Here you will find Pair of Linear Equations in Two Variables Class 10 Extra Questions Maths Chapter 3 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction.

## Extra Questions for Class 10 Maths Pair of Linear Equations in Two Variables with Answers Solutions

**Extra Questions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables with Solutions Answers**

### Pair of Linear Equations in Two Variables Class 10 Extra Questions Objective Type

Question 1.

If the pair of linear equations x – y – k = 0 and 6x – 2y – 3 = 0 represent an infinite solution, then the value of k is:

(a) k = 1

(b) k = 2

(c) k = 0

(d) No value of k

Answer:

(d) No value of k

Question 2.

If the pair of equations 6x + 5y = 4 and 12x + py = -8 has no solution, then the value of p is:

(a) 9

(b) 10

(c) 7

(d) 6

Answer:

(b) 10

Question 3.

The solution of the equation 2x + 3y = 18 and x – 2y = 2 is

(a) x = 2, y = 6

(b) x = 6, y = 2

(c) x = – 6, y = -2

(d) None of these

Answer:

(b) x = 6, y = 2

Question 4.

Find the values of x and y in the following equations:

x – 3y = 8 and 5x + 3y = 10

(a) x = 3, y = – \(\frac {5}{3}\)

(b) x = -3, y = \(\frac {5}{3}\)

(c) x = -3, y = – \(\frac {5}{3}\)

(d) None of these

Answer:

(a) x = 3, y = – \(\frac {5}{3}\)

### Pair of Linear Equations in Two Variables Class 10 Extra Questions Very Short Answer Type

Question 1.

Solve the following pair of linear equations by the substitution method.

(i) x + y = 14, x – y = 4

(ii) 3x – y = 3, 9x – 3y = 9

(iii) √2x + √3y = 0, √3x – √8y = 0

Solution:

(i) We have,

x + y = 14 …..(i)

and x – y = 4 ……(ii)

From Eq. (ii), y = x – 4 …(iii)

Substituting y from Eq. (iii) in Eq. (i), we get

x + x – 4 = 14

⇒ 2x = 18.

⇒ x = 9

On substituting x = 9 in Eq. (iii), we get

y = 9 – 4 = 5

⇒ y = 5

x = 9, y = 5

(ii) We have,

3x – y = 3 …(i)

and 9x – 3y = 9 …(ii)

From Eq. (i)y = 3x – 3 …. (iii)

On substituting y from Eq. (iii) in Eq. (ii), we get

9x – 3(3x – 3) = 9

⇒ 9 = 9

It is a true statement. Hence, every solution of Eq. (i) is a solution of Eq. (ii) and vice-versa.

On putting x = k in Eq. (i), we get

3k – y = 3 ⇒ y = 3k – 3

∴ x = k, y = 3k – 3 is a solution for every real k.

Hence, infinitely many solutions exist.

(iii) We have,

√2x + √3y = 0 …..(i)

and √3x – √8y = 0 …(ii)

From Eq. (ii),

√8y = √3x

⇒ y = \(\frac{\sqrt{3} x}{\sqrt{8}}\) …..(iii)

On substituting y from Eq. (iii) in Eq. (i), we get

⇒ √2 x √8x + 3x = 0

⇒ √16x + 3x = 0

⇒ 4x + 3x = 0

⇒ 7x = 0

⇒ x = 0

Putting x = 0 in Eq. (iii), y = 0

Question 2.

Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \(\frac {1}{2}\), if we only add 1 to the denominator. What is the fraction ?

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?

Solution:

(i) Let the fraction be \(\frac {x}{y}\)

According to the given conditions,

⇒ x + 1 = y – 1; 2x = y + 1 ……..(i)

⇒ x – y = -2 and 2x – y = 1 ……..(ii)

On subtracting Eq. (i) from Eq. (ii),

(2x – y) – (x – y) = 1 + 2

⇒ x = 3

On substituting x = 3 in Eq. (i),

3 – y = -2 ⇒ y = 5

Hence, the fraction is \(\frac {3}{5}\)

(ii) Let present age of Nuri = x years. Present age of Sonu = y years

According to the given conditions,

Five years ago,

x – 5 = 3(y – 5)

⇒ x – 3y = -10 …(i)

Ten years later,

x + 10 = 2(y + 10)

⇒ x – 2y = 10 …(ii)

On subtracting Eq. (i) from Eq. (ii),

(x – 2y) – (x – 3y) = 10 + 10

⇒ – 2y + 3y = 20

⇒ y = 20

From Eq. (ii), substituting y = 20, we get

⇒ x = 2y + 10 = 2 × 20 + 10

⇒ x = 50

Therefore, present age of Nuri = 50 years and present age of Sonu = 20 years

Question 3.

