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Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

10th Standard Maths 1 Practice Set 1.3 Chapter 1 Linear Equations in Two Variables Textbook Answers Maharashtra Board

Class 10 Maths Part 1 Practice Set 1.3 Chapter 1 Linear Equations in Two Variables Questions With Answers Maharashtra Board

Question 1.
Fill in the blanks with correct number.
Solution:

Question 2.
Find the values of following determinants.

Solution:

Question 3.
Solve the following simultaneous equations using Cramer’s rule.
i. 3x – 4y = 10 ; 4x + 3y = 5
ii. 4x + 3y – 4 = 0 ; 6x = 8 – 5y
iii. x + 2y = -1 ; 2x – 3y = 12
iv. 6x – 4y = -12 ; 8x – 3y = -2
v. 4m + 6n = 54 ; 3m + 2n = 28
vi. 2x + 3y = 2 ; x – \(\frac { y }{ 2 } \) = \(\frac { 1 }{ 2 } \)
Solution:
i. The given simultaneous equations are 3x – 4y = 10 …(i)
4x + 3y = 5 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a₁x + b₁y = c₁ and a₂x + b₂y = c₂, we get
a₁ = 3, b₁ = -4, c₁ = 10 and
a₂ = 4, b₂ = 3, c₂ = 5

∴ (x, y) = (2, -1) is the solution of the given simultaneous equations.

ii. The given simultaneous equations are
4x + 3y – 4 = 0
∴ 4x + 3y = 4 …(i)
6x = 8 – 5y
∴ 6x + 5y = 8 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a₁x + b₁y = c₁ and a₂x + b₂y = c₂, we get
a₁ = 4, b₁ = 3, c₁ = 4 and
a₂ = 6, b₂ = 5, c₂ = 8

∴ (x, y) = (-2, 4) is the solution of the given simultaneous equations.

iii. The given simultaneous equations are
x + 2y = -1 …(i)
2x – 3y = 12 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a₁x + b₁y = C₁ and a₂x + b₂y = c₂, we get
a₁ = 1, b₁ = 2, c₁ = -1 and
a₂ = 2, b₂ = -3, c₂ = 12

∴ (x, y) = (3, -2) is the solution of the given simultaneous equations.

iv. The given simultaneous equations are
6x – 4y = -12
∴ 3x – 2y = -6 …(i) [Dividing both sides by 2]
8x – 3y = -2 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a₁x + b₁y = c₁ and a₂x + b₂y = c₂, we get
a₁ = 3, b₁ = -2, c₁ = -6 and
a₂ = 8, b₂ = -3, c₂ = -2

∴ (x, y) = (2, 6) is the solution of the given simultaneous equations.

v. The given simultaneous equations are
4m + 6n = 54
2m + 3n = 27 …(i) [Dividing both sides by 2]
3m + 2n = 28 …(ii)
Equations (i) and (ii) are in am + bn = c form.
Comparing the given equations with
a₁m + b₁n = c₁ and a₂m + b₂n = c₂, we get
a₁ = 2, b₁ = 3, c₁ = 27 and
a₂ = 3, b₂ = 2, c₂ = 28


∴ (m, n) = (6, 5) is the solution of the given simultaneous equations.

vi. The given simultaneous equations are
2x + 3y = 2 …(i)
x = \(\frac { y }{ 2 } \) = \(\frac { 1 }{ 2 } \)
∴ 2x – y = 1 …(ii) [Multiplying both sides by 2]
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a₁x + b₁y = c₁ and a₂x + b₂y = c₂, we get
a₁ = 2, b₁ = 3, c₁ = 2 and
a₂ = 2, b₂ = -1, c₂ = 1

Question 1.
To solve the simultaneous equations by determinant method, fill in the blanks,
y + 2x – 19 = 0; 2x – 3y + 3 = 0 (Textbookpg.no. 14)
Solution:
Write the given equations in the form
ax + by = c.
2x + y = 19
2x – 3y = -3

Question 2.
Complete the following activity. (Textbook pg. no. 15)
Solution:

Question 3.
What is the nature of solution if D = 0? (Textbook pg. no. 16)
Solution:
If D = 0, i.e. a₁b₂ – b₁a₂ = 0, then the two simultaneous equations do not have a unique solution.
Examples:
i. 2x – 4y = 8 and x – 2y = 4
Here, a₁b₂ – b₁a₂ = (2)(-2) – (-4) (1)
= -4 + 4 = 0
Graphically, we can check that these two lines coincide and hence will have infinite solutions.

ii. 2x – y = -1 and 2x – y = -4
Here, a₁ b₂ – b₁ a₂ = (2)(-1) – (-1) (2)
= -2 + 2 = 0
Graphically, we can check that these two lines are parallel and hence they do not have a solution.

