SOLUTION: Hi, Every day Rachel walks 20 km at a certain speed. One day Rachel left 30 minutes late so she increased her speed by 25 kmh and as a result reached her destination 20 minutes ea

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Question 1035107: Hi,
Every day Rachel walks 20 km at a certain speed. One day Rachel left 30 minutes late so she increased her speed by 25 kmh and as a result reached her destination 20 minutes early. How long does she generally take to walk?
Thanks!
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39614) About Me  (Show Source):
You can put this solution on YOUR website!
Twenty minutes is %281%2F3%29hour. 
  
            RATE     TIME          DISTANCE

NORMAL        r        t             20

ONEDAY      r+25    t-1%2F2-1%2F3         20

Tricky part may be examining an expression for travel time on ONE DAY.
She left half hour later and arrival was one-third of hour early.

system%28rt=20%2C%28r%2B25%29%28t-5%2F6%29=20%29


Simplify the ONE DAY equation first.
Multiply left and right by 6.
%28r%2B25%29%286t-5%29=120
6rt%2B150t-5r-125=120
6rt%2B150t-5r=245

Use the NORMAL DAY equation to substitute for rt.
6%2A20%2B150t-5r=245
150t-5r=245-120
150t-5r=125
divide sides by 5.
30t-r=25

Now you have a simpler system of equations:
system%28rt=20%2C30t-r=25%29
and this should give you a quadratic equation to solve for either variable; and use it to find the other variable.

Answer by MathTherapy(10549) About Me  (Show Source):
You can put this solution on YOUR website!

Hi,
Every day Rachel walks 20 km at a certain speed. One day Rachel left 30 minutes late so she increased her speed by 25 kmh and as a result reached her destination 20 minutes early. How long does she generally take to walk?
Thanks! 
With T being normal walk-time, time taken on that day: T+-+5%2F6, leading to the following SPEED equation: 20%2FT+=+20%2F%28T+-+5%2F6%29+-+25. 

Solving for T gives a normal walk-time of: