Prasanna

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.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

Practice Set 4.3 Algebra 9th Std Maths Part 1 Answers Chapter 4 Ratio and Proportion

Ratio and Proportion Practice Set 4.3 Question 1.
If \(\frac { a }{ b }\) = \(\frac { 7 }{ 3 }\), then find the aIues of the following ratios.

Solution:

Ratio and Proportion Class 9 Practice Set 4.3 Question 2.
If \(\frac{15 a^{2}+4 b^{2}}{15 a^{2}-4 b^{2}}=\frac{47}{7}\), then find the value of the following ratios.

Solution:

Practice Set 4.3 Algebra 9th Question 3.
If \(\frac{3 a+7 b}{3 a-7 b}=\frac{4}{3}\)then find the value of the ratio \(\frac{3 a^{2}-7 b^{2}}{3 a^{2}+7 b^{2}}\).
Solution:

Class 9 Maths Chapter 4 Ratio And Proportion Practice Set 4.3 Question 4.
Solve the following equations.

Solution:

This equation is true for x = 0
∴ x = 0 is one of the solutions.
If x ≠ 0, then x² ≠ 0
∴ \(\frac { 1 }{ 12x – 20 }\) = \(\frac { 1 }{ 8x + 12 }\) … [Dividing both sides by x²]
∴ 8x + 12 = 12x – 20
∴ 12 + 20 = 12x – 8x
∴ 32 = 4x
∴ x = 8
∴ x = 0 or x = 8 are the solutions of the given equation.

∴ 21(x – 5) = 4(2x + 3)
∴ 21x – 105 = 8x + 12
∴ 21x – 8x = 12 + 105
∴ 13x = 117
∴ x = 9
∴ x = 9 ¡s the solution of the given equation.

∴ 9(4x + 1) = 25(x + 3)
∴36x + 925x + 75
∴ 36x – 25 = 75 – 9
∴11x = 66
∴ x = 6
∴ x = 6 is the solution of the given equation.


∴ 4(3x – 4) = 5(x + 1)
∴ 12x – 16 = 5x + 5
∴ 12x – 5x = 5 + 16
∴ 7x = 21
∴ x = 3
∴ x = 3 ¡s the solution of the given equation.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

Problem Set 4 Algebra 9th Std Maths Part 1 Answers Chapter 4 Ratio and Proportion

Question 1.
Select the appropriate alternative answer for the following questions.

i . If 6 : 5 = y : 20, then what will be the value of y?
(A) 15
(B) 24
(C) 18
(D) 22.5
Answer:
(B) 24

ii. What is the ratio of 1 mm to 1 cm ?
(A) 1 : 100
(B) 10: 1
(C) 1 : 10
(D) 100: 1
Answer:
(C) 1 : 10

iii. The ages of Jatin, Nitin and Mohasin are 16, 24 and 36 years respectively. What is the ratio of Nitin’s age to Mohasin’s age ?
(A) 3 : 2
(B) 2 : 3
(C) 4 : 3
(D) 3 : 4
Answer:
(B) 2 : 3

iv. 24 bananas were distributed between Shubham and Anil in the ratio 3 : 5, then how many bananas did Shubham get?
(A) 8
(B) 15
(C) 12
(D) 9
Answer:
(D) 9

v. What is the mean proportional of 4 and 25?
(A) 6
(B) 8
(C) 10
(D) 12
Answer:
(C) 10

Hints:
i . \(\frac{6}{5}=\frac{y}{20}\)
∴ \(\quad y=\frac{6 \times 20}{5}=24\)

ii. \(\frac{1 \mathrm{mm}}{1 \mathrm{cm}}=\frac{1 \mathrm{mm}}{10 \mathrm{mm}}=\frac{1}{10}=1 : 10\)

iii. \(\frac{\text { Age of Nitin }}{\text { Age of Mohasin }}=\frac{24}{36}=\frac{12 \times 2}{12 \times 3}\)
\(\frac { 2 }{ 3 }\) = 2 : 3

