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30 Jun 2021, 06:45
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
72% (01:58) correct
28%
(02:11)
wrong
based on 192
sessions
History
Date
Time
Result
Not Attempted Yet
If Bill traveled 15 miles per hour faster and it took him 1/3 less time, how fast was he going on the initial trip?
A. 15mph
B. 20mph
C. 25mph
D. 30mph
E. 45mph
A. 15mph
B. 20mph
C. 25mph
D. 30mph
E. 45mph
01 Jul 2021, 12:30
Kudos
Bookmarks
Bunuel
Solution :
Let Bill's speed in the initial trip be s.
Let the distance travelled be d.
Initial Trip
Distance travelled = d
Speed = s
Therefore, time = \(\frac{d}{s}\)
Second Trip
Distance travelled = d
Speed = s + 15 (since he travelled 15 miles/hour faster than his initial speed)
Therefore, time = \(\frac{d}{s + 15}\)
According to the problem,
Time taken in the second trip is \(\frac{1}{3}\) less than the time taken in the initial trip
- ⟹ \(\frac{d}{s + 15}\) = \(\frac{d}{s}\) - \(\frac{1}{3}\)*\(\frac{d}{s}\)
⟹ \(\frac{1}{s + 15}\) = \(\frac{1}{s}\) - \(\frac{1}{3}\)*\(\frac{1}{s}\)
⟹ \(\frac{1}{s + 15} = \frac{3 - 1}{3s}\)
⟹ \(\frac{1}{s + 15} = \frac{2}{3s}\)
⟹ \(3s = 2 (s + 15)\)
⟹ \(3s = 2s + 30\)
⟹ \(3s - 2s = 30\)
⟹ \(s = 30\)
Thus, the speed in the initial trip \(30mph\)
Hence, the correct answer is Option D.
01 Jul 2021, 19:57
Kudos
Bookmarks
Bunuel
Distance = Speed * Time
If the old speed is \(x\) miles/hour, new speed will be \(x+15\) miles/hour.
If he travelled \(30\) min(\(\frac{1}{2}\) hour) during the initial time, he would have taken \(\frac{2}{3}\)rds the time,
which is \(\frac{2}{3}*30 = 20\) min or \(\frac{2}{3}*\frac{1}{2} = \frac{1}{3}\) hour to travel the distance.
Since distances are equal, \(\frac{1}{2}*x = \frac{1}{3}*(x+15) \) -> \(3x = 2x + 30\) -> \(x = 30\) (Option D)
