Difference between revisions of "2023 AMC 8 Problems/Problem 9"

Revision as of 13:56, 6 August 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? $[asy] // Diagram by TheMathGuyd. Found cubic, so graph is perfect. import graph; size(8cm); int i; for(i=1; i<9; i=i+1) { draw((-0.2,2i-1)--(16.2,2i-1), mediumgrey); draw((2i-1,-0.2)--(2i-1,16.2), mediumgrey); draw((-0.2,2i)--(16.2,2i), grey); draw((2i,-0.2)--(2i,16.2), grey); } Label f; f.p=fontsize(6); xaxis(-0.5,17.8,Ticks(f, 2.0),Arrow()); yaxis(-0.5,17.8,Ticks(f, 2.0),Arrow()); real f(real x) { return -0.03125 x^(3) + 0.75x^(2) - 5.125 x + 14.5; } draw(graph(f,0,15.225),currentpen+1); real dpt=2; real ts=0.75; transform st=scale(ts); label(rotate(90)*st*"Elevation (meters)",(-dpt,8)); label(st*"Time (seconds)",(8,-dpt)); [/asy]$ $\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14$

Solution 1

We shall mark the time intervals in which Malaika's elevation is between $4$ and $7$ meters in red, as shown below: $[asy] // Diagram by TheMathGuyd. Found cubic, so graph is perfect. import graph; size(8cm); int i; for(i=1; i<9; i=i+1) { draw((-0.2,2i-1)--(16.2,2i-1), mediumgrey); draw((2i-1,-0.2)--(2i-1,16.2), mediumgrey); draw((-0.2,2i)--(16.2,2i), grey); draw((2i,-0.2)--(2i,16.2), grey); } Label f; f.p=fontsize(6); xaxis(-0.5,17.8,Ticks(f, 2.0),Arrow()); yaxis(-0.5,17.8,Ticks(f, 2.0),Arrow()); real f(real x) { return -0.03125 x^(3) + 0.75x^(2) - 5.125 x + 14.5; } draw(graph(f,0,15.225),currentpen+1); draw(graph(f,2,4)^^graph(f,6,10)^^graph(f,12,14),red+currentpen+2); real dpt=2; real ts=0.75; transform st=scale(ts); label(rotate(90)*st*"Elevation (meters)",(-dpt,8)); label(st*"Time (seconds)",(8,-dpt)); [/asy]$ The requested time intervals 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, Hmssoni82

Solution 2

Notice that the entire section between the $2$ second mark and the $14$ second mark is between the $4$ and $7$ feet elevation level except the $2$ seconds where she skis just under the $4$ feet mark and when she skis just above the $7$ feet mark, making the answer $14-2-2-2=\boxed{\textbf{(B)}\ 8}.$

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/wQd1lHtkVPU

~Education the Study of everything

Video Solution by SpreadTheMathLove

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

Video Solution by Magic Square

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

Video Solution by Interstigation

https://youtu.be/1bA7fD7Lg54?t=591

Video Solution by harungurcan

https://www.youtube.com/watch?v=oIGy79w1H8o&t=15s

~harungurcan

@@ Line 31: / Line 31: @@
 ==Solution 1==
-We mark the time intervals in which Malaika's elevation is between <math>4</math> and <math>7</math> meters in red, as shown below:
+We shall mark the time intervals in which Malaika's elevation is between <math>4</math> and <math>7</math> meters in red, as shown below:
 <asy>
 // Diagram by TheMathGuyd. Found cubic, so graph is perfect.
@@ Line 70: / Line 70: @@
 In total, Malaika spends <math>(4-2) + (10-6) + (14-12) = \boxed{\textbf{(B)}\ 8}</math> seconds at such elevation.
-~apex304, MRENTHUSIASM
+~apex304, MRENTHUSIASM, Hmssoni82
 ==Solution 2==

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