Haryana State Board HBSE 10th Class Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables Important Questions and Answers.

## Haryana Board 10th Class Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Short/Long Answer Type Questions

Question 1.

Solve the following pair of linear equation by graphical method : 4x – 5y = 20, and 3x + 5y = 15, with the help of this find the value of m while 4x + 3y = m.

Solution :

The given equations are:

4x – 5y = 20 …………(1)

And 3x + 5y = 15 …………..(2)

For representation of these equations graphically, we draw the graphs of these equations as follows:

4x – 5y = 20

y = \(\frac{20-4 x}{-5}\)

y = \(\frac{4 x-20}{5}\)

We put the different values of x in this equation, then we get different values of y and we prepare the table of x, y for the equaiton 4x – 5y = 20.

x | 5 | 10 | – 1 |

y | 0 | 4 | – 4.8 |

And 3x + 5y = 15

â‡’ y = \(\frac{15-3 x}{5}\)

We put the different values of x in this equation, then we get different values of y and we prepare the table of x, y for the equation 3x + 5y = 15.

x | 0 | 5 | 1 |

y | 3 | 0 | 2.4 |

Now, we plot the values of x hind y from tables 1 and 2 on the graph paper and we draw the graphs of the equations 1 and 2, those passes through these values.

Observe that we get two straight lines which intersect each other at point (5, 0). Hence, x = 5, y = 0 is the required solution put these values of x and y in the equation 4x + 3y = m, we get

4 Ã— 5 + 3 Ã— 0 = m

â‡’ 20 + 0 = m

â‡’ m = 20

Question 2.

Solve the following pair of linear equations by graphical method :

3x – 5y = – 1 and 2x – y = – 3. Thus find the value of A in the relation (x + y)^{2} = A

Solution :

The given equation are :

3x – 5y = – 1

And 2x – y = – 3 …………(2)

For representation of these equations graphically, we draw the graphs of these equations are follows:

3x – 5y = – 1

â‡’ y = \(\frac{-1-3 x}{-5}\)

â‡’ y = \(\frac{3 x+1}{5}\)

We put the different values of x in this equation, then we get different values of y and we prepare the table of x and y for the equation 3x – 5y = – 1.

x | 3 | 5 | 6 |

y | 2 | 3.2 | 3.8 |

And 2x – y = – 3

y = \(\frac{-3-2 x}{-1}\)

= 3 + 2x

We put the different values of x in this equation, then we get different values of y and we prepare the table of x and y for the equation

2x – y = – 3

x | 1 | 2 | – 2 |

y | 5 | 7 | – 1 |

Now, we plot the values of x and y from tables 1 and 2 on the graph paper and we draws the graphs of equations 1 and 2 those passes through these values.

Observe that we get two straight lines which intersects each other at point (-2, -1) Hence, x = -2, y = -1 is the required solution put the values of x and y in the equation (x + y)^{2} = A, we get

[- 2 + (-1)]^{2} = A

(-3) = A

A = 9

Question 4.

Solve the following pair of linear equations by graphical method 2x + y = 6 and 2x – y = 2. Thus find the value of P in the relation 6x + 7y = P.

Solution :

The given equations are :

2x + y = 6 ………..(1)

And 2x – y = 2 …………(2)

For representation of these equations graphically, we draw the graphs of these equations as follows

2x + y = 6

â‡’ y = 6 – 2x

We put the different values of x in this equation, then we get different galues of y and we prepare the table of x, y for the equation 2x + y = 6.

x | 1 | 2 | 0 |

y | 4 | 2 | 6 |

And 2x – y = 2

â‡’ y = \(\frac{2-2 x}{-1}\)

â‡’ y = 2x – 2

We put the different values of x in this equation, then we get different values of y and we prepare the table of x and y for the equation

2x – y = 2

x | 0 | 1 | 2 |

y | – 2 | 0 | 2 |

Now, we plot the values of x and y from tables 1 and 2 on the graph paper and we draw the graphs of the equations 1 and 2, those passes through these values.

