Pair of Linear Equations in Two Variables — Class 10 Maths practice questions

Question 1

A school canteen sells two items: samosas and juice cups. On Monday, 3 samosas and 2 juice cups together cost ₹46. On Tuesday, 5 samosas and 3 juice cups together cost ₹74.

Based on this situation, answer the following:
(i) If the cost of one samosa is ₹x and the cost of one juice cup is ₹y, write the pair of linear equations representing the above situation. (1 mark)
(ii) Find the cost of one samosa. (1 mark)
(iii) (A) Find the cost of one juice cup and verify your answer by substituting both values in either equation. (2 marks)
       OR
(iii) (B) On Wednesday, the canteen offers a combo: 2 samosas and 1 juice cup. A student has ₹35. Is ₹35 sufficient to buy the combo? Justify your answer using the values found above. (2 marks)

Accepted Answer

(i) According to the question,
3x + 2y = 46  ...(i)
5x + 3y = 74  ...(ii)
∴ The required pair of linear equations is 3x + 2y = 46 and 5x + 3y = 74.   [1 mark]

(ii) Multiplying equation (i) by 3 and equation (ii) by 2:
⟹ 9x + 6y = 138  ...(iii)
⟹ 10x + 6y = 148  ...(iv)

Subtracting (iii) from (iv):
⟹ (10x − 9x) + (6y − 6y) = 148 − 138
⟹ x = 10
∴ The cost of one samosa is ₹10.   [1 mark]

(iii)(A) Substituting x = 10 in equation (i):
3(10) + 2y = 46
⟹ 30 + 2y = 46
⟹ 2y = 16
⟹ y = 8
∴ The cost of one juice cup is ₹8.   [1 mark]

Verification — substituting x = 10, y = 8 in equation (i):
LHS = 3(10) + 2(8) = 30 + 16 = 46 = RHS ✓
∴ The values x = 10 and y = 8 are correct.   [1 mark]

OR

(iii)(B) Cost of the combo = cost of 2 samosas + cost of 1 juice cup
= 2x + y
= 2(10) + 8
= 20 + 8
= ₹28

Since ₹28 < ₹35,
∴ Yes, ₹35 is sufficient to buy the combo. The student will receive ₹35 − ₹28 = ₹7 as change.   [2 marks]

Question 2

A school canteen sells two items: a sandwich and a juice box. On Monday, 3 sandwiches and 2 juice boxes were sold for ₹74. On Tuesday, 1 sandwich and 4 juice boxes were sold for ₹62.

Based on the above information, answer the following questions:
(i) Write the pair of linear equations representing the above situation. [1 mark]
(ii) Find the cost of one sandwich. [1 mark]
(iii) Find the cost of one juice box. Also, find the total cost of buying 2 sandwiches and 3 juice boxes. [2 marks]

Accepted Answer

(i) Let the cost of one sandwich = ₹x and the cost of one juice box = ₹y.

According to the question,
3x + 2y = 74 ...(i)
x + 4y = 62 ...(ii)

∴ The required pair of linear equations is 3x + 2y = 74 and x + 4y = 62.

(ii) From equation (ii):
x = 62 − 4y ...(iii)

Substituting (iii) in equation (i):
⟹ 3(62 − 4y) + 2y = 74
⟹ 186 − 12y + 2y = 74
⟹ −10y = 74 − 186
⟹ −10y = −112
⟹ y = 11.2

Substituting y = 11.2 in (iii):
⟹ x = 62 − 4(11.2)
⟹ x = 62 − 44.8
⟹ x = 17.2

∴ The cost of one sandwich = ₹17.20

(iii) From part (ii), y = 11.20.

∴ The cost of one juice box = ₹11.20

Total cost of 2 sandwiches and 3 juice boxes:
⟹ 2x + 3y = 2(17.20) + 3(11.20)
⟹ = 34.40 + 33.60
⟹ = 68.00

∴ The cost of one juice box is ₹11.20 and the total cost of 2 sandwiches and 3 juice boxes is ₹68.00.

Question 3

A school canteen sells two items: a sandwich and a juice box. On Monday, Priya buys 2 sandwiches and 1 juice box and pays ₹90. On Tuesday, she buys 1 sandwich and 2 juice boxes and pays ₹75.

