Difference between revisions of "1950 AHSME Problems/Problem 28"
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==Solution== | ==Solution== | ||
| − | {{ | + | Let the speed of boy <math>A</math> be <math>a</math>, and the speed of boy <math>B</math> be <math>b</math>. Notice that <math>A</math> travels <math>4</math> miles per hour slower than boy <math>B</math>, so we can replace <math>b</math> with <math>a+4</math>. |
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| + | Now let us see the distances that the boys each travel. Boy <math>A</math> travels <math>60-12=48</math> miles, and boy <math>B</math> travels <math>60+12=72</math> miles. Now, we can use <math>d=rt</math> to make an equation, where we set the time to be equal: <cmath>\frac{48}{a}=\frac{72}{a+4}</cmath> | ||
| + | Cross-multiplying gives <math>48a+192=72a</math>. Isolating the variable <math>a</math>, we get the equation <math>24a=192</math>, so <math>a=\boxed{\textbf{(B) }8 \text{ mph}}</math>. | ||
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| + | ==Alternate Solution== | ||
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| + | Note that <math>A</math> travels <math>60-12=48</math> miles in the time it takes <math>B</math> to travel <math>60+12=72</math> miles. Thus, <math>B</math> travels <math>72-48=24</math> more miles in the given time, meaning <math>\frac{24\text{miles}}{4\text{miles}/\text{hour}} = 6 \text{hours}</math> have passed, as <math>B</math> goes <math>4</math> miles per hour faster. Thus, <math>A</math> travels <math>48</math> miles per <math>6</math> hours, or <math>8</math> miles per hour. | ||
==See Also== | ==See Also== | ||
| − | {{AHSME box|year=1950|num-b=27|num-a=29}} | + | {{AHSME 50p box|year=1950|num-b=27|num-a=29}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | [[Category:Rate Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:03, 9 June 2017
Problem
Two boys 
and 
start at the same time to ride from Port Jervis to Poughkeepsie, 
miles away. travels 
miles an hour slower than . reaches Poughkeepsie and at once turns back meeting 
miles from Poughkeepsie. The rate of was:
Solution
Let the speed of boy be 
, and the speed of boy be 
. Notice that travels miles per hour slower than boy , so we can replace with 
.
Now let us see the distances that the boys each travel. Boy travels 
miles, and boy travels 
miles. Now, we can use 
to make an equation, where we set the time to be equal: ![\[\frac{48}{a}=\frac{72}{a+4}\]](http://latex.artofproblemsolving.com/c/2/7/c2752908336a7fb7cdd1ed2658bdf150e0542ef6.png)
Cross-multiplying gives 
. Isolating the variable , we get the equation 
, so 
.
Alternate Solution
Note that travels miles in the time it takes to travel miles. Thus, travels 
more miles in the given time, meaning 
have passed, as goes miles per hour faster. Thus, travels 
miles per 
hours, or 
miles per hour.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 27 |
Followed by Problem 29 | |
| 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. 