Question 4.
What can you say about lines if common solution is not possible? (Textbook pg. no. 16)
Answer:
If the common solution is not possible, then the lines will either coincide or will be parallel to each other.

Class 10 Maths Digest

• Practice Set 1.1 Class 10 Answers
• Practice Set 1.2 Class 10 Answers
• Practice Set 1.3 Class 10 Answers
• Practice Set 1.4 Class 10 Answers
• Practice Set 1.5 Class 10 Answers
• Problem Set 1. Class 10 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 9.2 8th Std Maths Answers Solutions Chapter 9 Discount and Commission.

Discount and Commission Class 8 Maths Chapter 9 Practice Set 9.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 9.2 Chapter 9 Solutions Answers

Question 1.
John sold books worth Rs 4500 for a publisher. For this he received 15% commission. Complete the following activity to find the total commission John obtained.
Solution:
Selling price of the books = Rs 4500
Rate of commission = 15%
Commission obtained = 15% of selling price
\(=\frac{[15]}{[100]} \times[4500]\)
= 15 × 45
∴ Commission obtained = 675 Rupees.
∴ The total commission obtained by John is Rs 675.

Question 2.
Rafique sold flowers worth Rs 15,000 by giving 4% commission to the agent. Find the commission he paid. Find the amount received by Rafique.
Solution:
Here, selling price of flowers = Rs 15,000,
Rate of commission = 4%
i. Commission = 4% of selling price
= \(\frac { 4 }{ 100 }\) × 15,000
= 4 x 150
∴ Commission = Rs 600

ii. Amount received by Rafique = selling price – commission
= 15,000 – 600
= Rs 14,400
∴ Rafique paid Rs 600 as commission and the amount received by him was Rs 14,400.

Question 3.
A farmer sold food grains for Rs 9200 through an agent. The rate of commission was 2%. How much amount did the agent get ?
Solution:
Here, selling price of food grains = Rs 9200,
Rate of commission = 2%
Commission = 2% of selling price
= \(\frac { 2 }{ 100 }\) × 9200
= 2 × 92
= Rs 184
∴ The agent got a commission of Rs 184.

Question 4.
Umatai purchased following items from a Khadi – Bhandar.
i. 3 sarees for Rs 560 each.
ii. 6 bottles of honey for Rs 90 each.
On the purchase, she received a rebate of 12%. How much total amount did Umatai pay?
Solution:
Here, number of sarees = 3,
Price of each saree = Rs 560
∴ Cost of 3 sarees = 560 × 3
= Rs 1680 …(i)
Also, number of honey bottles = 6,
Price of each bottle = Rs 90
∴ Cost of 6 honey bottles = 90 × 6
= Rs 540
Total amount of purchase
= cost of 3 sarees + cost of 6 honey bottles
= 1680 + 540 … [From (i) and (ii)]
= Rs 2220 …(iii)
Rate of rebate = 12%
Rebate = 12% of total amount of purchase
= \(\frac { 12 }{ 100 }\) × 2220
= 12 × 22.20
= Rs 266.40 ..(iv)
Amount paid by Umatai
= Total amount of purchase – Rebate
= 2,220 – 266.40 … [From (iii) and (iv)]
= Rs 1953.60
∴ The total amount paid by Umatai is Rs 1953.60.

Question 5.
Use the given information and fill in the boxes with suitable numbers.
Smt. Deepanjali purchased a house for Rs 7,50,000 from Smt. Leelaben through an agent. Agent has charged 2 % brokerage from both of them.
Solution:
i. Smt. Deepanjali paid 7,50,000 × \(\frac { 2 }{ 100 }\)
= 7,500 × 2 = Rs 15,000 brokerage for purchasing the house.

ii. Smt. Leelaben paid brokerage of Rs 15,000

iii. Total brokerage received by the agent is = 15,000 + 15,000 = Rs 30,000

iv. The cost of house Smt. Deepanjali paid is = 7,50,000 + 15,000 = Rs 7,65,000

v. The selling price of house for Smt.Leelaben is = 7,50,000 – 15,000
= Rs 7,35,000

Std 8 Maths Digest

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

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 8.3 8th Std Maths Answers Solutions Chapter 8 Quadrilateral: Constructions and Types.

