Difference between revisions of "1954 AHSME Problems/Problem 30"
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==Solution== | ==Solution== | ||
Let <math>A</math> do <math>r_A</math> of the job per day, <math>B</math> do <math>r_B</math> of the job per day, and <math>C</math> do <math>r_C</math> of the job per day. These three quantities have unit <math>\frac{\text{job}}{\text{day}}</math>. Therefore our three conditions give us the three equations: | Let <math>A</math> do <math>r_A</math> of the job per day, <math>B</math> do <math>r_B</math> of the job per day, and <math>C</math> do <math>r_C</math> of the job per day. These three quantities have unit <math>\frac{\text{job}}{\text{day}}</math>. Therefore our three conditions give us the three equations: | ||
| − | \begin{align} | + | <cmath>\begin{align*} |
(2\text{ days})(r_A+r_B)&=1\text{ job},\nonumber\\ | (2\text{ days})(r_A+r_B)&=1\text{ job},\nonumber\\ | ||
(4\text{ days})(r_B+r_C)&=1\text{ job},\nonumber\\ | (4\text{ days})(r_B+r_C)&=1\text{ job},\nonumber\\ | ||
(2.4\text{ days})(r_C+r_A)&=1\text{ job}.\nonumber | (2.4\text{ days})(r_C+r_A)&=1\text{ job}.\nonumber | ||
| − | \end{align} | + | \end{align*}</cmath> |
We divide the three equations by the required constant so that the coefficients of the variables become 1: | We divide the three equations by the required constant so that the coefficients of the variables become 1: | ||
| − | \begin{align} | + | <cmath>\begin{align*} |
r_A+r_B&=\frac{1}{2}\cdot\frac{\text{job}}{\text{day}},\nonumber\\ | r_A+r_B&=\frac{1}{2}\cdot\frac{\text{job}}{\text{day}},\nonumber\\ | ||
r_B+r_C&=\frac{1}{4}\cdot\frac{\text{job}}{\text{day}},\nonumber\\ | r_B+r_C&=\frac{1}{4}\cdot\frac{\text{job}}{\text{day}},\nonumber\\ | ||
r_C+r_A&=\frac{5}{12}\cdot\frac{\text{job}}{\text{day}}.\nonumber | r_C+r_A&=\frac{5}{12}\cdot\frac{\text{job}}{\text{day}}.\nonumber | ||
| − | \end{align} | + | \end{align*}</cmath> |
If we add these three new equations together and divide the result by two, we obtain an equation with left-hand side <math>r_A+r_B+r_C</math>, so if we subtract <math>r_B+r_C</math> (the value of which we know) from both equations, we obtain the value of <math>r_A</math>, which is what we wish to determine anyways. So we add these three equations and divide by two: | If we add these three new equations together and divide the result by two, we obtain an equation with left-hand side <math>r_A+r_B+r_C</math>, so if we subtract <math>r_B+r_C</math> (the value of which we know) from both equations, we obtain the value of <math>r_A</math>, which is what we wish to determine anyways. So we add these three equations and divide by two: | ||
<cmath>r_A+r_B+r_C=\frac{1}{2}\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{5}{12}\right)\cdot\frac{\text{job}}{\text{day}}=\frac{7}{12}\cdot\frac{\text{job}}{\text{day}}.</cmath> | <cmath>r_A+r_B+r_C=\frac{1}{2}\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{5}{12}\right)\cdot\frac{\text{job}}{\text{day}}=\frac{7}{12}\cdot\frac{\text{job}}{\text{day}}.</cmath> | ||
Hence: | Hence: | ||
| − | \begin{align} | + | <cmath>\begin{align*} |
r_A &= (r_A+r_B+r_C)-(r_B+r_C)\nonumber\\ | r_A &= (r_A+r_B+r_C)-(r_B+r_C)\nonumber\\ | ||
&=\frac{7}{12}\cdot\frac{\text{job}}{\text{day}}-\frac{1}{4}\cdot\frac{\text{job}}{\text{day}}\nonumber\\ | &=\frac{7}{12}\cdot\frac{\text{job}}{\text{day}}-\frac{1}{4}\cdot\frac{\text{job}}{\text{day}}\nonumber\\ | ||
&=\frac{1}{3}\cdot\frac{\text{job}}{\text{day}}.\nonumber | &=\frac{1}{3}\cdot\frac{\text{job}}{\text{day}}.\nonumber | ||
| − | \end{align} | + | \end{align*}</cmath> |
This shows that <math>A</math> does one third of the job per day. Therefore, if <math>A</math> were to do the entire job himself, he would require <math>\boxed{\textbf{(B)}\ 3}</math> days. | This shows that <math>A</math> does one third of the job per day. Therefore, if <math>A</math> were to do the entire job himself, he would require <math>\boxed{\textbf{(B)}\ 3}</math> days. | ||
Latest revision as of 16:36, 18 July 2015
Problem
and 
together can do a job in 
days; and 
can do it in four days; and and in 
days.
The number of days required for A to do the job alone is:
Solution
Let do 
of the job per day, do 
of the job per day, and do 
of the job per day. These three quantities have unit 
. Therefore our three conditions give us the three equations:
We divide the three equations by the required constant so that the coefficients of the variables become 1:
If we add these three new equations together and divide the result by two, we obtain an equation with left-hand side 
, so if we subtract 
(the value of which we know) from both equations, we obtain the value of , which is what we wish to determine anyways. So we add these three equations and divide by two:
![\[r_A+r_B+r_C=\frac{1}{2}\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{5}{12}\right)\cdot\frac{\text{job}}{\text{day}}=\frac{7}{12}\cdot\frac{\text{job}}{\text{day}}.\]](http://latex.artofproblemsolving.com/c/8/4/c84e8dbd5827df45b177ffa8e4325dc8d51a33ab.png)
Hence:
This shows that does one third of the job per day. Therefore, if were to do the entire job himself, he would require 
days.
See Also
| 1954 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. 