Observe that we get two different straight lines which intersect each other at point (2, 2).

Hence, x = 2, y = 2 is the required solution. Put the values of x and y in the relation

6x + 7y = P we get

6 Ã— 2 + 7 Ã— 2 = P

â‡’ 12 + 14 = P

â‡’ P = 26

Question 5.

For ultarakhand Flood victims two reactions A and B of class 10 contributed â‚¹ 1500. If the contribution of class 10, A was â‚¹ 100 less than that of class 10 B, find the graphically the amount contributed by both reactions.

Solution :

Let contribution of A be â‚¹ x and contribution of B be â‚¹ y.

According to question

x + y = 1500 ………….. (1)

Contribution of A is â‚¹ 100 less than that of B

âˆ´ x + 100 = y ………… (2)

For representation of these equations graphically, we draw the graphs of these equations as follows:

x + y = 1500

â‡’ y = 1500 – x

We put the different values of x in this equation, then we get different values of y and we prepare the table of x, y for the equation

x + y = 1500

x | 400 | 500 | 700 |

y | 1100 | 1000 | 800 |

And x + 100 = y

â‡’ y = x + 100

We put the different values of x in this equation, then we get different value of y and we prepare the table of x, y for the equation

x + 100 = y

x | 400 | 500 | 300 |

y | 500 | 600 | 400 |

Now, we plot the values of x and y from tables 1 and 2 on the graph paper and we draw the graphs of the equations 1 and 2 those passes through these values.

Observe that we get two different straight lines which intersect each other at point (700, 800)

Hence,

x = â‚¹ 700,

y = â‚¹ 800

Therefore, contribution of A is â‚¹ 700 and contribution of B is â‚¹ 800.

Question 6.

If 2x + y = 23 and 4x – y = 19, find the value of (5y – 2x) and (\(\frac {y}{x}\)– 2)

Solution :

The given equations are :

2x + y = 23 ………(1)

4x – y = 19 ……..(2)

Adding equation (1) and (2) we get

6x = 42

â‡’ x = \(\frac {42}{6}\) = 7

Substituting the value of x in equation (1), we get

2 Ã— 7 + y = 23

â‡’ y = 23 – 14 = 9

So, x = 7, y = 9

Putting the values of x and y in 5y – 2x, we get 5 Ã— 9 – 2 Ã— 7 = 45 – 14 = 31.

And putting the values of x and y in the \(\frac {y}{x}\) – 2, we get

\(\frac {9}{7}\) – 2 = \(\frac{9-14}{7}=\frac{-5}{7}\)

Hence, 5y – 2x = 31 and \(\frac {y}{x}\) – 2 = – \(\frac {5}{7}\)

Question 7.

A father’s age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father.

Solution :

Let the age of father be x years and the sum of ages of his children be y years.

According to questions,

x = 3y …….. (1)

After 5 years

Age of father = (x + 5) years

And sum of ages of his children = (y + 10) years

According to the question,

x + 5 = 2(y + 10)

â‡’ x + 5 = 2y + 20

â‡’ x – 2y = 20 – 5

â‡’ x – 2y = 15 ……..(2)

From equation (1), substituting the value of x in the equation, we get

3y – 2y – 15

â‡’ y = 15 years

Putting the value of y in the equation we get

x = 3 Ã— 15 = 45 years

Hence, present age of father = 45 years.

Question 8.

Raghav scored 70 marks in a test, getting 4 marks for each right answer and losing 1 mark for each wrong answer. Had 5 marks been awarded for each correct answer and 2 marks been deducted for each wrong answer, then Raghav would have scored so marks. How many questions were there in the test.

Solution :

Let the number of right answers be x and number of wrong answers be y.

According to question,

4x – y = 70 ………… (1)

And 5x – 2y = 80 ………..(2)

Multiplying equation in (1) by 2 and substracting equ. (2) from equ. (1), we get

â‡’ x = \(\frac {60}{3}\) = 20

Substituting the value of x in the equ. (1) we

4 Ã— 20 – y = 70

â‡’ – y = 70 – 80 = – 10

â‡’ y = 10.

