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Mark biked from his house to his friend's house in how many

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Mark biked from his house to his friend's house in how many [#permalink]

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New post 25 Oct 2012, 20:44
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Mark biked from his house to his friend's house in how many hours?

(1) Mark bikes at an average speed of 72 blocks per hour.

(2) If Mark could bike an extra 8 blocks for each hour, he would have arrived at his friend's place 1 hour earlier

I got the correct answer (no spoilers!) but if I were to sit down and calculate the answer, how would I do it (I know purpose is not to calculate in DS, but it's really for practice, if anything).

Thanks again.
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Oct 2012, 04:26, edited 1 time in total.
Renamed the topic and edited the question.

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Re: Bike rates [#permalink]

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New post 25 Oct 2012, 21:17
elegan wrote:
Hello,

I'm looking at this question.

Mark biked from his house to his friend's house in how many hours?

1. Mark bikes at an average speed of 72 blocks per hour.

2. If Mark could bike an extra 8 blocks for each hour, he would have arrived at his friend's place 1 hour earlier

I got the correct answer (no spoilers!) but if I were to sit down and calculate the answer, how would I do it (I know purpose is not to calculate in DS, but it's really for practice, if anything).

Thanks again.


Mark bikes at an average speed of 72 blocks per hour.
Mark could bike an extra 8 blocks for each hour = 80 blocks per hour

LCM of 72 & 80 = 720. Time taken by Mark = 720/72 = 10 hrs.

Cheers!
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Re: Bike rates [#permalink]

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New post 25 Oct 2012, 23:15
why you took LCM?
to get a distance
can you explain in detial
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Re: Bike rates [#permalink]

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New post 25 Oct 2012, 23:41
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Aristocrat wrote:
why you took LCM?
to get a distance
can you explain in detial

There is no sense in taking LCM in such problems. if question said he saved 2 hrs riding his bike 15 blocks/hr faster, the solution would go haywire.

to solve such problems:

Let d be the distance and t the time taken in first case. thus t-1 is time taken in second case.

therefore from 1:
\(t = d/72\)
and from 2:
\(t-1 = d/80\)

combining these:
\(t-1 = 72t /80\)
=>\(80t -80 =72t\)
=>\(t= 10 hrs\)

Hope it helps.
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Re: Bike rates [#permalink]

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New post 26 Oct 2012, 01:01
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Another way of approaching this problem -

In all cases, if the speed increases by 1/x, the time taken decreases by 1/x+1 (as speed and time taken are inversely proportional to each other).

In this case the increase in speed is 1/9 (Mark's new speed is 72+8=80 blocks per hour, and the increase of 8 blocks per hour is 1/9th of his previous speed).

Therefore the corresponding decrease in time = 1/9+1 = 1/10. This represents 1/10th of the actual time taken.

1/10th of the actual time taken is given as 1 hour. Hence the total time taken is 10 hours.

Hope it helps.

Cheers!

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Re: Mark biked from his house to his friend's house in how many [#permalink]

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New post 01 Aug 2013, 13:08
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Mark biked from his house to his friend's house in how many hours?

Time = Distance/Speed

(1) Mark bikes at an average speed of 72 blocks per hour.
There is no information given about the distance Mark has to travel. All we know is that r=72
INSUFFICIENT

(2) If Mark could bike an extra 8 blocks for each hour, he would have arrived at his friend's place 1 hour earlier
t-1 = d/(r+8)
We can try and plug in various other distance/time/rate formulas to try and cancel out variables but it's unlikely that will leave us with only one given that there are three variables to plug in for.
INSUFFICIENT

1+2) r=72 and t-1 = d/(r+8)
We can plug in for r but we still have variables t and d left. d=r*t. If we were to substitute for d we could cancel out t leaving us with just t, as we are looking for the time it took him to bike to his friends house.
t-1 = d/(r+8)
t-1 = r*t/(72+8)
t-1 = 72*t/(80)
80(t-1) = 72t
80t-80=72t
8t=80
t=10
SUFFICIENT

(C)

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Re: Mark biked from his house to his friend's house in how many [#permalink]

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New post 24 Oct 2013, 21:41
WholeLottaLove wrote:
Mark biked from his house to his friend's house in how many hours?

Time = Distance/Speed

(1) Mark bikes at an average speed of 72 blocks per hour.
There is no information given about the distance Mark has to travel. All we know is that r=72
INSUFFICIENT

(2) If Mark could bike an extra 8 blocks for each hour, he would have arrived at his friend's place 1 hour earlier
t-1 = d/(r+8)
We can try and plug in various other distance/time/rate formulas to try and cancel out variables but it's unlikely that will leave us with only one given that there are three variables to plug in for.
INSUFFICIENT

1+2) r=72 and t-1 = d/(r+8)
We can plug in for r but we still have variables t and d left. d=r*t. If we were to substitute for d we could cancel out t leaving us with just t, as we are looking for the time it took him to bike to his friends house.
t-1 = d/(r+8)
t-1 = r*t/(72+8)
t-1 = 72*t/(80)
80(t-1) = 72t
80t-80=72t
8t=80
t=10
SUFFICIENT

(C)


if we take only second statement
(x+8)(t-1)=distance to his friends home=xt.................cant we find t from here ?
what am i doing wrong

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Re: Mark biked from his house to his friend's house in how many [#permalink]

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New post 25 Oct 2013, 01:59
tyagigar wrote:
WholeLottaLove wrote:
Mark biked from his house to his friend's house in how many hours?

Time = Distance/Speed

(1) Mark bikes at an average speed of 72 blocks per hour.
There is no information given about the distance Mark has to travel. All we know is that r=72
INSUFFICIENT

(2) If Mark could bike an extra 8 blocks for each hour, he would have arrived at his friend's place 1 hour earlier
t-1 = d/(r+8)
We can try and plug in various other distance/time/rate formulas to try and cancel out variables but it's unlikely that will leave us with only one given that there are three variables to plug in for.
INSUFFICIENT

1+2) r=72 and t-1 = d/(r+8)
We can plug in for r but we still have variables t and d left. d=r*t. If we were to substitute for d we could cancel out t leaving us with just t, as we are looking for the time it took him to bike to his friends house.
t-1 = d/(r+8)
t-1 = r*t/(72+8)
t-1 = 72*t/(80)
80(t-1) = 72t
80t-80=72t
8t=80
t=10
SUFFICIENT

(C)


if we take only second statement
(x+8)(t-1)=distance to his friends home=xt.................cant we find t from here ?
what am i doing wrong


Please try to solve and you'll get the answer yourself.
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Re: Mark biked from his house to his friend's house in how many [#permalink]

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New post 27 Oct 2013, 03:21
For st 1, don't we have to know that the rate is uniform during the entire distance to be able to compute the time?
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Mark biked from his house to his friend's house in how many [#permalink]

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New post 16 Oct 2016, 09:10
Let's assume that distance between the two houses is x blocks.

(1) With only avg. speed given, S = 72 blocks per hour, and no other information, it not possible to find the time. One equation, two variables \(t = \frac{x}{72}\)
INSUFFICIENT

(2) Given \(\frac{x}{S} = \frac{x}{(S+8)} + 1\), equation with two variables and no other constraints on x or/and S, it can not be solved.
INSUFFICIENT

However, combining (1) and (2), we can see that substituting value of S in (2) will give us x and then we can get time using \frac{x}{S}. (We don't even need to calculate the exact values.)
Hence, (1) and (2) together are SUFFICIENT.

C

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Mark biked from his house to his friend's house in how many   [#permalink] 16 Oct 2016, 09:10
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