Quadrilateral: Constructions and Types Class 8 Maths Chapter 8 Practice Set 8.3 Solutions Maharashtra Board

Std 8 Maths Practice Set 8.3 Chapter 8 Solutions Answers

Question 1.
Measures of opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of its each angle.
Solution:
Let ₹PQRS be the parallelogram.
m∠Q = (3x – 2)° and m∠S = (50 – x)°

m∠Q = m∠S
…..(i)
[Opposite angles of a parallelogram are congruent]
∴ 3x – 2 = 50 – x
∴ 3x + x = 50 + 2
∴ 4x = 52
∴ x = \(\frac { 52 }{ 4 }\)
∴ x = 13
Now, m∠Q = (3x – 2)°
= (3 x 13 – 2)° = (39 – 2)° = 37°
∴ m∠S = m∠Q = 37° …[From(i)]
m∠P + m∠Q = 180°
….[Adjacent angles of a parallelogram are supplementary]
∴ m∠P + 37° = 180°
∴ m∠P = 180° – 37° = 143°
∴ m∠R = m∠P = 143°
…..[Opposite angles of a parallelogram are congruent]
∴ The measures of the angles of the parallelogram are 37°, 143°, 37° and 143°.

Question 2.
Referring the given figure of a parallelogram, write the answers of questions given below.
i. If l(WZ) = 4.5 cm, then l(XY) = ?
ii. If l(YZ) = 8.2 cm, then l(XW) = ?
iii. If l(OX) = 2.5 cm, then l(OZ) = ?
iv. If l(WO) = 3.3 cm, then l(WY) = ?
v. If m∠WZY = 120°, then m∠WXY = ? and m∠XWZ = ?

Solution:
i. l(WZ) = 4.5 cm … [Given]
l(X Y) = l(WZ) ….[Opposite sides of a parallelogram are congrument ]
∴ l(X Y) = 4.5cm

ii. l(YZ) = 8.2 cm …[Given]
l(XW) = l(YZ)
…[Opposite sides of a parallelogram are congruent]
∴ l(XW) = 8.2cm … [Given]

iii. l(OX) = 2.5 cm …[Given]
l(OZ) = l(OX)
….[Diagonals of a parallelogram bisect each other]
∴ l(OZ) = 2.5cm

iv. l(WO) = 3.3 cm … [Given]
l(WO) = \(\frac { 1 }{ 2 }\) l(WY)
….[Diagonals of a parallelogram bisect each other]
∴ 3.3 = \(\frac { 1 }{ 2 }\) l(WY)
∴ 3.3 x 2 = l(WY)
∴ l(WY) = 6.6cm

v. m∠WZY =120° … [Given]
m∠WXY = m∠WZY
…..[Opposite angles of a parallelogram are congrument]
∴ m∠WXY = 120° …(i)
m∠XWZ + m∠WXY = 180°
….[Adjacent angles of a parallelogram are supplementary]
∴ m∠XWZ + 120° = 180° … [From (i)]
∴ m∠XWZ = 180°- 120°
∴ m∠XWZ = 60°

Question 3.
Construct a parallelogram ABCD such that l(BC) = 7 cm, m∠ABC = 40°, l(AB) = 3 cm
Solution:

Opposite sides of a parallelogram are congruent.
∴ l(AB) = l(CD) = 3cm
l(BC) = l(AD) = 7 cm

Question 4.
Ratio of consecutive angles of a quadrilateral is 1 : 2 : 3 : 4. Find the measure of its each angle. Write with reason, what type of a quadrilateral it is.
Solution:
Let ₹PQRS be the quadrilateral.
Ratio of consecutive angles of a quadrilateral is 1 : 2 : 3 : 4.
Let the common multiple be x.
∴m∠P = x°, m∠Q = 2x°, m∠R = 3x° and m∠S = 4x°
In ₹PQRS,
m∠P + m∠Q + m∠R + m∠S = 360°
…[Sum of the measures of the angles of a quadrilateral is 360°]
∴x° + 2x° + 3x° + 4x° = 360°
∴10 x° = 360°
∴x° = \(\frac { 360 }{ 10 }\)
∴x° = 36°
∴m∠P = x° = 36°

m∠Q = 2x° = 2 × 36° = 72°
m∠R = 3x° = 3 × 36° = 108° and
m∠S = 4x° = 4 × 36° = 144°
∴The measures of the angles of the quadrilateral are 36°, 72°, 108°, 144°.
Here, m∠P + m∠S = 36° + 144° = 180°
Since, interior angles are supplementary,
∴side PQ || side SR
m∠P + m∠Q = 36° + 72° = 108° ≠ 180°
∴side PS is not parallel to side QR.
Since, one pair of opposite sides of the given quadrilateral is parallel.
∴The given quadrilateral is a trapezium.