Hence, total number of questions = x + y = 20 + 10 = 30.

Question 9.

In a painting competition of a school a child made Indian national flag whose perimeter was 50 cm. Ita area will be decreased by 6 square em, if length is decreased by 3 cm and breadth is increased by 2 cm then find the dimension of flag.

Solution :

Let the length of the flag be x cm and its breadth be y cm,

so, its area = xy cm^{2} ………..(1)

According to condition (i)

2(x + y) = 50

â‡’ x + y = 25 …….(2)

If length is decreased by 3 cm. then new length is (x – 3) cm. And breadth is increased by 2 cm then new breadth is (y + 2) cm.

New area = (x – 3) (y + 2) cm^{2}

According to condition (ii)

(x – 3) (y + 2) = xy – 6 [Using equ. (1)]

â‡’ xy + 2x – 3y – 6 = xy – 6

â‡’ 2x – 3y = 0 ……(3)

So, the equation of pair is

x + y – 25 = 0 …….(2)

2x – 3y + 0 = 0 …….(3)

By cross multiplication method, we get

â‡’ x = 15 cm and y = 10 cm.

Hence, length of flag is 15 cm and its breadth is 10 cm.

Question 10.

Sumit is 3 times as old as his son. Five years later, he shall be two and half times as old as his son. How old is sumit at present age ?

Solution :

Let present age of sumit be x years and present age of his son be y years.

According to question

x = 3y

x – 3y = 0 …….. (1)

After 5 years later

Age of sumit = (x + 5) years

And age of his son = (y + 5) years

According to question,

x + 5 = 2\(\frac {1}{2}\)(y + 5)

x + 5 = \(\frac {5}{2}\)(y + 5)

2x + 10 – 5y + 25

2x – 5y = 25 – 10

2x – 5y = 15

Multiplying equation (1) by 2 and substracting eqn. (2) from equ. (1), we get

â‡’ y = 15

y = 15 Substituting the value of y in the equ. (1) we

x – 3 Ã— 15 = 0

â‡’ x = 45 years

Hence, present age of sumit = 45 years.

Question 11.

For what value of k, will the following pair of equations have infinitely many solutions :

2x + 3y = 7 and (k + 2) x – 3 (1 – k)y = 5k + 1.

Solution :

The given equations are :

2x + 3y – 7 = 0

And (k + 2)x – 3 (1 – k)y – (5k + 1) = 0

The condition for infinitely many solutions is :

â‡’ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

â‡’ \(\frac{2}{k+2}=\frac{3}{-3(1-k)}=\frac{-7}{-(5 k+1)}\)

â‡’ – 6 (1 – k) = 3k + 6

and – 10k – 2 = – 7k – 14

â‡’ – 6 + 6k = 3k + 6

and – 10k + 7k = – 14 + 2

â‡’ 3k = 12 and – 3k = – 12

â‡’ k = 4 and k = \(\frac {-12}{-3}\) = 4

Hence k = 4

Question 12.

For what values of m and n the following system of linear equations has infinitely many solutions:

3x + 4y = 12 and (m + n)x + 2 (m -n)y = 5m – 1.

Solution :

The given equations are :

3x + 4y – 12 = 0 ………(1)

(m + n)x + 2(m – n)y – (5m – 1) = 0

The condition for infinitely many solution is :

â‡’ 6m – 6n = 4m + 4n

and 15m – 3 = 12m + 12n

â‡’ 6m – 4m = 6n + 4n

and 15m – 12m – 12n = 3

2m = 10n and 3m – 12n = 3

â‡’ m = 5n and m – 4n = 1

Putting the value of m in (m – 4n = 1) we get

5n – 4n = 1

n = 1

Putting the value of n is m = 5n, we get

m = 5 Ã— 1 = 5

Hence, m = 5, n = 1.

