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# ps distance problem

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Joined: 24 Oct 2007
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26 Oct 2007, 10:49
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20. 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

any ideas on how to set up an equation?

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SVP
Joined: 29 Aug 2007
Posts: 2471

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26 Oct 2007, 11:10
20. 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

any ideas on how to set up an equation?

original: time = t, rate = r and distance = d = tr
after change: time = t+1, rate = r+5 and distance, d +70 = (t+1)(r+5)
d +70 = tr + 5t + r+5
70 = 5t + r+5

after change again: time = t+2, rate = r+10 and distance, d +x = (t+2)(r+10)
d +x = tr + 10t +2r+20
d +x = d + 10t +2r+10+ 10
x = 2 (5t +r+5) + 10
x = 2x70 + 10
x = 150

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Intern
Joined: 25 Apr 2007
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26 Oct 2007, 11:19
Lets say the motorist was on the road for 1 hour driving at a speed of X mph

Distance covered = X * 1 = X

If motorist had driven 1 hour longer at an average rate of 5 miles per hour faster , he would have covered 70 more miles

2(X + 5) = 70 + X
2X + 10 = 70 + X
X = 60 miles

So for the 2nd eqn new speed = 60 + 10 = 70mph and he travels for 3 hours which is 210 miles.

210 - 60 = 150 miles. Hence D.

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26 Oct 2007, 11:19
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# ps distance problem

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