Consider another situation. A science exhibition charges an entry fee that is different for adults and children. Here are the amount paid by two different groups.
Group A: 2 adult tickets and 3 child tickets for ₹600
Group B: 3 adult tickets and 2 child tickets for ₹700
How much do individual adult and child tickets cost?
If the cost of one adult ticket is ₹x and one child ticket is ₹y, then: 2x + 3y = 600, 3x + 2y = 700.
This gives us another pair of linear equations in two variables. A solution to these equations will give us the answer that we seek. Can you see why?
Step-by-step solution
Idea: Each group’s bill is one equation. The real prices must fit both bills at once, so they form the common solution of the pair. Here the coefficients are swapped, so adding and subtracting the equations is quickest.
- Why a solution answers the question: the real prices (x, y) make Group A’s bill 2x + 3y equal ₹600 and Group B’s bill 3x + 2y equal ₹700. A pair that satisfies both equations is exactly a pair of prices that matches both bills.1 mark
- Add the equations: 5x + 5y = 1300, so x + y = 260 … (3).½ mark
- Subtract the first from the second: x − y = 100 … (4).½ mark
- (3) + (4): 2x = 360, x = 180. Then y = 260 − 180 = 80. Adult ticket ₹180, child ticket ₹80.1 mark
Check: Group A: 2(180) + 3(80) = 360 + 240 = 600 ✓. Group B: 3(180) + 2(80) = 540 + 160 = 700 ✓.
Answer to write in the exam
Let adult ticket = ₹x, child ticket = ₹y
2x + 3y = 600 … (1); 3x + 2y = 700 … (2)
(1) + (2): 5x + 5y = 1300 ⇒ x + y = 260 … (3)
(2) − (1): x − y = 100 … (4)
(3) + (4): 2x = 360 ⇒ x = 180; y = 80
∴ Adult ticket = ₹180, child ticket = ₹80
Common mistakes that cost marks
- Solving only one equation. 2x + 3y = 600 alone has many solutions, e.g. (300, 0) or (0, 200); only the common solution fits both groups.
- After adding, forgetting to divide by 5 and writing x + y = 1300.
How this can come in the exam
A zoo charges different fees for adults and children. One family pays ₹520 for 3 adults and 2 children; another pays ₹480 for 2 adults and 3 children.
(i) Form the equations. (ii) Find each fee. (iii) How much would 1 adult and 4 children pay?
Show answer
(i) 3a + 2c = 520, 2a + 3c = 480 (1 mark). (ii) Add: a + c = 200; subtract: a − c = 40 ⇒ a = 120, c = 80 (2 marks). (iii) 120 + 320 = ₹440 (1 mark).Try one yourself
4 adult and 1 child ticket cost ₹460; 1 adult and 4 child tickets cost ₹340. Find each price.
Show answer
Add: 5(x + y) = 800 ⇒ x + y = 160; subtract: 3(x − y) = 120 ⇒ x − y = 40 ⇒ adult ₹100, child ₹60.
More questions like this
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