Difference between revisions of "1950 AHSME Problems/Problem 28"
Mrdavid445 (talk | contribs) (Created page with "==Problem== Two boys <math>A</math> and <math>B</math> start at the same time to ride from Port Jervis to Poughkeepsie, <math>60</math> miles away. <math>A</math> travels <math>...") |
(→Alternate Solution) |
||
(6 intermediate revisions by 4 users not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | |||
Two boys <math>A</math> and <math>B</math> start at the same time to ride from Port Jervis to Poughkeepsie, <math>60</math> miles away. <math>A</math> travels <math>4</math> miles an hour slower than <math>B</math>. <math>B</math> reaches Poughkeepsie and at once turns back meeting <math>A</math> <math>12</math> miles from Poughkeepsie. The rate of <math>A</math> was: | Two boys <math>A</math> and <math>B</math> start at the same time to ride from Port Jervis to Poughkeepsie, <math>60</math> miles away. <math>A</math> travels <math>4</math> miles an hour slower than <math>B</math>. <math>B</math> reaches Poughkeepsie and at once turns back meeting <math>A</math> <math>12</math> miles from Poughkeepsie. The rate of <math>A</math> was: | ||
Line 8: | Line 7: | ||
\textbf{(D)}\ 16\text{ mph} \qquad | \textbf{(D)}\ 16\text{ mph} \qquad | ||
\textbf{(E)}\ 20\text{ mph}</math> | \textbf{(E)}\ 20\text{ mph}</math> | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Alternate Solution== | ||
+ | |||
+ | 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== | ||
+ | {{AHSME 50p box|year=1950|num-b=27|num-a=29}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | [[Category:Rate Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:03, 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:
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.