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Given below are the

**Class 10 Maths** Problems for Linear equation

a) Concepts questions

b) Calculation problems

c) Multiple choice questions

d) Long answer questions

e) Fill in the blank's

**Question 1)** In a cyclic quadrilateral ABCD,Find the four angles.

a) $\angle A = (2x + 4)$, $\angle B = (y + 3)$, $\angle C = (2y + 10)$, $\angle D = (4x -5)$.

b) $\angle A = (2x - 1)$, $\angle B = (y + 5)$, $\angle C = (2y + 15)$ and $\angle D = (4x -7)$

Solution
a) $\angle A = (2x + 4)$, $\angle B = (y + 3)$, $\angle C = (2y + 10)$, $\angle D = (4x -5)$

In a cyclic quadrilateral, Opposite angles are supplementary.

$\angle A + \ angle C = 180^0$ and $\ angle B + \angle D = 180^0$

So 2x+4+2y+10=180 or x+y=83

y+3+4x-5=180 or y+4x=182

Solving the above equation by Substitution method

x=33 and y=50

So Angles are

70^{0},53^{0},110^{0},127 ^{0}

b) $\angle A = (2x - 1)$, $\angle B = (y + 5)$, $\angle C = (2y + 15)$ and $\angle D = (4x -7)$

Solving similarly, we get

65^{0}, 55^{0}, 115^{0}, 125^{0}

**Question 2)**Given below are three equations. Two of them have infinite solutions and two have a unique solution. State the two pairs:

3x - 2y = 4

6x + 2y = 4

9x -6y = 12

Solution
For Infinite solution:

We have $\frac{a_1}{a_2} = \frac {b_1}{b_2} = \frac {c_1}{c_2}$ is the equation for infinite solution.

Equation 1 and 3 is satisfying the condition

3x-2y=4; 9x-6y=12

3/9 = -2/-6 = 4/12

This also means equation 1 and equation 3 are same

Unique solution:

We have a1/a2 is not equal to b1/b2 is the equation for the unique solution.

Equation 1 and 2 and Equation 2 and 3 is satisfying the condition

3x-2y=4; 6x+2y=4 & 6x+2y=4; 9x-6y=1

3/6 is not equal to -2/2 & 6/9 is not equal to 2/-6.

**Question 3)** Find the values of a and b for which the following system of linear equations has infinite number of solutions:

2x + 3y = 7

2ax + (a + b)y = 28.

Solution
We have

$\frac{a_1}{a_2} = \frac {b_1}{b_2} = \frac {c_1}{c_2}$

$\frac{2}{2a} = \frac {3}{a+b} = \frac {7}{28}$

Solving this , we get a=4 and b=8

**Question 4)**Find the values of a and b for which the following system of linear equations has infinite number of solutions:

2x -3y = 7

(a + b)x - (a + b- 3)y = 4a + b.

Solution
We have

$\frac{a_1}{a_2} = \frac {b_1}{b_2} = \frac {c_1}{c_2}$

$\frac{2}{a+b} = \frac {-3}{a+b-3} = \frac {7}{4a+b}$

$\frac{2}{a+b} = \frac {-3}{a+b-3}$

a+b+6=0 -(1)

$\frac {-3}{a+b-3} = \frac {7}{4a+b}$

5a-4b+21=0 -(2)

Solving (1) and (2) by substitution method we get

a=-5,b=-1

**Question 5)** In a ABC, $\angle A = x$, $\angle B = (3x -2)$, $\angle C = y$ Also $\angle C - \angle B = 9$. Find the three angles.

Solution
In a triangle,sum of angles is equal to 360^{0}

So

x+ (3x-2) + y=180 or 4x+y=182 -(1)

Also given

$\angle C - \angle B = 9$ or y-(3x-2)=9 or y-3x=7 or y=7+3x

Substituting this in (1)

4x+ 7+3x=182

7x=175

x=25

So angles are

25,73, 82

**Question 6)** Find the value of k for which the system of equations x + 2y -3 = 0 and ky + 5x + 7 = 0 has a unique solution.

