HBSE 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Haryana State Board HBSE 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 1.
The ages of two Mends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution :
Let the present age of Ani be x years.
and present age of Biju be y years.
Therefore, Dharam’s age = 2x years
Cathy’s age = \(\frac{y}{2}\) years
According to question,
x – y = 3 …………………(1)
– x + y = 3 ………………….(2)
also 2x – \(\frac{y}{2}\) = 30
On multiplying both sides by 2
⇒ 4x – y = 60 …………….(3)
On subtracting equation (1) from (3)

On substituting the value of x in equation (1), we get
19 – y = 3
⇒ y = 19 – 3 = 16
So, x = 19, y = 16 is the required solution.
On adding equation (2) and (3)

On substituting the value of x in (2), we get
– 21 + y = 3 .
y = 3 + 21 = 24
So, another required solution is x = 21, y = 24
Hence, present age of Ani is 19 years and present age of Biju is 16 years or present age of Ani is 21 years and present age of Biju is 24 years.

Question 2.
One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital?
Solution:
Let the capital one has be ₹ x
and the capital other has be ₹ y
According to question,
x + 100 = 2(y – 100)
⇒ x + 100 = – 200
x – 2y = – 200 – 100
x – 2y = – 300 …………….(1)
and y + 10 = 6(x – 10)
y + 10 = 6x – 60
6x – y = 10 + 60
6x – y = 70 ……………..(2)
Multiplying equation (2) by 2, we get
12x – 2y = 140 ………………..(3)
Subtracting equation (3) from (1), we get

Substituting the value of x in equation (1), we get
40 – 2y = – 300
⇒ – 2y = – 300 – 40
⇒ – 2y = – 340
⇒ y = \(\frac{-340}{-2}\) = 170
So, x = 40, y = 170 is the required solution.
Hence, the capital of one be ₹ 40.
and the capital of other be ₹ 170.

Question 3.
A train covered a certain distance at a uniform speed. If the train would have been 10 km/hr faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 kmlhr, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:
Let the speed of a train be x km/hr
and time taken by a train be y hours.
Then distance covered by a train = xy km
Since, distance covered by train in each case is same.
Therefore, according to question,
(x + 10) (y – 2) = xy
[∵ Distance = Speed × Time]
⇒ xy – 2x + 10y – 20 = xy
⇒ – 2x + 10y – 20 = 0
⇒ – 2x + 10y = 20
and (x – 10) (y + 3) = xy
⇒ xy + 3x – 10y – 30 = xy
⇒ 3x – 10y – 30 = 0
⇒ 3x – 10y = 30
Adding equation (1) and equation (2), we get

Substituting the value of x in equation (2), we get
3 × 50 – 10y = 30
150 – 10y = 30
– 10y = 30 – 150 = – 120.
y = \(\frac{-120}{-10}\) = 12
So, x = 50, y = 12 is the required solution.
Hence, distance covered by a train = xy = 50 × 12 = 600 km.

Question 4.
The students of a class are made to stands in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less min a row, there would be 2 rows more. Find the number of students in the class.
Solution:
Let the number of rows be x and
number of students in each row be y.
Then total students in the class = xy
According to question,
(y + 3) (x – 1) = xy
⇒ xy – y + 3x – 3 = xy
⇒ 3x – y – 3 = 0
⇒ 3x – y = 3 ……………….(1)
and (y – 3) (x + 2) = xy
⇒ xy + 2y – 3x – 6 = xy
⇒ – 3x + 2y – 6 = 0
⇒ – 3x + 2y = 6 ………………..(2)
Adding equation (1) and equation (2), we get

Substituting the value of y in equation (1), we get
3x – 9 = 3
3x = 3 + 9= 12
x = \(\frac{12}{3}\) = 4
So, x = 4, y = 9 is the required solution.
Hence, total students in the class = xy = 4 × 9 = 36.

Question 5.
In a ∆ABC, ∠A = 3 ∠B = 2 (∠A + ∠B). Find the three angles.
Solution:
Let ∠A = x°, ∠B = y°
∵ ∠C = 3∠B (given)
∴ ∠C = 3y°
also 3∠B = 2(∠A + ∠B) (given)
⇒ 3y° = 2(x° + y°)
⇒ 3y = 2x° + 2y°
⇒ y° = 2x°
⇒ 2x°- y° = 0° …………………(1)
Since, the sum of ∠s of a triangle is 180°.
∠A + ∠B + ∠C = 180°
⇒ x° + y° + 3z° = 180°
⇒ x° + 4y° = 180° ………………(2)
Multiplying equation (1) by 4 and adding it to equation (2), we get

Substituting the value of x in equation (1), we get
2 × 20° – y° = 0°
⇒ 40° – y° = 0°
⇒ y° = 40°
So, x = 20°, y = 40° is the required solution.
Hence, ∠A = 20°,
∠B = 40°,
∠C = 3 × 40° = 120°.

Question 6.
Draw the graphs of the equations 5x -y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y-axis.
Solution:
We draw the graphs of given equa-tions as follows :
5x – y =5
⇒ y = 5x – 5
Putting the different values of x in above equation.
We get different values of y and we prepare the table of x, y.

and 3x – y =3
y = 3x – 3
Putting the different values of x in above equation.
We get different values of y and we prepare the table of x, y.

Plotting the points (0, – 5), (2, 5) and (3, 10) on a graph paper and draw a graph which passes through these points.
So, we obtain the graph of equation 5x – y = 5.
Again plotting the points (0, – 3), (2, 3) and (3, 6) on the same graph paper and draw a graph which passes through these points.
So, we obtain the graph of equation 3x – y = 3.

