CBSE Important Questions Class 10 Maths Ch3 Linear Equations 2025-26

The chapter Pair of Linear Equations in Two Variables is a key topic in Class 10 Maths Syllabus that combines algebra and geometry. It introduces students to systems of two linear equations and their solutions using different methods. 

The topic forms a foundation for higher-level algebra, coordinate geometry, and systems of equations in later classes. Since it involves both conceptual learning and numerical accuracy, regular practice is the best way to score full marks.

The Pair of Linear Equations in Two Variables Important Questions are designed to help you:

• Understand methods of solving equations systematically.
• Apply concepts to word problems and graphical representations.
• Strengthen accuracy and confidence in solving equations for board exams

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IMPORTANT QUESTIONS Class 10

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Ch3 Pair of Linear Equations In Two Variables Important Questions

1. Harsh correctly solved a pair of linear equations in two variables and found their only point of intersection as (3, -2). One of the lines was x - y = 5.

Which of the following could have been the other line?

1: 3x-3y = 15 

II: 2x-3y = 12 

III: 2x - 3y = 14

a. only I

b. only II

c. only I and II

d. only II and III

Soln. 

(c) only I and II

Explanation: 

To determine which of the given lines could be the other line, we check whether each line passes through the point of intersection (3,−2)(3, -2)(3,−2).

The point (3,−2)(3, -2)(3,−2) satisfies the equation x−y=5x - y = 5x−y=5, which we know is one of the lines.

Now, we will substitute x=3x = 3x=3 and y=−2y = -2y=−2 into each of the other equations and verify:

I: 3x−3y=15

Substitute x=3 and y=−2

3(3)−3(−2)

=9+6

=15

This equation is satisfied. So, Line I is valid.

II: 2x−3y=12

Substitute x=3 and y=−2

2(3)−3(−2)

=6+6

=12

This equation is also satisfied. So, Line II is valid.

III: 2x−3y=14

Substitute x=3 and y=−2:

2(3)−3(−2)

=6+6

=12

However, the equation is 2x−3y=14 and 12≠14. This line does not pass through (3,−2).

2. The ratio of a two-digit number and the sum of its digits is 7:1. How many such two-digit numbers are possible?

a. 1

b. 4

c. 9

d. (infinitely many)

Soln. 

(b) 4

Explanation: 

Let the two-digit number be N, where the tens digit is x, and the unit digit is y. Therefore:

N=10x+y

The sum of its digits is:

S=x+y

We are told that the ratio of the number and the sum of its digits is 7:17:17:1. Mathematically, this can be written as:

N/S=7  ⟹N=7S

Substitute N=10x+y and S=x+y into the equation N=7S

10x+y=7(x+y)

Simplify this equation:

10x+y=7

10x−7x=7y−y

3x=6y

x=2y

• x and y are digits, meaning x and y must be integers between 0 and 9.
• For x=2y, y must be such that x is also a single-digit number.

Let’s check possible values of yyy:

• If y=1, then x=2×1=2
• If y=2, then x=2×2=4
• If y=3, then x=2×3=6
• If y=4, then x=2×4=8
• If y≥5y, then x=2y would exceed 9, which is not allowed.

Thus, the valid pairs (x,y) are:

(2,1),(4,2),(6,3),(8,4)

Determine the corresponding two-digit numbers:

• For (2,1), the number is 10(2)+1=21
• For (4,2), the number is 10(4)+2=42
• For (6,3), the number is 10(6)+3=63
• For (8,4), the number is 10(8)+4=84

For each number, N/S=7

21/(2+1)=21/3=7

42/(4+2)=42/6=7

63/(6+3)=63/9=7

84/(8+4)=84/12=7

All satisfy the condition.

3. The two circles represent the ordered pairs, (a, b), which are solutions of the respective equations. The circles are divided into 3 regions P, Q and R as shown.

Write one ordered pair each belonging to P, Q and R. Show your work.

Soln.

