Difference between revisions of "2024 AMC 10A Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
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>. | 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> | + | |
| + | 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 | -i_am_suk_at_math_2 | ||
Revision as of 18:13, 8 November 2024
Contents
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 1
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
Solution 2
Compared to the first day 
, her scores change by 
, 
, 
, 
, and 
. So, the average is 
.
-mathfun2012
See Also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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. 
