It takes Tanya 50 minutes to drive to the country club. If the average

chetan2u

hi could u pls weigh in on this problem

Why one method works and the other doesn't is my favourite type of question. It's a good test of your fundamentals.

CounterSniper and royrijit1 allow me to reconcile the disparity between your respective approaches. rahulkashyap this might help you understand as well.

Let us first review the fundamentals required to solve this question:

Theory: If the same distance is covered twice, but with different speeds then the average speed for the total trip = 2(S1*S2)/(S1+S2)

Speed = Distance / Time , As long as one of the factors ( speed , time or distance ) is constant you can equate the other 2.

i.e for the same distance S1/S2 = T2/T1

for the same speed D1/T1 = D2/T2

for the same time D1/S1 = D2/S2


Solution:
Tanya's speed while travelling to the country club = S1

Tanya's speed while travelling back from the country club = S2

Average speed for the entire trip = 2(S1*S2)/(S1+S2)

As per the question, Average speed = 7/8 * S1

2(S1*S2)/(S1+S2) = 7/8 * S1 --> 16 * S2 = 7 * S1 + 7 * S2 --> 9*S2 = 7*S1

Since the distance, D, is constant for each trip --> S1/S2 = T2/T1 = 9/7

T2 = 50 * 9/7 = 64.28 mins ----Same as ---> T2 = 5/6 * 9/7 = 1.071 hrs


Lets us now review the approach used by CounterSniper.

d1=s1*(5/6)
d2=s1*(7/8)*t
s1*5/6=s1*7/8*t -------> (here lies the problem)

if you solve the above you will get approx 57 mins.

Note the average speed = 7/8 S1 and not the speed for the return trip (S2). The above method makes it seem as if you are equating the distance for each trip, but incorrectly using 7/8S1 as the speed for the second trip.

If you want to use the average speed to solve the question, using the method mentioned above, then multiply the distance by 2 as the average speed is calculated for the same distance covered twice.

2 * s1*5/6 = s1*7/8*t

t = 1.904 hrs approx = 114.2 mins approx (This is the time for covering the distance twice. ie going to and coming from the country club)

Time taken to return from the country club = 114 - 50 = 64 mins

NOTICE: if you just multiply the incorrect answer of 57 mins by 2 you get 114 mins.

E is the correct answer.

- Light Yagami

let t=time to drive home
d=one way distance
2d/(50+t)=7/8*d/50
t=450/7=64.3 minutes
answer e

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

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.)

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

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

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?

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)

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

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/

VeritasPrepKarishma

Got it, thanks!!

Hi VeritasPrepKarishma - Would you mind letting me know how you knew right away that 87.5% = 7/8? I know that 1/8 = .125 but didn't put together that 7x that would be 7/8. Is this something to be memorized as well?

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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