To identify one ordered pair each belonging to regions P, Q, and R, let’s break down the diagram:

1. Region P: Solutions to b=a+4 that do NOT satisfy b=−3a−4
2. Region Q: Solutions to b=−3a−4 that do NOT satisfy b=a+4.
3. Region R: Solutions that satisfy BOTH equations b=a+4 and b=−3a−4

To find the intersection, solve for aaa and bbb by equating the two equations:

b=a+4 and b=−3a−4.

Set the two equations equal to each other:

a+4=−3a−4

Solve for a:

a+3a=−4−4

⟹4a=−8

⟹a=−2.

Now substitute a=−2 into one of the equations to find b. Using b=a+4:

b=−2+4=2

Thus, the ordered pair (−2,2) satisfies both equations and lies in region R

Region P contains solutions to b=a+4 that do NOT satisfy b=−3a−4

Choose a value for a that does not satisfy b=−3a−4. Let a=0

Substitute a=0 into b=a+4:

b=0+4=4

Check if this pair (0,4) satisfies b=−3a−4:

b=−3(0)−4=−4.

Since b=4≠−4, the pair (0,4) belongs to region P.

Region Q contains solutions to b=−3a−4 that do NOT satisfy b=a+4.

Choose a value for a that does not satisfy b=a+4. Let a=0.

Substitute a=0 into b=−3a−4:

b=−3(0)−4=−4

Check if this pair (0,−4) satisfies b=a+4:

b=0+4=4.

Since b=−4≠4, the pair (0,−4) belongs to region Q

Region P: (0,4)

Region Q: (0,−4)

Region R: (−2,2)

4. Shown below is a graph representing straight lines l₁, l₂, and l₃, such that:

I is parallel to l₁, and l₃, intersects l₁, at exactly one point. 

The equation of l₁, is x + y = k, where k is a real number

Based on the above information, identify if the statements below are true or false. Justify your answer.

i) 2x + 2y = 2 k can be the equation of I₂

ii) (-x ) + y = k can be the equation of I₃

Soln.  

1. Line l₁: The equation is x+y=k, where k is a real number.

The slope of l₁  is −1 (since x+y=k can be rewritten as y=−x+ky).

2. Line l₂ (I is parallel to l₁​): If l₂ is parallel to l1. it must have the same slope as l₁​, which is −1.
3. Line l₃​ intersects l_1​ at exactly one point and is not parallel to l1​.

Now, let’s evaluate the statements.

‍i) 2x + 2y = 2 k can be the equation of I₂

_‍2(x+y)=2k⟹x+y=k

The equation x+y=k

x + y = k

x+y=k is identical to l₁.

l_2​ is parallel to l₁​, so it must have the same slope but should differ in the constant term to represent a distinct line.

Since 2x+2y=2k simplifies to x+y=k,

l₂ coincides with l₁​ rather than being a parallel, distinct line.

‍Conclusion: The statement is false because l₂​ coincides with l_1​ instead of being parallel and distinct.

‍ii) (-x ) + y = k can be the equation of I₃‍

The given equation is:−x+y=k

⟹y=x+k

The slope of this line is 1 (since y=x+k)

The slope of l₁ is −1,

so l₃​ (with slope 1) is not parallel to l₁​

‍Lines with slopes -1 and 1 intersect at exactly one point

‍Conclusion: The statement is true because l_3​ intersects l₁​ at exactly one point.  

5. Shown below is a pair of linear equations.

x + 0.999 y = 2.9990.999

x + y = 2.998

i) Without finding the values of x and y, prove that x - y = 1. 

ii) Find the values of x and y. Show your work.

Soln.  

i) Without finding the values of x and y, prove that x - y = 1. 

x+0.999y=2.999

0.999x+y=2.998

Subtract the second equation from the first:

(x+0.999y)−(0.999x+y)=2.999−2.998

x−0.999x+0.999y−y=0.001

(1−0.999)x+(0.999−1)y=0.001

0.001x−0.001y=0.001

x−y=1

Thus, we have proven that x−y=1

ii) Find the values of x and y. Show your work.

x=y+1

(y+1)+0.999y=2.999

y+1+0.999y=2.999

1.999y+1=2.999

1.999y=1.999

y=1

Substitute y=1 into x=y+1

x=1+1=2

x=2

y=1

6. At the Bengaluru marathon, the 42 km route is designed such that marathoners run in a straight line for 21 km, and return back along the same path in the opposite direction.

