Sam Loyd's Jack and Jill Puzzle

Level 2

Here is a pretty puzzle from Mother Goose's story of Jack and Jill's race for a pail of water.

The distance to the top of the hill was 440 yards, which is just a quarter of a mile.

Jack got to the top first and was 20 yards on the return trip when he met Jill whom he beat home by half a minute.

The record of the race is complicated by the runners being able to run down hill again one-half faster than they ran up, so you are asked to figure out Jack's time for the half mile run. (Give your answer in minutes.)

The answer is 6.3.

1 solution

Chew-Seong Cheong
Sep 16, 2017

Let the speeds of Jack and Jill to running uphill be $a$ and $b$ respectively.

Then, "Jack got to the top first and was 20 yards on the return trip when he met Jill" means:

$\begin{aligned} \frac {440}a + \frac {20}{\frac 32a} & = \frac {420}b \\ \frac {440}a + \frac {40}{3a} & = \frac {420}b \\ \frac {1360}{3a} & = \frac {420}b \\ \frac {1360}a & = \frac {1260}b \\ \implies b & = \frac {63}{68} a \end{aligned}$

And, "Jack beat Jill home by half a minute" means:

$\begin{aligned} 440 \left(\frac 1b + \frac 1{\frac 32b}\right) - 440 \left(\frac 1a + \frac 1{\frac 32a}\right) & = \frac 12 \\ 440 \times \frac 53 \left(\frac 1b - \frac 1a\right) & = \frac 12 \\ 440 \times \frac 53 \left(\frac {68}{63a} - \frac 1a\right) & = \frac 12 \\ 440 \times \frac 53 \times \frac 5 {63a} & = \frac 12 \\ \color{#3D99F6} 440 \times \frac 5{3a} & = \frac 12 \times \frac {63}5 & \small \color{#3D99F6} \text{Note that }440 \times \frac 5{3a} = t \text{ time for Jack} \\ \color{#3D99F6} t & = \boxed{6.3} & \small \color{#3D99F6} \text{to complete the half mile run.} \end{aligned}$

Sam Loyd's Jack and Jill Puzzle

The answer is 6.3.

1 solution

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