
By: drstevewu (offline) Tuesday, August 13 2013 @ 09:32 PM CDT (Read 946 times)



drstevewu 
Need help on this question: Town A and Town B are 600 km apart. At 9am, a van travelling at a uniform speed left Town A for Town B. At the same time, a car set off from Town B for Town A at a uniform speed that was 12 km/h faster than the van. Find the average speed of the car if the two vehicles met at 3pm.
Thank you.
Cheers,
Steve

Junior
Registered: 04/20/10 Posts: 31





By: aceacademy (offline) Wednesday, August 14 2013 @ 12:51 AM CDT



aceacademy 
Hi,
Here's one way to solve the question:
Both vehicles have travelled for 6 hours (9am  3pm) when they met.
In that time, the van would have travelled X km, and the car would have travelled 72 km more, ie, 12 km/h x 6 hrs.
So the car travelled (X+72) km.
Solving for X,
X + X + 72 = 600 km
X = 264 km
Hence, average speed of car is (264 + 72) km / 6 hrs = 56 km/h
Hope this helps.
Ace Academy

Regular Member
Registered: 03/26/13 Posts: 88





By: echeewh (offline) Wednesday, August 14 2013 @ 02:01 AM CDT



echeewh 
Hey <drstevewu>,
Another alternative offered to the above is as follows:
Let speed of Van be sV and speed of Car be sC, as both are unknown.
sV = sC  12
Time taken to meet up from opp direction (at the same time) = 600 ÷ (sC + sV)
= 600 ÷ (sC + sC  12)
= 600 ÷ (2sC  12)
Given time taken for both vehicles to meet is 6 hrs { 1500  0900 }, we have ...
6 = 600 ÷ (2sC  12)
6 × (2sC  12) = 600
12sC  72 = 600
sC = (600 + 72) ÷ 12
= 672 ÷ 12
= 56 km/h
===========
Trust this helps.
Do let me know again if there's further clarifications.
Cheers,
Edward

Active Member
Registered: 04/21/11 Posts: 627





By: drstevewu (offline) Wednesday, August 14 2013 @ 08:43 AM CDT



drstevewu 
Thank you guys for the working and solutions.
Cheers,
Steve..

Junior
Registered: 04/20/10 Posts: 31



