MP Board Class 10th Maths Solutions Chapter 3 Exercise 3.7

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.7 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board 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 1st condition : y – x = 3 …. (1)
∵ [Age of Ani’s father] = 2[Age of Ani] = 2x years

10th maths exercise 3.7 important questions

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

10th maths chapter 3 exerrcise 3.7 question answer

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

MP Board 10th maths chapter 3 Exercise 3.7 important question answer

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 1st friend = ₹ x,
and the capital of 2nd 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, 1st friend has ₹ 40 and the 2nd friend has ₹ 170.

10th maths exercise 3.7 solutions pdf 2022

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 1st condition: (x +10) × (y – 2) = xy
⇒ xy – 2x + 10y – 20 = xy
⇒ 2x – 10y + 20 = 0 … (1)
According to 2nd 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

Class 10th Maths Exercise 3.7 imp 2022 solutions mp board

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

MP Board Class 10th Maths Solutions

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

MP Board Class 10 Maths Chapter 3 exercise 3.7 solutions important

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°.

10th maths chapter 3 exercise 3.7 solutions pdf 10th maths 2022

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

x120
y05-5

and for equation 3x – y = 3, we get

x230
y36-3

Plotting the points (1, 0), (2, 5) and (0, -5), we get a straight line l1. Plotting the points (2, 3), (3, 6) and (0, -3), we get a straight line l2.

10th maths 2022 exercise 3.7 solutions

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) xa−yb= 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:

Class 10th maths chapter 3 important exercise 3.7 solutions,

10th maths exercise 3.7 Question 8.
ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.

ABCD is a 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)

MP Board Class 10th Maths Solutions 2022:

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