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A print shop charges exactly $0.35 to print one small flyer and exactly $0.55 to print one large flyer. A single order, containing at least one flyer of each size, cost exactly $7.20. What is the greatest possible number of flyers in the order?
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Correct answer
C
Let s be the number of small flyers and l the number of large flyers, with s at least 1 and l at least 1. Then 0.35s + 0.55l = 7.20. Multiply through by 20 to clear the decimals: 7s + 11l = 144.
Method 1: narrow l with modular arithmetic. Take 7s + 11l = 144 and reduce it modulo 7. The 7s term contributes 0, and 11 ≡ 4 and 144 ≡ 4 (mod 7), so 4l ≡ 4 (mod 7), which gives l ≡ 1 (mod 7). Since 11l ≤ 144 forces l ≤ 13, the only candidates are l = 1 and l = 8. For l = 1: s = (144 − 11)/7 = 133/7 = 19, a total of 20 flyers. For l = 8: s = (144 − 88)/7 = 56/7 = 8, a total of 16 flyers. The two valid orders contain 20 and 16 flyers, so the greatest possible number is 20. Check: 0.35(19) + 0.55(1) = 6.65 + 0.55 = 7.20. The answer is (C).
Method 2: trade large flyers for small ones. The small flyer is cheaper, so to fit more flyers into the same $7.20 you want as many small flyers as possible. Replacing 7 large flyers with 11 small flyers leaves the cost unchanged, because 7 × $0.55 = $3.85 = 11 × $0.35, while the head count rises by 11 − 7 = 4. So each such trade keeps the total at $7.20 and adds 4 flyers. Starting from the balanced order of 8 small and 8 large (which costs exactly $7.20), one trade drops the large flyers to 1 and raises the small flyers to 19, for 20 flyers. You cannot trade again without taking large flyers below 1, which the requirement of at least one of each forbids, so 20 is the maximum, confirming (C).
Why the other choices tempt: (A) 16 is the balanced order of 8 small and 8 large, a genuine valid order costing exactly $7.20, which is why it is tempting; but it is not the largest count, since shifting toward small flyers reaches 20. (B) 18 averages the two prices to $0.40 and divides 7.20 by 0.40, ignoring that the counts must be whole numbers of two distinct prices. (E) 24 makes the same averaging error in the other direction, using $0.30 per flyer. (D) 22 abandons both the whole-number and the at-least-one-large requirements and simply pads the count of cheap flyers.
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