Julius and Wilson are selling fruit for a school fundraiser.  Customers can buy small boxes of oranges and large boxes of oranges.  Julius sold 3 small boxes and 14 large boxes for a total of \$203.  Wilson sold 11 small boxes and 1 large boxes for a total of \$220.  Find the cost of one small box and one large box.

Let S and L be the cost in dollars of a small and large box respectively.

3S+14L=203 and 11S+L=220, so L=220-11S. (Note that L=11(20-S), so S cannot exceed 20, and if L>S it cannot exceed \$18.33. For example, if S=\$18, then L=\$22 and if S=\$19, L=\$11, cheaper than the small box!)

Therefore, by substitution: 3S+14(220-11S)=203.

3S+3080-154S=203; 2877=151S, S=2877/151=\$19.05 approx. L=220-11*19.05=\$10.42 approx.

S+L=\$29.47 approx. However, this seems wrong because the cost of a small box is bigger than the cost of a large box, so S<\$18.33 is violated. This suggest an error in the question if the calculations are correct.

If the question had, for example, contained \$2023 instead of \$203 the solution would be:

3S+14L=2023, 11S+L=220, so L=220-11S.

Substituting: 3S+14(220-11S)=2023, 3S+3080-154S=2023, 151S=1057, S=\$7 and L=\$143.

S+L=\$150, cost of one small and one large box.

A MORE LIKELY ANSWER AND CORRECTION

11S+11L=220, so S+L=20 and S=20-L.

Substituting for S: 3(20-L)+14L=203, 60-3L+14L=203, 11L=143 and L=\$13 and S=\$7. The small box of oranges costs \$7 and the large one costs \$13.

by Top Rated User (645k points)