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

(Video Solution 1 by SpreadTheMathLove)
Line 31: Line 31:
  
 
==Solution==
 
==Solution==
The time intervals in which Malaika's elevation is between <math>4</math> and <math>7</math> meters are:
+
We mark the time intervals in which Malaika's elevation is between <math>4</math> and <math>7</math> meters, as shown below:
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<asy>
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// Diagram by TheMathGuyd. Found cubic, so graph is perfect.
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import graph;
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size(8cm);
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int i;
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for(i=1; i<9; i=i+1)
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{
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draw((-0.2,2i-1)--(16.2,2i-1), mediumgrey);
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draw((2i-1,-0.2)--(2i-1,16.2), mediumgrey);
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draw((-0.2,2i)--(16.2,2i), grey);
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draw((2i,-0.2)--(2i,16.2), grey);
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}
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Label f;
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f.p=fontsize(6);
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xaxis(-0.5,17.8,Ticks(f, 2.0),Arrow());
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yaxis(-0.5,17.8,Ticks(f, 2.0),Arrow());
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real f(real x)
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{
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return -0.03125 x^(3) + 0.75x^(2) - 5.125 x + 14.5;
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}
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draw(graph(f,0,15.225),currentpen+1);
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draw(graph(f,2,4)^^graph(f,6,10)^^graph(f,12,14),red+currentpen+1.5);
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real dpt=2;
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real ts=0.75;
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transform st=scale(ts);
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label(rotate(90)*st*"Elevation (meters)",(-dpt,8));
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label(st*"Time (seconds)",(8,-dpt));
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</asy>
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The requested time intervals are:
  
 
* from the <math>2</math>nd to the <math>4</math>th seconds
 
* from the <math>2</math>nd to the <math>4</math>th seconds

Revision as of 04:18, 26 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? [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

We mark the time intervals in which Malaika's elevation is between $4$ and $7$ meters, 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+1.5); 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

Video Solution 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 (ProblemsAnswer KeyResources)
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

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