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2024 AMC 10A Problems/Problem 12: Difference between revisions

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~NSAoPS
~NSAoPS
==Video Solution(Faster computation)==
https://youtu.be/l3VrUsZkv8I


== Video Solution by Pi Academy ==
== Video Solution by Pi Academy ==

Revision as of 00:50, 15 May 2025

Problem

Zelda played the Adventures of Math game on August 1 and scored $1700$ points. She continued to play daily over the next $5$ 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 $1700 + 80 = 1780$ points.) What was Zelda's average score in points over the $6$ days?

$\textbf{(A)} 1700\qquad\textbf{(B)} 1702\qquad\textbf{(C)} 1703\qquad\textbf{(D)}1713\qquad\textbf{(E)} 1715$

Solution 1

Going through the table, we see her scores over the six days were: $1700$, $1700+80=1780$, $1780-90=1690$, $1690-10=1680$, $1680+60=1740$, and $1740-40=1700$.

Taking the average, we get $\frac{(1700+1780+1690+1680+1740+1700)}{6} = \boxed{\textbf{(E) } 1715}.$

-i_am_suk_at_math_2

Solution 2

Compared to the first day $(1700)$, her scores change by $+80$, $-10$, $-20$, $+40$, and $+0$. So, the average is $1700 + \frac{80-10-20+40+0}{6} = \boxed{\textbf{(E) }1715}$.

-mathfun2012

Solution 3

As the scores of each day are dependent on previous days, we get: $1700 + \dfrac{0\cdot6 + 80\cdot5 + (-90)\cdot4 + (-10)\cdot3 + 60\cdot2 + (-40)\cdot1}{6} = \boxed{\textbf{(E) }1715}$

~NSAoPS

Video Solution(Faster computation)

https://youtu.be/l3VrUsZkv8I

Video Solution by Pi Academy

https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM

Video Solution 1 by Power Solve

https://youtu.be/mfTDSXH9j2g

Video Solution by Daily Dose of Math

https://youtu.be/t8Dpj7dHZ3s

~Thesmartgreekmathdude

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

Video Solution by Just Math⚡

https://www.youtube.com/watch?v=o1Kiz_lZc90

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

These problems are copyrighted © by the Mathematical Association of America.

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