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PS - rate/distance/time increase twice

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PS - rate/distance/time increase twice

by ccassel » Mon Apr 18, 2011 12:03 pm
How would you explain the answer to this question?

If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average of 10 miles per hour faster on tht day?

A. 100
B. 120
C. 140
D. 150
E. 160

Answer D
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by force5 » Mon Apr 18, 2011 2:48 pm
this question can be attempted in many ways... first the algebra method.
Let the number of hours the motorist drives be x
Let his speed be y

Now in normal course he travels xy miles

(x+1)(y+5)=xy+70

=> 5x+y+5 = 70
5x +y = 65

Now if he travels 2 hours longer and 10 miles faster

The distance travelled = (x+2)(y+10)

= xy + 10x+2y+20 = xy+130+20 (substituting the value of 5x+y = 65 in 2(5x+y)+20
=xy+150

Hence he travels 150 miles extra

------------------------------------------
the other approach can be

He covered 70 miles more in an hour by driving 5 miles/hour faster.

Hence his speed current is 70 mile/hr
his original speed will be 65 miles/hr......WHY?? because he is traveling 5mile/hr faster).
Therefore his new speed will be 75miles/hr (given 10 mile/hr faster)
hence distance traveled will be 75*2 = 150


Hope it helps....
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by ccassel » Mon Apr 18, 2011 3:10 pm
Awsome. Thank you for the detailed explanation.

Cheers,
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