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Bunuel
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Time taken = 50 minutes = 5/6 hours
ones side distance = d1
speed = s1
d1=d2
d1=s1*(5/6)
d2=s1*(7/8)*t
s1*5/6=s1*7/8*t
t= 5*8/6*7=40/42= 20/21 = 95% = ~57 minutes
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Dear Bunuel, Is it E? I am getting approx 64.2 =~ 66 minutes


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let the speed at which he drives to club be s

then avg speed = (total distance)/(total time taken)

avg speed =0.875s

total distance = 2d (d being distance)

total time taken = 50 min + x ( x is the unknown to be found)

hence

0.875s= (2d)/(50 +x)

now we know that s=d/50

there fore

0.875(d/50)= (2d)/(50+x)

1.75/100= 2/50+x
1.75(50+x)=200
50+x=200/1.75
x=114.28-50== 64.28 min
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Bunuel
It takes Tanya 50 minutes to drive to the country club. If the average speed of the entire round trip to the club is 87.5% of the average speed on the way to the club, how many minutes approximately will it take Tanya to drive home from the country club?

A. 42 minutes.
B. 48 minutes.
C. 52 minutes.
D. 54 minutes.
E. 66 minutes.


When distances traveled on two legs of a journey are the same,

Avg Speed = \(\frac{2ab}{(a + b)}\)

\((\frac{7}{8})s = \frac{2*s *b}{(s + b)}\)
\(7(s + b) = 16b\)
\(b = (\frac{7}{9})s\)

So speed of going from club to home is (7/9)s
Time taken = 50*(9/7) = more than 63 mins

Answer (E)

Method 2: I would approximate here using the concept of average speed when same distance is covered at different speeds:
Speed on one leg is s and average is (7/8)s. If speed on the other leg were (6/8)s, the average would be slightly less than (7/8)s. So speed on the other leg must be a bit more than (6/8)s = (3/4)s

So time taken on the second leg would be a bit less than 50*(4/3) = 66.66

Answer (E)

(The option given is quite a bit off from the actual answer. That usually shouldn't be the case in an actual GMAT question.)
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Let the distance be d, so the round-trip distance=2d
Again, let he took x minutes to get back.
So, his up speed= d/50 and average speed=2d/x+50, which is 87.5% (7/8) of the up speed.

So, 2d/x+50=d/50*7/8
x=64+2/7 mins
So, ans E
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ok , I messed it up a bit as usual.
took 87.5% as the returning speed instead of the round trip.
though already solved,

average speed = 2ab/a+b (where a is the speed with which she drives to country club and b is while coming back)
as per the question.
2ab/a+b=(7/8)a
or,
a/b=9/7
now since , speed is inversely proportional to time
t1/t2=7/9 (where t1 = 50 min and t2= time taken at speed b )
or,
t2=50(9/7)= ~66 minutes
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Option E.

Since \(\frac{7}{8}\) is the average speed, then the inverse of it is the average time, \(\frac{8}{7}\). Calculate the average time by multiplying the given time: \(\frac{8}{7}*50 = ~57\). Therefore, since the average time is 57, the second time must be greater than 57 minutes. The only option for that is E.

VeritasPrepKarishma, Thoughts?
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Bunuel
It takes Tanya 50 minutes to drive to the country club. If the average speed of the entire round trip to the club is 87.5% of the average speed on the way to the club, how many minutes approximately will it take Tanya to drive home from the country club?

A. 42 minutes.
B. 48 minutes.
C. 52 minutes.
D. 54 minutes.
E. 66 minutes.

Somewhat different approach
Here's my 2 cents...

The average sped of the entire journey is \(\frac{7}{8}\)th of the original speed. If the speed had not changed, then the total journey time would have been 100 minutes. But it did change to \(\frac{7}{8}\)th . Therefore, the total time for the journey will be
\(\frac{8}{7}\)th of 100. We know that the first leg took 50 minutes. so the second leg took?

\(\frac{800}{7} - 50\)

= 64.2 ~ 66 (E)
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Bunuel
It takes Tanya 50 minutes to drive to the country club. If the average speed of the entire round trip to the club is 87.5% of the average speed on the way to the club, how many minutes approximately will it take Tanya to drive home from the country club?

A. 42 minutes.
B. 48 minutes.
C. 52 minutes.
D. 54 minutes.
E. 66 minutes.

We can let a = average speed of the round trip, b = average speed to the club, c = average speed back home and d = the one-way distance. We need to determine d/c.

We are given that d/b = 5/6 (recall that 50 minutes = 5/6 hour) and a = (7/8)b (recall that 87.5% = 7/8). Using the average speed for the round trip, we can create the following equation:

a = 2d/(d/b + d/c)

a = 2d/(5/6 + d/c)

5/6 + d/c = 2d/a

d/c = 2d/a - 5/6

Recall that a = (7/8)b, so substituting (7/8)b for a, we have:

d/c = 2d/[(7/8)b] - 5/6

d/c = 2(d/b)/(7/8) - 5/6

d/c = 2(5/6)/(7/8) - 5/6

d/c = 5/3 x 8/7 - 5/6

d/c = 40/21 - 5/6

d/c = 80/42 - 35/42 = 45/42 = 15/14 hr

We see that this is more than 1 hour or 60 minutes, and the only answer choice that is greater than 60 minutes is choice E (66 minutes) so E is the correct answer.

Answer: E
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Bunuel Hi!

Can you please have a look at my solution and tell me if i've done it correct? Suggestions from other members will be appreciates. Thanks!
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File comment: @Bunuel Hi!

Can you please have a look at my solution and tell me if i've done it correct? Suggestions from other members will be appreciates. Thanks!

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asfandabid
Bunuel Hi!

Can you please have a look at my solution and tell me if i've done it correct? Suggestions from other members will be appreciates. Thanks!

Note that this is not correct.
The average speed of x and y would be (x + y)/2 when they are maintained over the same stretch of TIME.
The average speed of x and y is 2xy/(x + y) when they are maintained over the same stretch of DISTANCE.

You can easily derive these from Average Speed = Total Distance / Total Time

Check this post for more: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/0 ... -the-gmat/
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VeritasPrepKarishma

Got it, thanks!!
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Ok I solved this a bit differently, may be a fluke!

Going:
T= D/S
50/60= D/100 --> assumed speed as 100

Returning:
T=D/S
t2 = D/s2 --> I don't know s2 here, but I know their avg speed is 87.5. Thus solving I get s2 as 75

Noe distance D is constant. Thus 50*100/60 (from going) is equal to t2*75 (from return)... This gives us t2 as 66.66666.... E

Not sure if this works always!

Bunuel the great, is the approach ok? Thanks.


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