Difference between revisions of "1952 AHSME Problems/Problem 48"
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\text{(E) } \frac{r+k}{t-k}</math> | \text{(E) } \frac{r+k}{t-k}</math> | ||
− | == Solution == | + | ==Solution== |
− | <math>\ | + | <asy> |
+ | pair A,B,C; | ||
+ | A=(0,0); B=(8,0); C=(4,1); | ||
+ | draw((A)--(B)); | ||
+ | label("$A$",A,S); label("$B$",B,SE); label("$k$",C,SW); | ||
+ | </asy> | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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 | ||
== See also == | == See also == |
Latest revision as of 00:27, 31 January 2023
Contents
[hide]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 | |
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All AHSME Problems and Solutions |
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