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Speed and distance

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Speed and distance

by naaga » Wed Feb 18, 2009 5:54 am
. 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 rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160


folks again I struck at this problem...
plz help me out ...




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by bluementor » Wed Feb 18, 2009 7:59 am
Actual speed = s
Actual time taken = t
Actual distance covered = d

d = s*t

1st Hypothetical situation: time taken = t + 1, average speed = s + 5
d + 70 = (s+5)(t+1)
st + 70 = st + 5t + s + 5
65 = s + 5t …. eqn 1

2nd hypothetical situation (the question): time taken = t + 2, average speed = s + 10

Total distance motorist would have covered = (s + 10)(t + 2)
= st + 10t + 2s + 20
= d + 2(5t + s) + 20 ….insert eqn 1 here
= d + 2(65) + 20
= d + 150

Additional mileage = 150 miles. Choose D.

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