Difference between revisions of "1994 AJHSME Problems/Problem 18"

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==Solution hi==
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==Solution==
 
When Mike is shopping at the mall, his distance doesn't change, so there should be a flat plateau shape on the graph. This rules out <math>C,D,E</math>. The portion of graph <math>A</math> in which the distance is changing is linear, inconsistent with how he changes speed from the city and the highway. The best representation of his travels is graph <math>\boxed{\text{(B)}}</math>.
 
When Mike is shopping at the mall, his distance doesn't change, so there should be a flat plateau shape on the graph. This rules out <math>C,D,E</math>. The portion of graph <math>A</math> in which the distance is changing is linear, inconsistent with how he changes speed from the city and the highway. The best representation of his travels is graph <math>\boxed{\text{(B)}}</math>.
  

Revision as of 12:19, 19 October 2016

Problem

Mike leaves home and drives slowly east through city traffic. When he reaches the highway he drives east more rapidly until he reaches the shopping mall where he stops. He shops at the mall for an hour. Mike returns home by the same route as he came, driving west rapidly along the highway and then slowly through city traffic. Each graph shows the distance from home on the vertical axis versus the time elapsed since leaving home on the horizontal axis. Which graph is the best representation of Mike's trip?

[asy] import graph; unitsize(12); real a(real x) {return ((x-15)^2)/2;} real b(real x) {return ((x-25)^2)/2;} real c(real x) {return ((x-30)^2 * (x-40)^2) * 8/625;} real d(real x) {return ((x-15)^2)*8/25-15;} real e(real x) {return ((x-25)^2)*8/25-15;} draw((0,9)--(0,0)--(11,0)); draw((15,9)--(15,0)--(26,0)); draw((30,9)--(30,0)--(41,0)); draw((0,-6)--(0,-15)--(11,-15)); draw((15,-6)--(15,-15)--(26,-15)); draw((0,0)--(3,8)--(7,8)--(10,0)); draw(graph(a,15,17)); draw((17,2)--(18,8)--(22,8)--(23,2)); draw(graph(b,23,25)); draw(graph(c,30,40)); draw((0,-15)--(5,-7)--(10,-15)); draw(graph(d,15,20)); draw(graph(e,20,25)); for (int k=0; k<3; ++k) { label("d",(15*k-1,8),N); label("i",(15*k-1,7),N); label("s",(15*k-1,6),N); label("t",(15*k-1,5),N); label("a",(15*k-1,4),N); label("n",(15*k-1,3),N); label("c",(15*k-1,2),N); label("e",(15*k-1,1),N); label("time",(15*k+8,0),S); } for (int k=0; k<2; ++k) { label("d",(15*k-1,8-15),N); label("i",(15*k-1,7-15),N); label("s",(15*k-1,6-15),N); label("t",(15*k-1,5-15),N); label("a",(15*k-1,4-15),N); label("n",(15*k-1,3-15),N); label("c",(15*k-1,2-15),N); label("e",(15*k-1,1-15),N); label("time",(15*k+8,0-15),S); } label("(A)",(5,9),N); label("(B)",(20,9),N); label("(C)",(35,9),N); label("(D)",(5,-6),N); label("(E)",(20,-6),N); [/asy]

Solution

When Mike is shopping at the mall, his distance doesn't change, so there should be a flat plateau shape on the graph. This rules out $C,D,E$. The portion of graph $A$ in which the distance is changing is linear, inconsistent with how he changes speed from the city and the highway. The best representation of his travels is graph $\boxed{\text{(B)}}$.

See Also

1994 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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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