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In 12 hours, how many times the hours and minutes hands of a clock will coincide with- each other?
Hour hand covers 360° in 12 hour or 720 minutes.
Also, Minute hand covers 360° in 60 minutes.
So, hour hand covers (1/2)° in 1 minute.
and minute hand covers 6° in 1 minute.
Let say, time in the clock when they coincide : hour hand moved H hour and minute hand moved M minutes.
So,
Angle between hands of a clock
When the minute hand is behind the hour hand, the angle between the two hands at M minutes past HH 'o clock
$$=(30H-6M)+(M/2)°$$
$$=(30H-(11M/2))°$$
When the minute hand is ahead of the hour hand, the angle between the two hands at M minutes past HH 'o clock
$$=((6M−30H)−M/2)°$$
$$=((11M/2)-30H)°$$
So, taking mod:
$$|60H-11M|=2\theta$$..........................(1)
So, hour hand and minute hand will coincide 11 times :
between 1 and 2 , 2 and 3 , 3 and 4, 4 and 5 , 5 and 6, 6 and 7 , 7 and 8 , 8 and 9, 9 and 10, 10 and 11, 11 and 12.
1 and 2 :
between 1 and 2 hour hand will pass 1 O'clock,
So,H=1 and angle should be 0.
using equation (1):$$60\times\ 1-11M=0\ or\ M=60\div11=5\frac{5\ }{11}.$$
So,between 1 and 2 , hour hand and minute hand will coincide at $$5\frac{5\ }{11}$$ minutes past 1.
2 and 3:
$$60\times2-11M=0\ or\ M=120\div11=10\frac{10\ }{11}.$$
They will coincide $$10\frac{10\ }{11}$$ minutes past 2.
3 and 4:
$$60\times3-11M=0\ or\ M=\frac{180}{11}=16\frac{4\ }{11}.$$
They will coincide $$16\frac{4\ }{11}$$ minutes past 3.
4 and 5:
$$60\times4-11M=0\ or\ M=\frac{240}{11}=21\frac{9\ }{11}.$$
They will coincide $$21\frac{9\ }{11}$$ minutes past 4.
5 and 6:
$$60\times5-11M=0\ or\ M=\frac{300}{11}=27\frac{3\ }{11}.$$
They will coincide $$27\frac{3\ }{11}$$ minutes past 5.
6 and 7:
$$60\times6-11M=0\ or\ M=\frac{360}{11}=32\ \frac{\ 8}{11}.$$
They will coincide $$32\ \frac{\ 8}{11}$$ minutes past 6.
7 and 8:
$$60\times7-11M=0\ or\ M=\frac{420}{11}=38\frac{2\ }{11}.$$
They will coincide $$38\frac{2\ }{11}$$ minutes past 7.
8 and 9:
$$60\times8-11M=0\ or\ M=\frac{480}{11}=43\frac{7\ }{11}.$$
They will coincide $$43\frac{7\ }{11}$$ minutes past 8.
9 and 10:
$$9\times60-11M=0\ or\ M=\frac{540}{11}=49\frac{1\ }{11}.$$
They will coincide $$49\frac{1\ }{11}$$ minutes past 9.
10 and 11:
$$60\times10-11M=0\ or\ M=\frac{600}{11}=54\frac{6\ }{11}.$$
They will coincide $$54\frac{6\ }{11}$$ minutes past 10.
11 and 12:
$$60\times11-11M=0\ or\ M=\frac{660}{11}=60.$$
So, last time they will coincide 60 minutes past 11O'clock which means at 12 noon.
So, Total 11 times they will coincide.
B is correct choice.
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