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Mark biked from his house to his friend's house in how many [#permalink]
25 Oct 2012, 19:44

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This post was BOOKMARKED

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Difficulty:

35% (medium)

Question Stats:

68% (01:59) correct
32% (01:01) wrong based on 232 sessions

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).

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! _________________

----------------------------------------------------------------------------------------- What you do TODAY is important because you're exchanging a day of your life for it! -----------------------------------------------------------------------------------------

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.

Re: Mark biked from his house to his friend's house in how many [#permalink]
01 Aug 2013, 12:08

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

Re: Mark biked from his house to his friend's house in how many [#permalink]
24 Oct 2013, 20:41

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

Re: Mark biked from his house to his friend's house in how many [#permalink]
25 Oct 2013, 00:59

Expert's post

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. _________________

Re: Mark biked from his house to his friend's house in how many [#permalink]
15 Jan 2015, 12:50

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