Question

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?

Answer 1

Answer:

Example: Simplify the ratio 6 : 10

The factors of 6 are: 1, 2, 3, 6.

The factors of 10 are: 1, 2, 5, 10.

Then the greatest common factor of 6 and 10 is 2.

Divide both terms by 2.

6 ÷ 2 = 3.

10 ÷ 2 = 5.

Rewrite the ratio using the results. The simplified ratio is 3 : 5.

6 : 10 = 3 : 5 in simplest form

Answer 2

Given:

Jack and Jill started walking at the same time

Jill arrived at the top ten minutes before Jack

If Jack had walked 50% faster and Jill had walked 50% slower, then they would have arrived at the top at the same time

To Find:

Time (t) taken by Jack to walk up to the top

Solution:

We know that Speed = Distance / Time

Let the speed of Jack be V 1 and the speed of Jill be V 2.

Let distance covered by Jack and Jill be s.

So, V 1 = s /t       - (1)

and V 2 = s / t-10      - (2)

According to the question, if Jack had walked 50% faster (V 1') and Jill had walked 50% slower (V 2'), then they would have arrived at the top at the same time.

⇒ V 1' = $V1 + \frac{50 }{100} V1$

= 3 X V1 /2

Similarly,

V 2' = V2 ( 1 - 1/2) = V2 / 2

Since they would have reached at the same time and would have covered the same distance s,

⇒ V 1' = V2'

or  3 V1 /2 = V2 / 2

or 3 V1 = V2     -(3)

Substituting the values of V1 and V2 in V3,

$3\frac{S}{t} = \frac{S}{t-10}$

Cross multiplying and solving,

3st - 30s = st

or 3t - 30 = t

or t = 15 minutes

Hence, Jack took 15 minutes to climb the hill.