NCERT Solutions for class 10 Maths Exercise 3.7 Chapter 3 Linear Equations In Two Variables helps the students strengthen their concepts. It carries solutions to the questions provided in the textbook and other miscellaneous exercises for self-practice. NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.7 carries solutions to the questions provided in the textbook page number 68.

NCERT Solutions for Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables is an important topic from the examination perspective and has been explained complete detail. Stepwise solutions are provided to all the questions with accurate answers.

NCERT Solutions Class 10 Maths is prepared by the subject experts after a lot of research to provide the best content possible. This will help the students score well in the examinations. This helps the students to analyze themselves for better practice.

NCERT Solutions can help the students strengthen their fundamentals. The more they practice the more they grasp and the better they perform in the examination.

### Download PDF for NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.7

## Access other exercise solutions of class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables

Exercise 3.1 Solutions– 3 Questions

Exercise 3.2 Solutions– 7 Questions

Exercise 3.3 Solutions– 3 Questions

Exercise 3.4 Solutions– 2 Questions

Exercise 3.5 Solutions– 4 Questions

Exercise 3.6 Solutions– 2 Questions

### Access Answers of Maths NCERT class 10 Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.7

**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:**

The age difference between Ani and Biju is 3 yrs.

Either Biju is 3 years older than that of Ani or Ani is 3 years older than Biju. From both the cases we find out that Ani’s father’s age is 30 yrs more than that of Cathy’s age.

Let the ages of Ani and Biju be A and B respectively.

Therefore, the age of Dharam = 2 x A = 2A yrs.

And the age of Biju sister Ann B/2 yrs

By using the information that is given,

Case (i)

When Ani is older than that of Biju by 3 yrs then A – B = 3 – – – – – – – – (1)

2A−B/2 = 30

4A – B = 60 – – – – – – – – – – – (2)

By subtracting the equations (1) and (2) we get,

3A = 60 – 3 = 57

A = 57/3 = 19

Therefore, the age of Ani = 19 yrs

And the age of Biju is 19 – 3 = 16 yrs.

Case (ii)

When Biju is older than Ani,

B – A = 3 – – – – – – – – – (1)

2A − B/2 = 30

4A – B = 60 – – – – – – – – – (2)

Adding the equation (1) and (2) we get,

3A = 63

A = 21

Therefore, the age of Ani is 21 yrs

And the age of Biju is 21 + 3 = 24 yrs.

**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 Sangam have Rs A with him and Reuben have Rs B with him.

Using the information that is given we get,

A + 100 = 2(B – 100) ⇒ A + 100 = 2B – 200

Or A – 2B = -300 – – – – – – – (1)

And

6(A – 10) = ( B + 10 )

Or 6A – 60 = B + 10

Or 6A – B = 70 – – – – – – (2)

When equation (2) is multiplied by 2 we get,

12A – 2B = 140 – – – – – – – (3)

When equation (1) is subtracted from equation (3) we get,

11A = 140 + 300

11A = 440

⇒ A = 440/11 = 40

Using A =40 in equation (1) we get,

40 – 2B = -300

40 + 300 = 2B

2B = 340

B = 170

Therefore, Sangam had Rs 40 and Reuben had 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.**

**Solution:**

Let the speed of the train be A km/hr and the time taken by the train to travel a distance be N hours and the distance to travel be X hours.