We observe that graphs of the given equations intersects each other at point A and at y-axis at points B and C respectively. So, the coordinates of the triangle so formed are A (1, 0), B (0, – 3) and C (0, – 5).

Question 7.
Solve the following pair of linear equations :
(i) px + qy = p – q
qx – py = p + q

(ii) ax + by = c
bx + ay = 1 + c

(iii) \(\frac{x}{a}-\frac{y}{b}\) = 0
ax + by = a² + b²

(iv) (a – b)x + (a + b)y = a² – 2ab – b²
(a + b)(x + y) = a² + b²

(v) 152x – 378y = – 74
– 378x + 152y = – 604
Solution :
(i) The given equations are :
px + qy = p – q …………………..(1)
qx – py = p + q ………………….(2)
Multiplying equation (1) by b and equation (2) by a and adding them, we get

Substituting the value of x in equation (1), we get

p × 1 + qy = p – q
qy = p – q – p
qy = – q
y = – \(\frac{q}{q}\) = – 1
Hence, x = 1, y = – 1 is the required solution.

(ii) The given equations are:
ax + by = c ………………..(1)
bx + ay = 1 + c
Multiplying equation (1) by b and equation (2) by a and subtracting them, we get

Substituting the value of y in equation (1), we get
ax + b × \(\left[\frac{c(a-b)+a}{a^2-b^2}\right]\) = c
ax + \(\frac{a b c-b^2 c+a b}{a^2-b^2}\) = c
⇒ a³x – ab²x + abc – b²c + ab = a²c – b²c
⇒ a(a² – b²)x = a²c – b²c – abc + b²c – ab
⇒ a(a² – b²)x = a²c – abc – ab
⇒ a(a² – b²)x = a(ac – bc – b)
⇒ a(a² – b²)x = a[c(a – b)- bi
⇒ x = \(\frac{a[c(a-b)-b]}{a\left(a^2-b^2\right)}\)
⇒ x = \(\frac{c(a-b)-b}{a^2-b^2}\)
Hence, x = \(\frac{c(a-b)-b}{a^2-b^2}\), y = \(\frac{c(a-b)+a}{a^2-b^2}\) the required solution.

(iii) The given equations are:
\(\frac{x}{a}-\frac{y}{b}\) = 0
bx – ay = 0 ……………….(1)
and ax + by = a² + b² …………………..(2)
Multiplying equation (1) by b and equation (2) by a and adding them, we get
b²x – aby = 0
a²x + aby = a(a² + b²)
(a² + b²) x = a (a² + b²)
⇒ x = \(\frac{a\left(a^2+b^2\right)}{\left(a^2+b^2\right)}\)
⇒ x = a
Substituting the value of x in equation (1), we get
b × a – ay = 0
– ay = – ab
y = \(\frac{-a b}{-a}\) = b
Hence, x = a, y = b is the required solution.

(iv) The given equations are
(a – b) x + (a + b) y = a² – 2ab – b²
(a + b) (x + y) = a² + b²
(a + b) x + (a + b) y = a² + b²
Multiplying equation (1) by (a + b) and equation (2) by (a – b) and subtracting them, we get

(a² + b² + 2ab – a² + b²) y = a³ – 2a²b – ab² + a²b – 2ab² – b³ – (a³ + ab² – ab – b³)
(2b² + 2ab) y = a³ – a²b – 3ab² – b³ – a³ – ab² + a²b + b³
2b(a + b)y = – 4ab²
y = \(\frac{-4 a b^2}{2 b(a+b)}=\frac{-2 a b}{a+b}\)
Substituting the value of y in equation (2), we get
(a + b)x + (a + b)\(\left(-\frac{2 a b}{a+b}\right)\) = a² + b²
(a + b)x – 2ab = a² + b²
(a + b)x = a² + b² + 2ab
= (a + b)x = (a + b)²
x = \(\frac{(a+b)^2}{(a+b)}\) = a + b
Hence, x = a + b, y = \(-\frac{2 a b}{a+b}\) is the required solution.

(v) The given equations are :
152x – 378y = – 74 ……………(1)
– 378x + 152y = – 604 …………..(2)
Adding equation (1) and equation (2), we get

⇒ x + y = 3 …………….(3)
Subtracting equation (2) from equation (1) we get

Substituting the value of x in equation (3), we get
2 + y = 3
y = 3 – 2 = 1
Hence, x = 2, y = 1 is the required solution.

Question 8.
ABCD is a cyclic quadrilateral (See Figure). Find the angles of the cyclic quadri lateral.

Solution:
Since, sum of opposite angles of cyclic quadrilateral is 180°.
∴ ∠A + ∠C = 180°
⇒ 4y° + 20 – 4x° = 180°
⇒ 4x° + 4y° = 180° – 20°
⇒ – 4x° + 4y° = 160°
⇒ – x° + y° = 40°
and ∠B +∠D = 180°
3y° – 5 – 7x° + 5 = 180°
– 7x° + 3y° = 180°
Multiplying equation (1) by 7 and subtracting equation (2) from it, we get

4y° = 100°
y° = \(\frac{100}{4}\) = 25°
Substituting the value ofy in equation (1), we get
– x° + 25 = 40°
– x° = 40° – 25° = 15°
x° = 15°
So, x° = – 15, y° = 25 is the required solution.
Hence,
∠A = 4° × 25° + 20° = 120°,
∠B = 3° × 25° – 5° = 70°,
∠C = – 4 × (- 15°) = 60°,
∠D = – 7 × (- 15°) + 5 = 110°.