NCERT Solutions for Class 10 Maths Exercise 3.7 Chapter 3 Pair of Linear Equations in Two Variables – FREE PDF Download
NCERT Class 10 Maths Ch 3 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter
3 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Pair of Linear Equations in Two Variables solutions will help you understand the chapter thoroughly.
NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables
1. The age 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.
Age of Dharam = 2x years and Age of Cathy = 
years
According to question, x – y = 3… (1)
And
4x – y = 60… (2)
Subtracting (1) from (2), we obtain:
3x = 60 – 3 = 57
x = Age of Ani = 19 years
Age of Biju = 19 – 3 = 16 years
Again, According to question, y – x = 3… (3)
And
4x – y = 60… (4)
Adding (3) and (4), we obtain:
3x = 63
x = 21
Age of Ani = 21 years
Age of Biju = 21 + 3 = 24 years
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]
According to the question,
x + 100 = 2(y – 100)
x + 100 = 2y – 200
x – 2y = – 300… (1)
Again, 6(x – 10) = (y + 10)
6x – 60 = y + 10
6x – y = 70… (2)
Multiplying equation (2) by 2, we obtain:
12x – 2y = 140… (3)
Subtracting equation (1) from equation (3), we obtain:
11x = 140 + 300
11x = 440
x = 40
Putting the value of x in equation (1), we obtain:
40 – 2y = –300
40 + 300 = 2y
2y = 340
y = 170
Thus, the two friends had Rs 40 and Rs 170 with them.
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.
Since Speed = 
… (1)
According to the question
[Since, xt=dxt=d]
……(2)[Using eq. (1)]
Again, 
……(3)[Using eq. (1)] [Since, xt=dxt=d]
Adding equations (2) and (3), we obtain:
x = 50
Substituting the value of x in equation (2), we obtain:
–100 + 10t = 20
10t = 120
⇒⇒ t = 12
From equation (1), we obtain:
d = xt = 
= 600
Thus, the distance covered by the train is 600 km.
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.
Total number of students in the class = Number of rows x Number of students in a row = xy
According to the question,
Total number of students = (x – 1) (y + 3)
xy = (x – 1) (y + 3)
xy = xy – y + 3x – 3
3x – y – 3 = 0
3x – y = 3… (1)
Total number of students = (x + 2) (y – 3)
xy = xy + 2y – 3x – 6
3x – 2y = –6… (2)
Subtracting equation (2) from (1), we obtain:
y = 9
Substituting the value of y in equation (1), we obtain:
3x – 9 = 3
3x = 9 + 3 = 12
x = 4
Number of rows = x = 4
Number of students in a row = y = 9
Hence, Total number of students in a class = xy = 
5. In a 
ABC, 
C = 3B = 2(A + B). Find three angles.
C = 3B = 2(A + B) Taking 3B = 2(A + B)
B = 2A
2A – B = 0 …….(1)
We know that the sum of the measures of all angles of a triangle is 180°.
A + B + C = 
A + B + 3B =
A + 4B = 
…….(2)
Multiplying equation (1) by 4, we obtain:
8A – 4B = 0 …….(3)
Adding equations (2) and (3), we get
9A =
A = 
From eq. (2), we get,
20∘+4∠B=180∘20∘+4∠B=180∘
B = 
AndC = 
Hence the measures of A, B and C are 
respectively.
6. Draw the graphs of the equations 
and 
Determine the co-ordinate of the vertices of the triangle formed by these lines and the 
axis.
Three solutions of this equation can be written in a table as follows:
It can be observed that the required triangle is ABC.
The coordinates of its vertices are A (1, 0), B (0, –3), C (0, –5).
7. Solve the following pair of linear equations:
(i)
(ii) 
(iii)
(iv) 
(v) 
… (1) … (2)
Multiplying equation (1) by p and equation (2) by q, we obtain:
… (3)
… (4)
Adding equations (3) and (4), we obtain:
= 1
Substituting the value of 
in equation (1), we obtain:
Hence the required solution is x = 1 and y = –1.
(ii)… (1)
… (2)
Multiplying equation (1) by a and equation (2) by b, we obtain:
… (3)
… (4)
Subtracting equation (4) from equation (3),
Substituting the value of x in equation (1), we obtain:
(iii)
……..(1)
……..(2)
Multiplying equation (1) and (2) by b and a respectively, we obtain:
……..(3)
……..(4)
Adding equations (3) and (4), we obtain:
Substituting the value of in equation (1), we obtain:
(iv) … (1)
……..(2)
Subtracting equation (2) from (1), we obtain:
Substituting the value of in equation (1), we obtain:
(v)152x – 378y = –74… (1)
–378x + 152y = –604 … (2)
Adding the equations (1) and (2), we obtain:
–226x – 226y = –678
x + y = 3 ………(3)
Subtracting the equation (2) from equation (1), we obtain:
530x – 530y = 530
x – y = 1 ……..(4)
Adding equations (3) and (4), we obtain:
2x = 4
x = 2
Substituting the value of x in equation (3), we obtain:
y = 1
8. ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.
. 
A + C =
………(1)
Also B + D =
………(2)
Multiplying equation (1) by 3, we obtain:
……….(3)
Adding equations (2) and (3), we obtain:
Substituting the value of in equation (1), we obtain:
A = 
B = 
C = 
D = 