Question 5.
Construct ₹BARC such that
l(BA) = l(BC) = 4.2 cm, l(AC) = 6.0 cm, l(AR) = l(CR) = 5.6 cm
Solution:

Question 6.
Construct ₹PQRS, such that l(PQ) = 3.5 cm, l(QR) = 5.6 cm, l(RS) = 3.5 cm, m∠Q = 110°, m∠R = 70°. If it is given that ₹PQRS is a parallelogram, which of the given information is unnecessary?
Solution:

1. Since, the opposite sides of a parallelogram are congruent.
   ∴ Either l(PQ) or l(SR) is required.
2. To construct a parallelogram lengths of adjacent sides and measure of one angle is required.
   ∴ Either l(PQ) and m∠Q or l(SR) and m∠R is the unnecessary information given in the question.

Maharashtra Board Class 8 Maths Chapter 8 Quadrilateral: Constructions and Types Practice Set 8.3 Intext Questions and Activities

Question 1.
Draw a parallelogram PQRS. Take two rulers of different widths, place one ruler horizontally and draw lines along its edges. Now place the other ruler in slant position over the lines drawn and draw lines along its edges. We get a parallelogram. Draw the diagonals of it and name the point of intersection as T.

1. Measure the opposite angles of the parallelogram.
2. Measure the lengths of opposite sides.
3. Measure the lengths of diagonals.
4. Measure the lengths of parts of the diagonals made by point T. (Textbook pg. no. 47)

Solution:
[Students should attempt the above activities on their own.]

Question 2.
In the given figure of ₹ABCD, verify with a divider that seg AB ≅ seg CB and seg AD ≅ seg CD. Similarly measure ∠BAD and ∠BCD and verify that they are congruent. (Textbook pg. no. 48)

Solution:
[Students should attempt the above activities on their own.]

Std 8 Maths Digest

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

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 7.3 8th Std Maths Answers Solutions Chapter 7 Variation.

Variation Class 8 Maths Chapter 7 Practice Set 7.3 Solutions Maharashtra Board

Std 8 Maths Practice Set 7.3 Chapter 7 Solutions Answers

Question 1.
Which of the following statements are of inverse variation?
i. Number of workers on a job and time taken by them to complete the job.
ii. Number of pipes of same size to fill a tank and the time taken by them to fill the tank.
iii. Petrol filled in the tank of a vehicle and its cost.
iv. Area of circle and its radius.
Solution:
i. Let, x represent number of workers on a job, and y represent time taken by workers to complete the job.
As the number of workers increases, the time required to complete the job decreases.
∴ \(x \propto \frac{1}{y}\)

ii. Let, n represent number of pipes of same size to fill a tank and t represent time taken by the pipes to fill the tank.
As the number of pipes increases, the time required to fill the tank decreases.
∴ \(\mathrm{n} \propto \frac{1}{\mathrm{t}}\)

iii. Let, p represent the quantity of petrol filled in a tank and c represent the cost of the petrol.
As the quantity of petrol in the tank increases, its cost increases.
∴ p ∝ c

iv. Let, A represent the area of the circle and r represent its radius.
As the area of circle increases, its radius increases.
∴ A ∝ r
∴ Statements (i) and (ii) are examples of inverse variation.

Question 2.
If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?
Solution:
Let, n represent the number of workers building the wall and t represent the time required.
Since, the number of workers varies inversely with the time required to build the wall.
∴ \(\mathrm{n} \propto \frac{1}{\mathrm{t}}\)
∴ \(\mathrm{n}=\mathrm{k} \times \frac{1}{\mathrm{t}}\)
where k is the constant of variation
∴ n × t = k …(i)
15 workers can build a wall in 48 hours,
i.e., when n = 15, t = 48
∴ Substituting n = 15 and t = 48 in (i), we get
n × t = k
∴ 15 × 48 = k
∴ k = 720
Substituting k = 720 in (i), we get
n × t = k
∴ n × t = 720 …(ii)
This is the equation of variation.
Now, we have to find number of workers required to do the same work in 30 hours.
i.e., when t = 30, n = ?
∴ Substituting t = 30 in (ii), we get
n × t = 720
∴ n × 30 = 720
∴ n = \(\frac { 720 }{ 30 }\)
∴ n = 24
∴ 24 workers will be required to build the wall in 30 hours.

