Difference between revisions of "1950 AHSME Problems/Problem 30"
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| − | ==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: | ||
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\textbf{(D)}\ 50 \qquad | \textbf{(D)}\ 50 \qquad | ||
\textbf{(E)}\ \text{None of these}</math> | \textbf{(E)}\ \text{None of these}</math> | ||
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| − | ==See Also== | + | == 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>. | ||
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| + | Expanding both equations and simplifying, we get <math>2g-30 = b</math> and <math>5b = g+210</math>. | ||
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| + | Substituting <math>b</math> for <math>2g-30</math>, we get <math>5(2g-30)=g+210</math>. Solving for <math>g</math>, we get <math>g = \boxed{\textbf{(A)}\ 40}</math>. | ||
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| + | == See Also == | ||
{{AHSME 50p box|year=1950|num-b=29|num-a=31}} | {{AHSME 50p box|year=1950|num-b=29|num-a=31}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:01, 12 October 2020
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 
and 
.
Substituting for 
, we get 
. Solving 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
