In this post, we will share NCERT Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.7. These solutions are based on new NCERT Syllabus.

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

Question 1.

The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Solution:

Let the age of Ani = x years

and the age of Biju’s = y years

Case I:

y > x

According to 1^{st} condition : y – x = 3 …. (1)

∵ [Age of Ani’s father] = 2[Age of Ani] = 2x years

Substituting the value of x in equation (1),

we get y – 21 = 3 ⇒ y = 3 + 21 = 24

∴ Age of Ani = 21 years

Age of Biju = 24 years

Substituting the value of y in equation (1),

we get 19 – y = 3 ⇒ y = 16

∴ Age of Ani = 19 years

Age of Biju = 16 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? [From the Bijaganita of Bhaskara II]

[Hint: x + 100 = 2(y – 100), y + 10 = 6(x – 10)]

Solution:

Let the capital of 1^{st} friend = ₹ x,

and the capital of 2^{nd} friend = ₹ y

According to the condition,

x + 100 = 2(y -100)

⇒ x + 100 – 2y + 200 = 0

⇒ x – 2y + 300 = 0 … (1)

Also, 6(x – 10) = y + 10 ⇒ 6x – y – 70 = 0 …. (2)

From (1), x = -300 + 2y (3)

Substituting the value of x in equation (2), we get

6[-300 + 2y] – y – 70 = 0

⇒ -1870 + 11y = 0

Now, Substituting the value of y in equation (3), we get, x = – 300 + 2y

= – 300 + 2(170) = – 300 + 340 = 40

Thus, 1^{st} friend has ₹ 40 and the 2^{nd} friend has ₹ 170.

Question 3.

A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Solution:

Let the actual speed of the train = x km/hr

and the actual time taken = y hours

Distance = speed × time

According to 1^{st} condition: (x +10) × (y – 2) = xy

⇒ xy – 2x + 10y – 20 = xy

⇒ 2x – 10y + 20 = 0 … (1)

According to 2^{nd} condition: (x -10) × (y + 3) = xy

⇒ xy + 1ox -10y – 30 = xy

⇒ 3x – 10y – 30 = 0 ….(2)

Using cross multiplication for solving (1) and (2), we get

Thus, the distance covered by the train = 50 × 12 km = 600 km.

Question 4.

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Solution:

Let the number of students = x

and the number of rows = y

∴ Number of students in each row

Question 5.

In a ∆ABC, ∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles.

Solution:

Sum of angles of a triangle = 180°

∴ ∠A + ∠B + ∠C = 180° … (1)

∵ ∠C = 3∠B = 2(∠A + ∠B) … (2)

From (1) and (2), we have ∠A + ∠B + 2 (∠A + ∠B) = 180°

⇒ ∠A + ∠B + 2∠A + 2∠B = 180°

⇒ ∠A + ∠B = 60° …. (3)

Also, ∠A + ∠B + 3∠B = 180°

⇒ ∠A + 4∠B = 180° ….(4)

Subtracting (3) from (4), we get

∠A + 4∠B – ∠A – ∠B = 180°- 60°

Substituting ∠B = 40° in (4) we get,

∠A + 4(40°) = 180°

⇒ ∠A = 180° – 160° = 20°

∴ ∠C = 3∠B = 3 × 40° = 120°

Thus, ∠A = 20°, ∠B = 40° and ∠C = 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 they-axis.

Solution:

To draw the graph of 5x – y = 5, we get

x | 1 | 2 | 0 |

y | 0 | 5 | -5 |

and for equation 3x – y = 3, we get

x | 2 | 3 | 0 |

y | 3 | 6 | -3 |

Plotting the points (1, 0), (2, 5) and (0, -5), we get a straight line l_{1}. Plotting the points (2, 3), (3, 6) and (0, -3), we get a straight line l_{2}.

From the figure, obviously, the vertices of the triangle formed are A(l, 0), B(0, -5) and C(0,-3).

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 = a^{2} + b^{2}

(iv) (a – b)x + (a + b)y = a^{2} – 2ab – b^{2},

(a + b)(x + y) = a^{2} + b^{2}

(v) 152x – 378y = -74, – 378x + 152y = -604

Solution:

Question 8.

ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.

Solution:

ABCD is a cyclic quadrilateral.

∴ ∠A + ∠C = 180° and

∠B + ∠D = 180°

⇒ [4y + 20] + [- 4x] = 180°

⇒ 4y – 4x + 20° -180° = 0

⇒ 4y – 4x – 160° = 0

⇒ y – x – 40° = 0 … (1)