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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>
  
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== Solution ==
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==Solution==
 
<asy>
 
<asy>
 
pair A,B,C;
 
pair A,B,C;
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</asy>
 
</asy>
  
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Solution 2
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==Solution 2==
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- I did this because this problem does not have any solution explained to us, rather just a diagram. This is not the official solution, it's just the way I solved it. Enjoy!
 
 
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
 
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
  

Latest revision as of 00:27, 31 January 2023

Problem

Two cyclists, $k$ miles apart, and starting at the same time, would be together in $r$ hours if they traveled in the same direction, but would pass each other in $t$ hours if they traveled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is:

$\text{(A) } \frac {r + t}{r - t} \qquad \text{(B) } \frac {r}{r - t} \qquad \text{(C) } \frac {r + t}{r} \qquad \text{(D) } \frac{r}{t}\qquad \text{(E) } \frac{r+k}{t-k}$

Solution

[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 $d = rt$. The times the problem gives us, $r$ and $t$ are kind of annoying, so I will just let $x$ and $y$ be $r$ and $t$, respectively. I will also let $r_{1}$ and $r_{2}$ being the rate of the riders, where $r_{1}>r_{2}$. We can then write our equations based off these numbers, so I got $k = (r_1-r_2)x$ and $k = (r_1+r_2)y$. Rearranging the equations and solving for $r_1$ and $r_2$, I got $r_1 = k/x + r_2$ (from first equation) and $r_1 = k/y - r_2$ (from second equation). Also, we have $r_2 = k/y - r_1$ and $r_2 = -(k/x) + r_1$. Adding the first two together and the last two together, I got $2r_1 = \frac{(xk+yk)}{xy}$ and $2r_2 = \frac{(xk - yk)}{xy}$. Finally, dividing $2r_1$ by $2r_2$ gives us $r_1/r_2 = \frac{(x+y)}{(x-y)}$. Substituting $x$ and $y$ for $r$ and $t$, I got $\boxed{\text{(A) } \frac {r + t}{r - t}}$ -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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