Difference between revisions of "1950 AHSME Problems/Problem 30"

(Created page with "==Problem== From a group of boys and girls, <math>15</math> girls leave. There are then left two boys for each girl. After this <math>45</math> boys leave. There are then <math>...")
 
(solution & see also sections)
Line 1: Line 1:
 
==Problem==
 
==Problem==
 
 
From a group of boys and girls, <math>15</math> girls leave. There are then left two boys for each girl. After this <math>45</math> boys leave. There are then <math>5</math> girls for each boy. The number of girls in the beginning was:
 
From a group of boys and girls, <math>15</math> girls leave. There are then left two boys for each girl. After this <math>45</math> boys leave. There are then <math>5</math> girls for each boy. The number of girls in the beginning was:
  
Line 8: Line 7:
 
\textbf{(D)}\ 50 \qquad
 
\textbf{(D)}\ 50 \qquad
 
\textbf{(E)}\ \text{None of these}</math>
 
\textbf{(E)}\ \text{None of these}</math>
 +
==Solution==
 +
{{solution}}
 +
 +
==See Also==
 +
{{AHSME box|year=1950|num-b=29|after=Last<br />Question}}
 +
 +
[[Category:Introductory Algebra Problems]]

Revision as of 13:19, 17 April 2012

Problem

From a group of boys and girls, $15$ girls leave. There are then left two boys for each girl. After this $45$ boys leave. There are then $5$ girls for each boy. The number of girls in the beginning was:

$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 43 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ \text{None of these}$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See Also

1950 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Last
Question
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 26 27 28 29 30
All AHSME Problems and Solutions