Henry started a trip into the country between 8 am and 9 am when the hand of clock were together, He arrived at his destination between 2 pm and 3 pm when the hands of the clock were exactly 180° apart. How long did he travel ?
A. 6 hours
B. 7 hours
C. 9 hours
D. 11 hours
Answer: Option A
Solution(By Examveda Team)
To be together between 8 am and 9 am, the minute hand has to gain 40 minutes spaces.55 minutes spaces are gained in 60 minutes.
40 minutes space are gained in $$\left( {\frac{{60}}{{55}} \times 40} \right)$$ minutes = $${\text{43}}\frac{7}{{11}}$$ minutes
So, Henry started his trip at $${\text{43}}\frac{7}{{11}}$$ minutes past 8 am.
Now, to be 180° apart, the hands must be 30 minutes spaces apart.
At 2 pm, they are 10 minutes spaces apart.
∴ The minute hand will have to gain (10 + 30) = 40 minutes spaces.
As calculate above, 40 minutes spaces are gained in $${\text{43}}\frac{7}{{11}}$$ minutes.
So, Henry's trip ended at $${\text{43}}\frac{7}{{11}}$$ minutes past 2 pm
∴ Duration of travel = Duration from $${\text{43}}\frac{7}{{11}}$$ minutes past 8 am to $${\text{43}}\frac{7}{{11}}$$ minutes past 2 pm = 6 hours
Related Questions on Clock
The reflex angle between the hands of a clock at 10.25 is:
A. 180º
B. $${\text{192}}{\frac{1}{2}^ \circ }$$
C. 195º
D. $${\text{197}}{\frac{1}{2}^ \circ }$$
A clock is started at noon. By 10 minutes past 5, the hour hand has turned through:
A. 145º
B. 150º
C. 155º
D. 160º
A. $$59\frac{7}{{12}}$$ min. past 3
B. 4 p.m.
C. $$58\frac{7}{{11}}$$ min. past 3
D. $$2\frac{3}{{11}}$$ min. past 4
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