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21 Apr 2020, 05:20
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Difficulty:
65%
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Question Stats:
69% (03:07) correct
31%
(03:05)
wrong
based on 193
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A car is moving from A to B at a constant speed. After 80 km, due to an engine malfunction, the speed of the car decreases to 4/5th of the original speed. Because of that, the car reaches B 1 hour and 24 minutes later than intended. Had the same malfunction happened after the car travelled 120 km, the car would have reached B 1 hour later than intended. What was the original speed of the car?
A. 15 Km/hr
B. 20 Km/hr
C. 25 Km/hr
D. 30 Km/hr
E. 40 Km/hr
Are You Up For the Challenge: 700 Level Questions
A. 15 Km/hr
B. 20 Km/hr
C. 25 Km/hr
D. 30 Km/hr
E. 40 Km/hr
Are You Up For the Challenge: 700 Level Questions
21 Apr 2020, 05:41
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Bunuel
A quick and logical approach would be..
Quote:
What is the difference in two scenarios...
So, if the vehicle travels 120-80 km by normal speed and not 4x/5, it should save 24 min..
Thus, \(\frac{40}{\frac{4x}{5}}-\frac{40}{x}=\frac{24}{60}........\)..
\(\frac{50}{x}-\frac{40}{x}=\frac{2}{5}........x=10*\frac{5}{2}=25\)
General Discussion
20 May 2020, 12:38
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[quote="Bunuel"]A car is moving from A to B at a constant speed. After 80 km, due to an engine malfunction, the speed of the car decreases to 4/5th of the original speed. Because of that, the car reaches B 1 hour and 24 minutes later than intended. Had the same malfunction happened after the bus travelled 120 km, the car would have reached B 1 hour later than intended. What was the original speed of the bus?
As speed reduced by 1/5; time should increase by 1/4 to cover the remaining distance
if t1 is the original time to cover the remaining distance in case 1 then, t1*1/4 = 1hr24min = 7/5 => t1=28/5 hrs
Similarly if t is the original time is t2 in case 2 then, t2 * 1/4 = 1 => t2 = 4 hrs;
Hence, the difference in distance in the two cases (=120-80) was to be covered in 28/5-4 hrs = 8/5 hrs
=> original speed = 40 / 8/5 = 25 km/hr
As speed reduced by 1/5; time should increase by 1/4 to cover the remaining distance
if t1 is the original time to cover the remaining distance in case 1 then, t1*1/4 = 1hr24min = 7/5 => t1=28/5 hrs
Similarly if t is the original time is t2 in case 2 then, t2 * 1/4 = 1 => t2 = 4 hrs;
Hence, the difference in distance in the two cases (=120-80) was to be covered in 28/5-4 hrs = 8/5 hrs
=> original speed = 40 / 8/5 = 25 km/hr
