NCERT Solutions for Class 10 Maths Exercise 3.7

NCERT Solutions for Class 10 Maths Exercise 3.7 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Maths Pair of Linear Equations in Two Variables Download as PDF

NCERT Solutions for Class 10 Maths 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.

Ans. Let the age of Ani and Biju be x years and y years respectively.

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

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NCERT Solutions for Class 10 Maths Exercise 3.7

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]

Ans. Let the money with the first person and second person be Rs x and Rs y respectively.
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.

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NCERT Solutions for Class 10 Maths Exercise 3.7

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.

Ans. Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.

Since Speed =

… (1)
According to the question

……(2)[Using eq. (1)]

Again,

……(3)[Using eq. (1)]

Adding equations (2) and (3), we obtain:
x = 50
Substituting the value of x in equation (2), we obtain:

–100 + 10t = 20

10t = 120t = 12

From equation (1), we obtain:

d = xt = = 600

Thus, the distance covered by the train is 600 km.

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NCERT Solutions for Class 10 Maths Exercise 3.7

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.

Ans. Let the number of rows be x and number of students in a row be y.
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 =

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NCERT Solutions for Class 10 Maths Exercise 3.7

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

Ans. 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,

B =

AndC =

Hence the measures of A, B and C are respectively.

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NCERT Solutions for Class 10 Maths Exercise 3.7

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.

Ans.

Three solutions of this equation can be written in a table as follows:

X’

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).

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NCERT Solutions for Class 10 Maths Exercise 3.7

7. Solve the following pair of linear equations:

(i)

(ii)

(iii)

(iv)

(v)

Ans. (i) … (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

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NCERT Solutions for Class 10 Maths Exercise 3.7

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

Ans. We know that the sum of the measures of opposite angles in a cyclic quadrilateral is .

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 =

NCERT Solutions for Class 10 Maths Exercise 3.7

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