = 70 – 40 = 30
∴ The speed of the cars starting from places A and B are 40 km/hr and 30 km/hr respectively.

Question 10.
The sum of a two digit number and the number obtained by interchanging its digits is 99. Find the number.
Solution:
Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.

According to the given condition,
the sum of a two digit number and the number
obtained by interchanging its digits is 99.
∴ 10y + x + 10x +y = 99
∴ 11x + 11y = 99
Dividing both sides by 11,
x + y = 9
If y = 1, then x = 8
If y = 2, then x = 7
If y = 3, then x = 6 and so on.
∴ The number can be 18, 27, 36, … etc.

Maharashtra Board Class 9 Maths Chapter 5 Linear Equations in Two Variables Practice Set 5 Intext Questions and Activities

Question 1.
On the glasses of following spectacles, write numbers such that (Textbook pg. no. 82)
i. Their sum is 42 and difference is 16.

ii. Their sum is 37 and difference is 11.

iii. Their sum is 54 and difference is 20.

iv. Their sum is … and difference is … .

Answer:
ii. x + y = 37 and x – y = 11
∴ x = 24, y = 13
iii. x +y = 54 and x – y = 20
∴ x = 37, y =17

Question 2.
There are instructions written near the arrows in the following diagram. From this information form suitable equations and write in the boxes indicated by arrows. Select any two equations from these boxes and find their solutions. Also verify the solutions. By taking one pair of equations at a time, how many pairs can be formed ? Discuss the solutions for these pairs. (Textbook pg. no. 92)

Answer:
Here, if we take a pair of any two equations, we get following 6 pairs.

1. equation (i) and (ii)
2. equation (i) and (iii)
3. equation (i) and (iv)
4. equation (ii) and (iii)
5. equation (ii) and (iv)
6. equation (iii) and (iv)

Solution of each pair given above is (21, 15).
Here, all four equations are of same rectangle. By solving any two equations simultaneously, we get length and breadth of the rectangle.

Question 3.
Find the function.

∴ Given function = \(\frac { 6 }{ 14 }\)
Verify the answer obtained. (Textbook pg. no. 92)
Answer:
For the fraction\(\frac { 6 }{ 14 }\), if the numerator is multiplied by 3 and 3 is subtracted from the denominator, we get fraction \(\frac { 18 }{ 11 }\).
Similarly, for the fraction \(\frac { 6 }{ 14 }\), if the numerator is increased by 8 and the denominator is doubled, we get fraction \(\frac { 1 }{ 2 }\).

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 6.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 6 Financial Planning.

Practice Set 6.1 Algebra 9th Std Maths Part 1 Answers Chapter 6 Financial Planning

Question 1.
Alka spends 90% of the money that she receives every month, and saves ₹ 120. How much money does she get monthly?
Solution:
Let Alka’s monthly income be ₹ x.
Alka spends 90% of the money that she receives every month.
∴ Amount spent by Alka = 90% of x
= \(\frac { 90 }{ 100 }\) × x = 0.9x 100
Now, Savings = Income – Expenditure
∴ 120 = x – 0.9x
∴120 = 0.1 x
∴ \(x=\frac{120}{0.1}=\frac{120 \times 10}{0.1 \times 10}\)
∴ x = 1200
Alka gets ₹ 1200 monthly.

Question 2.
Sumit borrowed a capital of ₹ 50,000 to start his food products business. In the first year he suffered a loss of 20%. He invested the remaining capital in a new sweets business and made a profit of 5%. How much was his profit or loss computed on his original capital ?
Solution:
Original capital borrowed by Sumit = ₹ 50000
Sumit suffered a loss of 20% in his food products business.
∴ Loss suffered in the first year = 20% of 50000
= \(\frac { 20 }{ 100 }\) × 50000
= ₹10000
Remaining capital = Original capital – loss suffered = 50000- 10000
= ₹ 40000
Sumit invested the remaining capital i.e. ₹ 40,000 in a new sweets business. He made a profit of 5%.
Profit in sweets business = 5% of 40000
= \(\frac { 5 }{ 100 }\) x 40000 100
= ₹ 2000
New capital with Sumit after the profit in new sweets business = 40000 + 2000 = ₹42000
Since, the new capital is less than the original capital, we can conclude that Sumit suffered a loss.
Total loss on original capital = Original capital – New capital
= 50000 – 42000 = ₹ 8000

∴ Sumit suffered a loss of 16% on the original capital.

