Review:

See this is as a classical relative rates problem – applying the formula RT = D.

- In order to bridge the distance between the two gears, we take the rate of the faster gear and subtract that with the slower gear.

Rp: 10/60 = 1/6. Rq = 10/60 = 1/4.

- The question asks us – when will Q have made exactly 6 more revolutions than gear P?

Rq – Rp = 3/6 = 1/2 .

1/2 is the rate at which Q makes more revolutions then P. So when will gear Q have made exactly 6 more revolutions than gear P?

- 1/2 * T = 6

- T = 6*2 = 12.

12 seconds. Remember to convert minutes into seconds prior to playing with the formulas; as they ask for how many seconds after the gears start rotating….

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