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By Titanz05 (Titanz05) on Friday, March 26, 2004 - 09:36 pm: Edit |

Ms.Clark drove from her home to the museum at an avg. speed of 40 miles per hour and returned home alone the same route at an avg. speed of 35 mph.if her total drving time for the trip was 2 hours, how many MINUTES did it take Ms.Clark to drive from her HOME to the MUSEUM?

a)70

b)60

c)56

d)45

e)40

can someone tell me how to logically set this problem up?

By Kewlkiwi102 (Kewlkiwi102) on Friday, March 26, 2004 - 10:11 pm: Edit |

Ahh...SAT math...fun stuff.

Ok here goes. (I think I did this right...)

The distance travelled one way is x miles.

So the time for the 1st part of the trip is x/40 hours. The time requiered for the second part of the trip is x/35 hours. (B/C you say x miles times 1h/40mi= x/40. then do the same thing for 35 mph)

So, the total time is 2 hours-->

x/40 + x/35 = 2

Cross multiply and solve for x, and you get

x=37.3333 miles one way.

Then, to get the time for the 1st part of her trip, you say

37.333 times 1hr/40mi=0.9333 hours

Convert to minutes

0.9333 hours times 60= 56 minutes.

So, the answer should be C.

By Jens (Jens) on Friday, March 26, 2004 - 10:37 pm: Edit |

d=rt, same distance back and forth so:

40t=35(120-t)

By Xiggi (Xiggi) on Friday, March 26, 2004 - 10:45 pm: Edit |

How to set this up logically?

After reading the problem, take a quick look at the proposed answers while reasoning:

She obviously drove faster to the museum, so she spent LESS time going than returning => eliminate 70 and 60 (longer and equal)

That leaves you 3 answers. Now look at the DIFFERENCE in rates (speed) => the difference between 40 and 35 is rather small at 5/75 or 1/15. This gives you a hint that the difference in time will be 1 hour +/- 1/15. Actually, it also gives you the correct answer, but that it not the point here. 1/15 of one hour is 4 minutes.

Would it be possible to spend only 45 or 40 minutes to drive to the museum? Hardly. This leaves you only answer C or 56 minutes as LOGICAL ANSWER.

If you are interested in the mathematical solution, here is one approach.

Apply the formula distance = rate x time. In this case we could set up that the time to go as x hours and the time to return as (2 hours -x hours). The formulas would yield:

d = 40x

d = 35 (2-x)

then you have 40x = 35(2-x) or

40x = 70 - 35x or

75x = 70 or

x = 70/75 or

x = 14/15 hours or

14/15 * 60 minutes or

56 minutes.

Another approach is to evaluate the difference in speeds, and realize that the difference in time has to equate it. The difference between 40 and 35 is 1/15th. Apply the same proportion to half the total time of 120 minutes and you obtain 60 X 14/15 to go or 56 minutes.

Those problem ALWAYS look a lot harder than they are. The answer can ALWAYS be guesstimated rather easily since it usually very close to the trick answer that is typically 1/2 of the total of the two speeds.

Hope this helps!

By Conker (Conker) on Saturday, March 27, 2004 - 04:31 am: Edit |

Just out of curiosity, what difficulty question is this?

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