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

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
(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
(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
(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
(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 = k2 and x + ky = 1 have infinitely many solutions ?
Solution:
Given equation are
kx + y = k2 ……(i)
and x + ky = 1 ……(ii)
have infinitely many solution

∴ k2 = 1 = k ⇒ ± 1
and k3 = 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 a1 = 3, b1 = 5, c1 = – 14
and a2 = 9, b2 = -10, c2 = -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)
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°.

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