Difference between revisions of "2024 AMC 10A Problems/Problem 12"
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==Solution== | ==Solution== | ||
| − | + | Going through the table, we see her scores over the six days were: <math>1700</math>, <math>1700+80=1780</math>, <math>1780-90=1690</math>, <math>1690-10=1680</math>, <math>1680+60=1740</math>, and <math>1740-40=1700</math>. | |
| + | Taking the average, we get <math>\frac{(1700+1780+1690+1680+1740+1700)}{6} = \boxed{\textbf{(E) } 1715}.</math> | ||
| + | |||
| + | -i_am_suk_at_math_2 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2024|ab=A|num-b=10|num-a=12}} | {{AMC10 box|year=2024|ab=A|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:02, 8 November 2024
Problem
Zelda played the Adventures of Math game on August 1 and scored 
points. She continued to play daily over the next 
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 
points.) What was Zelda's average score in points over the 
days?
Solution
Going through the table, we see her scores over the six days were: , 
, 
, 
, 
, and 
.
Taking the average, we get 
-i_am_suk_at_math_2
See Also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