iv. 3x + 5x = 24
∴ 8x = 24
∴ x = 3
∴ Number of bananas with Shubham = 3x = 9

v. x² = 4 x 25
∴ x² = 100
∴ x = 10

Question 2.
For the following numbers write the ratio of first number to second number in the reduced form. [1 Mark each]
i. 21,48
ii. 36,90
iii. 65,117
iv. 138,161
v. 114,133
Solution:
i. 21,48
\(\text { Ratio }=\frac{21}{48}=\frac{3 \times 7}{3 \times 16}=\frac{7}{16}=7 : 16\)
ii. 36,90
\(\text { Ratio }=\frac{36}{90}=\frac{18 \times 2}{18 \times 5}=\frac{2}{5}=2 : 5\)
iii. 65,117
\(\text { Ratio }=\frac{65}{117}=\frac{13 \times 5}{13 \times 9}=\frac{5}{9}=5 : 9\)
iv. 138,161
\(\text { Ratio }=\frac{138}{161}=\frac{23 \times 6}{23 \times 7}=\frac{6}{7}=6 : 7\)
v. 114,133
\(\text { Ratio }=\frac{114}{113}=\frac{19 \times 6}{19 \times 7}=\frac{6}{7}=6 : 7\)

Question 3.
Write the following ratios in the reduced form.
i. Radius to the diameter of a circle.
ii. The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm.
iii. The ratio of numbers denoting perimeter to area of a square, having side 4 cm.
Solution:
i. Radius to the diameter of a circle.
Let radius of the circle be r
then, diameter = 2 x radius = 2r
Ratio of radius to diameter of circle

∴ Ratio of radius to diameter of circle is 1 : 2.

ii. The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm.

Let □ ABCD be a rectangle.
In ∆ABC, ∠B = 90°
AC² = AB² + BC² … [Pythagoras theorem]
= 4² + 3² = 16 + 9
∴ AC² = 25
AC = 5 … [Taking square root on both side]
The ratio of diagonal to the length of a rectangle = \(\frac { AC }{ AB }\)
= \(\frac { 5 }{ 4 }\)
= 5 : 4
∴ The ratio of diagonal to the length of a rectangle is 5 : 4

iii. The ratio of numbers denoting perimeter to area of a square, having side 4 cm. side of square = 4cm
Perimeter of square = 4 x side = 4 x 4 = 16 cm
Area of square = (side)² = (4)² = 16 cm²
Ratio of numbers denoting perimeter to area of square

∴ The ratio of numbers denoting perimeter to area of a square is 1 : 1.

Question 4.
Check whether the following numbers are in continued proportion.
i. 2, 4, 8
ii. 1, 2, 3
iii. 9, 12, 16
iv. 3, 5, 8
Solution:
If a, b, c are in continued proportion then b² = ac.
i. 2, 4, 8
Let, a = 2, b = 4 and c = 8
Here, b² = 4² = 16
ac = 2 x 8 = 16
∴ b² = ac
∴ 2, 4,8 are in continued proportion.

ii. 1, 2, 3
Let, a = 1, b = 2 and c = 3
Here, b² = 2² = 4
ac = 1 x 3 = 3
∴ b² ≠ ac
∴ 1, 2,3 are not in continued proportion.

iii. 9, 12, 16
Let, a = 9, b = 12 and c = 16
Here, b² = 12² = 144
ac = 9 x 16 = 144
∴ b² = ac
∴ 9, 12, 16 are in continued proportion.

iv. 3, 5, 8
Let, a = 3, b = 5 and c = 8
Here, b² = 5² = 25
ac = 3 x 8 = 24
∴ b² ≠ ac
∴ 3, 5, 8 are not in continued proportion.

Question 5.
a, b, c are in continued proportion. If a = 3 and c = 27, then find b.
Solution:
a, b, c are in continued proportion. …[Given]
∴ b² = ac
∴ b² = 3 x 27 …[∵ a = 3 and c = 27]
∴ b² = 81
∴ b = 9 …[Taking square root of both sides]

Question 6.
Convert the following ratios into percentages.

Solution:
i. Let 37 : 500 = x%

∴ 37 : 500 = 7.4%

Question 7.
Write the ratio of first quantity to second quantity in the reduced form.
i. 1024 MB, 1.2 GB [(1024 MB = 1GB)]
ii. ₹ 17, ₹ 25 and 60 paise
iii. 5 dozen, 120 units
iv. 4 sq.m, 800 sq.cm
v. 1.5 kg, 2500 gm
Solution:
i. 1024 MB, 1.2 GB
1024 MB = 1 GB

ii. ₹ 17, ₹ 25 and 60 paise
₹ 17 = 17 x 100 paise = 1700 paise
₹ 25 and 60 paise = (25x 100) paise + 60 paise
= (2500 + 60) paise
= 2560 paise

iii. 5 dozen, 120 units
5 dozen = 5 x 12 units = 60 units

iv. 4 sq.m, 800 sq.cm
4 sq.m = 4 x 10000 sq.cm = 40000 sq.cm

v. 1.5 kg, 2500 gm
1.5 kg = 1.5 x 1000 gm = 1500gm

Question 8.
If \(\frac { a }{ b }\) = \(\frac { 2 }{ 3 }\), then find the values of the following expressions.

