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?
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.
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