# Maharashtra Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Problem Set 1

Maharashtra state Board 10th Standard Solutions Chapter 1 Linear Equations in two Variables – Here are all the MH Board solutions for 10th Standard maths Problem Set 1. This solution contains questions, answers, images, explanations of the complete Problem Set 1 titled Linear Equations in two Variables of maths taught in 10th Standard. If you are a student of 10th Standard who is using Maharashtra state Board Textbook to study maths, then you must come across Problem Set 1 Linear Equations in two Variables. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete Maharashtra Board Solutions for 10th Standard maths Chapter 1 Linear Equations in two Variables in one place.

## Maharashtra State Board Class 10 Maths Solutions Chapter 1 Linear Equations in Two Variables Problem Set 1

Choose correct alternative for each of the following questions.

Question 1.
To draw graph of 4x + 5y = 19, find y when x = 1.
(a) 4
(b) 3
(c) 2
(d) -3
(b)

Question 2.
For simultaneous equations in variables x and y, Dx = 49, Dy = – 63, D = 7 then what is x?
(a) 7
(b) -7
(c) $$\frac { 1 }{ 7 }$$
(d) $$\frac { -1 }{ 7 }$$
(a)

Question 3.
Find the value of

(a) -1
(b) -41
(c) 41
(d) 1
(d)

Question 4.
To solvex + y = 3; 3x – 2y – 4 = 0 by determinant method find D.
(a) 5
(b) 1
(c) -5
(d) -1
(c)

Question 5.
ax + by = c and mx + n y = d and an ≠ bm then these simultaneous equations have-
(a) Only one common solution
(b) No solution
(c) Infinite number of solutions
(d) Only two solutions.
(a)

Question 2.
Complete the following table to draw the graph of 2x – 6y = 3.

Question 3.
Solve the following simultaneous equations graphically.
i. 2x + 3y = 12 ; x – y = 1
ii. x – 3y = 1 ; 3x – 2y + 4 = 0
iii. 5x – 6y + 30 = 0; 5x + 4y – 20 = 0
iv. 3x – y – 2 = 0 ; 2x + y = 8
v. 3x + y= 10 ; x – y = 2
i. The given simultaneous equations are

The two lines interest at point (3,2).
∴ x = 3 and y = 2 is the solution of the simultaneous equations 2x + 3y = 12 and x – y = 1.

ii. The given simultaneous equations are

The two lines intersect at point (-2, -1).
∴ x = -2 and y = -1 is the solution of the simultaneous equations x – 3y = 1 and 3x – 2p + 4 = 0.

iii. The given simultaneous equations are

The two lines intersect at point (0, 5).
∴ x = 0 and y = 5 is the solution of the simultaneous equations 5x – 6y + 30 = 0 and 5x + 4y – 20 = 0.

iv. The given simultaneous equations are

The two lines intersect at point (2, 4).
∴ x = 2 and y = 4 is the solution of the simultaneous equations 3x – y – 2 = 0 and 2x + y = 8.

v. The given simultaneous equations are

The two lines intersect at point (3, 1).
∴ x = 3 and y = 1 is the solution of the simultaneous equations 3x + y = 10 and x – y = 2.

Question 4.
Find the values of each of the following determinants.

Solution:

Question 5.
Solve the following equations by Cramer’s method.

Solution:
i. The given simultaneous equations are
6x – 3y = -10 …(i)
3x + 5y – 8 = 0
∴ 3x + 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 = 6, b1 = -3, c1 = 10 and
a2 = 3, b2 = 5, c2 = 8

ii. The given simultaneous equations are
4m – 2n = -4 …(i)
4m + 3n = 16 …(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 = 4, b1 = -2, c1 = -4 and
a2 = 4, b2 = 3, c2 = 16

∴ (m, n) = (1, 4) is the solution of the given simultaneous equations.

iii. The given simultaneous equations are

iv. The given simultaneous equations are
7x + 3y = 15 …(i)
12y – 5x = 39
i.e. -5x + 12y = 39 …(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 = 7, b1 = 3, c1 = 15 and
a2 = -5, b2 = 12, c2 = 39

v. The given simultaneous equations are

∴ 4(x + y – 8) = 2(3x – y)
∴ 4x + 4y – 32 = 6x – 2y
∴ 6x – 4x – 2y – 4y = -32
∴ 2x – 6y = -32
∴ x – 3y = -16 …(ii)[Dividing 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 = 1, b1 = -1, c1 = -4 and
a2 = 1, b2 = -3, c2 = -16

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

Question 6.
Solve the following simultaneous equations:

i. The given simultaneous equations are

Subtracting equation (iv) from (iii), we get

∴ (x, y) = (6, – 4) is the solution of the given simultaneous equations.

