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Difference between revisions of "2025 AMC 8 Problems/Problem 20"

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==Problem==
 
Sarika, Dev, and Rajiv are sharing a large block of cheese. They take turns cutting off half of what remains and eating it: first Sarika eats half of the cheese, then Dev eats half of the remaining half, then Rajiv eats half of what remains, then back to Sarika, and so on. They stop when the cheese is too small to see. About what fraction of the original block of cheese does Sarika eat in total?
 
Sarika, Dev, and Rajiv are sharing a large block of cheese. They take turns cutting off half of what remains and eating it: first Sarika eats half of the cheese, then Dev eats half of the remaining half, then Rajiv eats half of what remains, then back to Sarika, and so on. They stop when the cheese is too small to see. About what fraction of the original block of cheese does Sarika eat in total?
  
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<math>\hspace*{5mm}\text{(A) } \frac{4}{7} \quad \text{(B) } \frac{3}{5} \quad \text{(C) } \frac{2}{3} \quad \text{(D) } \frac{3}{4} \quad \text{(E) } \frac{7}{8}</math>
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<math>\textbf{(A)}\ \frac{4}{7} \qquad \textbf{(B)}\ \frac{3}{5} \qquad \textbf{(C)}\ \frac{2}{3} \qquad \textbf{(D)}\ \frac{3}{4} \qquad \textbf{(E)}\ \frac{7}{8}</math>
  
 
==Video Solution==
 
==Video Solution==
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==Solution 1==
 
==Solution 1==
  
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WLOG, let the amount of total cheese be 1. Then Sarika eat <math>\dfrac{1/2}</math>, Dev eats <math>\dfrac{1}{4}</math>, Rajiv eats <math>\dfrac{1}{8}</math>, Sarika eats <math>\dfrac{1}{16}</math> and so on. After a couple for attempts, we see that Sarika eats cheese in an infinite geometric sequence with first term <math>\dfrac{1}{2}</math> and common ratio of <math>\dfrac{1}{8}</math>. Therefore, we use the infinite geometric sequence formula and get <cmath>\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{8}}}=\dfrac{\dfrac{1}{2}}{\dfrac{7}{8}}=\dfrac{4}{7}</cmath> To find how much Sarika eats, we just divide this by our original total and get <math>\dfrac{\dfrac{4}{7}}{1}=1</math>.  
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WLOG, let the amount of total cheese be <math>1</math>. Then Sarika eat <math>\dfrac{1}{2}</math>, Dev eats <math>\dfrac{1}{4}</math>, Rajiv eats <math>\dfrac{1}{8}</math>, Sarika eats <math>\dfrac{1}{16}</math> and so on. After a couple for attempts, we see that Sarika eats cheese in an infinite geometric sequence with first term <math>\dfrac{1}{2}</math> and common ratio of <math>\dfrac{1}{8}</math>. Therefore, we use the infinite geometric sequence formula and get <cmath>\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{8}}=\dfrac{\dfrac{1}{2}}{\dfrac{7}{8}}=\dfrac{4}{7}</cmath> To find how much Sarika eats, we just divide this by our original total and get <math>\dfrac{\dfrac{4}{7}}{1}=1</math>.  
  
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Therefore, Sarika eats <math>\frac{4}{7}</math>, <math>\boxed{\textbf{(A)}}</math> of the cheese.
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Therefore, Sarika eats <math>\frac{4}{7}</math> <math>\boxed{\textbf{(A)}}</math> of the cheese.
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~athreyay
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==Solution 2 (Estimation)==
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Sarika eats 1/2 of the original cheese, and then because the others eat 1/4 and 1/8, she eats 1/16 next, and then 1/128, and then so on. Since the values later are going to be too small to make a huge difference, we can use these 3 values. She ate (64 + 8 + 1)/128 = 73/128. We can replace the 73 with a 72 for now, so 72/128 = 9/16, which simplifies to around 56.25. Since there is a little bit more of the cheese to be accounted for, the amount that she eats will be around (A): 4/7
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~Sigmacuber
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==Video Solution 1 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=jTTcscvcQmI
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==Video Solution by Thinking Feet==
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https://youtu.be/PKMpTS6b988
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==See Also==
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{{AMC8 box|year=2025|num-b=19|num-a=21}}
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{{MAA Notice}}

Latest revision as of 00:36, 31 January 2025

Problem

Sarika, Dev, and Rajiv are sharing a large block of cheese. They take turns cutting off half of what remains and eating it: first Sarika eats half of the cheese, then Dev eats half of the remaining half, then Rajiv eats half of what remains, then back to Sarika, and so on. They stop when the cheese is too small to see. About what fraction of the original block of cheese does Sarika eat in total?

$\textbf{(A)}\ \frac{4}{7} \qquad \textbf{(B)}\ \frac{3}{5} \qquad \textbf{(C)}\ \frac{2}{3} \qquad \textbf{(D)}\ \frac{3}{4} \qquad \textbf{(E)}\ \frac{7}{8}$

Video Solution

Key Idea: Let $x$ be the fraction eaten by Sarika. Then Dev eats $\frac{x}{2}$ and Rajiv eats $\frac{x}{4}$. Hence $x + \frac{x}{2} + \frac{x}{4} = 1$ so solving for $x$ we get $x = \frac{4}{7}$, $\boxed{\textbf{(A)}}$.

Video Link: https://www.youtube.com/watch?v=VUR5VYabbrc

Solution 1

WLOG, let the amount of total cheese be $1$. Then Sarika eat $\dfrac{1}{2}$, Dev eats $\dfrac{1}{4}$, Rajiv eats $\dfrac{1}{8}$, Sarika eats $\dfrac{1}{16}$ and so on. After a couple for attempts, we see that Sarika eats cheese in an infinite geometric sequence with first term $\dfrac{1}{2}$ and common ratio of $\dfrac{1}{8}$. Therefore, we use the infinite geometric sequence formula and get \[\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{8}}=\dfrac{\dfrac{1}{2}}{\dfrac{7}{8}}=\dfrac{4}{7}\] To find how much Sarika eats, we just divide this by our original total and get $\dfrac{\dfrac{4}{7}}{1}=1$.

Therefore, Sarika eats $\frac{4}{7}$ $\boxed{\textbf{(A)}}$ of the cheese.

~athreyay

Solution 2 (Estimation)

Sarika eats 1/2 of the original cheese, and then because the others eat 1/4 and 1/8, she eats 1/16 next, and then 1/128, and then so on. Since the values later are going to be too small to make a huge difference, we can use these 3 values. She ate (64 + 8 + 1)/128 = 73/128. We can replace the 73 with a 72 for now, so 72/128 = 9/16, which simplifies to around 56.25. Since there is a little bit more of the cheese to be accounted for, the amount that she eats will be around (A): 4/7

~Sigmacuber

Video Solution 1 by SpreadTheMathLove

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

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

See Also

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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