System of Linear Equations: Coffee blend prices & loyalty discount — MI 2025 H2 Math Prelim Paper 1
What this question tests
Question
Three different blends of coffee beans — Blend A, Blend B and Blend C — are sold at both M Café and I Café. The usual selling prices of the blends are the same at both cafés. The total cost of buying one package each of Blend A, Blend B and Blend C is \$176. During a holiday sale, the two cafés offered the following discounts:
| Discounts given for each package | Total price | |||
|---|---|---|---|---|
| Café | Blend A | Blend B | Blend C | after the discount |
| M Café | 15% | 10% | 5% | $161.80 |
| I Café | 7% | 5% | 2% | $169.56 |
- Find the usual selling price of each package of Blend A, Blend B and Blend C coffee beans respectively.
Loyal customers of I Café receive an additional discount of \(x\)% on the usual selling price of the packages for Blend A and Blend B only. (This loyalty discount is calculated as \(x\)% of the usual price of each package and is subtracted from the sale prices above.)
I Café wants the total cost for its loyal customers to be less than the total cost at M Café when buying one of each package of the three blends.
- Form an inequality and find the smallest integer value of \(x\).
Show full worked solution▾
(i) Let \(x\), \(y\) and \(z\) be the usual selling price of Blend A, B and C coffee beans respectively. \[\begin{aligned} x + y + z &= 176 \quad\text{--- (1)}\\ 0.85x + 0.90y + 0.95z &= 161.80 \quad\text{--- (2)}\\ 0.93x + 0.95y + 0.98z &= 169.56 \quad\text{--- (3)} \end{aligned}\]
From GC:
\(x = \$32\), \(y = \$44\), \(z = \$100\).
Blend A \(= \$32\), Blend B \(= \$44\), Blend C \(= \$100\)
(ii) Price difference: \(169.56 - 161.80 = 7.76\).
To make I Café more attractive for loyal customers:
Additional loyalty discount \(>\) price difference. \[ \frac{x}{100}(32) + \frac{x}{100}(44) > 7.76 \] \[ x > 10.2\ldots \]
The smallest integer value of \(x\) is \(11\).