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2024 AMC 10A Problem 12

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2024 AMC 10A Problem 12AMC 10 Step by Step
Name:                             Date: October 28, 2026
Statistics & Data · difficulty 2 of 5 · about 2 min · position P11-15

Zelda played the Adventures of Math game on August 1 and scored 1,7001,700 points. She continued to play daily over the next 55 days. The bar chart below shows the daily change in her score compared to the day before. (For example, Zelda's score on August 2 was 1,700+80=1,7801,700 + 80 = 1,780 points.) What was Zelda's average score in points over the 66 days? figure

  1. A)17001700
  2. B)17021702
  3. C)17031703
  4. D)17131713
  5. E)17151715

Answer

E

Key insight

Turn the day-to-day changes into actual scores with a running total. Then average how far each score is above 1700 instead of adding six big numbers.

Solution

The chart shows how much the score changed from the day before, not the score itself. So first work out each day's actual score.

  1. Reading the bars for August 2 through 6, the changes are +80+80, −90-90, −10-10, +60+60, −40-40.
  2. Keep a running total starting from 17001700:
Aug 1Aug 2Aug 3Aug 4Aug 5Aug 6
170017001780178016901690168016801740174017001700
  1. Adding six four-digit numbers is slow, so look only at how far each score is from 17001700: 0,+80,−10,−20,+40,00, +80, -10, -20, +40, 0. These add to 9090.
  2. Share that 9090 over the 66 days: 90÷6=1590 \div 6 = 15. So on average the score is 1515 above 17001700.

The average score is 1700+15=17151700 + 15 = 1715.

The answer is (E) 1715\boxed{\textbf{(E)}\ 1715}.

Why this works

A chart of changes has to be turned into real values by adding up as you go, before you can average. When all the values are near one round number, average only the amounts above or below that number and add it back at the end; the arithmetic stays small.

The trap

Treating each bar as the score's distance from 1700 instead of the change from the day before, or averaging over only the five later days.

Concise solution

The bars give the day-to-day changes +80,−90,−10,+60,−40+80, -90, -10, +60, -40 for August 2 through 6. Accumulate them starting from 17001700:

Aug 1Aug 2Aug 3Aug 4Aug 5Aug 6
170017801690168017401700

Rather than adding six four-digit numbers, add the amounts above 17001700: 0+80−10−20+40+0=900 + 80 - 10 - 20 + 40 + 0 = 90. Spread over 66 days that is 1515 per day, so the average score is 1700+15=17151700 + 15 = 1715.

The answer is (E) 1715\boxed{\textbf{(E)}\ 1715}.

Review in the AoPS books

Techniques

Set up the equation/formula and compute; no special trick needed