Difference between revisions of "1968 AHSME Problems/Problem 32"
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== Solution == | == Solution == | ||
− | <math>\fbox{C}</math> | + | |
+ | Let the speed of <math>A</math> be <math>a</math> and the speed of <math>B</math> be <math>b</math>. The first time that <math>A</math> and <math>B</math> will be equidistant from <math>O</math>, <math>B</math> will have not yet reached <math>O</math>. Thus, after two minutes, <math>B</math>'s distance from <math>O</math> will be <math>500-2b</math>, and <math>A</math>'s distance from <math>O</math> will be <math>2a</math>. Setting these expressions equal to each other and dividing by 2, we see that <math>a=250-b</math>. | ||
+ | |||
+ | After another eight minutes (or after a total of ten minutes since <math>A</math> was at <math>O</math>), <math>A</math> and <math>B</math> will again be equidistant from <math>O</math>, but this time <math>B</math> will have passed <math>O</math>. The distance <math>A</math> will be from <math>O</math> is <math>10a</math>, and the distance <math>B</math> will be from <math>O</math> is <math>10b-500</math>. Setting these expressions equal to each other and dividing by 10, we see that <math>a=b-50</math>. | ||
+ | |||
+ | Adding the two equations that we have obtained above, we see that <math>2a=250-b+b-50</math>, and so <math>a=100</math>. Substituting this value of <math>a</math> into the second equation, we see that <math>100=b-50</math>, or <math>b=150</math>. Then, <math>\frac{a}{b}=\frac{100}{150}=\frac{2}{3}</math>, so the ratio of <math>A</math>'s speed to that of <math>B</math> is <math>\fbox{(C) 2:3}</math>. | ||
== See also == | == See also == | ||
− | {{AHSME box|year=1968|num-b=31|num-a=33}} | + | {{AHSME 35p box|year=1968|num-b=31|num-a=33}} |
[[Category: Intermediate Geometry Problems]] | [[Category: Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:01, 17 July 2024
Problem
and
move uniformly along two straight paths intersecting at right angles in point
. When
is at
,
is
yards short of
. In two minutes they are equidistant from
, and in
minutes more they are again equidistant from
. Then the ratio of
's speed to
's speed is:
Solution
Let the speed of be
and the speed of
be
. The first time that
and
will be equidistant from
,
will have not yet reached
. Thus, after two minutes,
's distance from
will be
, and
's distance from
will be
. Setting these expressions equal to each other and dividing by 2, we see that
.
After another eight minutes (or after a total of ten minutes since was at
),
and
will again be equidistant from
, but this time
will have passed
. The distance
will be from
is
, and the distance
will be from
is
. Setting these expressions equal to each other and dividing by 10, we see that
.
Adding the two equations that we have obtained above, we see that , and so
. Substituting this value of
into the second equation, we see that
, or
. Then,
, so the ratio of
's speed to that of
is
.
See also
1968 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 31 |
Followed by Problem 33 | |
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 | ||
All AHSME Problems and Solutions |
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