Given below are some everyday situations. Construct a pair of linear equations in two variables for each situation:
On two different days a family buys movie tickets and snack boxes. Let the cost of one movie ticket be ₹x and the cost of one snack box be ₹y. Frame a pair of linear equations in x and y to represent the following situations:
- (i) On the first day, the family buys 2 movie tickets and 3 snack boxes for ₹850.
- (ii) On the second day, the family buys 4 movie tickets and 1 snack box for ₹1100.
Answer: (i) 2x + 3y = 850 (ii) 4x + y = 1100.
Step-by-step solution
Idea: For each day: (number of tickets × price of a ticket) + (number of snack boxes × price of a box) = amount paid.
(i) On the first day, the family buys 2 movie tickets and 3 snack boxes for ₹850.
- Cost = 2 × x + 3 × y = 850, so 2x + 3y = 850.1 mark
2x + 3y = 850
(ii) On the second day, the family buys 4 movie tickets and 1 snack box for ₹1100.
- Cost = 4 × x + 1 × y = 1100, so 4x + y = 1100. Together, (i) and (ii) form the pair.1 mark
4x + y = 1100
2x + 3y = 850 and 4x + y = 1100.
Check: Solving the pair gives x = 245, y = 120: 490 + 360 = 850 ✓ and 980 + 120 = 1100 ✓, so the equations describe sensible prices.
Answer to write in the exam
(i)
Cost of 2 tickets + 3 snack boxes = ₹850
∴ 2x + 3y = 850
(ii)
Cost of 4 tickets + 1 snack box = ₹1100
∴ 4x + y = 1100
Common mistakes that cost marks
- Writing 3x + 2y = 850 by mixing up which number goes with tickets.
- Writing 4x + 1 = 1100 and dropping the variable of the single snack box.
How this can come in the exam
MCQ (1 mark)
5 notebooks and 2 pens cost ₹130. With a notebook ₹n and a pen ₹p, the equation is
- 2n + 5p = 130
- 5n + 2p = 130
- 7(n + p) = 130
- 5n − 2p = 130
Show answer
(B) 5n + 2p = 130
5 × n + 2 × p = 130.
Try one yourself
3 kg rice and 2 kg dal cost ₹390; 1 kg rice and 4 kg dal cost ₹530. Form the pair of equations.
Show answer
With rice ₹r/kg and dal ₹d/kg: 3r + 2d = 390 and r + 4d = 530.
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