AMC 10 Step by Step
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2007 AMC 10A Problem 1

P1-10~1 minPrint
Ratios, Percents & AveragesSet up the equation/formula and compute; no special trick needed

One ticket to a show costs $20\$20 at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickets using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?

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Key insight

Turn each discount into a price per ticket first: 25% off 20 dollars is 15 dollars, and 30% off is 14 dollars. Then multiply by the number of tickets.

Solution

Work out what one ticket costs for each person, then multiply by how many tickets they buy.

  1. Susan's ticket price. A 25%25\% discount on $20\$20 takes off 14\frac{1}{4} of 2020, which is 55. So each ticket costs her 20−5=1520 - 5 = 15 dollars.
  2. Susan's total. She buys 44 tickets: 4×15=604 \times 15 = 60 dollars.
  3. Pam's ticket price. A 30%30\% discount on $20\$20 takes off 0.3×20=60.3 \times 20 = 6. So each ticket costs her 20−6=1420 - 6 = 14 dollars.
  4. Pam's total. She buys 55 tickets: 5×14=705 \times 14 = 70 dollars.
  5. Difference. Pam pays 70−60=1070 - 60 = 10 dollars more.

The answer is (C) 10\boxed{\textbf{(C)}\ 10}.

Why this works

A discount changes the price of each ticket, so find the new price of one ticket before doing anything else. After that it is ordinary multiplication and subtraction. Pam gets the bigger discount, but she buys one more ticket, and the extra ticket costs more than her savings.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

A 25%25\% discount means Susan pays 75%75\% of full price on each ticket:
4⋅20⋅0.75=4⋅15=60 dollars.4 \cdot 20 \cdot 0.75 = 4 \cdot 15 = 60 \text{ dollars}.
A 30%30\% discount means Pam pays 70%70\% of full price on each ticket:
5⋅20⋅0.70=5⋅14=70 dollars.5 \cdot 20 \cdot 0.70 = 5 \cdot 14 = 70 \text{ dollars}.
Pam pays 70−60=1070 - 60 = 10 dollars more.

The answer is (C) 10\boxed{\textbf{(C)}\ 10}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site. Also appeared as 2007 12A #1.