Difference between revisions of "1952 AHSME Problems/Problem 48"
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draw((A)--(B)); | draw((A)--(B)); | ||
label("$A$",A,S); label("$B$",B,SE); label("$k$",C,SW); | label("$A$",A,S); label("$B$",B,SE); label("$k$",C,SW); | ||
+ | </asy> | ||
Solution 2 | Solution 2 | ||
− | We have a simple formula | + | We have a simple formula <math>d = rt</math>. The times the problem gives us, <math>r</math> and <math>t</math> are kind of annoying, so I will just let <math>x</math> and <math>y</math> be <math>r</math> and <math>t</math>, respectively. I will also let <math>r_{1}</math> and <math>r_{2}</math> being the rate of the riders, where <math>r_{1}>r_{2}</math>. We can then write our equations based off these numbers, so I got <math>k = (r_1-r_2)x</math> and <math>k = (r_1+r_2)y</math>. Rearranging the equations and solving for <math>r_1</math> and <math>r_2</math>, I got <math>r_1 = k/x + r_2</math> (from first equation) and <math>r_1 = k/y - r_2</math> (from second equation). Also, we have <math>r_2 = k/y - r_1</math> and <math>r_2 = -(k/x) + r_1</math>. Adding the first two together and the last two together, I got <math>2r_1 = \frac{(xk+yk)}{xy}</math> and <math>2r_2 = \frac{(xk - yk)}{xy}</math>. Finally, dividing <math>2r_1</math> by <math>2r_2</math> gives us <math>r_1/r_2 = \frac{(x+y)}{(x-y)}</math>. Substituting <math>x</math> and <math>y</math> for <math>r</math> and <math>t</math>, I got <math>\boxed{\text{(A) } \frac {r + t}{r - t}}</math> -DragonFish12345 with credits to @RJ5303707 for LATEX |
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== See also == | == See also == |
Revision as of 12:40, 14 September 2020
Problem
Two cyclists, miles apart, and starting at the same time, would be together in
hours if they traveled in the same direction, but would pass each other in
hours if they traveled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is:
Solution
Solution 2
We have a simple formula . The times the problem gives us,
and
are kind of annoying, so I will just let
and
be
and
, respectively. I will also let
and
being the rate of the riders, where
. We can then write our equations based off these numbers, so I got
and
. Rearranging the equations and solving for
and
, I got
(from first equation) and
(from second equation). Also, we have
and
. Adding the first two together and the last two together, I got
and
. Finally, dividing
by
gives us
. Substituting
and
for
and
, I got
-DragonFish12345 with credits to @RJ5303707 for LATEX
See also
1952 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 47 |
Followed by Problem 49 | |
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 |
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