In this page we have *NCERT Solutions for Class 10th Maths:Linear Equations* for
Exercise 3.7 on page 68. Hope you like them and do not forget to like , social_share
and comment at the end of the page.

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.

Let x and y be the ages of Ani and Biju

x -y =3 (Ani is older than Biju)

Or y -x =3 (Biju is older than Ani)

Case 1

x -y =3 -(1)

Now Dharam age =2x

Cathy Age = y/2

Now

2x – y/2 =30

Or 4x -y=60 – (2)

Subtracting (1) from (2)

3x=57

x=19 year

So, y=16 year

Case 2 Biju

y -x =3 -(3)

Now Dharam age =2x

Cathy Age = y/2

Now

2x – y/2 =30

Or 4x -y=60 – (4)

Adding (3) and (4)

3x=63

x=21 year

y=24 years

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)].

Let one have x Rs and other has y Rs, then

According to questions

x + 100 = 2(y – 100)

x-2y = -300 – (1)

y + 10 = 6(x – 10)

6x -y =70 -(2)

Multiplying (2) by 2 and subtracting from 1

11x =140 +300

x=40

Now substituting this in (1)

y=170

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.

Let x be the speed Km/h and y be the distance in Km and t (hour) be the time taken

Then

Speed = distance/time

x= y/t

or y=xt

Now Using the information that is given,

x+10= y/(t-2)

(x+10) (t-2) =d

xt -2x + 10 t-20 =d

Now d=xt, So

-2x+ 10 t=20 – (1)

x- 10= d/(t+3)

(x-10) (t+3) =d

xt-10t -30 +3x=d

3x-10t=30 --(2)

Adding (1) and (2)

x= 50 km/hr.

Substituting this value in (1)

t=12 hour

Now d=xt = 600 km

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.

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

Total number of students:

= Number of rows x Number of students in a row

=xy

Using the information that is given,

Total number of students = (x– 1) (y + 3)

xy = ( x – 1 )(y + 3)

xy= xy – y + 3x – 3

3x – y = 3 – – – – – – – – – – – – – (1)

Total Number of students = (x+ 2 ) ( y – 3 )

Or xy = xy + 2y – 3x – 6

Or 3x – 2y = -6 – – – – – – – – – (2)

Subtracting equation (2) from (1)

y=9

By using the equation (1) we get,

3x – 9 =3

3x = 9+3 = 12

x = 4

Number of rows, x= 4

Number of students in a row, y = 9

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

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

Given,

∠C = 3 ∠B = 2(∠B + ∠A)

3∠B = 2 ∠A+2 ∠B

∠B=2 ∠A

2∠A – ∠B= 0 -- (i)

We know, the sum of all the interior angles of a triangle is 180°.

Thus, ∠ A +∠B+ ∠C = 180°

∠A + ∠B +3 ∠B = 180°

∠A + 4 ∠B = 180 ----(ii)

Multiplying equation (i) by 4 and adding (ii),we get

9 ∠A = 180°

∠A = 20°

Using this in equation (ii), we get

20

∠B = 40°

3∠B =∠C

∠C = 3 x 40 = 120°

Therefore, ∠A = 20°

∠B=40°

∠C = 120°

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.

For 5x -y =5

y=5x -5

x |
0 |
1 |

y |
-5 |
0 |

y= 3x-3

x |
0 |
1 |

y |
-3 |
0 |

Graphical representation

From the above graph we can see that the triangle formed is $\Delta PQR$ (colored in red) by the lines and the y axis. Also, the coordinates of the vertices are P (1,0), Q (0, -5) and R (0, -3)

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) $152x - 378y = - 74$

$-378x + 152y = - 604$

i) px + qy = p – q -(1)

qx – py = p + q -(2)

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

p

x=1

Substituting this value in equation (1)

y=-1

ii) ax + by = c --(1)

bx + ay = 1 + c ---(2)

Multiplying a to equation (i) and b to equation (ii), and then subtracting (1) from (2) ,we get

a

x = [c(a-b) -b]/(a

Putting this value in equation (1), we get

a[c(a-b) -b]/(a

by = c - a[c(a-b) -b]/(a

by = (ca

x = [c(a-b) +a]/(a

iii)

(x/a) – (y/b) =0 => bx -ay=0 -(1)

ax+ by= a

Multiplying (1) by a and (2) by b and then subtracting (1) from (2)

a

y= b

Substituting this in (1)

x=a

iv)

(a + b) y + (a – b) x = a

(x + y)(a + b) = a

(a + b) y + (a + b) x = a

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

(a − b) x − (a + b) x = (a

x(a − b − a − b) = − 2ab − 2b

− 2bx = − 2b (b+a)

x = b + a

Substituting this value in equation (i), we get

(a + b)(a − b) +y (a + b) = a

a

(a + b) y = − 2ab

y=-2ab/(a+b)

152x − 378y = − 74

76x − 189y = − 37

x=(189y-37)/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)

189 × 37 + 302 × 76 = (189)

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

ABCD is a cyclic quadrilateral

Find the angles of the cyclic quadrilateral.

It is know that the sum of the opposite angles of a cyclic quadrilateral is 180

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°

- What is Linear equations
- |
- Linear equations Solutions
- |
- Graphical Representation of Linear equation in one and two variable
- |
- Steps to Draw the Given line on Cartesian plane
- |
- Simultaneous pair of Linear equation
- |
- Algebraic Solution of system of Linear equation
- |
- Simultaneous pair of Linear equation in Three Variable
- |
- Steps to solve the Linear equations

- NCERT Solutions Exercise 3.1
- |
- NCERT Solutions Exercise 3.2
- |
- NCERT Solutions Exercise 3.3,3.4,3.5
- NCERT Solutions Exercise 3.6
- |
- NCERT Solutions Exercise 3.7

- Important Questions
- |
- Linear equations Problems
- |
- Linear equation worksheet
- |
- Linear equation word problems
- |
- Linear equations graphical problems

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.

Thanks for visiting our website.

**DISCLOSURE:** THIS PAGE MAY CONTAIN AFFILIATE LINKS, MEANING I GET A COMMISSION IF YOU DECIDE TO MAKE A PURCHASE THROUGH MY LINKS, AT NO COST TO YOU. PLEASE READ MY **DISCLOSURE** FOR MORE INFO.