For What value of k, will the pair of linear equations kx + y = k^{2} and x + ky = 1 have infinitely many solutions ?

Solution:

Given equation are

kx + y = k^{2} ……(i)

and x + ky = 1 ……(ii)

have infinitely many solution

∴ k^{2} = 1 = k ⇒ ± 1

and k^{3} = 1 ⇒ k = 1

Hence k = 1

### Pair of Linear Equations in Two Variables Class 10 Extra Questions Short Answer Type

Question 1.

Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) \(\frac {4}{x}\) + 3y =14 and \(\frac {3}{x}\) – 4y = 23

(ii)

Solution:

(i) We have,

\(\frac {4}{x}\) + 3y =14 and \(\frac {3}{x}\) – 4y = 23

On putting \(\frac {1}{x}\) = X, we get

4x + 3y = 14 …..(i)

and 3x – 4y = 23 … (ii)

On multiplying Eq. (i) by 4 and Eq. (ii) by 3 and then adding, we get

16X + 9X = 4 × 14 + 3 × 23

⇒ 25X = 56 + 69

⇒ 25X = 125 = X = 5

Then, \(\frac {1}{x}\) = 5

⇒ x = \(\frac {1}{5}\) From Eq. (i), substituting x = 5, we get

4 × 5 + 3y = 14 = 3y = 14 – 20

⇒ 3y = -6 = y = -2

Hence, x = and y = -2

⇒ x – 1 = 3 and y – 2 = 3

⇒ x = 3 + 1

and y = 3 + 2

⇒ x = 4 and y = 5

Question 2.

Prove that the pair of linear equation \(\frac {22}{7}\)x + \(\frac {22}{7}\)y = 7 and y = 7 and 9x – 10y = 14 is coinsistent find its solution by method cross-multiplication.

Solution:

Given equation are

\(\frac {22}{7}\)x + \(\frac {22}{7}\)y = 7

⇒ 3x + 5y = 14 ….(i)

and 9x – 10y = 14 …(ii)

If \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\) then equation are coinsistance.

Here a_{1} = 3, b_{1} = 5, c_{1} = – 14

and a_{2} = 9, b_{2} = -10, c_{2} = -14

Question 3.

Solve the following pair of linear equations.

x ≠ 0, y ≠ 0

Solution:

Suppose \(\frac{1}{x-y}\) = a

and \(\frac{1}{x+y}\) = b

The reduced equation will be

15a + 22b = 5 ….(i)

and 40a + 55b = 13 (ii)

Multiply equation (i) by 8 and equation (ii) by 3, we get

120a + 176b = 40

and 120a + 165b = 39

on substracting 11b = 1

∴ x – y = 5 … (iv)

Solving Eq. (iii) and Eq. (iv) we get.

x = 8 and y = 3

Question 4.

The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2. Find the number.

Solution:

Let unit digit of the number bex and ten’s digit of the number be y

∴ Number = 10y + x

and Number obtain by reversing the digits is 10x + y

∴ By Ist condition,

(10y + x) + (10x + y) = 66

⇒ 11x + 11y = 66

∴ x + y = 6 ……(i)

By IInd condition, x – y = 2 (ii)

On adding (i) & (ii)

2x = 8

∴ x = 4

from equation (1), y = 6 – 4 = 2

Hence, required Number be 2 × 10 + 4 = 24

### Pair of Linear Equations in Two Variables Class 10 Extra Questions Short Answer Type 2

Question 1.

7 chairs and 4 tables for a classroom cost ₹ 7000 while 5 chairs and 6 tables cost ₹ 5080. Find the cost of each chair and that of each table.

Solution:

Let the cost of each chair be ₹ x and that of each table be ₹ y.

Then,

7x + 4y = 7000 …(i)

5x + 3y = 5080 …(ii)

On multiplying (i) by 3, (ii) by 4 and subtracting, we get:

(21x – 20x) = (21000 – 20320)

⇒ x = 680.

On substituting x = 680 in (i), we get:

(7000 – 4760) 4y = 7000

⇒ 4y = (7000 – 4760)

⇒ y = 560.

∴ cost of each chair = ₹ 680 and cost of each table = ₹ 560.

Question 2.

The sum of a two-digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.

Solution:

Let the tens and units digits of the required number be x and y respectively.

Then, the number = (10x + y).

The number obtained on reversing the digits = (10y + x).

∴ (10y + x) + (10x + y) = 99

= 11(x + y) = 99

⇒ x + y = 9.

Also, (x – y) = ± 3.

∴ x + y = 9 …..(i)

x – y = 3 …..(ii)

x + y = 9 ….(iii)

x – y = -3 …..(iv)

From (i) and (ii), we get: x = 6, y = 3.