Question 3.
120 bags of half litre milk can be filled by a machine within 3 minutes find the time to fill such 1800 bags?
Solution:
Let b represent the number of bags of half litre milk and t represent the time required to fill the bags.
Since, the number of bags and time required to fill the bags varies directly.
∴ b ∝ t
∴ b = kt …(i)
where k is the constant of variation.
Since, 120 bags can be filled in 3 minutes
i.e., when b = 120, t = 3
∴ Substituting b = 120 and t = 3 in (i), we get
b = kt
∴ 120 = k × 3
∴ k = \(\frac { 120 }{ 3 }\)
∴ k = 40
Substituting k = 40 in (i), we get
b = kt
∴ b = 40 t …(ii)
This is the equation of variation.
Now, we have to find time required to fill 1800 bags
∴ Substituting b = 1800 in (ii), we get
b = 40 t
∴ 1800 = 40 t
∴ t = \(\frac { 1800 }{ 40 }\)
∴ t = 45
∴ 1800 bags of half litre milk can be filled by the machine in 45 minutes.

Question 4.
A car with speed 60 km/hr takes 8 hours to travel some distance. What should be the increase in the speed if the same distance is
to be covered in \(7\frac { 1 }{ 2 }\) hours?
Solution:
Let v represent the speed of car in km/hr and t represent the time required.
Since, speed of a car varies inversely as the time required to cover a distance.
∴ \(v \propto \frac{1}{t}\)
∴ \(\mathbf{v}=\mathbf{k} \times \frac{1}{\mathbf{t}}\)
where, k is the constant of variation.
∴ v × t = k …(i)
Since, a car with speed 60 km/hr takes 8 hours to travel some distance.
i.e., when v = 60, t = 8
∴ Substituting v = 60 and t = 8 in (i), we get
v × t = k
∴ 60 × 8 = t
∴ k = 480
Substituting k = 480 in (i), we get
v × t = k
∴ v × t = 480 …(ii)
This is the equation of variation.
Now, we have to find speed of car if the same distance is to be covered in \(7\frac { 1 }{ 2 }\) hours.
i.e., when t = \(7\frac { 1 }{ 2 }\) = 7.5 , v = ?
∴ Substituting, t = 7.5 in (ii), we get
v × t = 480
∴ v × 7.5 = 480
\(v=\frac{480}{7.5}=\frac{4800}{75}\)
∴ v = 64
The speed of vehicle should be 64 km/hr to cover the same distance in 7.5 hours.
∴ The increase in speed = 64 – 60
= 4km/hr
∴ The increase in speed of the car is 4 km/hr.

Std 8 Maths Digest

• Practice Set 6.1 Class 8 Answers
• Practice Set 6.2 Class 8 Answers
• Practice Set 6.3 Class 8 Answers
• Practice Set 6.4 Class 8 Answers
• Practice Set 7.1 Class 8 Answers
• Practice Set 7.2 Class 8 Answers
• Practice Set 7.3 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 7.2 8th Std Maths Answers Solutions Chapter 7 Variation.

Variation Class 8 Maths Chapter 7 Practice Set 7.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 7.2 Chapter 7 Solutions Answers

Question 1.
The information about number of workers and number of days to complete a work is given in the following table. Complete the table.

Solution:
Let, n represent the number of workers and d represent the number of days required to complete a work.
Since, number of workers and number of days to complete a work are in inverse poportion.
∴ \(\mathbf{n} \propto \frac{1}{\mathrm{d}}\)
∴ \(\mathrm{n}=\mathrm{k} \times \frac{1}{\mathrm{d}}\)
where k is the constant of variation.
∴ n × d = k …(i)

i. When n = 30, d = 6
∴ Substituting n = 30 and d = 6 in (i), we get
n × d = k
∴ 30 × 6 = k
∴ k = 180
Substituting k = 180 in (i), we get
∴ n × d = k
∴ n × d = 180 …(ii)
This is the equation of variation

ii. When d = 12, n = 7
∴ Substituting d = 12 in (ii), we get
n × d = 180
∴ n × 12 = 180
∴ n = \(\frac { 180 }{ 12 }\)
∴ n = 15

iii. When n = 10, d = ?
∴ Substituting n = 10 in (ii), we get
n × d = 180
10 × d = 180
∴ d = \(\frac { 180 }{ 10 }\)
∴ d = 18

iv. When d = 36, n = ?
∴ Substituting d = 36 in (ii), we get
n × d = 180
∴ n × 36 = 180
∴ n = \(\frac { 180 }{ 36 }\)
∴ n = 5

