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20 Feb 2020, 13:20
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
67% (03:08) correct
33%
(03:18)
wrong
based on 217
sessions
History
Date
Time
Result
Not Attempted Yet
Arun drove from home to his hostel at 60 miles per hour. While returning home he drove half way along the same route at a speed of 25 miles per hour and then took a bypass road which increased his driving distance by 5 miles, but allowed him to drive at 50 miles per hour along this bypass road. If his return journey took 30 minutes more than his onward journey, then the total distance travelled by him is:
A. 55 miles
B. 60 miles
C. 65 miles
D. 70 miles
D. 76 miles
A. 55 miles
B. 60 miles
C. 65 miles
D. 70 miles
D. 76 miles
20 Feb 2020, 18:09
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Let’s say the the distance between home and hostel is S
(S/2 + 5 )/50 + ( S/2 )/ 25 = S/60 + 0.5
S= 30
So total distance is 2*S +5 =65
Posted from my mobile device
(S/2 + 5 )/50 + ( S/2 )/ 25 = S/60 + 0.5
S= 30
So total distance is 2*S +5 =65
Posted from my mobile device
21 Feb 2020, 00:31
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Solution
- • Let the distance between Arun’s home and his hostel be x miles.
• Time taken to travel his onward journey \(= \frac{x}{60}\) hours.
• Time taken to travel his return Journey \(= \frac{0.5*x}{25 }+ \frac{(0.5*x+5)}{50}\)
• It is given that his return journey took 30 minutes (or, \(\frac{1}{2}\) hours) more than his onward journey.
- o \(\frac{x}{60} + \frac{1}{2} = \frac{0.5*x}{25} + \frac{(0.5*x+5)}{50}\)
o Or, \(\frac{x}{60} + \frac{1}{2 }= \frac{x}{50} + \frac{x}{ 100} + \frac{5}{ 50}\)
o Or, \(\frac{x}{60} - \frac{3x}{100} = \frac{1}{10 }– \frac{1}{2}\)
o Or, \(\frac{{5x – 9x}}{300 }= \frac{-4}{10}\)
o Or, \(x = 30\)
