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If a motorist

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If a motorist

by ankit1383 » Fri Dec 19, 2008 1:43 am
If a motorist has 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
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by 720dreaming » Fri Dec 19, 2008 1:50 am
D. 70 more miles at 5 miles per hour faster means his real rate was 65. Since we know he only drove 1 additional hour.

At ten miles per hour faster, his new rate is 75. 75*2=150.
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by gmatman1 » Fri Dec 19, 2008 5:42 am
In my opinion the question is not formulated very clearly. I interpretted the question differently, which results in an unsolvable problem.

I assumed that the motorist drove the whole day at the new speed, instead of the last extra hour(s). This means there will be 3 variables. So there is one variable too much to solve the two equations shown below.

s = speed
t = time
d= distance

1. st = d
2. (s + 5)(t + 1) = d + 70
Last edited by gmatman1 on Fri Dec 19, 2008 10:29 am, edited 1 time in total.
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by mrsmarthi » Fri Dec 19, 2008 7:46 am
gmatman1 wrote:In my opinion the question is not formulated very clearly. I interpretted the question differently, which results in an unsolvable problem.

I assumed that the motorist drived the whole day at the new speed, instead of the last extra hour(s). This means there will be 3 variables. So there is one variable too much to solve the two equations shown below.

s = speed
t = time
d= distance

1. st = d
2. (s + 5)(t + 1) = d + 70
Let s be the speed, t be the time that he travelled actually. So distance travelled is

st = d -----> Eq1

Hypothetical speed when speed is 5 more and time is 1 hr additional then distance travelled is

(s+5)(t+1) = d + 70. ==> st + s + 5t + 5 = d+70

d + s + 5t = d + 65 (Substituting st = d from Eq 1)

==> s + 5t = 65 --> Eq 2.

Hupothetical situation 2 where speed is 10 miles greater and time is 2 hrs more then

(s + 10)(t + 2) = ?

==> st + 2s + 10t + 20 = ?

==> d + 130 + 20

==> d + 150.

So additional distance travelled would be 150 Miles.

IMO D.
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by trnguyen » Fri Dec 19, 2008 8:02 am
(r+5)(t+1) = 70+d

rt+r+5t+5 = 70 +d

rt+r+5t-d=65

d + r + 5t – d = 65

r + 5t = 65

(r+10)(t+2) = x+d

rt+2r+10t+20 = x +d

rt + 2r + 10t – d = x -20

d + 2r + 10t – d = x – 20

2r + 10t = x - 20

Multiply the first equation by 2: 2r + 10t = 130

130 = x – 20

X = 150
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by amitabhprasad » Fri Dec 19, 2008 9:29 am
Diff in velocity = +5
total time = t+1
Distance covered = 70
==> 70 = 5(t+1) ==> t = 13
if relative velocity = +10
extra time = +2
then distance = (13+2)*10 = 150
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by tritrantran » Fri Dec 19, 2008 10:22 am
amitabhprasad wrote:Diff in velocity = +5
total time = t+1
Distance covered = 70
==> 70 = 5(t+1) ==> t = 13
if relative velocity = +10
extra time = +2
then distance = (13+2)*10 = 150
I like this method. Clear and efficient!
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by gmatman1 » Fri Dec 19, 2008 10:28 am
I think 720dreaming uses an approach that is only valid if you assume that the extra distance (and change in speed) is made in the last hour(s).

The approach by mrsmarthi is based on the assumption that the motorist drives the whole day at the new speed, instead of the last extra hour(s).

How can it be then that both approaches result in the same answer? Or am I just confused?

By the way, I think mrsmarthi's method is the best.
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by gmatman1 » Tue Dec 23, 2008 8:37 am
Nobody? I would really appreciate some further explanation. :D
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