Question 2.
Find constant of variation and write equation of variation for every example given below:
i. \(p \propto \frac{1}{q}\) ; if p = 15 then q = 4.
ii. \(z \propto \frac{1}{w}\) ; when z = 2.5 then w = 24.
iii. \(s \propto \frac{1}{t^{2}}\) ; if s = 4 then t = 5.
iv. \(x \propto \frac{1}{\sqrt{y}}\) ; if x = 15 then y = 9.
Solution:
i. \(p \propto \frac{1}{q}\) …[Given]
∴ p = k × \(\frac { 1 }{ q }\)
where, k is the constant of variation.
∴ p × q = k …(i)
When p = 15, q = 4
∴ Substituting p = 15 and q = 4 in (i), we get
p × q = k
∴ 15 × 4 = k
∴ k = 60
Substituting k = 60 in (i), we get
p × q = k
∴ p × q = 60
This is the equation of variation.
∴ The constant of variation is 60 and the equation of variation is pq = 60.

ii. \(z \propto \frac{1}{w}\) …[Given]
∴ z = k × \(\frac { 1 }{ w }\)
where, k is the constant of variation,
∴ z × w = k …(i)
When z = 2.5, w = 24
∴ Substituting z = 2.5 and w = 24 in (i), we get
z × w = k
∴ 2.5 × 24 = k
∴ k = 60
Substituting k = 60 in (i), we get
z × w = k
∴ z × w = 60
This is the equation of variation.
∴ The constant of variation is 60 and the equation of variation is zw = 60.

iii. \(s \propto \frac{1}{t^{2}}\) …[Given]
∴ \(s=k \times \frac{1}{t^{2}}\)
where, k is the constant of variation,
∴ s × t² = k …(i)
When s = 4, t = 5
∴ Substituting, s = 4 and t = 5 in (i), we get
s × t² = k
∴ 4 × (5)² = k
∴ k = 4 × 25
∴ k = 100
Substituting k = 100 in (i), we get
s × t² = k
∴ s × t² = 100
This is the equation of variation.
∴ The constant of variation is 100 and the equation of variation is st² = 100.

iv. \(x \propto \frac{1}{\sqrt{y}}\) …[Given]
∴ \(x=\mathrm{k} \times \frac{1}{\sqrt{y}}\)
where, k is the constant of variation,
∴ x × √y = k …(i)
When x = 15, y = 9
∴ Substituting x = 15 and y = 9 in (i), we get
x × √y = k
∴ 15 × √9 = k
∴ k = 15 × 3
∴ k = 45
Substituting k = 45 in (i), we get
x × √y = k
∴ x × √y = 45 .
This is the equation of variation.
∴ The constant of variation is k = 45 and the equation of variation is x√y = 45.

Question 3.
The boxes are to be filled with apples in a heap. If 24 apples are put in a box then 27 boxes are needed. If 36 apples are filled in a box how many boxes will be needed?
Solution:
Let x represent the number of apples in each box and y represent the total number of boxes required.
The number of apples in each box are varying inversely with the total number of boxes.
∴ \(x \infty \frac{1}{y}\)
∴ \(x=k \times \frac{1}{y}\)
where, k is the constant of variation,
∴ x × y = k …(i)
If 24 apples are put in a box then 27 boxes are needed.
i.e., when x = 24, y = 27
∴ Substituting x = 24 and y = 27 in (i), we get
x × y = k
∴ 24 × 27 = k
∴ k = 648
Substituting k = 648 in (i), we get
x × y = k
∴ x × y = 648 …(ii)
This is the equation of variation.
Now, we have to find number of boxes needed
when, 36 apples are filled in each box.
i.e., when x = 36,y = ?
∴ Substituting x = 36 in (ii), we get
x × y = 648
∴ 36 × y = 648
∴ y = \(\frac { 648 }{ 36 }\)
∴ y = 18
∴ If 36 apples are filled in a box then 18 boxes are required.

Question 4.
Write the following statements using symbol of variation.

1. The wavelength of sound (l) and its frequency (f) are in inverse variation.
2. The intensity (I) of light varies inversely with the square of the distance (d) of a screen from the lamp.

Solution:

1. \(l \propto \frac{1}{\mathrm{f}}\)
2. \(\mathrm{I} \propto \frac{1}{\mathrm{d}^{2}}\)

Question 5.
\(x \propto \frac{1}{\sqrt{y}}\) and when x = 40 then y = 16. If x = 10, find y.
Solution:
\(x \propto \frac{1}{\sqrt{y}}\)
∴ \(x=\mathrm{k} \times \frac{1}{\sqrt{y}}\)
where, k is the constant of variation.
∴ x × √y = k …(i)
When x = 40, y = 16
∴ Substituting x = 40 andy = 16 in (i), we get
x × √y = k
∴ 40 × √16 = k
∴ k = 40 × 4
∴ k = 160
Substituting k = 160 in (i), we get
x × √y = k
∴ x × √y = 160 …(ii)
This is the equation of variation.
When x = 10,y = ?
∴ Substituting, x = 10 in (ii), we get
x × √y = 160
∴ 10 × √y = 160
∴ √y = \(\frac { 160 }{ 10 }\)
∴ √y = 16
∴ y = 256 … [Squaring both sides]