Speed of the train = Distance travelled by train / Time taken to travel that distance

A = N (distance)/ X (time)

Or, N = AX – – – – – – – – – – – (1)

Using the information that is given, we get:

(A+10) = X/(N-2)

(A + 10) (N – 2) = X

AN + 10N – 2A – 20 = X

By using the equation (1) we get,

– 2A + 10N = 20 – – – – – – – – – – (2)

(A-10) = X/(N+3)

(A – 10) (N + 3) = X

AN – 10N + 3A – 30 = X

By using the equation (1) we get,

3A – 10N = 30 – – – – – – – – – (3)

Adding equation (2) and equation (3) we get,

A = 50

Using the equation (2) we get,

(-2) x (50) + 10N = 20

-100 +10N = 20

=> 10N = 120

N = 12hours

From the equation (1) we get,

Distance travelled by the train, X = AN

= 50 x 12

= 600 km

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

**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 rows be A and the number of students in a row be B.

Total number of students = Number of rows x Number of students in a row

=AB

Using the information, that is given,

**First Condition:**

Total number of students = (A – 1) ( B + 3)

Or AB = ( A – 1 )(B + 3) = AB – B + 3A – 3

Or 3A – B – 3 = 0

Or 3A – Y = 3 – – – – – – – – – – – – – (1)

**Second condition:**

Total Number of students = (A + 2 ) ( B – 3 )

Or AB = AB + 2B – 3A – 6

Or 3A – 2B = -6 – – – – – – – – – (2)

When equation (2) is subtracted from (1)

(3A – B) – (3A – 2B) = 3 – (-6)

-B + 2B = 3 + 6B = 9

By using the equation (1) we get,

3A – 9 =3

3A = 9+3 = 12

A = 4

Number of rows, A = 4

Number of students in a row, B = 9

Number of total students in a class = AB = 4 x 9 = 36

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

**Solution:**

Given,

∠C = 3 ∠B = 2(∠B + ∠A)

∠B = 2 ∠A+2 ∠B

∠B=2 ∠A

∠A – ∠B= 0- – – – – – – – – – – – (i)

We know, the sum of all the interior angles of a triangle is 180^{O}.

Thus, ∠ A +∠B+ ∠C = 180^{O}

∠A + ∠B +3 ∠B = 180^{O}

∠A + 4 ∠B = 180^{O}– – – – – – – – – – – – – – -(ii)

Multiplying 4 to equation (i) , we get

8 ∠A – 4 ∠B = 0- – – – – – – – – – – – (iii)

Adding equations (iii) and (ii) we get

9 ∠A = 180^{O}

∠A = 20^{O}

Using this in equation (ii), we get

20^{O}+ 4∠B = 180^{O}

∠B = 40^{O}

3∠B =∠C

∠C = 3 x 40 = 120^{O}

Therefore, ∠A = 20^{O}

∠B=40^{O}

∠C = 120^{O}

**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.**

**Solutions: **

Given,

5x – y = 5

=> y = 5x – 5

Its solution table will be.

Also given,3x – y = 3

y = 3x – 3

The graphical representation of these lines will be as follows:

From the above graph we can see that the triangle formed is ∆ABC by the lines and the y axis. Also the coordinates of the vertices are A(1,0) , C(0,-5) and B(0,-3).

**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) x/a – 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**

**Solutions:**

**(i) px + qy = p – q……………(i)**

**qx – py = p + q……………….(ii)**

Multiplying p to equation (1) and q to equation (2), we get

p^{2}x + pqy = p^{2} − pq ………… (iii)

q^{2}x − pqy = pq + q^{2} ………… (iv)

Adding equation (iii) and equation (iv),we get

p^{2}x + q^{2} x = p^{2} + q^{2}

(p^{2} + q^{2} ) x = p^{2} + q^{2}

x = (p^{2} + q^{2})/ p^{2} + q^{2 }= 1

From equation (i), we get

p(1) + qy = p – q

qy = p-q-p

qy = -q

y = -1

**(ii) ax + by= c…………………(i)**

**bx + ay = 1+ c………… ..(ii)**

Multiplying a to equation (i) and b to equation (ii), we obtain

a^{2}x + aby = ac ………………… (iii)

b^{2}x + aby = b + bc…………… (iv)

Subtracting equation (iv) from equation (iii),

(a^{2} – b^{2}) x = ac − bc– b

x = (ac − bc– b)/ (a^{2} – b^{2})

x = c(a-b) –b / (a^{2}+b^{2})

