Difference between revisions of "1971 AHSME Problems/Problem 2"
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<math>\textbf{(A) }fb^2\qquad \textbf{(B) }b/f^2\qquad \textbf{(C) }f^2/b\qquad \textbf{(D) }b^2/f\qquad \textbf{(E) }f/b^2</math> | <math>\textbf{(A) }fb^2\qquad \textbf{(B) }b/f^2\qquad \textbf{(C) }f^2/b\qquad \textbf{(D) }b^2/f\qquad \textbf{(E) }f/b^2</math> | ||
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== Solution == | == Solution == | ||
| − | We can use a modified version of the equation <math>\text{Distance} = \text{Rate} \times {\text{Time}}</math>, which is <math>\text{Work Done} = \text{Rate of Work} \times{ \text{Time Worked}}</math>. In this case, the work done is the number of bricks laid, the rate of work is the number of men working, and the time worked is the number of days. With these definitions, we see from the first equation that <math>f = b \cdot c</math>. If we let <math>X</math> be the number of days it will take <math>c</math> men working at the same rate to <math>b</math> bricks, then we have the equation <math>b = c \cdot x</math>. So, <math>x = \frac{b}{c}</math>. The first equation says that <math>\frac{f}{c^2} = \frac{b}{c}</math>, which leads to <math>x = \frac{f}{c^2}</math>. This doesn't match any of our answer choices though, so we have to fiddle around a bit before we realize that <math>c = \frac{f}{b}</math>, a substitution we can make to see that <math>x = \frac{b^2}{f}</math>. Thus, the answer is \boxed{\ | + | We can use a modified version of the equation <math>\text{Distance} = \text{Rate} \times {\text{Time}}</math>, which is <math>\text{Work Done} = \text{Rate of Work} \times{ \text{Time Worked}}</math>. In this case, the work done is the number of bricks laid, the rate of work is the number of men working, and the time worked is the number of days. With these definitions, we see from the first equation that <math>f = b \cdot c</math>. If we let <math>X</math> be the number of days it will take <math>c</math> men working at the same rate to <math>b</math> bricks, then we have the equation <math>b = c \cdot x</math>. So, <math>x = \frac{b}{c}</math>. The first equation says that <math>\frac{f}{c^2} = \frac{b}{c}</math>, which leads to <math>x = \frac{f}{c^2}</math>. This doesn't match any of our answer choices though, so we have to fiddle around a bit before we realize that <math>c = \frac{f}{b}</math>, a substitution we can make to see that <math>x = \frac{b^2}{f}</math>. Thus, the answer is <math>\boxed{\textbf{(D) }b^2/f}</math>. |
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| + | == See Also == | ||
| + | {{AHSME 35p box|year=1971|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 08:30, 1 August 2024
Problem
If 
men take 
days to lay 
bricks, then the number of days it will take men working at the same rate to lay bricks, is
Solution
We can use a modified version of the equation 
, which is 
. In this case, the work done is the number of bricks laid, the rate of work is the number of men working, and the time worked is the number of days. With these definitions, we see from the first equation that 
. If we let 
be the number of days it will take men working at the same rate to bricks, then we have the equation 
. So, 
. The first equation says that 
, which leads to 
. This doesn't match any of our answer choices though, so we have to fiddle around a bit before we realize that 
, a substitution we can make to see that 
. Thus, the answer is 
.
See Also
| 1971 AHSC (Problems • Answer Key • Resources) | ||
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
