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A Man walked from his house to office at 5kmph and got 20 mi

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A Man walked from his house to office at 5kmph and got 20 mi [#permalink]

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New post 24 Aug 2013, 22:06
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Question Stats:

72% (02:42) correct 28% (02:32) wrong based on 204 sessions

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A Man walked from his house to office at 5kmph and got 20 minutes late. if he had travelled at 7.5kmph, he would have reached 12 minutes early. The distance from his house to office is?

A) 4Km
B) 16Km
C) 8Km
D) 9Km
E) 12Km
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Re: A Man walked from his house to office at 5kmph and got 20 mi [#permalink]

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New post 24 Aug 2013, 22:22
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rrsnathan wrote:
A Man walked from his house to office at 5kmph and got 20 minutes late. if he had travelled at 7.5kmph, he would have reached 12 minutes early. The distance from his house to office is?
A) 4Km
B) 16Km
C) 8Km
D) 9Km
E) 12Km


Let the original distance and time be D & t
Case 1 - If man travels at 5kmph, he reaches late 20 min or 1/3 hr
We can write the same as
\(\frac{D}{5}\) = t + \(\frac{1}{3}\) -----(1)

Case 2 - If man travels at 7.5kmph, he reaches early 12 min or 1/5 hr
We can write the same as
\(\frac{D}{7.5}\) = t - \(\frac{1}{5}\) -----(2)

Solving both we get D = 8
Answer C
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Re: A Man walked from his house to office at 5kmph and got 20 mi [#permalink]

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New post 25 Aug 2013, 06:21
1
rrsnathan wrote:
A Man walked from his house to office at 5kmph and got 20 minutes late. if he had travelled at 7.5kmph, he would have reached 12 minutes early. The distance from his house to office is?

A) 4Km
B) 16Km
C) 8Km
D) 9Km
E) 12Km


Discussed here: time-speed-and-distance-simplified-150163.html
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Re: A Man walked from his house to office at 5kmph and got 20 mi [#permalink]

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New post 25 Aug 2013, 19:23
2
rrsnathan wrote:
A Man walked from his house to office at 5kmph and got 20 minutes late. if he had travelled at 7.5kmph, he would have reached 12 minutes early. The distance from his house to office is?

A) 4Km
B) 16Km
C) 8Km
D) 9Km
E) 12Km


1 Distance traveled is the same in both the cases
2 5 * (t + 1/3) = 7.5 * (t- 1/5)
3 t=19/15 hrs
4 Substituting the value of t in the LHS of (2), Distance = 5 *( 19/15 +1/3) = 8 km

Substituting the value of t in RHS of (2) will also give the same answer
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Re: A Man walked from his house to office at 5kmph and got 20 mi [#permalink]

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New post 11 Jan 2014, 08:17
rrsnathan wrote:
A Man walked from his house to office at 5kmph and got 20 minutes late. if he had travelled at 7.5kmph, he would have reached 12 minutes early. The distance from his house to office is?

A) 4Km
B) 16Km
C) 8Km
D) 9Km
E) 12Km


I'm not sure if this method is valid but please comment

d/5 = t+20
d/7.5 = t-12

So 5t + 100 = 7.5t - 90

t = (190)(2)/5 = 78

So distance (78)(5) + 100 /60

C

I divided by 60 to go back to hours cause I was working with minutes before

Hope it helps
Cheers!
J :)
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A Man walked from his house to office at 5kmph and got 20 mi [#permalink]

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New post 27 Aug 2015, 02:32
jlgdr wrote:

I'm not sure if this method is valid but please comment

d/5 = t+20
d/7.5 = t-12

So 5t + 100 = 7.5t - 90

t = (190)(2)/5 = 78

So distance (78)(5) + 100 /60

C

I divided by 60 to go back to hours cause I was working with minutes before

Hope it helps
Cheers!
J :)


I did it the same way, however t = (190*2)/5 should read = 76 not 78.

(76*5 + 100) / 60 = 8
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Re: A Man walked from his house to office at 5kmph and got 20 mi [#permalink]

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New post 24 Nov 2016, 10:50
1
rrsnathan wrote:
A Man walked from his house to office at 5kmph and got 20 minutes late. if he had travelled at 7.5kmph, he would have reached 12 minutes early. The distance from his house to office is?

A) 4Km
B) 16Km
C) 8Km
D) 9Km
E) 12Km


Distance is constant

\(\frac{5* ( t + 20 )}{60} = \frac{( t - 12 )*30}{4*60}\)

\(20( t + 20 ) = ( t - 12 )30\)

\(20t + 400 = 30t - 360\)

\(760 = 10t\)

So, \(t = 76\)

Thus distance = \(\frac{5( 76 + 20 )}{60}\)

Hence distace = 8 km

Answer will be (C) 8 Km

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Re: A Man walked from his house to office at 5kmph and got 20 mi [#permalink]

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New post 24 Nov 2016, 13:48
rrsnathan wrote:
A Man walked from his house to office at 5kmph and got 20 minutes late. if he had travelled at 7.5kmph, he would have reached 12 minutes early. The distance from his house to office is?

A) 4Km
B) 16Km
C) 8Km
D) 9Km
E) 12Km


d/5-8/15=d/7.5
d=8 km
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Re: A Man walked from his house to office at 5kmph and got 20 mi [#permalink]

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New post 23 Jan 2018, 08:02
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Re: A Man walked from his house to office at 5kmph and got 20 mi   [#permalink] 23 Jan 2018, 08:02
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