A marathoner, running against the wind, covered the first half of the marathon in 2 hours. Then, he covered the second half, running with the wind, in 1.5 hours.

Assume that the marathoner ran at a constant speed and that the wind speed and direction did not change throughout the marathon.

(Note: The figure is not to scale.)

Calculate the speed (in km/h) of the marathoner and the wind. Show your steps.

Soln.  

Let:

• v_m = speed of the marathoner in still air (in km/h),
• v_w = speed of the wind (in km/h).

When running against the wind:

• The marathoner's effective speed = v_m−v_w.

When running with the wind:

• The marathoner's effective speed = v_m+v_w

Speed=Distance/Time

Distance for each half = 21 km(since the total route is 42 km),

Time for the first half (against the wind) = 2 hours,

Time for the second half (with the wind) = 1.5 hours

For the first half (against the wind):

v_m−v_w. =21/2​

=10.5km/h

For the second half (with the wind):

v_m+v_w=21/21.5

=14km/h

Add the two equations to eliminate v_m+v_w

(v_m​−v_w​)+(v_m​+v_w​)=10.5+14

2v_m​=24.5

⟹v_m​=24.5/2​

=12.25km/h

(v_m​+v_w​)-(v_m​-v_w​)=14+10.5

2v_w​=3.5

⟹ v_w​=3.5/2

=1.75km/h.

7. A new intra-city transportation startup has employed both taxis and auto rickshaws. The night fare for a taxi is Rs 9 for the first half kilometre and Rs 15 per kilometre thereafter, while the night fare for an auto rickshaw is Rs 20 for the first one kilometre and Rs 13 per kilometre thereafter.

i) Express the night fare structure for taxis and auto rickshaws in the form of linear equations. Use f as the fare (in Rs) and d as the distance travelled (in km).

ii) At what distance is the night fare for a taxi and an auto rickshaw equal? Show your steps.

Soln.  

i) Express the night fare structure for taxis and auto rickshaws in the form of linear equations. Use f as the fare (in Rs) and d as the distance travelled (in km).

Let f represent the fare (in Rs) and d represent the distance traveled (in km)

For Taxis:

• For the first half kilometer (d≤0.5), the fare is Rs 9.
• Beyond the first half kilometer (d>0.5), the fare increases at Rs 15 per km.

To represent this:

When d≤0.5:

f=9

When d>0.5:

The additional distance beyond 0.5 km is d−0.5. The fare for this distance is 15×(d−0.5)

Therefore, the total fare is:

f=9+15(d−0.5)

f=9+15d−7.5

⟹f=1.5+15d.

Thus, the fare structure for taxis is:

f=  { 9 if d≤0.5 }

 =  { 1.5 + 15d  if d > 0.5 }

For Auto Rickshaws:

• For the first kilometer (d≤1), the fare is Rs 20.
• Beyond the first kilometer (d>1), the fare increases at Rs 13 per km.

To represent this:

When d≤1:

f=20

When d>1:

The additional distance beyond 1 km is d−1. The fare for this distance is 13×(d−1). Therefore, the total fare is:

f=20+13(d−1)

f=20+13d−13

⟹f=7+13d

Thus, the fare structure for auto rickshaws is:

f=  { 20 if d≤1 }

 =  { 7 + 13d  if d > 1} 

ii) At what distance is the night fare for a taxi and an auto rickshaw equal? Show your steps.

To find the distance at which the fares for taxis and auto rickshaws are equal, we consider the fare equations for d>1, since for smaller distances, the fares are fixed and not linear.

For taxis:

f=1.5+15d

For auto rickshaws:

f=7+13d.

