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03 May 2024, 01:30
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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:
55%
(hard)
Question Stats:
66% (01:46) correct
34%
(01:40)
wrong
based on 289
sessions
History
Date
Time
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Not Attempted Yet
Train X leaves Los Angeles at 10:00AM and travels East at a constant speed of x miles per hour. If another Train Y leaves Los Angeles at 11:30AM and travels East along the same tracks at a constant speed of y, then at what time will Train Y catch Train X ?
(1) y = 4x/3
(2) x = 60 miles per hour
(1) y = 4x/3
(2) x = 60 miles per hour
03 May 2024, 04:18
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As train X leaves \(1.5\) hours before train Y and travels at \(x\) mph, then train X will be \(1.5x\) miles ahead when Y begins its trip.
As Y is travelling at a faster speed of \(y\) mph, their relative speed will be \(y - x\).
This means that \(y\) will pass \(x\) after (formula for time = distance/speed): \(\frac{1.5x}{y-x}\)
(1) y = 4x/3
Plugging this into \(\frac{1.5x}{y-x}\) gives:
\(\frac{1.5x}{\frac{4}{3}x-x}\)
\(\frac{1.5x}{\frac{1}{3}x}\)
\(4.5\) hours.
SUFFICIENT
(2) x = 60 miles per hour
Plugging this into \(\frac{1.5x}{y-x}\) gives:
\(\frac{90}{y-60}\)
As we know nothing about \(y\) we cannot solve the question with only this info.
INSUFFICIENT
ANSWER A
As Y is travelling at a faster speed of \(y\) mph, their relative speed will be \(y - x\).
This means that \(y\) will pass \(x\) after (formula for time = distance/speed): \(\frac{1.5x}{y-x}\)
(1) y = 4x/3
Plugging this into \(\frac{1.5x}{y-x}\) gives:
\(\frac{1.5x}{\frac{4}{3}x-x}\)
\(\frac{1.5x}{\frac{1}{3}x}\)
\(4.5\) hours.
SUFFICIENT
(2) x = 60 miles per hour
Plugging this into \(\frac{1.5x}{y-x}\) gives:
\(\frac{90}{y-60}\)
As we know nothing about \(y\) we cannot solve the question with only this info.
INSUFFICIENT
ANSWER A
General Discussion
04 May 2024, 12:01
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Nidzo
Nice approach....I will solve it this way:
Background: Since X left before Y, total time that X will spend at the point when they both catch up will be greater by 1.5hrs
So we can let time spent by Y at point of catch up be T, the. X will spend (T+1.5)hr.
At the catch up point, both parties must have travelled the same distance. We can put this in math form;
xT=y(T+1.5).....eq1
statement
1. 4x/3=y
Eq 1 above can be written as
xT=4x/3(T+1.5)
*Since x cannot be zero ( this can be infer from free info)*- This is a simple but powerful mindset that I think we should always keep i mind when working with equation like this.
We can cancel out x in both side.
Thus, T can be solved for. Stm 1 is sufficient
2.
x=60
inserting this into eq 1 above, we have:
60T=y(T+1.5)
We have 2 unknown, hence T can not be solved for...Not sufficient
P.S D=S×T
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