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Difference between revisions of "2023 AMC 8 Problems/Problem 9"
 
(39 intermediate revisions by 18 users not shown)
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==Problem==  
 
==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?
+
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 <math>4</math> and <math>7</math> 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>
 +
<math>\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14</math>
 +
 
 +
==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:
 +
<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 <math>2</math>nd to the <math>4</math>th seconds
 +
 
 +
* from the <math>6</math>th to the <math>10</math>th seconds
 +
 
 +
* from the <math>12</math>th to the <math>14</math>th seconds
 +
 
 +
In total, Malaika spends <math>(4-2) + (10-6) + (14-12) = \boxed{\textbf{(B)}\ 8}</math> seconds at such elevation.
 +
 
 +
~apex304, MRENTHUSIASM
 +
 
 +
==Solution 2==
 +
Notice that the entire section between the <math>2</math> second mark and the <math>14</math> second mark is between the <math>4</math> and <math>7</math> feet elevation level except the <math>2</math> seconds where she skis just under the <math>4</math> feet mark and when she skis just above the <math>7</math> feet mark, making the answer <math>14-2-2-2=\boxed{\textbf{(B)}\ 8}.</math>
 +
 
 +
==Video Solution by CoolMathProblmes==
 +
https://youtu.be/Pf93RGtKo1I?feature=shared&t=615
 +
 
 +
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 +
https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=499
 +
~hsnacademy
 +
 
 +
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/wQd1lHtkVPU
 +
 
 +
~Education the Study of everything
 +
 
 +
==Video Solution by Math-X (Let's first Understand the question)==
 +
https://youtu.be/Ku_c1YHnLt0?si=_0SCHsHavl1dJJpP&t=1364
 +
 
 +
~Math-X
 +
 
 +
== Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=lfyg5ZMV0gg
 +
 
 +
https://www.youtube.com/watch?v=TAa6jarbATE
  
−
[[Image:2023 AMC 8-9.png|thumb|center|300px]]
+
==Video Solution by Magic Square==
 +
https://youtu.be/-N46BeEKaCQ?t=4903
  
−
<math>\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14</math>
+
==Video Solution by Interstigation==
 +
https://youtu.be/DBqko2xATxs&t=786
 +
 
 +
==Video Solution by harungurcan==
 +
https://www.youtube.com/watch?v=oIGy79w1H8o&t=15s
 +
 
 +
~harungurcan
 +
 
 +
==Video Solution by Dr. David==
 +
https://youtu.be/MTYUJ1wg2q0
  
−
==Solution==
+
==Video Solution by WhyMath==
 +
https://youtu.be/cHyJB7oXgxk
  
−
From our graph we can read the graph to see that the answer is <math>2+4+2 = \boxed {{(B)}8}</math>
+
==See Also==
−
-apex304
+
{{AMC8 box|year=2023|num-b=8|num-a=10}}
 +
{{MAA Notice}}

Latest revision as of 11:22, 12 January 2025

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 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

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 by CoolMathProblmes

https://youtu.be/Pf93RGtKo1I?feature=shared&t=615

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=499 ~hsnacademy

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/wQd1lHtkVPU

~Education the Study of everything

Video Solution by Math-X (Let's first Understand the question)

https://youtu.be/Ku_c1YHnLt0?si=_0SCHsHavl1dJJpP&t=1364

~Math-X

Video Solution by SpreadTheMathLove

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

https://www.youtube.com/watch?v=TAa6jarbATE

Video Solution by Magic Square

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

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=786

Video Solution by harungurcan

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

~harungurcan

Video Solution by Dr. David

https://youtu.be/MTYUJ1wg2q0

Video Solution by WhyMath

https://youtu.be/cHyJB7oXgxk

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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