Based on this situation, answer the following:
(i) Write a pair of linear equations representing the above situation, where x is the price (in ₹) of one sandwich and y is the price (in ₹) of one juice box. [1 mark]
(ii) Find the price of one sandwich and one juice box. [2 marks]
(iii) On Wednesday, Priya's friend Ravi buys 3 sandwiches and 3 juice boxes. How much does Ravi pay in total? [1 mark]

Accepted Answer

(i) According to the question,
2x + y = 90   ...(i)
x + 2y = 75   ...(ii)
∴ The required pair of linear equations is 2x + y = 90 and x + 2y = 75.
[1 mark]

(ii) From equation (i):
y = 90 − 2x   ...(iii)

Substituting ...(iii) into equation (ii):
x + 2(90 − 2x) = 75
⟹ x + 180 − 4x = 75
⟹ −3x = 75 − 180
⟹ −3x = −105
⟹ x = 35

Substituting x = 35 into ...(iii):
y = 90 − 2(35)
⟹ y = 90 − 70
⟹ y = 20

∴ Price of one sandwich = ₹35 and price of one juice box = ₹20.
[2 marks]

(iii) Total amount paid by Ravi for 3 sandwiches and 3 juice boxes:
= 3x + 3y
= 3(35) + 3(20)
= 105 + 60
∴ Ravi pays ₹165 in total.
[1 mark]

Question 4

A school canteen sells two items: a sandwich and a juice box. On Monday, Priya buys 2 sandwiches and 1 juice box and pays ₹70. On Tuesday, Rohan buys 1 sandwich and 2 juice boxes and pays ₹80.

Based on this situation, answer the following questions:

(i) If the price of one sandwich is ₹x and the price of one juice box is ₹y, write the pair of linear equations representing the above situation. [1 mark]

(ii) Find the price of one sandwich and one juice box by solving the pair of equations. [2 marks]

(iii) On Wednesday, the canteen offers a combo: 1 sandwich + 1 juice box at a discount of ₹5 from the total price. What will a student pay for this combo? [1 mark]

Accepted Answer

(i) According to the question,

2x + y = 70  ...(i)
x + 2y = 80  ...(ii)

∴ The required pair of linear equations is 2x + y = 70 and x + 2y = 80.

[Award 1 mark for both equations correctly stated]

---

(ii) From equation (i):
y = 70 − 2x  ...(iii)

Substituting (iii) in equation (ii):
x + 2(70 − 2x) = 80
⟹ x + 140 − 4x = 80
⟹ −3x = 80 − 140
⟹ −3x = −60
⟹ x = 20

Substituting x = 20 in (iii):
y = 70 − 2(20) = 70 − 40 = 30

∴ Price of one sandwich = ₹20 and price of one juice box = ₹30.

[Award 1 mark for correct method/substitution step; 1 mark for correct values of x and y with conclusion]

---

(iii) Total price of 1 sandwich + 1 juice box = x + y = 20 + 30 = ₹50

Discount offered = ₹5

∴ Amount paid for the combo = 50 − 5 = ₹45.

[Award 1 mark for correct final answer with unit]

Question 5

A school canteen sells two items: a sandwich and a juice box. One day, a group of students placed two orders:
• Order 1: 3 sandwiches and 2 juice boxes cost ₹ 74.
• Order 2: 5 sandwiches and 3 juice boxes cost ₹ 121.

Based on the above information, answer the following questions:
(i) Form a pair of linear equations in two variables to represent the situation. [1 mark]
(ii) Find the price of one sandwich and the price of one juice box. [2 marks]
(iii) The next day, the canteen offered a combo deal: any number of sandwiches at ₹ 2 less per sandwich and juice boxes at ₹ 1 more per juice box. A student bought 4 sandwiches and 5 juice boxes using the combo prices. How much did the student pay? OR
(iii) If a student has ₹ 200 and wants to buy as many sandwiches as possible (buying only whole sandwiches, with no juice boxes), how many sandwiches can the student buy, and how much money will be left over? [2 marks]

Accepted Answer

Let the price of one sandwich = ₹ x and the price of one juice box = ₹ y.

(i) According to the question,
3x + 2y = 74  ...(i)
5x + 3y = 121  ...(ii)
∴ The required pair of linear equations is 3x + 2y = 74 and 5x + 3y = 121.

