Jack and Jill went up and down a hill. They started together from the bottom,…

2024

Jack and Jill went up and down a hill. They started together from the bottom, and while returning, Jack met Jill again 20 miles from the top. Jack completed the whole race 1 minute ahead of Jill. If the hill is 440 miles high and each person's downhill speed is 1.5 times their own uphill speed, how long did Jack take to complete the race?

  1. A.

    12.6

  2. B.

    16.5

  3. C.

    11.3

  4. D.

    17.8

Show answer & explanation

Correct answer: A

For a journey covering a fixed distance, time = distance divided by speed. When two travellers meet at the same instant (a meeting-point problem), the total time each has spent up to that instant is equal, so equating their individual distance-by-speed sums gives one equation. A stated difference between their overall finish times gives a second, independent equation. Two such equations let us solve for the two unknown speeds.

Let Jack's uphill speed = x and Jill's uphill speed = y (in miles per minute). Each person's downhill speed is 1.5 times their own uphill speed: Jack's downhill speed = 1.5x, Jill's downhill speed = 1.5y.

  1. Meeting condition: they meet 20 miles from the top while Jack is returning, so Jack has covered 440 miles up plus 20 miles down = 460 miles, while Jill (still climbing) has covered 440 minus 20 = 420 miles. Since they meet at the same instant, their elapsed times are equal: 440/x + 20/(1.5x) = 420/y.

  2. Combine the left side over a common denominator: 440/x + 20/(1.5x) = 680/(1.5x). So 680/(1.5x) = 420/y, which gives 680y = 630x, i.e. y = 63x/68. Call this equation (1).

  3. Finish condition: Jack's total race time (440 up + 440 down) is 1 minute less than Jill's total race time: 440/x + 440/(1.5x) = 440/y + 440/(1.5y) minus 1.

  4. Each side simplifies to (2200/3) divided by that person's uphill speed, since 440/x + 440/(1.5x) = 440 times (1 + 2/3) divided by x = (2200/3)/x. So the equation becomes (2200/3)/x = (2200/3)/y minus 1, i.e. (2200/3) times (1/y minus 1/x) = 1. Call this equation (2).

  5. Substitute y = 63x/68 from (1) into (2): 1/y minus 1/x = 68/(63x) minus 1/x = 5/(63x). So (2200/3) times 5/(63x) = 1, which gives x = 11000/189.

  6. Jack's total time is t = (2200/3)/x = (2200/3) times (189/11000) = 12.6 minutes.

Cross-check: from y = 63x/68, y = 11000/204, so Jill's total time = (2200/3)/y = 13.6 minutes, exactly 1 minute more than Jack's 12.6 minutes, confirming both the meeting condition and the finish condition are satisfied.

Hence Jack takes 12.6 minutes to complete the race.

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