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21 Apr 2020, 05:18
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
64% (03:22) correct
36%
(03:20)
wrong
based on 167
sessions
History
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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 is the distance between A and B?
A. 120 km
B. 140 km
C. 220 km
D. 240 km
E. 250 km
Are You Up For the Challenge: 700 Level Questions
A. 120 km
B. 140 km
C. 220 km
D. 240 km
E. 250 km
Are You Up For the Challenge: 700 Level Questions
22 Apr 2020, 00:27
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Let the normal speed of the car be 5x kmph. Due to the malfunction, its speed becomes 4x kmph.
Let’s draw some line diagrams to represent the situations described in the question.
22nd April 2020 - Reply 6.jpg [ 45.72 KiB | Viewed 5642 times ]
The first line diagram represents the reference case where the car reached its destination on time since it travelled the entire distance at its normal speed. Let’s call this car C.
The second line diagram represents the first situation described in the question – that of the car breaking down after 80 km. In this case, the car reaches its destination 84 mins later than the reference case. Let’s call this car \(C_1\).
The third line diagram represents the second situation described in the question – that of the car breaking down after 120 km. In this case, the car reaches its destination 60 mins later than the reference case. Let’s call this car \(C_2\).
When we compare the second and the third line diagrams, we see that the car \(C_2\) saved 24 minutes of time only because it travelled an extra 40 km at its normal speed. If we consider the two times as \(T_1\) and \(T_2\),
\(\frac{T_2 }{ T_1}\) = \(\frac{S_1 }{ S_2}\) = \(\frac{4}{5}\).
This means \(T_2\) = \(\frac{4}{5}\) \(T_1\); this essentially means that the car \(C_2\) saved some time in the third case since it travelled at its original speed. The time saved is \(\frac{1}{5}\) T1. This can be equated to 24 minutes since it is the actual time saved.
\(\frac{1}{5}\) \(T_1\) = 24 which gives us \(T_1\) = 120 minutes.
This means \(C_1\) took 2 hours to travel 40 km which tells us that speed of \(C_1\) = 20kmph. But, this represents \(\frac{4}{5}\) th of its original speed. Therefore, original speed = 25 kmph.
Let the total distance be X km. Then,
Time taken by C = \(\frac{X }{25}\)
Time taken by \(C_1\) = \(\frac{80 }{ 25}\) + \(\frac{(X-80) }{ 20}\)
Time taken by \(C_2\) = \(\frac{120 }{ 25}\) + \(\frac{(X-120) }{ 20}\)
Time taken by \(C_1\) = Time taken by C + \(\frac{7}{5}\) hours
\(\frac{80 }{ 25}\) + \(\frac{(X-80) }{ 20}\) = \(\frac{X }{ 25}\) + \(\frac{7}{5}\).
Solving the above equation, we get X = 220 km.
The correct answer option is C.
Hope that helps!
Let’s draw some line diagrams to represent the situations described in the question.
Attachment:
22nd April 2020 - Reply 6.jpg [ 45.72 KiB | Viewed 5642 times ]
The first line diagram represents the reference case where the car reached its destination on time since it travelled the entire distance at its normal speed. Let’s call this car C.
The second line diagram represents the first situation described in the question – that of the car breaking down after 80 km. In this case, the car reaches its destination 84 mins later than the reference case. Let’s call this car \(C_1\).
The third line diagram represents the second situation described in the question – that of the car breaking down after 120 km. In this case, the car reaches its destination 60 mins later than the reference case. Let’s call this car \(C_2\).
When we compare the second and the third line diagrams, we see that the car \(C_2\) saved 24 minutes of time only because it travelled an extra 40 km at its normal speed. If we consider the two times as \(T_1\) and \(T_2\),
\(\frac{T_2 }{ T_1}\) = \(\frac{S_1 }{ S_2}\) = \(\frac{4}{5}\).
This means \(T_2\) = \(\frac{4}{5}\) \(T_1\); this essentially means that the car \(C_2\) saved some time in the third case since it travelled at its original speed. The time saved is \(\frac{1}{5}\) T1. This can be equated to 24 minutes since it is the actual time saved.
\(\frac{1}{5}\) \(T_1\) = 24 which gives us \(T_1\) = 120 minutes.
This means \(C_1\) took 2 hours to travel 40 km which tells us that speed of \(C_1\) = 20kmph. But, this represents \(\frac{4}{5}\) th of its original speed. Therefore, original speed = 25 kmph.
Let the total distance be X km. Then,
Time taken by C = \(\frac{X }{25}\)
Time taken by \(C_1\) = \(\frac{80 }{ 25}\) + \(\frac{(X-80) }{ 20}\)
Time taken by \(C_2\) = \(\frac{120 }{ 25}\) + \(\frac{(X-120) }{ 20}\)
Time taken by \(C_1\) = Time taken by C + \(\frac{7}{5}\) hours
\(\frac{80 }{ 25}\) + \(\frac{(X-80) }{ 20}\) = \(\frac{X }{ 25}\) + \(\frac{7}{5}\).
Solving the above equation, we get X = 220 km.
The correct answer option is C.
Hope that helps!
23 Apr 2020, 00:39
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Attachment:
Engine malfunction.PNG [ 52.76 KiB | Viewed 5386 times ]
In scenario 1 where the engine malfunctions after 80 km, the car reaches its destination 1 hour and 24 minutes (i.e \(\frac{7}{5}\) hours) later than originally anticipated.
In scenario 2 where the engine malfunctions after 120 km, the car reaches its destination 1 hour later than originally anticipated.
Implying, the car saves 24 minutes if it travels the 40 km (120km - 80km) at its original speed (assumed to be 'x') rather than the reduced speed ( given as '4/5x')
Time (t) = Distance (d) / Speed (S)
Therefore;
\(\frac{40}{4/5x}\) - \(\frac{40}{x}\) = 24 minutes = \(\frac{2}{5}\) hours
Solving this gives us x = 25 km/hr. Implying that after the engine malfunction, reduced speed = \(\frac{4}{5}\)*25 = 20km/hr
Now, can use time equation in either of the scenarios to calculate distance
1.
\(\frac{80}{25}\) + \(\frac{(d-80)}{20}\) = \(\frac{d}{25}\) + \(\frac{7}{8}\) OR
2.
\(\frac{120}{25}\) + \(\frac{(d-120)}{20}\) = \(\frac{d}{25}\) + 1
to solve for 'd' which works out to 220 km