From (iii) and (iv), we get: x = 3, y = 6.

Hence, the required number is 63 or 36.

Question 3.

The monthly incomes of A and B are in the ratio 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves ₹ 5000 per month, find the monthly income of each.

Solution:

Let the monthly incomes of A and B be ₹ 8x and ₹ 7x respectively, and let their expenditures be ₹ 19y and ₹ 16y respectively.

Then, A’s monthly savings = ₹ (8x – 19y).

And, B’s monthly savings = ₹ (7x – 16y).

But, the monthly saving of each is ₹ 5000.

∴ 8x – 19y = 5000 …..(i)

7x – 16y = 5000…..(ii)

Multiplying (ii) by 19, (i) by 16 and subtracting the results, we get:

(19 × 7 – 16 × 8)x = (19 × 5000 – 16 × 5000)

⇒ (133 – 128)x = (19 – 16) 5000

⇒ 5x = 15000

⇒ x = 3000.

∴ A’s monthly income = ₹ (8x) = ₹ (8 × 3000) = ₹ 24000.

And, B’s monthly income = ₹ (7x) = ₹(7 × 3000) = ₹ 21000.

Question 4.

A two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.

Solution:

Let the tens and units digits of the required number be x and y respectively.

Then, xy = 14.

Required number= (10x + y).

Number obtained on reversing its digits = (10y + x).

∴ (10x + y) + 45 = (10y + x)

⇒ 9(y – x) = 45

⇒ y – x = 5 …(i)

Now, (y + x)^{2} – (y – x)^{2} = 4xy

= y + x = 9 …(ii)

[∴ digits are never negative)

2y = 14 ⇒ y = 7.

Putting y = 7 in (ii), we get:

7 + x = 9 ⇒ x = (9 – 7) = 2.

x = 2 and y = 7.

Hence, the required number is 27.

Question 5.

The sum of the numerator and the denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes \(\frac {1}{2}\). Find the fraction.

Solution:

Let the required fraction be \(\frac {x}{y}\).

Then,

∴ x + y = 12 ……(i)

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

⇒ 2x = y + 3

⇒ 2x – y = 3 …(ii)

Adding (i) and (ii), we get:

3x = 15 x = 5.

Putting x = 5 in (i), we get:

5 + y = 12 y = (12 – 5) = 7.

Thus, x = 5 and y = 7.

Hence, the required fraction is \(\frac {5}{7}\).

Question 6.

Five year ago, A was thrice as old as B and ten years later A shall be twice as old as B. What are the present ages of A and B ?

Solution:

Let the present ages of B and A be x years and y years respectively. Then,

B’s age 5 years ago = (x – 5 ) years and A’s age 5 years ago = (y – 5) years.

∴ (y – 5) = 3(x – 5) = 3x – y = 10 …(i)

B’s age 10 years hence = (x + 10) years.

A’s age 10 years hence = (y + 10) years.

∴ (y + 10) = 2(x + 10) = 2x – y = -10 (ü)

On subtracting (ii) from (i), we get:

x = 20.

Putting x = 20 in (i), we get:

(3 x 20) – y = 10 = y = (60 – 10) = 50.

∴ x = 20 and y = 50.

Hence, B’s present age = 20 years and A’s present age = 50 years.

Question 7.

If the length of a rectangle is reduced by 5 units and its breadth is increased by 2 units, then the area of the rectangle is reduced by 80 sq units. However, if we increase its length by 10 units and decrease the breadth by 5 units, its area is increased by 50 sq units. Find the length and breadth of the rectangle.

Solution:

Let the length and breadth of the rectangle be x units and y units respectively.

Then, area of the rectangle = xy sq units

Case I: When the length is reduced by 5 units and the breadth is increased by 2 units.

Then,new length= (x – 5) units

andnew breadth = (y + 2) units.

∴ new area = (x – 5)(y + 2) sq units.

∴ xy – (x – 5)(y + 2) = 80

⇒ 5y – 2x = 70…(i)

Case II: When the length is increased by 10 units and the breadth is decreased by 5 units.

Then, new length= (x + 10) units

andnew breadth = (y – 5) units.

∴ new area = (x + 10)(y – 5) sq units.

∴ (x + 10)(y – 5) – xy = 50

⇒ 10y – 5x = 100

⇒ 2y – x = 20 …….(ii)

On multiplying (ii) by 2 and subtracting esult from (i), we get:

y = 30.

Putting y = 30 in (ii), we get:

(2 × 30) – x = 20 ⇒ 60 – x = 20

⇒ x = (60 – 20) = 40.

∴ x = 40 and y = 30.

Hence, length = 40 units and breadth = 30 units.

Question 8.