Solution:

Question 9.
If a, b, c, d are in proportion, then prove that

Solution:
a, b, c are in continued proportion. …[Given]

Question 10.
If a, b, c are ¡n continued proportion, then prove that

Solution:
a, b, c are in continued proportion. … [Given]
∴ \(\frac { a }{ b }\) = \(\frac { b }{ c }\)
Let \(\frac { a }{ b }\) = \(\frac { b }{ c }\) = k
∴ b = ck
∴ a = bk
= (ck)k . .. [From (j)]
a = ck² …(ii)

Question 11.
Solve:
\( \frac{12 x^{2}+18 x+42}{18 x^{2}+12 x+58}=\frac{2 x+3}{3 x+2}\)
Solution:


∴ 29(2x + 3) = 21 (3x + 2)
∴ 5 + 87= 63x + 42
∴ 87 – 42 = 63x – 58x
∴ 45 = 5x
∴ x = 9
∴ x = 9 is the solution of the given equation.

Question 12.
If \( \frac{2 x-3 y}{3 z+y}=\frac{z-y}{z-x}=\frac{x+3 z}{2 y-3 x}\) ,then prove that every ratio = \(\frac { x }{ y }\).
Solution:

Question 13.
If \(\frac{b y+c z}{b^{2}+c^{2}}=\frac{c z+a x}{c^{2}+a^{2}}=\frac{a x+b y}{a^{2}+b^{2}}\), then prove \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
Solution:

Question 1.
Take 5 pieces of card paper. Write the following statements, one on each paper.

a, b, c, d are positive numbers and \(\frac{a}{b}=\frac{c}{d}\) is given. Which of the above statements are true or false, write at the back of each card, if false explain why. (Textbook pg. no. 70)
Answer:
i. True
ii. True
iii. False
Here, numerator and denominator are multiplied by two different numbers a and b.
iv. False
Here, different numbers a and b are subtracted from numerator and denominator.
v. True

Question 2.
In the following activity, the values of a and b can be changed. That is by changing a : b we can create many examples. Teachers should give lot of practice to the students and encourage them to construct their own examples. (Textbook pg. no. 70)

Question 3.
Observe the political map of India from a Geography text book. Study the scale of this map.
From the given scale find the straight line distances between various cities like
i. New Delhi to Bengaluru
ii. Mumbai to Kolkata
iii. Jaipur to Bhuvaneshvar. (Textbook pg. no. 77)
[Students should attempt the above activity on their own.]

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 8.2 8th Std Maths Answers Solutions Chapter 8 Quadrilateral: Constructions and Types.

Practice Set 8.2 8th Std Maths Answers Chapter 8 Quadrilateral: Constructions and Types

Question 1.
Draw a rectangle ABCD such that l(AB) = 6.0 cm and l(BC) = 4.5 cm.
Solution:

Question 2.
Draw a square WXYZ with side 5.2 cm.
Solution:

Question 3.
Draw a rhombus KLMN such that its side is 4 cm and m∠K = 75°.
Solution:

Question 4.
If diagonal of a rectangle is 26 cm and one side is 24 cm, find the other side.
Solution:

Let ₹ABCD be the rectangle.
l(BC) = 24cm, l(AC) = 26cm
In ∆ABC,
m∠ABC = 90° …[Angle of a rectangle]
∴[l(AC)]² = [l(AB)]2 + [l(BC)]²
…[Pythagoras theorem]
∴ (26 )² = [l(AB)]² + (24)²
∴(26)² – (24)² = [l(AB)]²
∴(26 + 24) (26 – 24) = [l(AB)]²
…[∵ a² – b² = (a + b)(a – b)]
∴50 x 2 = [l(AB)]²
∴100 = [l(AB)]²
i.e. [l(AB)]² = 100
∴l(AB) = √100
…[Taking square root of both sides]
∴l(AB) =10 cm
∴The length of the other side is 10 cm.

