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Difference between revisions of "2001 AMC 8 Problems/Problem 17"

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==Video Solution==
 
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https://www.youtube.com/shorts/3vlKd6yf3rU Soo, DRMS, NM
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https://www.youtube.com/watch?v=S8hyVeObqUs Soo, DRMS, NM
  
 
==See Also==
 
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{{AMC8 box|year=2001|num-b=16|num-a=18}}
 
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Latest revision as of 17:16, 10 January 2024

Problem

For the game show Who Wants To Be A Millionaire?, the dollar values of each question are shown in the following table (where K = 1000).

\[\begin{tabular}{rccccccccccccccc} \text{Question} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\ \text{Value} & 100 & 200 & 300 & 500 & 1\text{K} & 2\text{K} & 4\text{K} & 8\text{K} & 16\text{K} & 32\text{K} & 64\text{K} & 125\text{K} & 250\text{K} & 500\text{K} & 1000\text{K} \end{tabular}\]

Between which two questions is the percent increase of the value the smallest?

$\text{(A)}\ \text{From 1 to 2} \qquad \text{(B)}\ \text{From 2 to 3} \qquad \text{(C)}\ \text{From 3 to 4} \qquad \text{(D)}\ \text{From 11 to 12} \qquad \text{(E)}\ \text{From 14 to 15}$

Solution

Notice that in two of the increases, the dollar amount doubles. The increases in which this is not true is $2$ to $3$, $3$ to $4$, and $11$ to $12$. We can disregard $11$ to $12$ since that increase is almost $2$ times. The increase from $2$ to $3$ is $\frac{300-200}{200}=\frac{1}{2}$, so it's only multiplied by a factor of $1.5$. The increase from $3$ to $4$ is $\frac{500-300}{300}=\frac{2}{3}$, so it is multiplied by a factor of $1.\bar{6}$. Therefore, the smallest percent increase is $\text{From 2 to 3}, \boxed{\text{B}}$

Video Solution

https://www.youtube.com/watch?v=S8hyVeObqUs Soo, DRMS, NM

See Also

2001 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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

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