Question 6.
x varies inversely as y, when x = 15 then y = 10, if x = 20, then y = ?
Solution:
Given that,
\(x \propto \frac{1}{\sqrt{y}}\)
∴ \(x=\mathrm{k} \times \frac{1}{\sqrt{y}}\)
where, k is the constant of variation.
∴ x × y = k …(i)
When x = 15, y = 10
∴ Substituting, x = 15 and y = 10 in (i), we get
x × y = k
∴ 15 × 10 = k
∴ k = 150
Substituting, k = 150 in (i), we get
x × y = k
∴ x × y = 150 …(ii)
This is the equation of variation.
When x = 20, y = ?
∴ substituting x = 20 in (ii), we get
x × y = 150
∴ 20 × y = 150
∴ y = \(\frac { 150 }{ 20 }\)
∴ y = 7.5

Std 8 Maths Digest

• Practice Set 6.1 Class 8 Answers
• Practice Set 6.2 Class 8 Answers
• Practice Set 6.3 Class 8 Answers
• Practice Set 6.4 Class 8 Answers
• Practice Set 7.1 Class 8 Answers
• Practice Set 7.2 Class 8 Answers
• Practice Set 7.3 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 9.1 8th Std Maths Answers Solutions Chapter 9 Discount and Commission.

Discount and Commission Class 8 Maths Chapter 9 Practice Set 9.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 9.1 Chapter 9 Solutions Answers

8th Standard Maths Practice Set 9.1 Question 1. If marked price = Rs 1700, selling price = Rs 1540, then find the discount.
Solution:
Here, Marked price = Rs 1700,
selling price = Rs 1540
Selling price = Marked price – Discount
∴ 1540 = 1700 – Discount
∴ Discount = 1700 – 1540
= Rs 160
∴ The amount of discount is Rs 160.

Discount and Commission Practice Set 9.1 Question 2. If marked price Rs 990 and percentage of discount is 10, then find the selling price.
Solution:
Here, marked price = Rs 990,
discount = 10%
Let the percentage of discount be x
∴ x = 10%
i. Discount

= Rs 99

ii. Selling price = Marked price – Discount
= 990 – 99
= Rs 891
∴ The selling price is Rs 891.

Practice Set 9.1 Question 3. If selling price Rs 900, discount is 20%, then find the marked price.
Solution:
Here, selling price = Rs 900, discount = 20%
Let the marked price be Rs 100
Since, the discount given = 20%
∴ Amount of discount = Rs 20
∴ Selling price = 100 – 20 – Rs 80
Let actual marked price be Rs x
∴ For marked price of Rs x, selling price is Rs 900
\(\frac{80}{100}=\frac{900}{x}\)
∴ 80 × x = 100 × 900
∴ \(x=\frac{100 \times 900}{80}\)
∴ x = Rs 1125
∴ The marked price is Rs 1125.

Discount and Commission Std 8 Question 4. The marked price of the fan is Rs 3000. Shopkeeper gave 12% discount on it. Find the total discount and selling price of the fan.
Solution:
Here, Marked price = Rs 3000, discount = 12%
Let the percentage of discount be x.
∴ x = 12%
i. Discount

= 30 × 12
= Rs 360

ii. Selling price = Marked price – Discount
= 3000 – 360
= Rs 2640
∴ The total discount is Rs 360 and the selling price of the fan is Rs 2640.

Discount and Commission 8th Standard Question 5. The marked price of a mixer is Rs 2300. A customer purchased it for Rs 1955. Find percentage of discount offered to the customer.
Solution:
Here, marked price = Rs 2300,
selling price = Rs 1955
i. Selling price = Marked price – Discount
∴ 1955 = 2300 – Discount
∴ Discount = 2300 – 1955
= Rs 345

ii. Let the percentage of discount be x


∴ x = 15%
∴ The percentage of discount offered to the customer is 15%.

Question 6.
A shopkeeper gives 11% discount on a television set, hence the cost price of it is Rs 22,250. Then find the marked price of the television set.
Solution:
Here, selling price = Rs 22,250, discount = 11%
Let marked price be Rs 100
Since, the discount given = 11%
∴ Amount of discount = Rs 11
∴ Selling price = 100 – 11 = Rs 89
∴ Let actual marked price be Rs x
∴ For marked price of Rs x, selling price is Rs 22,250

∴ x = Rs 25,000
∴ The marked price of the television set is Rs 25,000.