Question 5.
Lengths of diagonals of a rhombus ABCD are 16 cm and 12 cm. Find the side and perimeter of the rhombus.
Solution:
In rhombus ABCD,

l(AC) = 16 cm and l(BD) = 12 cm.
Let the diagonals of rhombus ABCD intersect at point O.
l(AO) = \(\frac { 1 }{ 2 }\) l(AC)
…[Diagonals of a rhombus bisect each other]
∴l(AO) = \(\frac { 1 }{ 2 }\) × 16
∴l(AO) = 8 cm
Also, l(DO) = \(\frac { 1 }{ 2 }\) l(BD)
…[Diagonals of a rhombus bisect each other]
∴l(DO) = \(\frac { 1 }{ 2 }\) × 12
∴l(DO) = 6 cm
In ∆DOA,
m∠DOA = 90°
..[Diagonals of a rhombus are perpendicular to each other]
[l(AD)]² = [l(AO)]² + [l(DO)]²
…[Pythagoras theorem]

= (8)² + (6)²
= 64 + 36
∴[l(AD)]² = 100
∴l(AD) = √100
… [Taking square root of both sides]
∴l(AD) = 10 cm
∴l(AB) = l(BC) = l(CD) = l(AD) = 10 cm
…[Sides of a rhombus are congruent]
Perimeter of rhombus ABCD
= l(AB) + l(BC) + l(CD) + l(AD)
= 10+10+10+10
= 40 cm
∴The side and perimeter of the rhombus are 10 cm and 40 cm respectively.

Question 6.
Find the length of diagonal of a square with side 8 cm.
Solution:
Let ₹XYWZ be the square of side 8cm.

seg XW is a diagonal.
In ∆ XYW,
m∠XYW = 90°
… [Angle of a square]
∴ [l(XW)]² = [l(XY)]² + [l(YW)]²
…[Pythagoras theorem]
= (8)² + (8)²
= 64 + 64
∴ [l(XW)]² = 128
∴ l(XW) = √128
…[Taking square root of both sides]
= √64 × 2
= 8 √2 cm
∴ The length of the diagonal of the square is 8 √2 cm.

Question 7.
Measure of one angle of a rhombus is 50°, find the measures of remaining three angles.
Solution:
Let ₹ABCD be the rhombus.

m∠A = 50°
m∠C = m∠A
….[Opposite angles of a rhombus are congruent]
∴ m∠C = 50°
Also, m∠D = m∠B …(i)
….[Opposite angles of a rhombus are congruent]
In ₹ABCD,
m∠A + m∠B + m∠C + m∠D = 360°
….[Sum of the measures of the angles of a quadrilateral is 360°]
∴ 50° + m∠B + 50° + m∠D = 360°
∴ m∠B + m∠D + 100° = 360°
∴ m∠B + m∠D = 360° – 100°
∴ m∠B + m∠B = 260° …[From (i)]
∴ 2m∠B = 260°
∴ m∠B = \(\frac { 260 }{ 2 }\)
∴ m∠B = 130°
∴ m∠D = m∠B = 130° …[From (i)]
∴ The measures of the remaining angles of the rhombus are 130°, 50° and 130°.

Maharashtra Board Class 8 Maths Chapter 8 Quadrilateral: Constructions and Types Practice Set 8.2 Intext Questions and Activities

Question 1.
Construct a rectangle PQRS by taking two convenient adjacent sides. Name the point of intersection of diagonals as T. Using divider and ruler, measure the following lengths.
i. lengths of opposite sides, seg QR and seg PS.
ii. lengths of seg PQ and seg SR.
iii. lengths of diagonals PR and QS.
iv. lengths of seg PT and seg TR, which are parts of the diagonal PR.
v. lengths of seg QT and seg TS, which are parts of the diagonal QS.
Observe the measures. Discuss about the measures obtained by your classmates. (Textbook pg. no. 44)
Solution:
Draw a rectangle PQRS such that, l(PQ) = 3 cm and l(QR) = 4 cm.

Steps of construction:
i. As shown in the rough figure, draw seg QR of length 4 cm.
ii. Placing the centre of the protractor at point Q, draw ray QW making an angle of 90° with seg QR.
iii. By taking a distance of 3 cm on the compass and placing it at point Q, draw an arc on ray QW. Name the point as P.
iv. Draw ray PV and ray RU making an angle of 90° with seg PQ and seg QR respectively.
v. Name the point of intersection of ray PV and ray RU as S.
₹PQRS is the required rectangle.