(ii) From equation (i):
3x + 2y = 74
⟹ 2y = 74 − 3x
⟹ y = (74 − 3x) / 2  ...(iii)

Substituting (iii) in equation (ii):
5x + 3 · [(74 − 3x) / 2] = 121
⟹ 10x + 3(74 − 3x) = 242
⟹ 10x + 222 − 9x = 242
⟹ x = 242 − 222
⟹ x = 20

Substituting x = 20 in (iii):
y = (74 − 3 × 20) / 2 = (74 − 60) / 2 = 14 / 2 = 7

∴ Price of one sandwich = ₹ 20 and price of one juice box = ₹ 7.

(iii) [Main option]
Combo prices: price of one sandwich = ₹ (20 − 2) = ₹ 18; price of one juice box = ₹ (7 + 1) = ₹ 8.

Cost for 4 sandwiches and 5 juice boxes at combo prices:
= 4 × 18 + 5 × 8
= 72 + 40
= 112

∴ The student paid ₹ 112.

[OR option]
Price of one sandwich = ₹ 20 (from part (ii)).

Maximum number of sandwiches the student can buy:
Let the number of sandwiches = n.
According to the question, 20n ≤ 200
⟹ n ≤ 200 / 20
⟹ n ≤ 10

∴ Maximum number of whole sandwiches = 10.
Money spent = 10 × 20 = ₹ 200.
Money left over = 200 − 200 = ₹ 0.

∴ The student can buy 10 sandwiches and ₹ 0 will be left over.

Question 6

A school canteen sells two types of meal combos: Combo A (a sandwich + a juice) and Combo B (a burger + a juice). On Monday, 30 students bought Combo A and 20 students bought Combo B, and the canteen collected ₹2,200. On Tuesday, 15 students bought Combo A and 35 students bought Combo B, and the canteen collected ₹2,350.

Based on the above situation, answer the following:
(i) If the price of Combo A is ₹x and the price of Combo B is ₹y, write the pair of linear equations representing the above situation.
(ii) Find the price of each combo by solving the equations algebraically (elimination method).
(iii) A student says: "On Wednesday, if 25 students buy Combo A and 25 students buy Combo B, the canteen will collect exactly ₹2,500." Verify whether this statement is correct, showing full working. OR (iii) On Thursday, the canteen wants to collect exactly ₹3,000. If the number of students buying Combo B is twice the number buying Combo A, find how many students bought each combo.

Accepted Answer

(i) Let the price of Combo A = ₹x and the price of Combo B = ₹y.

According to the question,
30x + 20y = 2200  ...(i)
15x + 35y = 2350  ...(ii)

∴ The required pair of linear equations is:
30x + 20y = 2200 and 15x + 35y = 2350.

[1 mark]

---

(ii) From equations (i) and (ii), applying Elimination Method:

Dividing equation (i) by 10:
3x + 2y = 220  ...(iii)

Dividing equation (ii) by 5:
3x + 7y = 470  ...(iv)

Subtracting (iii) from (iv):
(3x + 7y) − (3x + 2y) = 470 − 220
⟹ 5y = 250
⟹ y = 50

Substituting y = 50 in equation (iii):
3x + 2(50) = 220
⟹ 3x + 100 = 220
⟹ 3x = 120
⟹ x = 40

∴ Price of Combo A = ₹40 and Price of Combo B = ₹50.

[1 mark]

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(iii) [MAIN OPTION]

If 25 students buy Combo A and 25 students buy Combo B on Wednesday, the total collection would be:

Total = 25x + 25y
⟹ Total = 25(40) + 25(50)
⟹ Total = 1000 + 1250
⟹ Total = ₹2,250

∵ ₹2,250 ≠ ₹2,500

∴ The student's statement is INCORRECT. The canteen will collect ₹2,250, not ₹2,500.

[2 marks]

---

(iii) [OR OPTION]

Let the number of students buying Combo A on Thursday = n.
According to the question, the number of students buying Combo B = 2n.

According to the question,
n × 40 + 2n × 50 = 3000
⟹ 40n + 100n = 3000
⟹ 140n = 3000
⟹ n = 3000/140
⟹ n = 150/7

∵ n = 150/7 is not a whole number, and the number of students must be a natural number, this scenario is not possible with exact whole-number students.

∴ The canteen cannot collect exactly ₹3,000 under the given condition using whole numbers of students.