Solution
For unique solution

$\frac {a_1}{a_2} \neq \frac {b_1}{b_2}$

Subsituting the values

$\frac {1}{5} \neq \frac {2}{k}$

$k\neq10$

**Question 7)** For what value of a the system of linear equations 2x + 3y = 7 and (a -1)x + (a + 1)y = 3a - 1 represent parallel lines.

Solution
For parallel lines

$\frac{a_1}{a_2} = \frac {b_1}{b_2} = \frac {c_1}{c_2}$

$\frac{2}{a-1} = \frac {3}{a+1} = \frac {7}{3a-1}$

or a=5

**Question 8)** If the lines x + 2y + 7 = 0 and 2x + ky + 18 = 0 intersect at a point, then find the value of k.

Solution
For unique solution

$\frac {a_1}{a_2} \neq \frac {b_1}{b_2}$

Subsituting the values

$\frac {1}{2} \neq \frac {2}{k}$

$k\neq4$

**Question 9)** Is x = 5, y = -5 a solution of the linear equation 3x + 2y - 5 = 0?

Solution
3(5) +2(-5) -5=0

So x=5 and y=-5 is solution of the equation

**Question 10)** The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. Find the present ages of the son and the father.

Solution
let Father age is y and son age is x

First condition

y=6x

Second condition

y+4=4(x+4)

So we have

6x+4=4x+16

x=6 and y=36

**Question 11)** Replace p by an appropriate number so that the following system of equations has a unique solution.

2x + 3y = 5

px +9y = 12

**Question 12)** For what value of k, the following pair of linear equations has infinitely many solutions?

i) kx + 4y - (k + 8) = 0

4x + ky + 4 = 0

ii) 2x + 3y = 4

(k + 2)x + 6y = 3k + 2

Solution
For infinite many solutions

$\frac{a_1}{a_2} = \frac {b_1}{b_2} = \frac {c_1}{c_2}$

i)

$\frac{k}{4} = \frac {4}{k} = \frac {-(k+8)}{4}$

Taking first two

k^{2}=16

or k=+4 or -4

for k=4, the condition becomes

$\frac{4}{4} = \frac {4}{4} \neq \frac {-12}{4}$

So it is not correct

For k=-4, the condition becomes

$\frac{-4}{4} = \frac {4}{-4} = \frac {-4}{4}$

So k=-4 is the solution

ii)
$\frac{2}{k+2} = \frac {3}{6} = \frac {4}{3k+2}$

Taking First two

k+2=4

or k=2

So condition becomes

$\frac{2}{4} = \frac {3}{6} = \frac {4}{8}$

So it is the correct solution

**Question 13)** For what value of k, will the following system of equations have no solutions?

i) (3k + 1) x + 3y = 2

(k

^{2} + 1) x + (k - 2) y = 5

ii)3x + y = 1

(2k - 1) x + (k - 1) y = 2k + 1

iii)
2x + ky = 11

5x -7y = 5

**Question 14)** For what value of k, does the pair of equations given below has a unique solution?

2x + ky = 6, 4x + 6y = 0

**Question 15)** Find the value of k for which the following system of linear equations has infinite solutions:

i) x + (k + 1) y = 5

(k + 1) x + 9y = 8k – 1

ii)
8x + 5y = 9

kx + 10y = 18

iii) 2x + 3y = k

(k – 1) x + (k + 2) y = 3k

**Question 16)** For what value of α, the system of equations

αx + 3y = α – 3

12x + αy = α

will have no solution?

**Question 17)** Find the value of k for which the system

kx + 2y = 5

3x + y = 1

has (i) a unique solution, and (ii) no solution.

**Question 18)** Find the value of k for which the system

2x + ky = 1

3x - 5y = 7

has (i) a unique solution, and (ii) no solution.

**Question 19)**find the value of a and b for which the following system of equations has infinitely many solutions:

2x - (2a + 5) y = 5

(2b + 1) x – 9y = 15

**Question 20)** Find the value of a for which the following system of equations has infinitely many solutions:

2x + 3y - 7 = 0

(a – 1) x + (a + 1) y = (3a – 1)

Answer

16) -6

17) k

≠6 , k=6

18) k≠ -10/3 , k=10/3

19) (-1, 5/2)

20) a=5

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