8th Std Maths Discount and Commission Question 7. After offering discount of 10% on marked price, a customer gets total discount of Rs 17. To find the cost price for the customer, fill in the following boxes with appropriate numbers and complete the activity.
Solution:
Suppose, marked price of the item = 100 rupees Therefore, for customer that item costs 100 – 10 = 90 rupees.
Hence, when the discount is [10] then the selling price is [90] rupees.
Suppose when the discount is [17] rupees, the selling price is x rupees.

∴ The customer will get the item for Rs 153.

Question 8.
A shopkeeper decides to sell a certain item at a certain price. He tags the price on the item by increasing the decided price by 25%. While selling the item, he offers 20% discount. Find how many more or less percent he gets on the decided price.
Solution:
Here, price increase = 25%,
discount offered = 20%
Let the decided price be Rs 100
∴ Increase in price = Rs 25
∴ Shopkeeper marks the price = 100 + 25
= Rs 125
∴ marked price = Rs 125
Let the percentage of discount be x
∴ x = 20%

∴ Selling price = Marked price – Discount
= 125 – 25
= Rs 100
∴ If the decided price is Rs 100, then shopkeeper gets Rs 100.
∴ The shopkeeper gets neither more nor less than the decided price i.e. he gets 0% more / less.

Maharashtra Board Class 8 Maths Chapter 9 Discount and Commission Practice Set 9.1 Intext Questions and Activities

Question 1.
Write the appropriate numbers in the following boxes. (Textbook pg. no. 51)

1. \(\frac { 12 }{ 100 }=\) __ percent = __%
2. 47% = __
3. 86% = __
4. 4% of 300 = 300 × __ = __
5. 15% of 1700 = 1700 × __= __

Solution:

1. \(\frac { 12 }{ 100 }=\) 12 percent = 12%
2. 47% = \(\frac { 47 }{ 100 }\)
3. 86% = \(\frac { 86 }{ 100 }\)
4. 4% of 300 = 300 × \(\frac { 4 }{ 100 }\) = 12
5. 15% of 1700 = 1700 × \(\frac { 15 }{ 100 }\) = 255

Question 2.
You may have seen advertisements like ‘Monsoon Sale’, ‘Stock Clearance Sale’ etc offering different discount. In such a sale, a discount is offered on various goods. Generally in the month of July, sales of clothes are declared. Find and discuss the purpose of such sales. (Textbook pg. no. 51)
Solution:
(Students should attempt the above activity on their own)

Maharashtra Board Class 8 Maths Solutions

Std 8 Maths Digest

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

Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Practice Set 7 Answers Solutions Chapter 1 Geometrical Constructions.

Geometrical Constructions Class 7 Maths Chapter 1 Practice Set 7 Solutions Maharashtra Board

Std 7 Maths Practice Set 7 Solutions Answers

Question 1.
Some angles are given below. Using the symbol of congruence write the names of the pairs of congruent angles in these figures.

Solution:
i. ∠AOC ≅ ∠PQR
ii. ∠DOC ≅ ∠LMN
iii. ∠AOB ≅ ∠BOC ≅ ∠RST

Maharashtra Board Class 7 Maths Chapter 1 Geometrical Constructions Practice Set 7 Intext Questions and Activities

Question 1.
Observe the given angles and write the names of those having equal measures.
(Textbook pg. no. 8 and 9)

Solution:
i. ∠ABC and ∠SPM
ii. ∠NIT and ∠SRI
iii. ∠PTQ and ∠RTS

Question 2.
Observe the image shown in the adjacent figure and answer the following questions. (Textbook pg. no. 9)

1. What time does this clock show?
2. What is the measure of the angle between its two hands?
3. At which other times is the angle between the hands congruent with this angle?

Solution:

1. 3 o’ clock.
2. 90°.
3. 9 o’ clock.

Question 3.
Get bangles of different sizes but equal thickness and find the congruent ones among them. (Textbook pg. no. 10)
Solution:
[Students should attempt the above activities on their own.]

Question 4.
Find congruent circles in your surroundings. (Textbook pg. no. 10)
Solution:
[Students should attempt the above activities on their own.]

Question 5.
Take some round bowls and plates. Place their edges one upon the other to find pairs of congruent edges. (Textbook pg. no. 10)
Solution:
[Students should attempt the above activities on their own.]

Std 7 Maths Digest

• Practice Set 1 Class 7 Answers
• Practice Set 2 Class 7 Answers
• Practice Set 3 Class 7 Answers
• Practice Set 4 Class 7 Answers
• Practice Set 5 Class 7 Answers
• Practice Set 6 Class 7 Answers
• Practice Set 7 Class 7 Answers