[2 marks]

Question 7

A school canteen sells two items: a sandwich and a juice box. On Monday, Priya buys 2 sandwiches and 1 juice box and pays ₹70. On Tuesday, Ravi buys 1 sandwich and 2 juice boxes and pays ₹65.

Based on the above information, answer the following questions:
(i) Write the pair of linear equations representing the above situation. [1 mark]
(ii) Check whether this pair of equations is consistent or not. [1 mark]
(iii) Find the price of one sandwich and one juice box.  [2 marks]
  OR
(iii) If the canteen offers a combo of 3 sandwiches and 3 juice boxes, find the total cost of the combo. [2 marks]

Accepted Answer

(i) Let the price of one sandwich = ₹x and the price of one juice box = ₹y.

According to the question,
2x + y = 70   ...(i)
x + 2y = 65   ...(ii)

∴ The required pair of linear equations is 2x + y = 70 and x + 2y = 65.

(ii) Comparing 2x + y = 70 with a₁x + b₁y + c₁ = 0:
a₁ = 2, b₁ = 1, c₁ = −70.
Comparing x + 2y = 65 with a₂x + b₂y + c₂ = 0:
a₂ = 1, b₂ = 2, c₂ = −65.

a₁/a₂ = 2/1 = 2;   b₁/b₂ = 1/2.

⟹ a₁/a₂ ≠ b₁/b₂

∴ The pair of linear equations is consistent (the lines intersect at exactly one point).

(iii) From equation (i):  y = 70 − 2x   ...(iii)

Substituting (iii) in equation (ii):
x + 2(70 − 2x) = 65
⟹ x + 140 − 4x = 65
⟹ −3x = 65 − 140
⟹ −3x = −75
⟹ x = 25

Substituting x = 25 in (iii):
y = 70 − 2(25) = 70 − 50 = 20

∴ The price of one sandwich is ₹25 and the price of one juice box is ₹20.

OR

(iii) Adding equations (i) and (ii):
(2x + y) + (x + 2y) = 70 + 65
⟹ 3x + 3y = 135

∴ The total cost of a combo of 3 sandwiches and 3 juice boxes = 3x + 3y = ₹135.

Question 8

A school canteen offers two combo meals: Combo A (1 sandwich + 2 cold drinks) and Combo B (2 sandwiches + 1 cold drink). During the lunch break, a group of students made the following purchases:

• Group 1 bought 3 Combo A meals and 2 Combo B meals and paid ₹410 in total.
• Group 2 bought 2 Combo A meals and 3 Combo B meals and paid ₹390 in total.

Let the cost of one sandwich be ₹x and the cost of one cold drink be ₹y.

(i) Write the pair of linear equations representing the above situation. [1 mark]

(ii) Find the cost of one sandwich and one cold drink by solving the pair of equations algebraically. [2 marks]

(iii) On a particular day, the canteen offers a 'Student Special': any combination of 5 items (sandwiches and/or cold drinks) for a flat price of ₹110. Determine whether this flat price is cheaper or more expensive than buying the same 5 items at the individual prices found in part (ii), if a student picks 3 sandwiches and 2 cold drinks. Justify your answer. [1 mark]

Accepted Answer

Part (i) [1 mark]

Combo A costs: (x + 2y) and Combo B costs: (2x + y).

According to the question,

3(x + 2y) + 2(2x + y) = 410
⟹ 3x + 6y + 4x + 2y = 410
⟹ 7x + 8y = 410   ...(i)

2(x + 2y) + 3(2x + y) = 390
⟹ 2x + 4y + 6x + 3y = 390
⟹ 8x + 7y = 390   ...(ii)

∴ The required pair of linear equations is:
7x + 8y = 410   ...(i)
8x + 7y = 390   ...(ii)

─────────────────────────────────────
Part (ii) [2 marks]

Adding equations (i) and (ii):
7x + 8y + 8x + 7y = 410 + 390
⟹ 15x + 15y = 800
⟹ x + y = 800/15

[Examiner note: Award marks for correct method; re-examine arithmetic route below which gives integer solution and matches question intent.]

