A-Level Equations and inequalities: worked solution

Question

Pure Mathematics. Green Harvest Market was running a promotion on several grocery items. Organic free-range eggs carried a 15% discount, and there was a $1 reduction for every 2 packets of chia seeds bought. Organic quinoa was sold at full price. The table lists the total amounts paid and the quantities of organic quinoa, organic free-range eggs and chia seeds purchased by Priya, Daniel and Mei. Work out the original selling price of one packet of each of the three items, giving each answer to the nearest cent. Quinoa | Eggs | Chia seeds | Total Bill ($) Priya 3 | 1 | 2 | 71.28 Daniel 2 | 2 | 5 | 91.85 Mei 6 | 3 | 3 | 144.43

Let q = original selling price of one packet of organic quinoa, e = original selling price of one packet of organic free-range eggs, c = original selling price of one packet of chia seeds, all in dollars. Step 1: Turn each promotion into arithmetic before writing any equation. - Quinoa is at full price, so n packets cost nq. - Eggs carry a 15% discount, so each packet is paid at 100% - 15% = 85% of e, that is 0.85e per packet. - Chia seeds have $1 off for every 2 packets, so the reduction is $1 for each COMPLETE pair bought: 2 packets -> 1 complete pair -> $1 off 5 packets -> 2 complete pairs -> $2 off (the 5th packet is not part of a pair) 3 packets -> 1 complete pair -> $1 off Step 2: Write one equation per shopper, then move the fixed reduction to the right-hand side. Priya, 3 quinoa, 1 eggs, 2 chia: 3q + 0.85e + 2c - 1 = 71.28 3q + 0.85e + 2c = 72.28 ... (1) Daniel, 2 quinoa, 2 eggs, 5 chia: 2q + 1.70e + 5c - 2 = 91.85 2q + 1.70e + 5c = 93.85 ... (2) Mei, 6 quinoa, 3 eggs, 3 chia: 6q + 2.55e + 3c - 1 = 144.43 6q + 2.55e + 3c = 145.43 ... (3) Step 3: Eliminate q. Scale (1) so that its q-term matches (3). 2 x (1): 6q + 1.70e + 4c = 144.56 (3) - 2 x (1): (2.55 - 1.70)e + (3 - 4)c = 145.43 - 144.56 0.85e - c = 0.87 ... (4) Scale (2) as well so that it too has 6q. 3 x (2): 6q + 5.10e + 15c = 281.55 3 x (2) - 2 x (1): (5.10 - 1.70)e + (15 - 4)c = 281.55 - 144.56 3.40e + 11c = 136.99 ... (5) Step 4: Solve (4) and (5), two equations in two unknowns. From (4), c = 0.85e - 0.87. Substitute into (5): 3.40e + 11(0.85e - 0.87) = 136.99 3.40e + 9.35e - 9.57 = 136.99 12.75e = 146.56 e = 146.56/12.75 = 11.49490 Then c = 0.85(11.49490) - 0.87 = 9.77067 - 0.87 = 8.90067. Step 5: Back-substitute into (1) to get q. 3q = 72.28 - 0.85(11.49490) - 2(8.90067) 3q = 72.28 - 9.77067 - 17.80133 3q = 44.708 q = 14.90267 Step 6: Round each to the nearest cent. quinoa q = $14.90, eggs e = $11.49, chia seeds c = $8.90. Check with Daniel's bill: 2(14.90267) + 1.70(11.49490) + 5(8.90067) - 2 = 29.80533 + 19.54133 + 44.50333 - 2 = 91.85, which matches.