From equation (i), we obtain

ax +by = c

a{c(a−b)−b)/ (a^{2} – b^{2})} +by=c

ac(a−b)−ab/ (a^{2} – b^{2})+by=c

by=c–ac(a−b)−ab/(a^{2} – b^{2})

by=abc – b^{2 }c+ab/a^{2}-b^{2}

^{y = c(a-b)+a/}a^{2}-b^{2}

**(iii) x/a – y/b = 0**

**ax + by = a ^{2} + b^{2}**

x/a – y/b = 0

=> bx − ay = 0 ……. (i)

ax + by = a^{2} + b^{2} …….. (ii)

Multiplying a and b to equation (i) and (ii) respectively, we get

b^{2}x − aby = 0 …………… (iii)

a^{2}x + aby = a ^{3} + ab^{3} …… (iv)

Adding equations (iii) and (iv), we get

b^{2}x + a^{2}x = a ^{3} + ab^{2}

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

Using equation (i), we get

b(a) − ay = 0

ab − ay = 0

ay = ab,

y = b

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

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

(a + b) y + (a – b) x = a^{2}− 2ab − b^{2} …………… (i)

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

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

Subtracting equation (ii) from equation (i), we get

(a − b) x − (a + b) x = (a ^{2} − 2ab − b ^{2}) − (a^{2} + b^{2})

x(a − b − a − b) = − 2ab − 2b^{2}

− 2bx = − 2b (b + a)

x = b + a

Substituting this value in equation (i), we get

(a + b)(a − b) +y (a + b) = a^{2}− 2ab – b^{2}

a^{2} − b^{2} + y(a + b) = a^{2}− 2ab – b^{2}

(a + b) y = − 2ab

y = -2ab/(a+b)

**(v) 152x − 378y = − 74**

**76x − 189y = − 37**

x =(189y-137)/76……………..…(i)

− 378x + 152y = − 604

− 189x + 76y = − 302 ………….. (ii)

Using the value of x in equation (ii), we get

−189(189y−37/76)+76y=−302

− (189)^{2}y + 189 × 37 + (76)^{2} y = − 302 × 76

189 × 37 + 302 × 76 = (189)^{2} y − (76)^{2}y

6993 + 22952 = (189 − 76) (189 + 76) y

29945 = (113) (265) y

y = 1

Using equation (i), we get

x = (189-37)/76

x = 152/76 = 2

**8. ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.**

**Solution:**

It is known that the sum of the opposite angles of a cyclic quadrilateral is 180^{o}

Thus, we have

∠C +∠A = 180

4y + 20− 4x = 180

− 4x + 4y = 160

x − y = − 40 ……………(1)

And, ∠B + ∠D = 180

3y − 5 − 7x + 5 = 180

− 7x + 3y = 180 ………..(2)

Multiplying 3 to equation (1), we get

3x − 3y = − 120 ………(3)

Adding equation (2) to equation (3), we get

− 7x + 3x = 180 – 120

− 4x = 60

x = −15

Substituting this value in equation (i), we get

x − y = − 40

-y−15 = − 40

y = 40-15

= 25

∠A = 4y + 20 = 20+4(25) = 120°

∠B = 3y − 5 = − 5+3(25) = 70°

∠C = − 4x = − 4(− 15) = 60°

∠D = 5-7x

∠D= 5− 7(−15) = 110°

Hence, all the angles are measured.

The last exercise of the chapter contains 8 questions the solutions to which are provided in NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.7. This has been created by subject experts. The stepwise solutions to these questions will help the students in case of any doubts.

### Key Features of NCERT Solutions Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.7

- The questions are solved by the subject experts
- The answers to the questions are accurate.
- The solutions to all the questions in the textbook are provided here.
- NCERT Solutions for Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables Exercise 3.7 can help the students to score well in the examinations.