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

Revision as of 00:35, 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 (B): 4/7

~Soupboy0

Video Solution 1 by SpreadTheMathLove

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

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

@@ Line 18: / Line 18: @@
 ~athreyay
-==Solution 2 (If you forgot the infinite geometric series formula)==
+==Solution 2 (Estimation)==
-At first, Sarika eats <math>\frac{1}{2}</math> of the cheese, and then after <math>2</math> more people eat, there is <math>\frac{1}{8}</math> of the cheese remaining, so Sarika eats <math>\frac{1}{16}</math> of the cheese. Continuing this pattern until a reasonable amount, we get new fractions of <math>\frac{1}{128}</math> and <math>\frac{1}{1024}</math>. Adding these fractions together yields <math>\frac{1}{2}+\frac{1}{16}+\frac{1}{128}+\frac{1}{1024} = \frac{512+64+8+1}{1024} = \frac{585}{1024}</math>. Approximating this answer yields about <math>0.5713</math> which is about <math>\frac{4}{7}</math> <math>\boxed{\textbf{(A)}}</math> of the cheese.
+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 (B): 4/7
 ~Soupboy0

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
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