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Carl drove from his home to the beach at an average speed of

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Carl drove from his home to the beach at an average speed of [#permalink] New post 16 Nov 2003, 18:20
Carl drove from his home to the beach at an average speed of 80 km/hr and returned home by the same route at an average speed of 70km/hr. If the trip home took 1/2 hour longer than the trip to the beach, how many km did Carl drive each way?

A)350
B)345
C)320
D)280
E)240


i always spend a lot of time on these types of problems....any shortcuts?
quickest way to solve these types of problems?


s


Answer is:
D
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 [#permalink] New post 16 Nov 2003, 18:32
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Distance will be the same... so, D1 = D2
80*t = 70*(t+1/2)
gives t=7/2
distance=80*7/2 = 280

:)
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 [#permalink] New post 17 Nov 2003, 07:15
Yep, agree on D. I'd also agree with dj's explanation. That is a very simple and short way of doing this. The key is recognizing how to set up the equation.
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 [#permalink] New post 17 Nov 2003, 10:15
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See if the following helps

I think you can solve such problems in lesser time, if you always follow one approach consistently. You can always solve any such problem by using the Distance formula. The moment you see distances/times or speeds put down the formula and try to fit in given information.

Distance= Speed * Time.

Distance - Is being asked so don't know - D
Speed - Given 80
Time - don't know so T.
D= 80*T

Wait.. speed is given twice. So put down the equation again.
Distance - D2
Speed - 70
Time - don't know so T2
D2 = 70 * T2

What all information do we have ?
1) The distance is the same. So D=D2
2) The time taken to return is 1/2 hr more than the time taken to go to the beach => T + 1/2 = T2

Therefore D=80*T = 70*(T+1/2) => T= 3.5 => D= 80*3.5= 280. D is the answer.

Usually we take more time to solve when we don't get an exact sense of the question or if we are torn between two ore more methods to solve. This usually happens if you can sense the possibility of easy solution but are not quite clear about what it is.

Probably the intuitive method would be -> In half an hour at 70 km/hr Carl would have travelled 35 k.m. At a difference in speed of 10 km/hr how much time would it take to fall behind by 35 km? 35/10 = 3.5 hrs. So the distance travelled in these 3.5 hrs = 3.5 * 80
(Think of somebody else starting from the other side at the same time (and the speed of 10km/hr) Carl starts his return. In the time Carl would take to go to the beach, our guy will meet Carl and would have travelled 35 km. Therefore, the time taken by him =35/10 which will also be equal to the time taken by Carl to go to the beach.)

But it's difficult to rely on being able to use such intuitive approaches during the test. Such approaches tend to be problem specific and if you are under pressure, you may find it difficult to think clearly enough to get such solutions. I think it would be better to go to the test with some straight forward methods for such straight questions and leave the intuitive thinking for the really tough questions where you may not have the easier alternative.

Or, you can solve few speed/distance problems before the test. This usually helps.
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 [#permalink] New post 17 Nov 2003, 10:20
pawargmat--

Considering how much time I waste on problems like these, I think your advice is probably very good.

Any other suggestions?
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 [#permalink] New post 17 Nov 2003, 11:43
pawargmat....thank you for that explanation...i think that's just what i needed....
  [#permalink] 17 Nov 2003, 11:43
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