Love to Take the Metro?

Let the time taken with 50 mph be T min

Then time taken with 80 mph will be 5/8 X T min

And the time taken with 60 mph will be 5/6 X T min

Given that T - 5/8 X T = 30 min

i.e. 3/8 X T = 30 min

T = 80 min

Time taken with 60 mph is 5/6 X T = 5/6 X 80 = 200/3 = 66 min 40 sec

I guess this is missing in your options.

let time supposed to enter (in mins) = t and distance (in miles) = d

so

d=50/60*(t+20)
d=80/60*(t-10)

d=d
5t+100=8t-80
180 = 3t
t=60 mins supposed to enter

now let x be ans

5(60+20) = 6(60+x)
5/6(80) - 60 = x
400/6 - 360/6 = x
x=40/6=20/3

which isn't a valid answer choice??

m&m wrote:let time supposed to enter (in mins) = t and distance (in miles) = d

so

d=50/60*(t+20)
d=80/60*(t-10)

d=d
5t+100=8t-80
180 = 3t
t=60 mins supposed to enter

now let x be ans

5(60+20) = 6(60+x)
5/6(80) - 60 = x
400/6 - 360/6 = x
x=40/6=20/3

which isn't a valid answer choice??

You Nailed it. Answer choice is D 20/3. Somehow the 3 did not copy. Good job.

A train goes from town A to town B. If it averages 50 mph (miles per hour) then it will be 20 minutes late, and if it averages 80 mph then it will be 10 minutes early. If it goes 60 mph it will be (in minutes)

(a) early by 3
(b) early by 7/3
(c) late by 5
(d) late by 15/2
(e) late by 20/3


These are the correct answer choices.

Let the time taken with 50 mph be T min

Then time taken with 80 mph will be 5/8 X T min

And the time taken with 60 mph will be 5/6 X T min

Given that T - 5/8 X T = 30 min

i.e. 3/8 X T = 30 min

T = 80 min (which mean normal time taken is 60 min)

Time taken with 60 mph is 5/6 X T = 5/6 X 80 = 200/3 min

So late by 200/3 - 60 = 20/3 minutes

dtweah wrote:A train goes from town A to town B. If it averages 50 mph (miles per hour) then it will be 20 minutes late, and if it averages 80 mph then it will be 10 minutes early. If it goes 60 mph it will be (in minutes)

(a) early by 3
(b) early by 7/3
(c) late by 5
(d) late by 15/2
(e) late by 20/3


These are the correct answer choices.

OA is E.
Let d miles be the distance between towns and t the hours if the train is on time. Then d = 50(t + 1/3) = 80(t – 1/6). Solving gives t = 1and hence d = 200/3 miles. Solving 60(1 + x) = 200/3 gives x = 1/9 hour = 20/3 minutes.