Set the fares equal to each other:

1.5+15d=7+13d

15d−13d=7−1.5.

2d=5.5.

d= 5.5/2

d= 2.75km

8. Arvind owns a dry fruits store. He sells cashew nuts at Rs 600/kg and pistachio nuts at Rs 750/kg.

A customer asks for a mixture of cashew nuts and pistachio nuts with the following conditions:

both the items should together weigh 500 g.

both the items should together cost Rs 360.

i) If Arvind packs x kg of cashew nuts and y kg of pistachio nuts for the customer, frame the equations that represent the given context.

ii) Find the weights of cashew nuts and pistachio nuts that Arvind packed for the customer.

Show your work.

Soln.  

i) If Arvind packs x kg of cashew nuts and y kg of pistachio nuts for the customer, frame the equations that represent the given context.

Let:

x = weight of cashew nuts (in kg),

y = weight of pistachio nuts (in kg).

The given conditions are:

Both items together weigh 500 g. Since 500 g = 0.5 kg, we write:

x+y=0.5

Both items together cost Rs 360:

Cashew nuts cost Rs 600/kg

Pistachio nuts cost Rs 750/kg

total cost is:

600x+750y=360

Thus, the two equations are:

x+y=0.5,

600x+750y=360

ii) Find the weights of cashew nuts and pistachio nuts that Arvind packed for the customer.

From x+y=0.5, we get:

y=0.5−x.

Substitute y=0.5−x into 600x+750y=360:

600x+750(0.5−x)=360

600x+750(0.5)−750x=360.

600x+375−750x=360

−150x+375=360

−150x=−15

x= (−15/−150)

=0.1

y=0.5−0.1

=0.4

• The weight of cashew nuts is 0.1 kg.
• The weight of pistachio nuts is 0.4 kg.

Answer the questions based on the given information.

The total cost of snowden ice cream parlour is divided into fixed cost ( x ) and variable cost ( y ). Fixed cost is the cost that the ice cream parlour has to incur even at zero level of production and variable cost is the cost that will be directly proportional to each unit of ice cream sold.

The parlour launched a new flavour of ice cream and wanted to find the fixed and variable cost associated with it. They found that their total cost for that flavour was Rs 27500 after selling 150 units and Rs 32500 after selling 250 units.

9. (i) Frame the equations that represent the total cost incurred by snowden ice cream parlour for the new flavour in terms of fixed and variable costs.

9. (ii) Find the fixed cost incurred by the ice cream parlour for the new flavour. Show your work.

9. (iii) Find the variable cost per unit incurred by the ice cream parlour for the new flavour. Show your work.

Soln.  

(i) Frame the equations that represent the total cost incurred by snowden ice cream parlour for the new flavour in terms of fixed and variable costs.

Let:

• xx = fixed cost (in Rs),
• y = variable cost per unit (in Rs),
• C = total cost,
• n = number of ice cream units sold.

The total cost is the sum of the fixed cost and the variable cost:

C=x+y⋅n

From the given data:

1. When n=150, C=27500.
2. When n=250, C=32500

Using the general equation C=x+y⋅n, we can write two equations:

x+150y=27500 (Equation 1)

x+250y=32500 (Equation 2)

(ii) Find the fixed cost incurred by the ice cream parlour for the new flavour. Show your work.

We will solve the two equations to find x (the fixed cost).

(x+250y)−(x+150y)=32500−27500

x−x+250y−150y=5000

100y=5000

y= 5000/100

y= 50

The variable cost per unit is y=50 Rs/unit

From Equation 1:

x+150y=27500.

Substitute y=50:

x+150(50)=27500

x+7500=27500

x=27500−7500

=20000

The fixed cost is x=20000 

(iii) Find the variable cost per unit incurred by the ice cream parlour for the new flavour. Show your work.

(x+250y)−(x+150y)=32500−27500

x−x+250y−150y=5000

100y=5000

y= 5000/100

y= 50

The variable cost per unit is y=50 Rs/unit

10. Ananya had red, blue and yellow marbles in the ratio 4:5:3. She gave all her red marbles and some blue marbles to Neha. The ratio of the number of blue marbles and yellow marbles left with Ananya was 7:9.