From the figure,
i. l(QR) = l(PS) = 4 cm
ii. l(PQ) = l(SR) = 3 cm
iii. l(PR) = l(QS) = 5 cm
iv. l(PT) = l(TR) = 2.5 cm
v. l(QT) = l(TS) = 2.5 cm

From the above measures, we can say that for any rectangle,
i. Opposite sides are congruent.
ii. Diagonals are congruent.
iii. Diagonals bisect each other.

Question 2.
Draw a square by taking convenient length of side. Name the point of intersection of its diagonals as E. Using the apparatus in a compass box, measure the following lengths.
i. lengths of diagonal AC and diagonal BD.
ii. lengths of two parts of each diagonal made by point E.
iii. all the angles made at the point E.
iv. parts of each angle of the square made by each diagonal, (e.g. ∠ADB and ∠CDB).
Observe the measures. Also observe the measures obtained by your classmates and discuss about them. (Textbook pg. no. 44)
Solution:
Draw a square ABCD such that its side is 5cm

Steps of construction:
i. As shown in the rough figure, draw seg BC of length 5 cm.
ii. Placing the centre of the protractor at point B, draw ray BP making an angle of 90° with seg BC.
iii. By taking a distance of 5 cm on the compass and placing it at point B, draw an arc on ray BP. Name the point as A.
iv. Placing the centre of the protractor at point C, draw ray CQ making an angle of 90° with seg BC.
v. By taking a distance of 5 cm on the compass and placing it at point C, draw an arc on ray CQ. Name the point as D.
vi. Draw seg AD.
₹ABCD is the required square.

From the figure,
i. l(AC) = l(BD) ≅ 7cm
ii. l(AE) = l(EC) ≅ 3.5cm,
l(BE) = l(ED) ≅ 3.5cm
iii. m∠AED = m∠BEC = m∠CED = m∠BEA = 90°
iv. Angles made by diagonal AC:
m∠BAC = m∠DAC = 45°
m∠BCA = m∠DCA = 45°
Angles made by diagonal BD:
m∠ABD = m∠CBD = 45°
m∠ADB = m∠CDB = 45°

From the above measures, we can say that for any square,
i. Diagonals are congruent.
ii. Diagonals bisect each other.
iii. Diagonals are perpendicular to each other.
iv. Diagonals bisect the opposite angles.

Question 3.
Draw a rhombus EFGH by taking convenient length of side and convenient measure of an angle.
Draw its diagonals and name their point of Intersection as M.
i. Measure the opposite angles of the quadrilateral and angles at the point M.
ii. Measure the two parts of every angle made by the diagonal.
iii. Measure the lengths of both diagonals. Measure the two parts of diagonals made by point M.
Observe the measures. Also observe the measures obtained by your classmates and discuss about them. (Textbook pg. no. 45)
Solution:
Draw a rhombus EFGH such that its side is 5 cm and m∠F = 60°.

Steps of construction:
i. As shown in the rough figure, draw seg FG of length 5 cm.
ii. Placing the centre of the protractor at point F, draw ray FX making an angle 60° with seg FG.
iii. By taking a distance of 5 cm on the compass and placing it at point F, draw an arc on ray FX. Name the point as E.
iv. By taking a distance of 5 cm on the compass and placing it at point E and point G, draw arcs. Name the point of intersection of arcs as H. ₹EFGH is the required rhombus.

From the figure,
i. Opposite angles:
m∠EFG = m∠GHE = 60°,
m∠FEH = m∠HGF = 120°
Angles at the point M:
m∠EMF = m∠FMG = m∠GMH = m∠HME = 90°

ii. Angles made by diagonal FH:
m∠EFH = m∠GFH = 30° m∠EHF = m∠GHF = 30°
Angles made by diagonal EG:
m∠FEG = m∠HEG = 60° m∠FGE = m∠HGE = 60°

iii. l(FH) ≈ 8.6 cm
l(EG) = 5 cm
l(FM) = l(HM) ≈ 4.3 cm
l(EM) = l(GM) ≈ 2.5 cm

From the above measures, we can say that for any rhombus,
i. Opposite angles are congruent.
ii. Diagonals bisect the opposite angles.
iii. Diagonals bisect each other and they are perpendicular to each other.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

Practice Set 2.4 Algebra 9th Std Maths Part 1 Answers Chapter 2 Real Numbers

Question 1.
Multiply.

Solution:

Question 2.
Rationalize the denominator.

Solution: