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# One day, Rocky walked from his home to his office at three fourths of

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Intern
Joined: 05 May 2018
Posts: 9
One day, Rocky walked from his home to his office at three fourths of  [#permalink]

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08 Aug 2019, 03:11
5
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Difficulty:

85% (hard)

Question Stats:

48% (03:11) correct 52% (03:18) wrong based on 33 sessions

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One day, Rocky walked from his home to his office at three fourths of his usual speed. When he reached midway, he realised that he was 10 minutes late at that point. He, then, increased his speed by 25% and completed the remaining journey. Find the time (in minutes) taken by Rocky to reach his office that day.

A) 64
B) 72
C) 80
D) 60
E) 48
Director
Joined: 19 Oct 2018
Posts: 772
Location: India
Re: One day, Rocky walked from his home to his office at three fourths of  [#permalink]

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08 Aug 2019, 15:09
2
Let the total distance between his home and office= 2x
His usual speed= v

$$\frac{x}{0.75v} - \frac{x}{v}=10$$
x/v=30

As he took 10 minutes more to cover his first half of his journey, time taken to cover first half= 30+10=40

His speed to cover the remaining journey= $$\frac{3v}{4}*\frac{5}{4}$$= $$\frac{15v}{16}$$

time taken to cover second half= $$\frac{16}{15}*(\frac{x}{v})$$=$$\frac{16}{15}*30$$= 32 mins

Total time taken= 40+32=72 mins

MrVarghese wrote:
One day, Rocky walked from his home to his office at three fourths of his usual speed. When he reached midway, he realised that he was 10 minutes late at that point. He, then, increased his speed by 25% and completed the remaining journey. Find the time (in minutes) taken by Rocky to reach his office that day.

A) 64
B) 72
C) 80
D) 60
E) 48
Director
Joined: 12 Feb 2015
Posts: 886
One day, Rocky walked from his home to his office at three fourths of  [#permalink]

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13 Aug 2019, 22:17
MrVarghese wrote:
One day, Rocky walked from his home to his office at three fourths of his usual speed. When he reached midway, he realised that he was 10 minutes late at that point. He, then, increased his speed by 25% and completed the remaining journey. Find the time (in minutes) taken by Rocky to reach his office that day.

A) 64
B) 72
C) 80
D) 60
E) 48

You can solve this question simply by jotting down what is given and proceed step by step.

Assume the distance between home and office as "D" and the normal time taken as "T"

Usual speed is D/T but Rocky walked from his home to his office at three fourths of his usual speed; Therefore, the revised speed for the first is 3D/4T. Accordingly one can calculate the time taken in terms of T to cover D/2 at respective speeds. The time taken to cover half of the distance at usual speed is T/2 and at revised speed is 2T/3.

The difference between the time taken is 10 minutes and if you solve for T you get T = 60 minutes. [time taken to cover first half distance is 40 minutes.

For the second half of the distance Rocky increases the speed by 25%. Therefore, 3D/4T*5/4=15D/16T.

D/2 divided by 15D/16T and substituting T = 60 minutes we get the time taken to cover second half as 32 minutes.

Total time is 40 + 32 = 72 minutes. Option B is the correct ans.
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One day, Rocky walked from his home to his office at three fourths of   [#permalink] 13 Aug 2019, 22:17
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# One day, Rocky walked from his home to his office at three fourths of

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