Subtracting equation (ii) from equation (i):
(7x + 8y) − (8x + 7y) = 410 − 390
⟹ −x + y = 20
⟹ y − x = 20   ...(iii)

Adding equations (i) and (ii):
15x + 15y = 800
⟹ x + y = 800/15 = 160/3   ...(iv)

From (iii): y = x + 20

Substituting in (iv):
x + (x + 20) = 160/3
⟹ 2x + 20 = 160/3
⟹ 2x = 160/3 − 20 = 160/3 − 60/3 = 100/3
⟹ x = 50/3

∴ y = 50/3 + 20 = 50/3 + 60/3 = 110/3

Verification in (i): 7(50/3) + 8(110/3) = 350/3 + 880/3 = 1230/3 = 410 ✓
Verification in (ii): 8(50/3) + 7(110/3) = 400/3 + 770/3 = 1170/3 = 390 ✓

∴ Cost of one sandwich = ₹50/3 ≈ ₹16.67 and cost of one cold drink = ₹110/3 ≈ ₹36.67.

[Full marks awarded for correct algebraic method with verification even if values are non-integer; ECF applies in part (iii).]

─────────────────────────────────────
Part (iii) [1 mark]

Using values from part (ii): x = ₹50/3, y = ₹110/3.

Cost of 3 sandwiches and 2 cold drinks at individual prices:
= 3x + 2y
= 3(50/3) + 2(110/3)
= 50 + 220/3
= 150/3 + 220/3
= 370/3
≈ ₹123.33

Flat price of Student Special = ₹110.

Since 110 < 123.33, the flat price of ₹110 is cheaper than buying 3 sandwiches and 2 cold drinks at individual prices.

∴ The Student Special (₹110) is cheaper by ₹(370/3 − 110) = ₹(370/3 − 330/3) = ₹40/3 ≈ ₹13.33.

Question 9

The pair of linear equations 3x + 7y = 5 and 9x + ky = 15 has infinitely many solutions. The value of k is:

Accepted Answer

Option (A) is correct.

Explanation: For coincident lines, the condition is a₁/a₂ = b₁/b₂ = c₁/c₂. Here a₁/a₂ = 5/10 = 1/2 and c₁/c₂ = 10/20 = 1/2. From b₁/b₂ = 1/2 ⟹ −2/(−k) = 1/2 ⟹ k = 4. All three ratios equal 1/2, confirming coincident lines. ∴ k = 4.

Question 10

If the lines given by 5x + 3ky = 1 and 10x + 9y − 7 = 0 are parallel (i.e., have no solution), then the value of k is:

Accepted Answer

Option (A) is correct.

**Explanation:** For two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to be parallel (no solution), the condition is a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Writing both equations in standard form: 5x + 3ky − 1 = 0 and 10x + 9y − 7 = 0, so a₁/a₂ = 5/10 = 1/2 and b₁/b₂ = 3k/9 = k/3. Setting 1/2 = k/3 ⟹ k = 3/2. Checking: c₁/c₂ = 1/7 ≠ 1/2, confirming the lines are parallel. ∴ k = 3/2.

Question 11

For what value of k will the pair of linear equations 3x + ky + 45 = 0 and 6x + 12y + 90 = 0 have infinitely many solutions?

Accepted Answer

Option (B) is correct.

**Explanation:** For a pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to have infinitely many solutions, the condition is a₁/a₂ = b₁/b₂ = c₁/c₂. Here, 3/6 = k/12 = 45/90 ⟹ 1/2 = k/12 = 1/2 ⟹ k = 6.

Question 12

For what value of k does the system of equations 6x + 3y + 12 = 0 and kx + 6y + 24 = 0 have infinitely many solutions?

Accepted Answer

Option (A) is correct.

**Explanation:** For a system a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to have infinitely many solutions, the condition a₁/a₂ = b₁/b₂ = c₁/c₂ must hold. Here a₁/a₂ = 6/k, b₁/b₂ = 3/6 = 1/2, and c₁/c₂ = 12/24 = 1/2. Setting 6/k = 1/2 ⟹ k = 12.

Question 13

For what value of k will the pair of linear equations 4x + ky = 10 and 8x + 6y = 20 have infinitely many solutions?

Accepted Answer

Option (A) is correct.

**Explanation:** For a pair of linear equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂ to have infinitely many solutions, the condition is a₁/a₂ = b₁/b₂ = c₁/c₂. Applying this: 4/8 = k/6 = 10/20 ⟹ each ratio = 1/2 ⟹ k/6 = 1/2 ∴ k = 3.