If Ananya gave 20 marbles to Neha, how many of them are red marbles? Show your work.

Soln.  

Let the total number of marbles be x.

Red marbles = 4/12x

=1/3x

Blue marbles = 5/12x

Yellow marbles = 3/12x

=1/4x

Ananya gave all her red marbles and some blue marbles to Neha. The number of blue marbles left and yellow marbles with Ananya is in the ratio 7:9.

Let the number of blue marbles Ananya gave to Neha be b. Then:

Blue marbles left with Ananya = 5/12x−b,

Yellow marbles left with Ananya = 3/12x (unchanged).

Blue marbles left​/Yellow marbles = 7/9

(5/12x−b)/(3/12x) = 7/9

(5/12x−b)/(3/12x) = (5x-12b)/3x

(5x-12b)/3x = 7/9

9(5x−12b)=7(3x)

45x−108b=21x

24x=108b.

b= 24x/108

b= x/4.5

The total marbles given to Neha is 20:

1/3x + x/4.5 = 20

3x/9 + 2x/9= 20

5x/9 = 20

x=20× (9/5)

x= 36

The number of red marbles is: 1/3x 

= ⅓ x (36)

= 12

The number of red marbles Ananya gave to Neha is 12

Some More Important Question Answer of Class 10th Maths

Q1. Solve by substitution: x + y = 10, 2x – y = 1.

Ans. From first: y = 10 – x.
Substitute in second: 2x – (10 – x) = 1 ⇒ 2x – 10 + x = 1 ⇒ 3x = 11 ⇒ x = 11/3.
Then y = 10 – 11/3 = 19/3.
Answer: (11/3, 19/3)

Q2. Solve by elimination: 3x + 2y = 11, 2x – 3y = –4.

Ans. Multiply first by 3: 9x + 6y = 33.
Multiply second by 2: 4x – 6y = –8.
Add: 13x = 25 ⇒ x = 25/13.
Substitute: 3(25/13) + 2y = 11 ⇒ 75/13 + 2y = 11 ⇒ 2y = 68/13 ⇒ y = 34/13.
Answer: (25/13, 34/13)

Q3. The sum of two numbers is 30. Their difference is 4. Find the numbers.

Ans. x + y = 30, x – y = 4.
Add: 2x = 34 ⇒ x = 17. Then y = 13.
Answer: 17 and 13

Q4. A two-digit number has tens digit 3 more than units digit. The sum of the number and its reverse is 121. Find the number.

Ans. Let tens = x, units = y. Then x = y + 3.
Number + reverse = 11(x + y) = 121 ⇒ x + y = 11.
So (y + 3) + y = 11 ⇒ 2y = 8 ⇒ y = 4, x = 7.
Number = 74.
Answer: 74

Q5. One number is twice the other. Their sum is 36. Find the numbers.

Ans. Let smaller = y, larger = x = 2y.
x + y = 36 ⇒ 2y + y = 36 ⇒ y = 12, x = 24.
Answer: 24 and 12

Q6. Find k if 2x + 3y = 7 and 4x + ky = 14 have infinitely many solutions.

Ans. For infinitely many solutions: 2/4 = 3/k = 7/14 ⇒ 1/2.
So 3/k = 1/2 ⇒ k = 6.
Answer: k = 6

Q7. Find k if x + 2y = 3 and 5x + ky = 7 have no solution.

Ans. Condition: a1/a2 = b1/b2 ≠ c1/c2.
Here 1/5 = 2/k ⇒ k = 10.
Check: 3/7 ≠ 1/5.
Answer: k = 10

Q8. Solve: 7x – 3y = 5, 3x + 2y = 11.