Question 3.
Nikhil spent 5% of his monthly income on his children’s education, invested 14% in shares, deposited 3% in a bank and used 40% for his daily expenses. He was left with a balance of ₹ 19,000. What was his income that month?
Solution:
Let the monthly income of Nikhil be ₹ x.
Nikhil invested 14% in shares and deposited 3% in a bank.
∴ Total investment = (14% + 3%) of x
= 17% of x
= \(\frac { 17 }{ 100 }\) × x
= 0.1 7 x
Nikhil spent 5% on his children’s education and used 40% for his daily expenses.

∴ Total expenditure = (5% + 40%) of x
= 45% of x
= \(\frac { 45 }{ 100 }\) × x
= 0.45x
Amount left with Nikhil = 19,000
Amount left with Nikhil = Income – (Total investment + Total expenditure)
∴ 19000 = x – (0.17x + 0.45x)
∴ 19000 = x – 0.62x ,
∴ 19000 = 0.38x
∴ \(x=\frac{19000}{0.38}=\frac{19000 \times 100}{0.38 \times 100}=\frac{1900000}{38}\)
= 50000
∴ The monthly income of Nikhil is ₹ 50000.

Question 4.
Mr. Sayyad kept ₹ 40,000 in a bank at 8% compound interest for 2 years. Mr. Fernandes invested ₹ 1,20,000 in a mutual fund for 2 years. After 2 years, Mr. Fernandes got ₹ 1,92,000. Whose investment turned out to be more profitable?
Solution:
Mr. Sayyad:
Mr. Sayyad kept ₹ 40,000 in a bank at 8% compound interest for 2 years P = ₹ 40000, R = 8%, n = 2 years
∴ Compound interest (I)
= Amount (A) – Principal (P)

= 40000 [(1 +0.08)² – 1]
= 40000 [(1.08)² – 1]
= 40000(1.1664 – 1)
= 40000 (0.1664)
= ₹ 6656
∴ Mr. Sayyad’s percentage of profit Interest

Mr. Fernandes:
Amount invested by Mr. Fernandes in mutual fund = ₹ 120000
Amount received by Mr. Fernandes after 2 years = ₹ 192000
∴ Profit earned by Mr. Fernandes
= Amount received – Amount invested
= 192000- 120000
= ₹72000
∴ Mr. Fernandes percentage of profit Profit earned

= 60%
From (i) and (ii),
Investment of Mr. Fernandes turned out to be more profitable.

Question 5.
Sameera spent 90% of her income and donated 3% for socially useful causes. If she was left with ₹ 1750 at the end of the month, what was her actual income ?
Solution:
Let the actual income of Sameera be ₹ x.
Sameera spent 90% of her income and donated 3%.
∴ Sameera’s total expenditure
= (3% + 90%) of x
= 93% of x
= \(\frac { 93 }{ 100 }\) × x
= 0.93x
Now, Savings = Income – Expenditure
∴ 1750 = x-0.93x
∴ 1750 = 0.07x
\(x=\frac{1750}{0.07}=\frac{1750 \times 100}{0.07 \times 100}=\frac{175000}{7}=25000\)
∴ The actual income of Sameera is ₹ 25000.

Maharashtra Board Class 9 Maths Chapter 6 Financial Planning Practice Set 6.1 Intext Questions and Activities

Question 1.
Amita invested some part of ₹ 35000 at 4% and the rest at 5% interest for one year. Altogether her gain was ₹ 1530. Find out the amounts she had invested at the two different rates. Write your answer in words. (Textbook pg. no. 97)
Solution:
Let the amount invested at the rate of 4% and 5% be ₹ x and ₹ y respectively.
According to the first condition, total amount invested = ₹ 35000
∴ x + y = 35000 …(i)
According to the second condition,
total interest received at 4% and 5% is ₹ 1530.
∴ 4 % of x + 5 % of y = 1530
∴ \(\frac { 4 }{ 100 }\) x x + \(\frac { 5 }{ 100 }\) x y = 1530
∴ 4x + 5y = 153000 …(ii)
Multiplying equation (i) by 4, we get
4x + 4y = 140000 …(iii)
Subtracting equation (iii) from (ii),

Substituting y = 13000 in equation (i),
x + 13000 = 35000
∴ x = 35000 – 13000 = 22000

∴ Amita invested ₹ 22000 at the rate of 4% and ₹ 13000 at the rate of 5%.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 5.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 5 Linear Equations in Two Variables.