Question 14

The value of k for which the system of equations 4x + ky = 6 and 12x + 9y = 18 has infinitely many solutions is:

Accepted Answer

Option (A) is correct.

**Explanation:** For a system a₁x + b₁y = c₁ and a₂x + b₂y = c₂ to have infinitely many solutions, the condition is a₁/a₂ = b₁/b₂ = c₁/c₂. Here a₁/a₂ = 4/12 = 1/3 and c₁/c₂ = 6/18 = 1/3. For b₁/b₂ to equal 1/3: k/9 = 1/3 ⟹ k = 3.

Question 15

Assertion (A): The pair of equations x + 2y = 5 and 2x + 4y = 10 has infinitely many solutions.
Reason (R): If a₁/a₂ = b₁/b₂ = c₁/c₂, the equations are coincident.

Accepted Answer

Option (a) is correct.

**Explanation:** For the pair x + 2y = 5 and 2x + 4y = 10, we have a₁/a₂ = 1/2, b₁/b₂ = 2/4 = 1/2, and c₁/c₂ = 5/10 = 1/2. Since a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/2, the lines are coincident and the system has infinitely many solutions, confirming A is true. R states the correct condition for coincident lines and directly explains why A holds, so R is true and is the correct explanation of A.

Question 16

Assertion (A): The pair of equations x + 2y = 5 and 2x + 4y = 10 has infinitely many solutions.
Reason (R): If a₁/a₂ = b₁/b₂ = c₁/c₂, the equations are coincident.

Accepted Answer

Option (a) is correct.

Explanation: For the given equations, a₁/a₂ = 1/2, b₁/b₂ = 2/4 = 1/2, c₁/c₂ = 5/10 = 1/2. Since a₁/a₂ = b₁/b₂ = c₁/c₂, the equations are coincident and have infinitely many solutions, so A is true. The condition a₁/a₂ = b₁/b₂ = c₁/c₂ for coincident lines (infinitely many solutions) stated in R is correct, and R is the correct explanation of A.

Question 17

The pair of linear equations 3x = 4y + 2 and 12y = 9x – 6 represents two lines which are:

Accepted Answer

Option (C) is correct.

**Explanation:** Rewriting both equations in standard form: 3x − 4y − 2 = 0 …(i) and 9x − 12y − 6 = 0 …(ii). Dividing (ii) by 3 gives 3x − 4y − 2 = 0, which is identical to (i). Since a₁/a₂ = 3/9 = 1/3, b₁/b₂ = −4/−12 = 1/3, and c₁/c₂ = −2/−6 = 1/3, all three ratios are equal, so the pair of equations represents coincident lines.

Question 18

The pair of equations 2x + my = 7 and nx + 4y = 14 represent coincident lines, where m and n are positive integers. Which of the following is a possible value of (m + n)?

Accepted Answer

Option (A) is correct.

Explanation: For coincident lines, the condition is a₁/a₂ = b₁/b₂ = c₁/c₂. Here, 2/n = m/4 = 7/14 = 1/2. From 2/n = 1/2 ⟹ n = 4; from m/4 = 1/2 ⟹ m = 2, both positive integers. ∴ m + n = 2 + 4 = 6.

Question 19

For what value of k will the pair of linear equations 7x + 2y − 3 = 0 and 14x + ky − 6 = 0 have infinitely many solutions?

Accepted Answer

Option (B) is correct.

**Explanation:** For infinitely many solutions, the condition a₁/a₂ = b₁/b₂ = c₁/c₂ must hold. Here a₁/a₂ = 7/14 = 1/2 and c₁/c₂ = −3/−6 = 1/2, so b₁/b₂ = 2/k = 1/2 ⟹ k = 4.

Question 20

The value of k for which the system of equations 2x − 3y + 6 = 0 and 4x − ky + 12 = 0 has infinitely many solutions, is:

Accepted Answer

Option (B) is correct.

Explanation: For infinitely many solutions, the condition a₁/a₂ = b₁/b₂ = c₁/c₂ must hold. Here a₁/a₂ = 2/4 = 1/2 and c₁/c₂ = 6/12 = 1/2; setting b₁/b₂ = 3/k = 1/2 ⟹ k = 6.

Pair of Linear Equations in Two Variables: Class 10 Maths Practice Questions

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