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Difference between revisions of "2014 AMC 8 Problems/Problem 7"

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<math>\textbf{(A) }3 : 4\qquad\textbf{(B) }4 : 3\qquad\textbf{(C) }3 : 2\qquad\textbf{(D) }7 : 4\qquad \textbf{(E) }2 : 1</math>
 
<math>\textbf{(A) }3 : 4\qquad\textbf{(B) }4 : 3\qquad\textbf{(C) }3 : 2\qquad\textbf{(D) }7 : 4\qquad \textbf{(E) }2 : 1</math>
 
==Solution==
 
==Solution==
−
We can set up an equation with <math>x</math> being the number of girls in the class. The number of boys in the class is equal to <math>x-4</math>. Since the total number of students is equal to <math>28</math>, we get <math>x+x-4=28</math>. Solving this equation, we get <math>x=16</math>. There are <math>16-4=12</math> boys in our class, and our answer is <math>16:12=\boxed{4:3}</math>. So the answer is <math>\boxed{B}</math>
+
We can set up an equation with <math>x</math> being the number of girls in the class. The number of boys in the class is equal to <math>x-4</math>. Since the total number of students is equal to <math>28</math>, we get <math>x+x-4=28</math>. Solving this equation, we get <math>x=16</math>. There are <math>16-4=12</math> boys in our class, and our answer is <math>16:12=\boxed{\textbf{(B)}~4:3}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=6|num-a=8}}
 
{{AMC8 box|year=2014|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:10, 2 December 2015

Problem

There are four more girls than boys in Ms. Raub's class of $28$ students. What is the ratio of number of girls to the number of boys in her class?

$\textbf{(A) }3 : 4\qquad\textbf{(B) }4 : 3\qquad\textbf{(C) }3 : 2\qquad\textbf{(D) }7 : 4\qquad \textbf{(E) }2 : 1$

Solution

We can set up an equation with $x$ being the number of girls in the class. The number of boys in the class is equal to $x-4$. Since the total number of students is equal to $28$, we get $x+x-4=28$. Solving this equation, we get $x=16$. There are $16-4=12$ boys in our class, and our answer is $16:12=\boxed{\textbf{(B)}~4:3}$.

See Also

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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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