Practice Set 5.2 Algebra 9th Std Maths Part 1 Answers Chapter 5 Linear Equations in Two Variables

Question 1.
In an envelope there are some ₹5 notes and some ₹10 notes. Total amount of these notes together is ₹350. Number of ₹5 notes are less by 10 than twice the number of ₹10 notes. Then find the number of ₹5 and ₹10 notes.
Solution:
Let the number of ₹5 notes be ‘x’ and the number of ₹10 notes be ‘y’
Total amount of x notes of ₹ 5 = ₹ 5x
Total amount ofy notes of ₹ 10 = ₹ 10y
∴ Total amount = 5x + 10y
According to the first condition,
total amount of the notes together is ₹350.
∴ 5x + 10y = 350 …(i)
According to the second condition,
Number of ₹ 5 notes are less by 10 than twice the number of ₹ 10 notes.
∴ x = 2y – 10
∴ x – 2y = -10 …..(ii)
Multiplying equation (ii) by 5,
5x – 10y = -50 …(iii)
Adding equations (i) and (iii),
5x + 10y =350
+ 5x – 10y = -50
10x =300
∴ x = \(\frac { 300 }{ 10 }\)
∴ x = 30
Substituting x = 30 in equation (ii),
x – 2y = -10
30 – 2y = -10
∴ 30 + 10 = 2y
∴ 40 = 2y
∴ y = \(\frac { 40 }{ 2 }\)
∴ y = 20
There are 30 notes of ₹ 5 and 20 notes of ₹ 10 in the envelope.

Question 2.
The denominator of a fraction is 1 less than twice its numerator. If 1 is added to numerator and denominator respectively, the ratio of numerator to denominator is 3 : 5. Find the fraction.
Solution:
Let the numerator of the fraction be ‘x’ and its denominator be ‘y’.
Then, the required fraction is \(\frac { x }{ y }\) .
According to the first condition,
the denominator is 1 less than twice its numerator.
∴ y = 2x – 1
∴ 2x – y = 1 …(i)
According to the second condition,
if 1 is added to the numerator and the denominator, the ratio of numerator to denominator is 3 : 5.
∴ \(\frac { x+1 }{ y+1 }\) = \(\frac { 3 }{ 5 }\)
∴ y + 1 = 5
∴ 5(x + 1) = 3(y + 1)
∴ 5x + 5 = 3y + 3
∴ 5x – 3y = 3 – 5
∴ 5x – 3y = -2 ……(ii)
Multiplying equation (i) by 3,
6x – 3y = 3 …(iii)
Subtracting equation (ii) from (iii),

Substituting x = 5 in equation (i),
∴ 2x – y = 1
∴ 2(5) – y = 1
∴ 10 – y = 1
∴ y= 10 – 1 =9
∴ The required fraction is \(\frac { 5 }{ 9 }\).

Question 3.
The sum of ages of Priyanka and Deepika is 34 years. Priyanka is elder to Deepika by 6 years. Then find their present ages.
Solution:
Let the present age of Priyanka be ‘x’ years and that of Deepika be ‘y’ years.
According to the first condition,
Priyanka’s age + Deepika’s age = 34 years
∴ x + y = 34 …(i)
According to the second condition,
Priyanka is elder to Deepika by 6 years.
∴ x =y + 6
∴ x – y = 6 …..(ii)
Adding equations (i) and (ii),

∴ x = 20
Substituting x = 20 in equation (i),
x + y = 34
∴ 20 + y = 34
∴ y = 34 -20= 14
∴ The present age of Priyanka is 20 years and that of Deepika is 14 years.

