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Difference between revisions of "2023 AMC 8 Problems/Problem 9"
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<math>\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14</math>
 
<math>\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14</math>
  
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==Solution==
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==Solution 1==
 
The time intervals in which Malaika's elevation is between <math>4</math> and <math>7</math> meters are:
 
The time intervals in which Malaika's elevation is between <math>4</math> and <math>7</math> meters are:
  
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~apex304, MRENTHUSIASM
 
~apex304, MRENTHUSIASM
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 +
==Solution 2==
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 +
We count the number of seconds and get <math>\boxed{\textbf{(B)}\ 8}</math>
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 +
Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]]
  
 
== Video Solution 1 by SpreadTheMathLove==  
 
== Video Solution 1 by SpreadTheMathLove==  

Revision as of 12:16, 25 January 2023

Problem

Malaika is skiing on a mountain. The graph below shows her elevation, in meters, above the base of the mountain as she skis along a trail. In total, how many seconds does she spend at an elevation between $4$ and $7$ meters?

2023 AMC 8-9.png

$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14$

Solution 1

The time intervals in which Malaika's elevation is between $4$ and $7$ meters are:

  • from the $2$nd to the $4$th seconds
  • from the $6$th to the $10$th seconds
  • from the $12$th to the $14$th seconds

In total, Malaika spends $(4-2) + (10-6) + (14-12) = \boxed{\textbf{(B)}\ 8}$ seconds at such elevation.

~apex304, MRENTHUSIASM

Solution 2

We count the number of seconds and get $\boxed{\textbf{(B)}\ 8}$

Solution by ILoveMath31415926535

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=lfyg5ZMV0gg

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=4903

See Also

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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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