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Mathematics

Where will the hour hand of a clock stop if it

(i) starts at 10 and makes 12\dfrac{1}{2} of a revolution, clockwise?

(ii) starts at 4 and makes 14\dfrac{1}{4} of a revolution, clockwise?

(iii) starts at 4 and makes 34\dfrac{3}{4} of a revolution, clockwise?

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Answer

One complete revolution of the hour hand = 12 hours.

(i) 12\dfrac{1}{2} of a revolution = 12×12\dfrac{1}{2} \times 12 = 6 hours

Starting from 10 and moving 6 hours clockwise: 10 → 11 → 12 → 1 → 2 → 3 → 4

∴ The hour hand will stop at 4.

(ii) 14\dfrac{1}{4} of a revolution = 14×12\dfrac{1}{4} \times 12 = 3 hours

Starting from 4 and moving 3 hours clockwise: 4 → 5 → 6 → 7

∴ The hour hand will stop at 7.

(iii) 34\dfrac{3}{4} of a revolution = 34×12\dfrac{3}{4} \times 12 = 9 hours

Starting from 4 and moving 9 hours clockwise: 4 → 5 → 6 → 7 → 8 → 9 → 10 → 11 → 12 → 1

∴ The hour hand will stop at 1.

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