Question 4.
The total number of lions and peacocks in a certain zoo is 50. The total number of their legs is 140. Then find the number of lions and peacocks in the zoo.
Solution:
Let the number of lions in the zoo be ‘x’ and the number of peacocks be ‘y’.
According to the first condition,
the total number of lions and peacocks is 50.
∴ x + y = 50 …(i)
Lion has 4 legs and Peacock has 2 legs.
According to the second condition,
the total number of their legs is 140.
∴ 4x + 2y = 140
Dividing both sides by 2,
2x + y = 70 …(ii)
Subtracting equation (i) from (ii),

Substituting x = 20 in equation (i),
x + y = 50
∴ 20 + y = 50
∴ y = 50 – 20 = 30
∴ The number of lions and peacocks in the zoo are 20 and 30 respectively.

Question 5.
Sanjay gets fixed monthly income. Every year there is a certain increment in his salary. After 4 years, his monthly salary was ₹ 4500 and after 10 years his monthly salary became ₹ 5400, then find his original salary and yearly increment.
Solution:
Let the original salary of Sanjay be ₹ ‘x’ and his yearly increment be ₹ ‘y’.
According to the first condition, after 4 years his monthly salary was ₹ 4500
∴ x + 4y = 4500 …..(i)
According to the second condition,
after 10 years his monthly salary became ₹ 5400
∴ x + 10y = 5400 …(ii)
Subtracting equation (i) from (ii),

∴ y = 150
Substituting y = 150 in equation (i),
x + 4y = 4500
∴ x +4(150) = 4500
∴ x + 600 = 4500
∴ x = 4500 – 600 = 3900
∴ The original salary of Sanjay is ₹ 3900 and his yearly increment is ₹ 150.

Question 6.
The price of 3 chairs and 2 tables is ₹ 4500 and price of 5 chairs and 3 tables is ₹ 7000, then find the price of 2 chairs and 2 tables.
Solution:
Let the price of one chair be ₹ ‘x’ and that of one table be ₹ ‘y’.
According to the first condition,
the price of 3 chairs and 2 tables is ₹ 4500
∴ 3x + 2y = 4500 ,..(i)
According to the second condition, the price of 5 chairs and 3 tables is ? 7000
∴ 5x + 3y = 7000 …(ii)
Multiplying equation (i) by 3,
9x + 6y = 13500 ….(iii)
Multiplying equation (ii) by 2,
10x + 6y= 14000 …(iv)
Subtracting equation (iii) from (iv),

Substituting x = 500 in equation (i),
3x + 2y = 4500
∴ 3(500)+ 2y = 4500
∴ 1500 + 2y = 4500
∴ 2y = 4500- 1500
∴ 2y = 3000
∴ y = \(\frac { 3000 }{ 2 }\)
∴ y = 1500
∴ Price of 2 chairs and 2 tables = 2x + 2y
= 2(500)+ 2(1500)
= 1000 + 3000 = ₹ 4000
∴ The price of 2 chairs and 2 tables is ₹ 4000.

Question 7.
The sum of the digits in a two-digit number is 9. The number obtained by interchanging the digits exceeds the original number by 27. Find the two-digit number.
Solution:
Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.

According to the first condition.
the sum of the digits in a two-digit number is 9
x + y = 9 …(i)
According to the second condition,
the number obtained by interchanging the digits exceeds the original number by 27
∴ 10x + y = 10y + x + 27
∴ 10x – x + y – 10y = 27
∴ 9x – 9y = 27
Dividing both sides by 9,
x – y = 3 …….(ii)
Adding equations (i) and (ii),

∴ x = 6
Substituting x = 6 in equation (i),
x + y = 9
∴ 6 + y = 9
∴ y = 9 – 6 = 3
∴ Original number = 10y + x = 10(3)+ 6
= 30 + 6 = 36
∴ The two digit number is 36.

Question 8.
In ∆ABC, the measure of ∠A is equal to the sum of the measures of ∠B and ∠C. Also the ratio of measures of ∠B and ∠C is 4 : 5. Then find the measures of angles of the triangle.
Solution:
Let the measure of ∠B be ‘x°’ and that of ∠C be ‘y°’.
According to the first condition,
m∠A = m∠B + m∠C
∴ m∠A = x° + y°
In AABC,
m∠A + m∠B + m∠C = 180° …[Sum of the measures of the angles of a triangle is 180°]
∴ x + y + x + y = 180 ,
∴ 2x + 2y = 180
Dividing both sides by 2,
x + y = 90 …(i)
According to the second condition,
the ratio of the measures of ∠B and ∠C is 4 : 5.
∴ \(\frac { x }{ y }\) = \(\frac { 4 }{ 5 }\)
∴ 5x = 4y
∴ 5x – 4y = 0 …….(ii)
Multiplying equation (i) by 4,
4x + 4y = 360 …(iii)
Adding equations (ii) and (iii),

