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 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

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.

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

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