Fill in the Blanks

Question 1.

If \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) the pair of equations a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0 is …………

Solution :

Inconsistent

Question 2.

The graph of a pair of linear equations in two variables is represented by two ……………. lines.

Solution :

Straight

Question 3.

Every solution of a linear equation in two variables is a ………. on the line representing it.

Solution :

Point

Question 4.

A pair of linear equation in two variables can be represented and solved by the algebraic and ………….. methods.

Solution :

Graphical

Question 5.

If the lines are ……….. then the pair of linear equations in two variables has no solution.

Solution :

Parallel

Question 6.

The general form of a pair of ……………… equations in two variables x and y is a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 where \(a_1^2+b_1^2\) â‰ 0, \(a_2^2+b_2^2\) â‰ 0.

Solution :

Linear.

Multiple Choice Questions

Question 1.

The value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 is inconsistent is :

(a) – \(\frac {14}{3}\)

(b) \(\frac {2}{5}\)

(c) 5

(d) 10

Solution :

(d) 10

Given equation are :

x + 2y – 3 = 0 And

5x + ky + 7 = 0

The condition for inconsistant is:

\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

Taking first two order, we get

\(\frac{a_1}{a_2}=\frac{b_1}{b_2}\)

â‡’ \(\frac{1}{5}=\frac{2}{k}\)

â‡’ k = 10

Question 2.

The value of k for which the system of equations.

2x + 3y = 5

4x + ky = 10 has infinite number of solutions, is :

(a) 3

(b) 6

(c) 0

(d) 1

Solution :

(b) 6

For inifinite number of solutions

\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

\(\frac{2}{4}=\frac{3}{k}=\frac{5}{10}\)

â‡’ k = \(\frac {12}{2}\) = 6

So correct option (b).

Question 3.

The value of k for which the given system of equations has a unique solution is :

kx + 2y – 5 = 0

x + 3y – 2 = 0

(a) â‰ \(\frac {3}{2}\)

(b) â‰ \(\frac {2}{3}\)

(c) â‰ \(\frac {2}{3}\)

(d) â‰ \(\frac {3}{2}\)

Solution :

(b) â‰ \(\frac {2}{3}\)

kx + 2y – 5 = 0

x + 3y – 2 = 0

For unique solution

\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

â‡’ \(\frac{k}{1} \neq \frac{2}{3}\)

â‡’ k â‰ \(\frac {2}{3}\)

So correct option (b).

Question 4.

The value of k for which the system of equations

2x + 4y = 5

6x + ky = 9 has no solution, is :

(a) 10

(b) 9

(c) 12

(d) 11.

Solution :

(c) 12

2x + 4y = 5

6x + ky = 9

For no solution

\(\frac{a_1}{a_2}=\frac{b_1}{b_2}\)

â‡’ \(\frac{2}{6}=\frac{4}{k}\)

â‡’ k = \(\frac{6 \times 4}{2}\) = 12

So correct option (c).

Question 5.

The values of a and b for which the following system of equations

2x – 3y = 7

(a + b)x – (a + b – 3)y = 4a + b has infinitely many solutions, are :

(a) a = -5, b = 2

(b) a = -2, b = -5

(c) a = -5, b = -1

(d) a = -1, b = -5.

Solution :

(c) a = -5, b = -1

2x – 3y = 7

(a + b)x – (a + b – 3)y

= 4a + b

For inifinitily many solutions 2 -3

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

By solving a = -5, b = – 1 so correct choice is (c).

Question 6.

For which values of P, will the lines represented by the following pair of linear equations be parallel 3x – y – 5 = 0, 6x – 2y – P = 0.

(a) all real values expect 10

(b) 10

(c) \(\frac {5}{2}\)

(d) \(\frac {1}{2}\)

Solution :

(a) all real values expect 10

The given equation are :

3x – y – 5 = 0 …………….(1)

6x – 2y – P = 0 …………….(2)

The condition for parallel lines is

So, P can have any real values other than 10.