∴ x = 40
Substituting x = 40 in equation (i),
x + y = 90
∴ 40 + y = 90
∴ y = 90 – 40
∴ y = 50
∴ m∠A = x° + y° = 40° + 50° = 90°
∴ The measures of ∠A, ∠B and ∠C are 90°, 40°, and 50° respectively.

Question 9.
Divide a rope of length 560 cm into 2 parts such that twice the length of the smaller part is equal to \(\frac { 1 }{ 3 }\) of the larger part. Then find the length of the larger part.
Solution:
Let the length of the smaller part of the rope be ‘x’ cm and that of the larger part be ‘y’ cm.
According to the first condition,
total length of the rope is 560 cm.
∴ x + y = 560 …(i)
Twice the length of the smaller part = 2x
\(\frac { 1 }{ 3 }\)rd length of the larger part = \(\frac { 1 }{ 3 }\)y
According to the second condition,
2x = \(\frac { 1 }{ 3 }\) 3
∴ 6x = y
∴ 6x – y = 0 ……(ii)
Adding equations (i) and (ii),

∴ x = 80
Substituting x = 80 in equation (ii),
6x – y = 0
∴ 6(80) – y = 0
∴ 480 – y = 0
∴ y = 480
∴ The length of the larger part of the rope is 480 cm.

Question 10.
In a competitive examination, there were 60 questions. The correct answer would carry 2 marks, and for incorrect answer 1 mark would be subtracted. Yashwant had attempted all the questions and he got total 90 marks. Then how many questions he got wrong?
Solution:
Let us suppose that Yashwant got ‘x’ questions right and ‘y’ questions wrong.
According to the first condition, total number of questions in the examination are 60.
∴ x + y = 60 …(i)
Yashwant got 2 marks for each correct answer and 1 mark was deducted for each wrong answer.
∴ He got 2x – y marks.
According to the second condition,
he got 90 marks.
2x – y = 90 … (ii)
Adding equations (i) and (ii),

∴ x = 50
Substituting x = 50 in equation (i),
50 + y = 60
∴ y = 60 – 50 = 10
∴ Yashwant got 10 questions wrong.

Maharashtra Board Class 9 Maths Chapter 5 Linear Equations in Two Variables Practice Set 5.2 Intext Questions and Activities

Question 1.
The population of a certain town was 50,000. In a year, male population was increased by 5% and female population was increased by 3%. Now the population became 52020. Then what was the number of males and females in the previous year? (Textbook pg. no. 89)
Solution:
Step 1: Read the given word problem carefully and try to understand it.

Step 2: Make assumptions using two variables x and y.
Let the number of males in previous year be
‘x’ and the number of females be ‘y’.

Step 3: From the given information, form mathematical statements using the above variables.
According to the first condition,
the total population of town was 50,000.
∴ x + y = 50000 …(i)
Male population increased by 5%.
∴ Number of males = x + 5% of x , 5

Female population increased by 3%.
∴ Number of females = y + 3% of y

According to the second condition,
in a year population became 52020
∴ \(\frac{105}{100} x+\frac{103}{100} y=52020\)
∴ 105 x + 103 y = 5202000 …(ii)
Multiplying equation (i) by 103,
103 x + 103 y = 5150000 …(iii)

Step 4: Here, we use elimination method.
Subtracting equation (iii) from (ii),

∴ x = 26000
Substituting x = 26000 in equation (i),
∴ 26000 + y = 50000
∴ y = 50000 – 26000
∴ y = 24000
∴ Number of males = x = 26000
∴ Number of females = y = 24000

Step 5: Write the answer.
The number of males and females in the previous year were 26,000 and 24,000 respectively.

Step 6: Verify your result using smart check.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 6.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 6 Financial Planning.

Practice Set 6.2 Algebra 9th Std Maths Part 1 Answers Chapter 6 Financial Planning

Question 1.
Observe the table given below. Check and decide, whether the individuals have to pay income tax.

Solution:
i. Miss Nikita’s age = 27 years < 60 years
Miss Nikita’s income = ₹ 2,34,000
Miss Nikita’s income is below the basic
exemption limit of ₹ 2,50,000.
∴ Miss Nikita will not have to pay income tax.

ii. Mr. Kulkarni’s age 36 years < 60 years
Mr. Kulkarni’s income = ₹3,27,000
Mr. Kulkarni’s income is above the basic exemption Limit of ₹2,50,000.
∴ Mr. Kulkarni will have to pay income tax.

iii. Miss Mehta’s age = 44 years < 60 years Miss Mehta’s income = ₹5.82,000
Miss Mehta’s income is above the basic exemption limit of ₹2,50,000.
∴ Miss Mehta will have to pay income tax.

iv. Mr. Bajaj’s age = 64 years (Age 60 to 80 years)
Mr. Bajaj’s income = ₹8,40,000
Mr. Bajaj’s income is above the basic exemption Limit of ₹3,00,000.
∴ Mr. Bajaj will have to pay income tax.

v. Mr. Desilva’s age = 81 years > 80 years
Mr. Desilva’s income = ₹4,50,000
Mr. Desilva’s income is below the basic exemption limit of ₹ 5,00.000.
∴ Mr. Desilva will not have to pay income tax.

Question 2.
Mr. Kartarsingh (age 48 years) works in a private company. His monthly income after deduction of allowances is ₹ 42,000 and every month he contributes ₹ 3000 to GPF. He has also bought ₹ 15,000 worth of NSC (National Savings Certificate) and donated ₹ 12,000 to the PM’s Relief Fund. Compute his income tax.
Solution:
Mr. Kartarsingh’s monthly income = ₹ 42,000
Mr. Kartarsingh’s yearly income = 42,000 x 12 = ₹ 5,04,000

Mr. Kartarsingh’s investment
= GPF + NSC
= (3000 x 12)+ 15,000
= 36,000 + 15,000
= ₹ 51,000

Donation to PM’s relief fund = ₹ 12, 000
∴ Taxable income
= yearly income – (investment + donation)
= 5,04,000 – (51,000 + 12,000)
= 5,04,000 – 63,000 = ₹ 4,41,000
Mr. Kartarsingh income falls in the slab 2,50,001 to 5,00,000.
∴ Income tax = 5% of (Taxable income – 250000) = 5% of (4,41,000 – 2,50,000)
= \(\frac { 5 }{ 100 }\) x 1,91,000 100
= ₹ 9550
Education cess = 2% of income tax
= \(\frac { 2 }{ 100 }\) x 9550
= 191
Secondary and Higher Education cess = 1% of income tax
= \(\frac { 1 }{ 100 }\) x 9550 100
= 95.50
Total income tax = Income tax + Education cess + Secondary and higher education cess
= 9550 + 191 + 95.50
= ₹ 9836.50
∴ Mr. Kartarsingh’s income tax is ₹ 9836.50

Maharashtra Board Class 9 Maths Chapter 6 Financial Planning Practice Set 6.2 Intext Questions and Activities

Question 1.
Use Table I given above and write the appropriate amount/figure in the boxes for the example given below. (Textbook pg. no. 102)
Mr. Mehta’s annual income is ₹4,50,000
i. If he does not have any savings by which he can claim deductions from his income, to which slab does his taxable income belong ? ______
ii, What is the amount on which he will have to pay income tax and at what percent rate? on ₹ _______
percentage _______
iii. On what amount will the cess be levied? _______
Answer:
1. ₹2,50,001 to ₹5,00,000
ii. 5% of (4,50,000 – 2,50,000)
i.e. 5% of ₹2,00,000
iii. income tax = 5% of 2,00,000
= \(\frac { 5 }{ 100 }\) x 2,00,000
= ₹10,000
∴ Education cess and Secondary and higher education cess will be levied on the income tax i.e., on ₹10,000.

