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

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.

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!

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.

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?

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

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

1) Speed = 72 blocks / hour & time taken = x NOT SUFFICIENT
2) new Speed = y+8 blocks / hour & time taken = x-1 NOT SUFFICIENT

(1+2)
Speed = 72 blocks / hour & time taken = x
New Speed = 80 blocks / hour & time taken = x-1
distance is same : 72x=80(x-1) =>8x=80 => x=10

ANSWER C

time = distance / speed

(1) We are told Mike bikes at an average speed of 72 blocks per hour:

time = distance / 72

Not sufficient to determine time because we don't know the distance.

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

time - 1 = distance / 80

Not sufficient to determine time because we don't know the distance traveled.

(1&2)

time - 1 = distance / 80
80*t - 80 = distance

We know from statement 1 that 72*t = distance

80*t - 80 = 72*t
t = 10

Sufficient.

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