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The Oddest Distane Speed Problem

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The Oddest Distane Speed Problem

by wilderness » Sun Jul 13, 2008 8:04 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


I am told that the answer is 150. But I am not sure and more importantly how ?
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by ildude02 » Sun Jul 13, 2008 8:25 am
rt = OrigD - 1st

(t+1)(r+5) = OrigD +70; - 2nd

((t+2)(r+10) = NewD; - 3rd

We are asked for, NewD - OrgD;

rt + r +5t + 5 = rt +70; combining 1 and 2
=> r + 5t = 65;

Substituing 1 and 2 in equation 3, rt + 10t + 2r +20 = NewD;

OrgD + 2(r +5t) +20 = NewD;
NewD - OrgD = 2(r+5t) +20; you can easily substitue 65 for r + 5t to get 150.
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by raunekk » Sun Jul 13, 2008 8:27 am
Let speed =r,distance=d,time=t.

we get 2 equations by d=rt,

(r+5)(1+h)-rh=70,
i.e r+5h=65 (d=rt).

Also,
(r+10)(h+2)-rh=x (x =distance travelled)
2(r+5h)+20=x

As, r+5h=65,substituting

we get x=150.

thanks.
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by wilderness » Sun Jul 13, 2008 8:57 am
thnaks both of u
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by reachac » Sun Jul 13, 2008 9:19 am
I solve it this way.

The extra 5mile/hr speed for the usual time say 't' + the extra 1 hour at this incresed speed is used to cover 70 miles.

so 5*(t+1)= 70
t=13.

similarly in the second case

10(t+2)=????
or 10*(13+2)=150....
Last edited by reachac on Sun Jul 13, 2008 9:35 am, edited 2 times in total.
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by ildude02 » Sun Jul 13, 2008 9:31 am
Didn't get how you got the 10 mile? I assume it's 70.
reachac wrote:I solve it this way.

The extra 5mile/hr speed for the usual time say 't' + the extra 1 hour at this incresed speed is used to cover 10 miles.

so 5*(t+1)= 10
t=13.

similarly in the second case

10(t+2)=????
or 10*(13+2)=150....
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by reachac » Sun Jul 13, 2008 9:34 am
Typo dude I meant 70 only.

Sorry for the confusion.
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by ildude02 » Sun Jul 13, 2008 9:37 am
I didn't look at the value of the "t" when I posted that response, I was kinda figuring out may be somehow 10 might be a possibility :)

BTW, it's a real good approach.
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by parallel_chase » Sun Jul 13, 2008 12:29 pm
ildude02 wrote:I didn't look at the value of the "t" when I posted that response, I was kinda figuring out may be somehow 10 might be a possibility :)

BTW, it's a real good approach.
I have done it in a simpler way. I know simple is a relative term. What is simple for me may not be simple for another person. Anyways here is my explanation.


Everything is already given to us.

The motorist drives 70 miles more, 5 hours faster and 1 hr longer.

let x be the original rate.

x+5 = 70/1 (R= D/T)

x = 65

The motorist this time drives 10 hours faster, and 2 hours longer.

65 + 10 = D / 2 => D= 75*2 => 150


Let me know if you have any doubts in the above explanation.
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