Question 2.
Use table lito carry out the following activity.
Mr. Pandit is 75 years old. Last year his annual income was ₹ 13,25,000. How much is his taxable income? How much tax does he have to pay? (Textbook pg. no. 103)
Solution:
Mr. Pandit’s age = 75 years (Age 60 to 80 years)
Mr. Pandit’s income is more than 10,00,000.
According to the table,
Income tax = ₹ 1,10,000 + 30 % of (taxable income – 10,00,000)
Taxable income – 10,00,000 = 13,25,000 – 10,00,000 = 3,25,000
In addition, on ₹ 3,25,000 rupees he has to pay 30% income tax.
3,25,000 x \(\frac { 30 }{ 100 }\) = ₹ 97500
Therefore, his total income tax amounts to 1,10,000 + 97,500 ₹ 207500
Besides this, education cess willi be 2% of income tax 207500 x \(\frac { 2 }{ 100 }\) = ₹ 4150.
A secondary and higher education cess at 1% of income tax = 207500 x \(\frac { 1 }{ 100 }\) = ₹ 2075.
∴ Total income tax = Income tax + education cess + secondary and higher education cess
= 207500 + 4150 + 2075
= ₹2,13,725

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 5.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 5 Linear Equations in Two Variables.

Practice Set 5.1 Algebra 9th Std Maths Part 1 Answers Chapter 5 Linear Equations in Two Variables

Question 1.
By using variables x and y form any five linear equations in two variables.
Answer:
The general form of a linear equation in two variables x and y is ax + by + c = 0,
where a, b, c are real numbers and a ≠ 0, b ≠ 0.
Five linear equations in two variables are as follows:
i. 3x + 4y – 12 = 0
ii. 3x – 4y + 12 = 0
iii. 5x + 5y – 6 = 0
iv. 7x + 12y – 11 = 0
v. x – y + 5 = 0

Question 2.
Write five solutions of the equation x + y = 1.
Answer:
i. x = 1, y = 6
ii. x = -1, y = 8
iii. x = 5, y = 2
iv. x = 0, y = 7
v. x = 10, y = -3

Question 3.
Solve the following sets of simultaneous equations.
i. x + y = 4 ; 2x – 5y = 1
ii. 2x + y = 5 ; 3x – y = 5
iii. 3x – 5y = 16; x – 3y= 8
iv. 2y – x = 0; 10x + 15y = 105
v. 2x + 3y + 4 = 0; x – 5y = 11
vi. 2x – 7y = 7; 3x + y = 22
Solution:
i. Substitution Method:
x + y = 4
∴ x = 4 – y …(i)
2x – 5y = 1 ……(ii)
Substituting x = 4 – y in equation (ii),
2(4 – y) – 5y = 1
∴ 8 – 2y – 5y = 1
∴ 8 – 7y = 1
∴ 8 – 1 = 7y
∴ 7 = 7y
∴ y = \(\frac { 7 }{ 7 }\)
∴ y = 1
Substituting y = 1 in equation (i),
x = 4 – 1 = 3
∴ (3,1) is the solution of the given equations.

Alternate method:
Elimination Method:
x + y = 4 …(i)
2x – 5y = 1 ……(ii)
Multiplying equation (i) by 5,
5x + 5y = 20 … (iii)
Adding equations (ii) and (iii),
2x – 5y = 1
+ 5x + 5y = 20
7 = 21
∴ x = \(\frac { 21 }{ 7 }\)
∴ x = 3
Substituting x = 3 in equation (i),
3 + y = 4
∴ y = 4 – 3 = 1
(3,1) is the solution of the given equations.

ii. 2x + y = 5 …(i)
3x – y = 5 …(ii)
Adding equations (i) and (ii),
2x + y = 5
+ 3x – y = 5
5x = 10
∴ x = \(\frac { 10 }{ 5 }\)
∴ x = 2
Substituting x = 2 in equation (i),
2(2) + y = 5
4 + y = 5
∴ y = 5 – 4 = 1
∴ (2, 1) is the solution of the given equations.

iii. 3x – 5y = 16 …(i)
x – 3y = 8
∴x = 8 + 3y …..(ii)
Substituting x = 8 + 3y in equation (i),
3(8 + 3y) – 5y = 16
24 + 9y- 5y = 16
∴4y= 16 – 24
∴ 4y = -8
∴ y = \(\frac { -8 }{ 4 }\)
y = -2
Substituting y = -2 in equation (ii),
x = 8 + 3 (-2)
∴ x = 8 – 6 = 2
∴ (2, -2) is the solution of the given equations.

