# Where will the hand of a clock stop if it(a) starts at 12 and makes $\frac{1}{2}$ of a revolution, clockwise?(b) starts at 2 and makes $\frac{1}{2}$ of a revolution, clockwise?(c) starts at 5 and makes $\frac{1}{4}$ of a revolution, clockwise?(d) starts at 5 and makes $\frac{3}{4}$ of a revolution, clockwise?

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To do:

We have to find the number at which the hand of the clock stops in each case.

Solution:

(a) Number of hours in 1 revolution $= 12$ hours

Number of hours in $\frac{1}{2}$ revolution $= 12 \times \frac{1}{2}$

$= 6$ hours

So, the hand will move 6 hours and stop at 6.

(b) Number of hours in 1 revolution $= 12$ hours

Number of hours in $\frac{1}{2}$ revolution $= 12 \times \frac{1}{2}$

$= 6$ hours

The final position of the hand of the clock $=2+6=8$

So, the hand will move for 6 hours and stop at 8.

(c) Number of hours in 1 revolution $= 12$ hours

Number of hours in $\frac{1}{4}$ revolution $= 12 \times \frac{1}{4}$

$= 3$ hours

The final position of the hand of the clock $=5+3=8$

So, the hand will move 5 hours and stop at 8.

(d) Number of hours in 1 revolution $= 12$ hours

Number of hours in $\frac{3}{4}$ revolution $= 12 \times \frac{3}{4}$

$= 9$ hours

The final position of the hand of the clock $=5+9=14$

$14=12+2$

The clock hand will cross the 12 mark and reach the 2 mark

So, the hand will move 9 hours and stop at 2.

Updated on 10-Oct-2022 13:32:11