Ans. Multiply first by 2: 14x – 6y = 10.
Multiply second by 3: 9x + 6y = 33.
Add: 23x = 43 ⇒ x = 43/23.
Substitute: 3(43/23) + 2y = 11 ⇒ 2y = 124/23 ⇒ y = 62/23.
Answer: (43/23, 62/23)

Q9. The sum of ages of father and son is 45. Five years ago my father had 4 sons. Find present ages.

Ans. F + S = 45.
F – 5 = 4(S – 5). ⇒ F = 4S – 15.
Substitute: 4S – 15 + S = 45 ⇒ 5S = 60 ⇒ S = 12, F = 33.
Answer: Father 33 years, Son 12 years

Q10. A two-digit number is divisible by the sum of digits and the quotient is 4. Digits differ by 2. Find the number.

Ans. 10x + y = 4(x + y). ⇒ 10x + y = 4x + 4y ⇒ 6x = 3y ⇒ y = 2x.
|x – y| = 2 ⇒ |x – 2x| = 2 ⇒ x = 2 ⇒ y = 4.
Number = 24.
Answer: 24

Q11. Solve: x + y = 5, 2x – y = 4.

Ans. x + y = 5, 2x – y = 4.
Add: 3x = 9 ⇒ x = 3, y = 2.
Answer: (3, 2)

Q12. Solve: 5x – 4y = –2, –7x + 6y = 10.

Ans. Multiply first by 3: 15x – 12y = –6.
Multiply second by 2: –14x + 12y = 20.
Add: x = 14.
Substitute: 5(14) – 4y = –2 ⇒ –4y = –72 ⇒ y = 18.
Answer: (14, 18)

Q13. Solve: 9x + 7y = 59, 4x + 3y = 26.

Ans. Multiply first by 3: 27x + 21y = 177.
Multiply second by 7: 28x + 21y = 182.
Subtract: –x = –5 ⇒ x = 5.
Then 4(5) + 3y = 26 ⇒ 3y = 6 ⇒ y = 2.
Answer: (5, 2)

Q14. Solve: 3x + 4y = 10, 2x – y = 1.

Ans. From second: y = 2x – 1.
Substitute: 3x + 8x – 4 = 10 ⇒ 11x = 14 ⇒ x = 14/11.
y = 28/11 – 1 = 17/11.
Answer: (14/11, 17/11)

Q15. Solve: 2x + 3y = 12, 3x + 2y = 13.

Ans. Multiply first by 3: 6x + 9y = 36.
Multiply second by 2: 6x + 4y = 26.
Subtract: 5y = 10 ⇒ y = 2.
Then 2x + 6 = 12 ⇒ x = 3.
Answer: (3, 2)

Q16. Solve: x/2 + y/3 = 5, x/3 + y/2 = 6.

Ans. Multiply by 6: 3x + 2y = 30, 2x + 3y = 36.
Multiply first by 3: 9x + 6y = 90.
Multiply second by 2: 4x + 6y = 72.
Subtract: 5x = 18 ⇒ x = 18/5.
Then 3(18/5) + 2y = 30 ⇒ 2y = 96/5 ⇒ y = 48/5.
Answer: (18/5, 48/5)

Q17. Solve: 4x + 5y = 19, 7x – 2y = 20.

Ans. Multiply first by 2: 8x + 10y = 38.
Multiply second by 5: 35x – 10y = 100.
Add: 43x = 138 ⇒ x = 138/43.
Substitute: 4x + 5y = 19 ⇒ y = 31/25.
Answer: (138/43, 31/25)

Q18. Solve: 5x + y = 11, x + 5y = 7.

Ans. Multiply first by 5: 25x + 5y = 55.
Subtract second: 24x = 48 ⇒ x = 2.
Substitute: 5(2) + y = 11 ⇒ y = 1.
Answer: (2, 1)

Q19. The larger of two numbers is three times smaller. Sum is 60. Find numbers.

Ans. Let smaller = y, larger = 3y.
3y + y = 60 ⇒ 4y = 60 ⇒ y = 15, larger = 45.
Answer: 45 and 15

Q20. The sum of two numbers is 100. If the larger is decreased by 10 and the smaller increased by 6, the larger becomes three times the smaller. Find the numbers.

