Math, asked by tahauddin007, 7 months ago

A man walks 12 km at a certain speed and then 8 km at 1 km / hour faster. If the total time is 6 hours,
find his original speed?

Answers

Answered by Anonymous
23

\huge\pink{Answer}

Let r = the rate the man walked

then (r+.5) = 1 hr less rate

and (r-1) = 3 hr more rate

let t = time he actually walked

then (t-1) = time taken at the faster speed

and (t+3) = time at the slower speed

Distance for all three situations is the same (dist = rate * time):

So, rt = (r+.5)(t-1)

rt = rt - r + .5t - .5

rt - rt + r - .5t = -.5

r - .5t = -.5 (1)

Also, for the slower speed case,

rt = (r-1)(t+3)

rt = rt + 3r -t - 3

rt - rt - 3r + t = -3

-3r + t = -3 (2)

multiply the 1st equation by 2 and add to the above equation

2r - t = -1

-3r + t = -3

---------------eliminates t

-r = -4

r = 4 km/h is the actual walking rate

Find t using: r -.5t = -.5

4 - .5t = -.5

-.5t = -.5 - 4

-.5t = - 4.5

t = 9 hrs actual time

Hence, Dist = r*t

4 * 9 = 36 km

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Answered by PaapiPandit
0

Answer:

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