8 men and 12 boys can finish a piece of work in 5 days, while 6 men and 8 boys can finish it in 7 days. Find the time taken by 1 man alone and that by 1 boy alone to finish the work…

Solution:

Suppose 1 man alone can finish the work in x days and 1 boy alone can finish it in y days.

Then, 1 man’s 1 day’s work = \(\frac {1}{x}\)

And, 1 boy’s 1 day’s work = \(\frac {1}{y}\)

8 men and 12 boys can finish the work in 5 days

⇒ (8 men’s 1 day’s work) + (12 boys’ 1 day’s work) = \(\frac {1}{5}\)

⇒ \(\frac {8}{x}\) + \(\frac {12}{y}\) = \(\frac {1}{5}\),

⇒ 8υ + 12υ = \(\frac {1}{5}\),

[Where \(\frac {1}{x}\) = υ and \(\frac {1}{y}\) = υ] …..(i)

Again, 6 men and 8 boys can finish the work in 7 days

⇒ (6 men’s 1 day’s work) + (8 boys’ 1 day’s work) = \(\frac {1}{7}\)

∴ one man alone can finish the work in 70 days,

and one boy alone can finish the work in 140 days.

Question 9.

A boat goes 16 km upstream and 24 km downstream in 6 hours. Also, it covers 12 km upstream and 36 km downstream at the same time. Find the speed of the boat in still water and that of the stream.

Solution:

Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr.

Then, speed upstream = (x – y) km/hr

and speed downstream = (x + y) km/hr.

Time taken to cover 16 km upstream

On adding (v) and (vi), we get: 2x = 16 ⇒ x = 8.

On subtracting (vi) from (v), we get:

2y = 8 y = 4.

∴ speed of the boat in still water

= 8 km/hr.

And, speed of the stream = 4 km/hr.

Question 10.

90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid Solution: Find the quantity of each type of acid to be mixed to form the mixture.

Solution:

Let the given solutions be labelled as A and B respectively.

Let x litres of A be mixed with y litres of B. Then,

x + y = 21

Quantity of acid in x litres of A = (90% of x)

⇒ 90x + 97y – 1995 …….(ii)

Multiplying (i) by 90 and subtracting the result from (ii), we get:

7y = 105 ⇒ y = 15.

Putting y = 15 in (i), we get:

x + 15 = 21 = x = 6.

∴ x = 6 and y = 15.

So, 6 litres of 90% solution is mixed with 15 litres of 97% Solution.

Question 11.

On selling a tea-set at 5% loss and a lemon-set at 15% gain, a shopkeeper gains ₹ 84. However, if he sells the tea-set at 5% gain and the lemon-set at 10% gain, he gains ₹ 104. Find the price of the tea-set and that of the lemon-set paid by the shopkeeper.

Solution:

Let the CP of the tea-set and the lemon-set be ₹ x and ₹ y respectively.

On adding (i) and (ii), we get:

5y = 3760 ⇒ y = 752.

Putting y = 752 in (ii), we get:

x + (2 × 752) = 2080

x = (2080 – 1504) = 576.

∴ x = 576 and y = ₹ 752.

Hence, CP of the tea-set = ₹ 576

and CP of the lemon-set = ₹ 752.

Question 12.

A chemist has one solution containing 50% acid and a second one containing 25% acid. How much of each should be used to make 10 litres of a 40% acid solution ?

Solution:

Let x liters of 50% solution be mixed with y liters of 25% Solution. Then,

x + y = 10 and 50% of x + 25% of y = 40% of 10

⇒ x + y 10 and \(\frac {50}{100}\) × x + \(\frac {25}{100}\) × y = \(\frac {40}{100}\) × 10

⇒ x + y 10 and \(\frac {x}{2}\) + \(\frac {y}{4}\) = 4

⇒ x + y = 10 …(i)

and 2x + y = 16 …(ii)

On solving (i) and (ii), we get: x = 6 and y = 4.

∴ 6 litres of 50% solution is to be mixed with 4 litres of 25% Solution.

Question 13.

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

Solution:

Let ∠A = x° and ∠B = y°. Then,

∠C = 3∠B = (3y)°.

Now, ∠A + ∠B + ∠C = 180°

⇒ x + y + 3y = 180°

x + 4y = 180 ……(i)

Also, ∠C = 2(∠A + ∠B)

⇒ 3y = 2(x + y)

⇒ 2x – y = 0 ……(ii)

Multiplying (ii) by 4 and adding the result to (i), we get:

9x = 180 ⇒ x = 20.

Putting x = 20 in (i), we get:

20 + 4y = 180 ⇒ 4y = 160

⇒ y = \(\frac {160}{4}\) = 40.

∴ x = 20 and y = 40.

∴ ∠A = 20°, ∠B = 40°

and ∠C = (3 x 40)° = 120°.