Ans.

1. Let the larger number be xxx and the smaller be yyy.
   Given: x+y=100x + y = 100x+y=100 …(1).
2. After changes: Larger → x−10x - 10x−10, Smaller → y+6y + 6y+6.
   Condition: x−10=3(y+6)x - 10 = 3(y + 6)x−10=3(y+6) …(2).
3. Expand (2): x−10=3y+18x - 10 = 3y + 18x−10=3y+18 ⇒ x=3y+28x = 3y + 28x=3y+28.
4. Substitute in (1): 3y+28+y=1003y + 28 + y = 1003y+28+y=100.
   ⇒ 4y+28=1004y + 28 = 1004y+28=100.
   ⇒ 4y=724y = 724y=72.
   ⇒ y=18y = 18y=18.
5. From (1): x+18=100x + 18 = 100x+18=100 ⇒ x=82x = 82x=82.
6. Check: Sum = 82 + 18 = 100 ✓
   After change: 82 – 10 = 72; 18 + 6 = 24; and 72 = 3 × 24 ✓

Answer: The numbers are 82 and 18.

Q21. A factory makes tables and chairs. Each table needs 3 units of wood, each chair 2 units. In one day 88 items were made, using 210 units of wood. Find tables and chairs.

Ans. T + C = 88.
3T + 2C = 210.
C = 88 – T.
3T + 2(88 – T) = 210 ⇒ 3T + 176 – 2T = 210 ⇒ T = 34, C = 54.
Answer: 34 tables, 54 chairs

Q22. System: x + 2y = 11, 3x + ky = 33. For what k infinite/unique solutions?

Ans. Multiply first by 3: 3x + 6y = 33.
Compare with 3x + ky = 33.
For infinite: k = 6.
Unique: k ≠ 6.
Answer: Infinite for k = 6, unique for k ≠ 6

Q23. A bag contains 20p and 25p coins, 50 in all, total value ₹11.25. Find coins.

Ans.      
x + y = 50.
20x + 25y = 1125 (in paise).
y = 50 – x.
20x + 25(50 – x) = 1125 ⇒ –5x + 1250 = 1125 ⇒ –5x = –125 ⇒ x = 25, y = 25.
Answer: 25 coins of each type

Q24. A rectangle has length 4 m more than breadth. Perimeter = 56 m. Find dimensions.

Ans. Let breadth = b, length = b + 4.
Perimeter: 2(l + b) = 56 ⇒ l + b = 28.
b + 4 + b = 28 ⇒ 2b = 24 ⇒ b = 12, l = 16.
Answer: Length 16 m, Breadth 12 m

Extra Practice – For you to ace your board  examinations!

Q1. The sum of the digits of a two-digit number is 9. If the digits are reversed, the new number is 27 more than the original. Find the number.

Q2. A person buys 2 pens and 3 pencils for ₹38, and another person buys 3 pens and 2 pencils for ₹44. Find the cost of a pen and a pencil.

Q3. The sum of a two-digit number and the number obtained by reversing its digits is 99. If the difference of digits is 3, find the number.

Q4. A boat goes 12 km downstream in 4 hours and comes back the same distance upstream in 6 hours. Find the speed of the boat in still water and the speed of the stream.

Q5. Solve for x and y: 2x+3y=11,  3x−2y=42x + 3y = 11,\; 3x - 2y = 42x+3y=11,3x−2y=4.

Q6. A two-digit number is such that the product of digits is 24. If we add 18 to the number, the digits get interchanged. Find the number.

Q7. The sum of the ages of a father and son is 50 years. Ten years ago, the father’s age was three times that of his son. Find their present ages.

Q8. The larger of two numbers is 5 more than twice the smaller. If their sum is 65, find the numbers.

Q9. Find the value of k for which the system has infinitely many solutions: 3x−y=7,  6x−2y=k3x - y = 7,\; 6x - 2y = k3x−y=7,6x−2y=k.

