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jill and her brother jack both walked up a hill. they started at the same time, but jill arrived at the top ten minutes before jack. while resting at the top, jack calculated that if he had walked 50% faster and jill had walked 50% slower, then they would have arrived at the top at the same time. how long did jack actually take to walk up to the top of the hill?

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Oladayoademosuidn

It took Jack 15 minutes to walk up to the top of the hill.

Let v = Jill's speed and v' = Jack's speed. Let t = time it takes Jill to walk up to the top of the hill, since Jill arrives 10 minutes earlier, Jack's time, t' = t + 10.

Since the distance covered by both Jack and Jill is the same, we have that

vt = v't'

vt = v'(t + 10)

Since Jack calculated that if he walked 50 % slower and Jill 50 % faster, they would arrived at the top at the same time, Jack's new speed is V = v' + 0.5v'. (since there is a 50 % increase in speed).

Also, Jill's new speed is V' = v - 0.5v (since there is a 50 % decrease in speed)

Since they both arrive at the top at the same time, T, and the distance covered is equal, we have that VT = V'T

(v' + 0.5v')T = (v - 0.5v)T

1.5v' = 0.5v

v'/v = 1.5/0.5

v'/v = 3

From (1)

vt = v'(t + 10)

(v'/v)t = t + 10

3t = t + 10

3t - t = 10

2t = 10

t = 5 minutes

Since Jack's time t' = t + 10, substituting t = 5 into the equation, we have

t' = t + 10

= 5 + 10

= 15 minutes

So, it took Jack 15 minutes to walk up to the top of the hill.

Learn more about time of travel here:

https://brainly.com/question/12190190

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