ii. The given simultaneous equations are

Adding equations (v) and (vi), we get

iii. The given simultaneous equations are

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

iv. The given simultaneous equations are

∴ Equations (i) and (ii) become 7q – 2p = 5 …(iii)
8q + 7p = 15 …(iv)
Multiplying equation (iii) by 7, we get
49q – 14p = 35 …(v)
Multiplying equation (iv) by 2, we get
16q + 14p = 30 …(vi)
Adding equations (v) and (vi), we get

Substituting q = 1 in equation (iv), we get
8(1) + 7p = 15
∴ 8 + 7p = 15
∴ 7p = 15 – 8
∴ 7p = 7
∴ p = $$\frac { 7 }{ 7 }$$ = 1
∴ (P, q) = (1,1)
Resubstituting the values of p and q, we get
1 = $$\frac { 1 }{ x }$$ and 1 = $$\frac { 1 }{ y }$$
∴ x = 1 and y = 1
∴ (x, y) = (1, 1) is the solution of the given simultaneous equations.

v. The given simultaneous equations are

Resubstituting the values of p and q, we get

∴ 3x + 4y = 10 …(v)
and 2x – 3y = 1 …(vi)
Multiplying equation (v) by 3, we get
9x + 12y = 30 …(vii)
Multiplying equation (vi) by 4, we get
8x – 12y = 4 …(viii)
Adding equations (vii) and (viii), we get

Substituting x = 2 in equation (v), we get
3(2) + 4y = 10
⇒ 6 + 4y = 10
⇒ 4y = 10 – 6
⇒ y = 4/4 = 1
∴ y = 1
∴ (x, y) = (2, 1) is the solution of the given simultaneous equations.

Question 7.
Solve the following word problems, i. A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit’s place is 3 more than the digit in the ten’s place. Find the original number.
Solution:
Let the digit in unit’s place be x
and that in the ten’s place be y.

ii. Kantabai bought 1 $$\frac { 1 }{ 2 }$$ kg tea and 5 kg sugar from a shop. She paid ₹ 50 as return fare for rickshaw. Total expense was ₹ 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So, next month she placed the order online for 2 kg tea and 7 kg sugar. She paid ₹ 880 for that. Find the rate of sugar and tea per kg.
Solution:
Let the rate of tea be ₹ x per kg and that of sugar be ₹ y per kg.
According to the first condition,
cost of 1 $$\frac { 1 }{ 2 }$$ kg tea + cost of 5 kg sugar + fare for rickshaw = total expense

According to the second condition,
cost of 2 kg tea + cost of 7 kg sugar = total expense
2x + 7y = 880 …(ii)
Multiplying equation (i) by 2, we get
6x + 20y = 2600 …(iii)
Multiplying equation (ii) by 3, we get
6x + 21y = 2640 …(iv)
Subtracting equation (iii) from (iv), we get

∴ The rate of tea is ₹ 300 per kg and that of sugar is ₹ 40 per kg.

iii. To find number of notes that Anushka had, complete the following activity.

Solution:
Anushka had x notes of ₹ 100 and y notes of ₹ 50.
According to the first condition,
100x + 50y = 2500
∴ 2x + y = 50 …(i) [Dividing both sides by 50]
According to the second condition,
100y + 50x = 2000
∴ 2y + x = 40 … [Dividing both sides by 50]
i.e. x + 2y = 40
∴ 2x + 4y = 80 …(ii) [Multiplying both sides by 2]
Subtracting equation (i) from (ii), we get

∴ Anushka had 20 notes of ₹ 100 and 10 notes of ₹ 50.

iv. Sum of the present ages of Manish and Savita is 31, Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.
Solution:
Let the present ages of Manish and Savita be x years and y years respectively.
According to the first condition,
x + y = 31 …(i)
3 years ago,
Manish’s age = (x – 3) years
Savita’s age = (y – 3) years
According to the second condition,
(x – 3) = 4 (y – 3)
∴ x – 3 = 4y – 12
∴ x – 4y = -12 + 3
∴ x – 4y = -9 …(ii)
Subtracting equation (ii) from (i), we get

∴ x = 31 – 8
∴ x = 23
The present ages of Manish and Savita are 23 years and 8 years respectively.

v. In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is ₹ 720. Find daily wages of skilled and unskilled workers.
Solution:
Let the daily wages of skilled workers be ₹ x
that of unskilled workers be ₹ y.
According to the first condition,

∴ The daily wages of skilled workers is ₹ 450 and that of unskilled workers is ₹ 270.

vi. Places A and B are 30 km apart and they are on a straight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A), Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.
Solution:
Let the speeds of Hamid and Joseph be x km/hr andy km/hr respectively.
Distance travelled by Hamid in 20 minutes

According to the first condition,

∴ The speeds of Hamid and Joseph 50 km/hr and 40 km/hr respectively.