A swimming club sells single-visit passes at \(\pounds x\) and 10-visit passes at \(\pounds y\). On Monday, 8 single-visit and 3 ten-visit passes brought in...
Question 1 Report
A swimming club sells single-visit passes at \(\pounds x\) and 10-visit passes at \(\pounds y\). On Monday, 8 single-visit and 3 ten-visit passes brought in £166. On Tuesday, 5 single-visit and 2 ten-visit passes brought in £109.
- Form two equations in \(x\) and \(y\). (1)
- Solve these equations simultaneously to find \(x\) and \(y\). (4)
- A member buys one 10-visit pass instead of paying for 10 single visits. Work out how much the member saves. (2)
Answer Details
Two days of pass sales, each combining single-visit and 10-visit pass counts, give simultaneous equations in the two prices, and the resulting prices then let a saving from buying the discounted pass be calculated directly.
(a) Letting \(x\) be the single-visit pass price and \(y\) the 10-visit pass price:
\[8x+3y=166 \qquad 5x+2y=109\] [1 mark](b) Multiplying the first equation by 2 and the second by 3 so the \(y\) coefficients match: \(16x+6y=332\) and \(15x+6y=327\). Subtracting eliminates \(y\):
\[(16x+6y)-(15x+6y)=332-327\] \[x=5\]Substituting back into the second original equation: \(5(5)+2y=109\), so \(2y=84\), giving \(y=42\). So a single-visit pass costs \(\pounds5\) and a 10-visit pass costs \(\pounds42\). [4 marks]
(c) Ten single visits paid individually would cost \(10\times5=\pounds50\). Buying the 10-visit pass instead costs \(\pounds42\), so the member saves:
\[50-42=\pounds8\] [2 marks]Choosing multipliers (2 and 3) that make the \(y\) coefficients equal is what allows elimination in one subtraction; a different pair of multipliers, matching the \(x\) coefficients instead, would work just as well and give the same solution.
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