Daily Math Puzzle: 2026-06-05

$132 — First note the cost per ride for each option: single $12, 3‑bundle $30 → $10 per ride, 5‑bundle $45 → $9 per ride. To minimise cost we want as many 5‑bundles as possible without exceeding 14 rides. Try 2 bundles of 5 rides = 10 rides. We still need 4 rides. The cheapest way to get 4 rides is a 3‑bundle (cost $30) plus one single ticket (cost $12). Total cost = 2×$45 + $30 + $12 = $90 + $30 + $12 = $132. Check other combinations: - 1×5‑bundle + 3×3‑bundles = 5 + 9 = 14 rides, cost = $45 + 3×$30 = $135. - 4×3‑bundles + 2 singles = 12 + 2 = 14 rides, cost = 4×$30 + 2×$12 = $144. - Any use of more singles only raises the total because singles are the most expensive per ride. Thus the cheapest possible total is $132, which corresponds to choice A.