Difference between revisions of "2007 AMC 8 Problems/Problem 2"

(Problem)
 
Line 42: Line 42:
  
 
The answer is <math> \boxed{\textbf{(E)}\ \dfrac{5}{2}} </math>
 
The answer is <math> \boxed{\textbf{(E)}\ \dfrac{5}{2}} </math>
 +
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=omFpSGMWhFc
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2007|num-b=1|num-a=3}}
 
{{AMC8 box|year=2007|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:51, 2 July 2024

Problem

$650$ students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?

[asy] size(200); defaultpen(linewidth(0.7)); defaultpen(fontsize(8)); draw(origin--(0,250)); int i; for(i=0; i<6; i=i+1) { draw((0,50*i)--(5,50*i)); } filldraw((25,0)--(75,0)--(75,150)--(25,150)--cycle, gray, black); filldraw((75,0)--(125,0)--(125,100)--(75,100)--cycle, gray, black); filldraw((125,0)--(175,0)--(175,150)--(125,150)--cycle, gray, black); filldraw((225,0)--(175,0)--(175,250)--(225,250)--cycle, gray, black); label("$50$", (0,50), W); label("$100$", (0,100), W); label("$150$", (0,150), W); label("$200$", (0,200), W); label("$250$", (0,250), W); label(rotate(90)*"Lasagna", (50,0), S); label(rotate(90)*"Manicotti", (100,0), S); label(rotate(90)*"Ravioli", (150,0), S); label(rotate(90)*"Spaghetti", (200,0), S); label(rotate(90)*"$\mbox{Number of People}$", (-40,140), W); [/asy]

$\mathrm{(A)} \frac{2}{5} \qquad \mathrm{(B)} \frac{1}{2} \qquad \mathrm{(C)} \frac{5}{4} \qquad \mathrm{(D)} \frac{5}{3} \qquad \mathrm{(E)} \frac{5}{2}$

Solution

The answer is $\dfrac{\text{number of students who preferred spaghetti}}{\text{number of students who preferred manicotti}}$

So,

$\frac{250}{100}$

Simplify,

$\frac{5}{2}$

The answer is $\boxed{\textbf{(E)}\ \dfrac{5}{2}}$

Video Solution by SpreadTheMathLove

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

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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