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Difference between revisions of "1961 AHSME Problems/Problem 37"

(Created page with "In racing over a given distance <math>d</math> at uniform speed, <math>A</math> can beat <math>B</math> by 20 yards, <math>B</math> can beat <math>C</math> by 10 yards, and <math...")
 
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In racing over a given distance <math>d</math> at uniform speed, <math>A</math> can beat <math>B</math> by 20 yards, <math>B</math> can beat <math>C</math> by 10 yards, and <math>A</math> can beat <math>C</math> by 28 yards. Then <math>d</math>, in yards, equals:
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== Problem ==
  
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In racing over a distance <math>d</math> at uniform speed, <math>A</math> can beat <math>B</math> by <math>20</math> yards, <math>B</math> can beat <math>C</math> by <math>10</math> yards,
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and <math>A</math> can beat <math>C</math> by <math>28</math> yards. Then <math>d</math>, in yards, equals:
  
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<math> \textbf{(A)}\ \textrm{not determined by the given information} \qquad \textbf{(B)}\ 58 \qquad \textbf{(C)}\ 100 \qquad \textbf{(D)}\ 116 \qquad \textbf{(E)}\ 120</math>
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<math>\textbf{(A)}\ \text{Not determined by the given information} \qquad  
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\textbf{(B)}\ 58\qquad
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\textbf{(C)}\ 100\qquad
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\textbf{(D)}\ 116\qquad
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\textbf{(E)}\ 120</math>
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==Solution==
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Let <math>a</math> be speed of <math>A</math>, <math>b</math> be speed of <math>B</math>, and <math>c</math> be speed of <math>C</math>.
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Person <math>A</math> finished the track in <math>\frac{d}{a}</math> minutes, so <math>B</math> traveled <math>\frac{db}{a}</math> yards at the same time.  Since <math>B</math> is <math>20</math> yards from the finish line, the first equation is
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<cmath>\frac{db}{a} + 20 = d</cmath>
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Using similar steps, the second and third equation are, respectively,
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<cmath>\frac{dc}{b} + 10 = d</cmath>
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<cmath>\frac{dc}{a} + 28 = d</cmath>
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Get rid of the denominator in all three equations by multiplying both sides by the denominator in each equation.
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<cmath>db + 20a = da</cmath>
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<cmath>dc + 10b = db</cmath>
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<cmath>dc + 28a = da</cmath>
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These equations can be rearranged to get the following.
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<cmath>(d-20)a = db</cmath>
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<cmath>(d-10)b = dc</cmath>
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<cmath>(d-28)a = dc</cmath>
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Solving for <math>b</math> in the first equation yields <math>\frac{(d-20)a}{d} = b</math>.  Substituting for <math>b</math> and solving for <math>c</math> in the second equation yields <math>\frac{(d-10)(d-20)a}{d^2} = c</math>.  Substituting for <math>c</math> in the third equation yields
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<cmath>(d-28)a = \frac{(d-10)(d-20)a}{d}</cmath>
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<cmath>d^2 - 28d = d^2 - 30d + 200</cmath>
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Solve the equation to get <math>d = 100</math>.  Thus, the track is <math>100</math> yards long, which is answer choice <math>\boxed{\textbf{(C)}}</math>.
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== Solution 2 ==
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Let <math>s_a</math>, <math>s_b</math>, <math>s_c</math> be the speeds of <math>A</math>, <math>B</math>, <math>C</math>, respectively and let <math>t_a</math>, <math>t_b</math>, <math>t_c</math> be their respective times needed, such that
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<cmath>d=s_at_a=s_bt_b=s_ct_b</cmath>
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Putting each pair of them into a race gets the <math>3</math> equations
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<cmath>s_at_a-s_bt_a=20</cmath>
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<cmath>s_bt_b-s_ct_b=10</cmath>
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<cmath>s_at_a-s_ct_a=28</cmath>
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Subtract the first equation from the third yields
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<cmath>s_bt_a-s_ct_a=t_a(s_b-s_c)=8</cmath>
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Divide the equation by the second original equation yields
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<cmath>\frac{t_a}{t_b}=\frac{4}{5}</cmath>
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Since <math>s_at_a=s_bt_b</math> we have
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<cmath>\frac{s_a}{s_b}=\frac{5}{4}</cmath>
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Thus from the first equation we have
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<cmath>s_at_a-s_bt_a=s_at_a-\frac{4}{5}s_at_a=20\Longrightarrow d=s_at_a=100 \boxed{C}</cmath>
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~ Nafer
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==See Also==
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{{AHSME 40p box|year=1961|num-b=36|num-a=38}}
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 +
[[Category:Intermediate Algebra Problems]]
 +
 
 +
{{MAA Notice}}

Latest revision as of 17:30, 23 December 2019

Problem

In racing over a distance $d$ at uniform speed, $A$ can beat $B$ by $20$ yards, $B$ can beat $C$ by $10$ yards, and $A$ can beat $C$ by $28$ yards. Then $d$, in yards, equals:

$\textbf{(A)}\ \text{Not determined by the given information} \qquad \textbf{(B)}\ 58\qquad \textbf{(C)}\ 100\qquad \textbf{(D)}\ 116\qquad \textbf{(E)}\ 120$

Solution

Let $a$ be speed of $A$, $b$ be speed of $B$, and $c$ be speed of $C$.

Person $A$ finished the track in $\frac{d}{a}$ minutes, so $B$ traveled $\frac{db}{a}$ yards at the same time. Since $B$ is $20$ yards from the finish line, the first equation is \[\frac{db}{a} + 20 = d\] Using similar steps, the second and third equation are, respectively, \[\frac{dc}{b} + 10 = d\] \[\frac{dc}{a} + 28 = d\] Get rid of the denominator in all three equations by multiplying both sides by the denominator in each equation. \[db + 20a = da\] \[dc + 10b = db\] \[dc + 28a = da\] These equations can be rearranged to get the following. \[(d-20)a = db\] \[(d-10)b = dc\] \[(d-28)a = dc\]

Solving for $b$ in the first equation yields $\frac{(d-20)a}{d} = b$. Substituting for $b$ and solving for $c$ in the second equation yields $\frac{(d-10)(d-20)a}{d^2} = c$. Substituting for $c$ in the third equation yields \[(d-28)a = \frac{(d-10)(d-20)a}{d}\] \[d^2 - 28d = d^2 - 30d + 200\] Solve the equation to get $d = 100$. Thus, the track is $100$ yards long, which is answer choice $\boxed{\textbf{(C)}}$.


Solution 2

Let $s_a$, $s_b$, $s_c$ be the speeds of $A$, $B$, $C$, respectively and let $t_a$, $t_b$, $t_c$ be their respective times needed, such that \[d=s_at_a=s_bt_b=s_ct_b\] Putting each pair of them into a race gets the $3$ equations \[s_at_a-s_bt_a=20\] \[s_bt_b-s_ct_b=10\] \[s_at_a-s_ct_a=28\] Subtract the first equation from the third yields \[s_bt_a-s_ct_a=t_a(s_b-s_c)=8\] Divide the equation by the second original equation yields \[\frac{t_a}{t_b}=\frac{4}{5}\] Since $s_at_a=s_bt_b$ we have \[\frac{s_a}{s_b}=\frac{5}{4}\] Thus from the first equation we have \[s_at_a-s_bt_a=s_at_a-\frac{4}{5}s_at_a=20\Longrightarrow d=s_at_a=100 \boxed{C}\]

~ Nafer

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 36 		Followed by
Problem 38
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
All AHSME Problems and Solutions

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