The Oddest Distane Speed Problem

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.

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.

thnaks both of u

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....

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....

Typo dude I meant 70 only.

Sorry for the confusion.

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.

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.