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Re: A driver completed the first 20 miles of a 40-mile trip at an average
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12 Apr 2018, 21:18
12 Apr 2018, 21:18
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Expert Reply
asfandabid
VeritasPrepKarishma
I figured out. Sorry!
It's like:
t1/t2=(x-60)/(60-50)
Where t1= 2/5 & t2= 20/x
The equation looks something like this after solving a bit
2x/5*20=(x-60)/10
x=5(x-60)
X=5x-300
4x=300
x= 75
Option D
I figured out. Sorry!
It's like:
t1/t2=(x-60)/(60-50)
Where t1= 2/5 & t2= 20/x
The equation looks something like this after solving a bit
2x/5*20=(x-60)/10
x=5(x-60)
X=5x-300
4x=300
x= 75
Option D
Though you have applied the scale method correctly, I would like to point out that the best method to use here is the formula 2ab/(a+b)
When one drives at speed a for half the distance and at speed b for the other half of the distance, the average speed is given by 2ab/(a+b)
60 = 2*50*b/(50 + b)
b = 3000/40 = 75
Check this post for these formulas: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/0 ... -the-gmat/
A driver completed the first 20 miles of a 40-mile trip at an average
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29 Jun 2021, 23:26
29 Jun 2021, 23:26
1
Kudos
Average speed when a body covers m part of the trip at speed x and the second n part of the trip at speed y:
average speed = (m+n)xy / (my+nx)
60 = (20+20)50x / (20*50 + 20x)
60 = 2000x / 1000 + 20x
60 000 = 800x
x = 75
Questions of speed, distance, work and time are the few types of questions where I really prefer to use memorized formulas.
average speed = (m+n)xy / (my+nx)
60 = (20+20)50x / (20*50 + 20x)
60 = 2000x / 1000 + 20x
60 000 = 800x
x = 75
Questions of speed, distance, work and time are the few types of questions where I really prefer to use memorized formulas.
Re: A driver completed the first 20 miles of a 40-mile trip at an average
[#permalink]
18 Aug 2024, 20:19
18 Aug 2024, 20:19
I solved this in 40 seconds so sharing a quick solution
When distances are same and you have travelled with u rate for x distance and y rate for another x distance avg speed is 2uv/(u+v)
We can see that the distances are same here - so that shouldnt be an issue for formula
Let v be the speed of the second leg of the journey
2(50)(v)/(50+v)= 60
Solve to get v 75
When distances are same and you have travelled with u rate for x distance and y rate for another x distance avg speed is 2uv/(u+v)
We can see that the distances are same here - so that shouldnt be an issue for formula
Let v be the speed of the second leg of the journey
2(50)(v)/(50+v)= 60
Solve to get v 75
