Difference between revisions of "1950 AHSME Problems/Problem 30"
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\textbf{(E)}\ \text{None of these}</math> | \textbf{(E)}\ \text{None of these}</math> | ||
==Solution== | ==Solution== | ||
| − | {{ | + | Let us represent the number of boys <math>b</math>, and the number of girls <math>g</math>. |
| + | |||
| + | From the first sentence, we get that <math>2(g-15)=b</math> | ||
| + | |||
| + | From the second sentence, we get <math>5(b-45)=g-15</math> | ||
| + | |||
| + | Expanding both equations and simplifying, we get <cmath>2g-30=b</cmath> <cmath>5b=g+210</cmath> | ||
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| + | Substituting <math>b</math> for <math>2g-30</math>, we get <math>5(2g-30)=g+210\implies 10g-150=g+210\implies 9g=360\implies g=\boxed{\textbf{(A)}\ 40}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 20:54, 9 April 2013
Problem
From a group of boys and girls, 
girls leave. There are then left two boys for each girl. After this 
boys leave. There are then 
girls for each boy. The number of girls in the beginning was:
Solution
Let us represent the number of boys 
, and the number of girls 
.
From the first sentence, we get that 
From the second sentence, we get 
Expanding both equations and simplifying, we get ![\[2g-30=b\]](http://latex.artofproblemsolving.com/8/a/a/8aadbd45ef82e84dddd3ee5125b49d4320c2bdb2.png)
![\[5b=g+210\]](http://latex.artofproblemsolving.com/7/6/2/76227cd896a879efd159e465189b855bd667035c.png)
Substituting for 
, we get 
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Problem 31 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
