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A person went out between $4$pm and $5$pm to chat with her friend and returned between $5$pm and $6$pm. On her return, she found that the hour-hand and the minute-hand of her (well-functioning) clock had just exchanged their positions with respect to their earlier positions at the time of her leaving. The person must have gone out to chat at

  1. