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[[Category:Introductory Algebra Problems]]
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Latest revision as of 20:22, 3 July 2013

Problem

A bar graph shows the number of hamburgers sold by a fast food chain each season. However, the bar indicating the number sold during the winter is covered by a smudge. If exactly $25\%$ of the chain's hamburgers are sold in the fall, how many million hamburgers are sold in the winter?

[asy] size(250); void bargraph(real X, real Y, real ymin, real ymax, real ystep, real tickwidth, string yformat, Label LX, Label LY, Label[] LLX, real[] height,pen p=nullpen) { draw((0,0)--(0,Y),EndArrow); draw((0,0)--(X,0),EndArrow); label(LX,(X,0),plain.SE,fontsize(9)); label(LY,(0,Y),plain.NW,fontsize(9)); real yscale=Y/(ymax+ystep); for(real y=ymin; y<ymax; y+=ystep) { draw((-tickwidth,yscale*y)--(0,yscale*y)); label(format(yformat,y),(-tickwidth,yscale*y),plain.W,fontsize(9)); } int n=LLX.length; real xscale=X/(2*n+2); for(int i=0;i<n;++i) { real x=xscale*(2*i+1); path P=(x,0)--(x,height[i]*yscale)--(x+xscale,height[i]*yscale)--(x+xscale,0)--cycle; fill(P,p); draw(P); label(LLX[i],(x+xscale/2),plain.S,fontsize(10)); } for(int i=0;i<n;++i) draw((0,height[i]*yscale)--(X,height[i]*yscale),dashed); } string yf="%#.1f"; Label[] LX={"Spring","Summer","Fall","Winter"}; for(int i=0;i<LX.length;++i) LX[i]=rotate(90)*LX[i]; real[] H={4.5,5,4,4}; bargraph(60,50,1,5.1,0.5,2,yf,"season","hamburgers (millions)",LX,H,yellow); fill(ellipse((45,30),7,10),brown); [/asy]

$\text{(A)}\ 2.5 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 3.5 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 4.5$

Solution

What we want to find is the number of hamburgers sold in the winter. Since we don't know what it is, let's call it $x$. From the graph, we know that in Spring, 4.5 million hamburgers were sold, in the Summer was 5 million and in the Fall was 4 million. We know that the number of hamburgers sold in Fall is exactly $\frac{1}{4}$ of the total number of hamburgers sold, so we can say that...

$4 \times \text{Fall} = \text{Spring} + \text{Winter} + \text{Fall} + \text{Summer}$

$4 \times 4 = 4.5 + 4 + x + 5$

$16 = x + 13.5$

$2.5 = x$

The answer is 2.5, or $\boxed{\text{A}}$

See Also

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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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