iv. 2y – x = 0
∴ x = 2y …(i)
10x + 15y = 105 …(ii)
Substituting x = 2y in equation (ii),
10(2y) + 15y = 105
∴ 20y + 15y = 105
∴ 35y = 105
∴ y = \(\frac { 105 }{ 35 }\)
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y
∴ x = 2(3) = 6
∴ (6, 3) is the solution of the given equations.

v. 2x + 3y + 4 = 0 …(i)
x – 5y = 11
∴x = 11 + 5y …(ii)
Substituting x = 11 + 5y in equation (i),
2(11 +5y) + 3y + 4 = 0
∴ 22 + 10y + 3y + 4 = 0
∴ 13y + 26 = 0
∴ 13y = -26
∴y = \(\frac { -26 }{ 13 }\)
∴ y = -2
Substituting y = -2 in equation (ii),
x = 11 + 5y
∴ x = 11 + 5(-2)
∴ x = 11 – 10 = 1
∴ (1, -2) is the solution of the given equations.

vi. 2x – 7y = 7 …(i)
3x + y = 22
∴ y = 22 – 3x ……(ii)
Substituting y = 22 – 3x in equation (i),
2x – 7(22 – 3x) = 7
∴ 2x – 154 + 21x = 7
∴ 23x = 7 + 154
∴ 23x = 161
∴ x = \(\frac { 161 }{ 23 }\)
∴ x = 7
Substituting x = 7 in equation (ii),
y = 22 – 3x
∴ y = 22 – 3(7)
∴ 7 = 22 -21= 1
∴ (7, 1) is the solution of the given equations.

Question 1.
Solve the following equations. (Textbook pg. no. 80)
i. m + 3 = 5
ii. 3y + 8 = 22
iii. \(\frac { x }{ 3 }\) = 2
iv. 2p = p + \(\frac { 4 }{ 9 }\)
Solution:
i. m + 3 = 5
m = 5 – 3
∴m = 2

ii. 3y + 8 = 22
∴ 3y = 22 – 8
∴ 3y = 14
∴ y = \(\frac { 14 }{ 9 }\)

iii. \(\frac { x }{ 3 }\) = 2
∴ x = 2 × 3
∴ x = 6

iv. 2p = p + \(\frac { 4 }{ 9 }\)
∴ 2p – p = \(\frac { 4 }{ 9 }\)
∴ p = \(\frac { 4 }{ 9 }\)

Question 2.
Which number should be added to 5 to obtain 14? (Textbook pg. no. 80)
Solution:
x + 5 = 14
∴ x = 14 – 5
x = 9
∴ 9 + 5 = 14

Question 3.
Which number should be subtracted from 8 to obtain 2? (Textbook pg. no. 80)
Solution:
8 – y = 2
∴ y = 8 – 2
∴ y = 6
∴ 8 – 6 = 2

Question 4.
x + y = 5 and 2x + 2y= 10 are two equations in two variables. Find live different solutions of x + y = 5, verify whether same solutions satisfy the equation 2x + 2y = 10 also. Observe both equations. Find the condition where two equations in two variables have all solutions in common. (Textbook pg. no. 82)
Solution:
Five solutions of x + y = 5 are given below:
(1,4), (2, 3), (3, 2), (4,1), (0, 5)
The above solutions also satisfy the equation 2x + 2y = 10.
∴ x + y = 5 …[Dividing both sides by 2]
∴ If the two equations are the same, then the two equations in two variables have all solutions common.

Question 5.
3x – 4y – 15 = 0 and y + x + 2 = 0. Can these equations be solved by eliminating x ? Is the solution same? (Textbook pg. no. 84)
Solution:
3x – 4y – 15 = 0
∴ 3x – 4y = 15 …(i)
y + x + 2 = 0
∴ x + y = -2 ……(ii)
Multiplying equation (ii) by 3,
3x + 3y = -6 …(iii)
Subtracting equation (iii) from (i),

∴ y = -3
Substituting y = -3 in equation (ii),
∴ x – 3 = -2
∴ x = – 2 + 3
∴ x = 1
∴ (x, y) = ( 1, -3)
Yes, the given equations can be solved by eliminating x. Also, the solution will remain the same.