I'm starting to despair with the difficulty of the maths regarding stage 1. How hard are the problems?? I keep finding different types of S/D/T problems which I am not familiar with and all sorts of arduousness. Is anybody capable of telling me if, for instance, the problems copied below are more difficult than the ones in the real test or if easier??. Thankss loadss!, I'm in complete despair..
A cargo plane flew to the manteinance facility and back. It took one hour less time to get there than it did to get back. The average speed on the trip was 220 mph. The average speed on the way back was 200 mph. How many hours did the trip there take? (SOL: 10 hours).
For this question let the distance travelled for each leg equal D. Let the time taken to get there be (t-1) and the time taken to get back as t (total time would be 2t-1).
The average speed of the trip is 220mph and the average speed on the way back is 200mph. From this info you can make two equations.
(1) D = 200t
(2) 2D = 220(2t-1)
(these were made using simple distance = speed * time)
Doubling (1) gives 2D = 400t to make equation (3)
You can now equate (2) and (3) so that:
400t = 220(2t-1) Expanding gives:
400t = 440t - 220 Re-arranging gives:
40t = 220 so t = 5.5
But you know that the total time for the trip is 2t-1 so:
(2*5.5)-1 = 10h
Aircraft A left airport and flew south. Aircraft B left three hours later flying at 42mph faster in effort to catch up to A. After two hours B finally caught up. Find A's average speed. (SOL: 28mph)
For questions like this you should write out everything you know about each AC, and what they have in common.
In this question the value they have in common is their distance which we will call D again.
AC A:
All we know is that AC A has been flying for five hours total at an average speed S, as B left 3 hours after and caught up after 2 hours. So all you can say is that:
D = 5xS
AC B
AC B has been flying for a total of two hours at a speed which is 42 mph faster than AC A.
D = 2(S+42)
These two can be equated (they both equal D) to give:
5S = 2(S+42) Expanding gives:
5S = 2S + 84 Re-arranging gives:
3S = 84
So the average speed is 28 mph
Plane B left Chicago flying due east at a speed of 400 mph. Plane A left Chicago 2 hours later, flying in the same direction at a speed of 600 mph. If plane A is 400 miles behind plane B now, how long has plane B traveled? (SOL 4h)
Plane B:
Plane B has allready travelled 400*2 = 800 miles
So its distance for any time after can be written as:
D = 800 + 400t (1)
Plane A
Plane A has travells at 600mph so its distance can be written as:
D = 600t
The qustion asks you to find the time at when place A is 400 miles behind B, so its distance will be 400 less (D-400) which means that:
D - 400 = 600t Re-arranged to give:
D = 400 + 600t Which can be equated to (1) to give:
800 + 400t = 400 + 600t re-arranged to give:
200t = 400
So t = 2 hours
The plane has allready been travelling for 2 hours so 2+2 = 4 hours
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Hope this helped, if I have made any mistakes others could point them out and lemme know if you have any other questions.