Q10. The area of a rectangle is 180 m². If its length is 5 m more than its breadth, find its length and breadth.

Benefits of Practising These Class 10 Pair of Linear Equations Questions

here are some of the benefits of practising these class 10 pair of linear equations questions:

• Improved Conceptual Understanding: Strengthens clarity on the relationship between equations and lines.
• Better Problem-Solving Skills: Increases accuracy in algebraic methods and real-life applications.
• Exam-Oriented Practice: Covers all types of questions from CBSE sample papers and previous years.
• Enhanced Logical Thinking: Builds confidence in forming and interpreting equations.
• High Scoring Chapter: Straightforward calculations make it one of the most reliable chapters to score full marks.

How to Ace These Class 10 Pair of Linear Equations Important Questions

This chapter becomes easy when you understand the meaning of a “solution” and the relationship between equations. Here’s a clear plan to prepare:

Step 1: Revise the General Form of Linear Equations

Understand the standard form of a pair of linear equations as:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
Learn how the coefficients (a₁, b₁, c₁, a₂, b₂, c₂) determine whether the lines intersect, are parallel, or coincide.

Step 2: Understand the Graphical Method

Learn to plot both equations on a coordinate plane.

• If the lines intersect, the system has one unique solution.
• If the lines are parallel, there is no solution.
• If the lines coincide, there are infinitely many solutions. Practise drawing graphs neatly with proper scales and labeling.

Step 3: Learn the Algebraic Methods of Solving Equations

Study all four algebraic methods step by step:

1. Substitution Method – Replace one variable and solve.
2. Elimination Method – Add or subtract equations to eliminate a variable.
3. Cross Multiplication Method – Use the cross multiplication formula for direct solution.
4. Comparison Method – Compare coefficients to find relationships between equations.

Step 4: Study Consistency of Equations

Understand the conditions for consistency and types of solutions based on the ratio of coefficients:

• a₁/a₂ ≠ b₁/b₂ → one solution
• a₁/a₂ = b₁/b₂ ≠ c₁/c₂ → no solution
• a₁/a₂ = b₁/b₂ = c₁/c₂ → infinitely many solutions

Step 5: Solve Word Problems

Practise application-based questions on speed, distance, age, and geometry. These are common in exams and test your ability to translate real-life situations into equations.

Step 6: Double-Check Calculations

Since most errors come from arithmetic steps, carefully check each step when simplifying or substituting values.

Tips for Solving Linear Equations in Two Variables

Acing the topic of linear equations in two variables requires practice and a good understanding of different solving techniques. Going through the NCERT Book class 10 maths can also help you build clarity before attempting these questions.Here are some tips to help you:

• Carefully read the question to identify what is being asked and what the variables represent.
• Depending on the nature of the problem, choose the most efficient method (substitution, elimination, or graphical) to solve the equations.
• Always substitute your solutions back into the original equations to verify that they satisfy both equations.
• Being able to accurately plot equations on a graph can help you visualize the solution and understand the relationship between the variables.
• Practice converting word problems into algebraic equations. This skill is essential for solving real-life problems using linear equations.

FAQs

Q1. How many marks are usually allotted to this chapter in Class 10 board exams?

Ans. This chapter generally carries 6 to 8 marks, including one word problem and one numerical or reasoning question.

Q2. Which topics are most important for exams?

Ans. Algebraic methods (especially elimination and substitution), conditions for consistency, and word problems based on speed, age, and geometry are most important.

Q3. How can I easily remember the cross multiplication method?

Ans. Use the pattern:
x / (b₁c₂ – b₂c₁) = y / (c₁a₂ – c₂a₁) = 1 / (a₁b₂ – a₂b₁)
Practise it several times until it feels natural.

Q4. Are graphical questions asked frequently?

Ans. Yes, you may be asked to draw graphs to show the number of solutions or interpret given graphs, so practise graph plotting neatly.

Q5. How can I prepare for word problems effectively?

Ans. Write the problem in words, define your variables clearly, form equations systematically, and then use the elimination or substitution method to solve.