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profileganesh
asked 2015-07-12 20:18:34 
A man goes out between $5$ p.m and $6$ p.m. When he comes back in between $6$ p.m and $7$ p.m, he observes that the two hands of clock have interchanged their positions. Find when the man did go out.
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profileRaj
commented 2015-12-20 13:24:06 
@yusuf ansari, see answer given by Dev. you can find the variables $a$ and $b$ by solving $(3)$ and $(4)$ and get his leaving time and time of return. Then calculate the duration he was out.
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profileyusuf ansari
commented 2015-11-15 15:49:27 
Sir if we need to find out for what duration he was out or time between his leaving and coming back, how will we calculate that.
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profileJaved Khan
answered 2015-07-13 19:28:42 
A general formula can be derived along the lines of Dev's explanation. Suppose the man goes out $x$ minutes past $H$ (with minutes hand ahead of the hour hand) and the hands of the clock interchange positions by minutes hand and hour hand together travelling $60$ minute spaces (like the case given in the question), then $x = \dfrac{720+780H}{143}$

In this case, $x=\dfrac{720+780×5}{143}=\dfrac{420}{13}=32\dfrac{4}{13}$

i.e., he left $32\dfrac{4}{13}$ minutes past $5$ pm
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profiledev
answered 2015-07-13 18:29:55 
Minute hand moves $360°$ in $60$ minutes.
=> Minute hand moves $6°$ in $1$ minute$~~\cdots(1)$

Hour hand moves $360°$ in $12$ hours.
=> Hour hand moves $30°$ in $1$ hour.
=> Hour hand moves $\dfrac{1}{2}°$ in $1$ minute $~~\cdots(2)$

if the time is $h:m \quad (0\le h \lt 12\text{ and }0 \le m \lt 60)$
position of the hour hand $=h×30°+m×\dfrac{1}{2}°$
position of the minute hand $=m×6°$

Let the man goes out between $5:a$ and comes back at $6:b$

When the time is $5:a,$
position of the hour hand $=5×30°+a×\dfrac{1}{2}°=150°+a×\dfrac{1}{2}°$
position of the minute hand $=a×6°$

When the time is $6:b,$
position of the hour hand $=6×30°+b×\dfrac{1}{2}°=180°+b×\dfrac{1}{2}°$
position of the minute hand $=b×6°$

Since the hands interchange positions in $5:a$ and $6:b,$

$a×6°=180°+b×\dfrac{1}{2}°~~\cdots(1)\\
b×6°=150°+a×\dfrac{1}{2}°~~\cdots(2)$

Two equations and two variables. Solve $(1)$ and $(2)$ for $a$

$(1)/6° \implies a=30+\dfrac{b}{12}~~\cdots(3)\\
(2)/6° \implies b=25+\dfrac{a}{12}~~\cdots(4)$

Substitute the value of $b$ from $(4)$ in $(3)$
$a=30+\dfrac{25+\left(\dfrac{a}{12}\right)}{12}\\
\Rightarrow \text{a} = 30 + \dfrac{25}{12}+\dfrac{\text{a}}{144}\\
\Rightarrow \dfrac{143a}{144}=\dfrac{385}{12}\\
\Rightarrow a=\dfrac{144×385}{143×12}=\dfrac{12×385}{143}=\dfrac{420}{13}=32\dfrac{4}{13}$

Therefore the time the man went out at $32\dfrac{4}{13}$ minutes past $5$ pm
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profileRaj
answered 2015-07-13 16:52:12 
Answer : $32\dfrac{4}{13}$ minutes past $5$ pm



Let the man leaves $x$ minutes past $5.$ When he returns, the positions of the hands of the clock have been interchanged. This means, minutes hand and hour hand together has travelled $60$ minute spaces.

Speeds of hour hand : Speed of the  minute hand $=1:12$

Distance travelled by hour hand when the positions are interchanged
$=60×\dfrac{1}{13} =\dfrac{60}{13}$ minute spaces.

This means, when the man went out, the minute hand was ahead of the hour hand by $\dfrac{60}{13}$ minute spaces.

We know that at $5$ pm, hour hand is ahead by $25$ minute spaces

i.e., During the time interval from $5$ pm to the time at which man goes out, minute hand gained $\dfrac{60}{13}+25=\dfrac{385}{13}$ minute spaces over hour hand.

In $60$ minutes, minute hand gains $55$ minute spaces over the hour hand
=> In $\dfrac{60}{55}$ minute, minute hand gains $1$ minute spaces over the hour hand

To gain $\dfrac{385}{13}$ minute spaces, time needed
$=\dfrac{385}{13}×\dfrac{60}{55}=\dfrac{420}{13}=32\dfrac{4}{13}$ minutes

Hence, he went out at $32\dfrac{4}{13}$ minutes past $5$ pm
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