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25 Mar 2025, 00:30
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
67% (01:45) correct
33%
(01:59)
wrong
based on 174
sessions
History
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Not Attempted Yet
Exactly one straight road connects Albertville and Bowton, which are 90 miles apart. If Sierra leaves Albertville headed for Bowton at the same time that Dylan leaves Bowton headed for Albertville, and each travels along the road at a constant rate, how far will Sierra have traveled when they meet?
(1) Sierra travels at 3x miles per hour and Dylan travels at 4x miles per hour.
(2) Sierra travels at 5 miles per hour slower than Dylan does.
(1) Sierra travels at 3x miles per hour and Dylan travels at 4x miles per hour.
(2) Sierra travels at 5 miles per hour slower than Dylan does.
25 Mar 2025, 00:49
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Let Sierra’s speed be \(s\) miles per hour and Dylan’s speed be \(d\) miles per hour.
The total distance between Albertville and Bowton is 90 miles.
Since they start at the same time and meet at some point, they travel for the same duration \(t\).
Thus, their combined distance covered is:
\( s t + d t = 90 \)
\( t (s + d) = 90 \)
\( t = \frac{90}{s + d} \)
We need to determine \( s t \), the distance Sierra travels before they meet.
Statement (1): Sierra travels at \(3x\) mph, and Dylan at \(4x\) mph.
Thus,
\( s = 3x \) and \( d = 4x \)
\( s + d = 3x + 4x = 7x \)
\( t = \frac{90}{7x} \)
Sierra’s distance:
\( s t = 3x \times \frac{90}{7x} = \frac{270}{7} \) miles.
Sufficient
Statement (2): Sierra travels 5 mph slower than Dylan.
Thus,
\( s = d - 5 \)
But we do not have the actual values for \( s \) or \( d \), so we cannot determine \( s t \) uniquely.
Not sufficient
Option A
The total distance between Albertville and Bowton is 90 miles.
Since they start at the same time and meet at some point, they travel for the same duration \(t\).
Thus, their combined distance covered is:
\( s t + d t = 90 \)
\( t (s + d) = 90 \)
\( t = \frac{90}{s + d} \)
We need to determine \( s t \), the distance Sierra travels before they meet.
Statement (1): Sierra travels at \(3x\) mph, and Dylan at \(4x\) mph.
Thus,
\( s = 3x \) and \( d = 4x \)
\( s + d = 3x + 4x = 7x \)
\( t = \frac{90}{7x} \)
Sierra’s distance:
\( s t = 3x \times \frac{90}{7x} = \frac{270}{7} \) miles.
Sufficient
Statement (2): Sierra travels 5 mph slower than Dylan.
Thus,
\( s = d - 5 \)
But we do not have the actual values for \( s \) or \( d \), so we cannot determine \( s t \) uniquely.
Not sufficient
Option A
25 Mar 2025, 05:55
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Bunuel
The distance between A and B is 90 miles.
S leaves from A, and D leaves from B at the same time. Each travel at a constant rates.
Statement 1:
(1) Sierra travels at 3x miles per hour and Dylan travels at 4x miles per hour.
Speed ratio is proportional to distance ratio when time is constant.
Speed ratio S : D = 3x : 4x = 3:4
Hence, S covers 3/7 * 90 = 270/7 miles. And D covers 4/7*90 = 360/7 miles at the time of meeting.
A is Sufficient
Statement 2:
(2) Sierra travels at 5 miles per hour slower than Dylan does.
S travels at speed D-5 mph.
D travels at speed D mph.
with unknown variables. statement 2 is NOT SUFFICIENT.
HENCE, OPTION A
