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# If a motorist had driven 1 hour longer on a certain day and

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Manager
Joined: 23 Mar 2008
Posts: 217

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If a motorist had driven 1 hour longer on a certain day and [#permalink]

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20 Sep 2008, 10:34
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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 that 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

Kudos [?]: 201 [0], given: 0

Manager
Joined: 21 Aug 2008
Posts: 199

Kudos [?]: 18 [0], given: 0

Re: PS: motorist [#permalink]

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20 Sep 2008, 10:48
puma wrote:
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 that 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

D = R x T (where, D = distance, R = speed, and T = time) --> A
D + 70 = (T+1) x (R + 5) --> B

from A and B, we get D = (T+1)x(R+5) - 70 = RT = 5T+R-70=0 --> 5T+R = 70

question is asking, 2 hours longer and 10mph faster...So,

D + x = (T+2) x (R+10) = TR + 10T + 2R + 20 = D + 2(5T + R) + 20
since 5T + R = 70,

D + x = D + 140 + 20
x = 160

Thus, ANS is E.

Kudos [?]: 18 [0], given: 0

VP
Joined: 05 Jul 2008
Posts: 1402

Kudos [?]: 437 [0], given: 1

Re: PS: motorist [#permalink]

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20 Sep 2008, 14:40
Richardson wrote:
puma wrote:
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 that 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

D = R x T (where, D = distance, R = speed, and T = time) --> A
D + 70 = (T+1) x (R + 5) --> B

from A and B, we get D = (T+1)x(R+5) - 70 = RT = 5T+R-70=0 --> 5T+R = 70

question is asking, 2 hours longer and 10mph faster...So,

D + x = (T+2) x (R+10) = TR + 10T + 2R + 20 = D + 2(5T + R) + 20
since 5T + R = 70,

D + x = D + 140 + 20
x = 160

Thus, ANS is E.

150 is my answer. You missed subtracting 5 from 70 . Its 5T+R+5=70

Kudos [?]: 437 [0], given: 1

Manager
Joined: 28 Aug 2008
Posts: 101

Kudos [?]: 43 [0], given: 0

Re: PS: motorist [#permalink]

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21 Sep 2008, 13:58
75*2 = 150

70-5= 65, +10 =75, *2 = 150

Kudos [?]: 43 [0], given: 0

Re: PS: motorist   [#permalink] 21 Sep 2008, 13:58
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# If a motorist had driven 1 hour longer on a certain day and

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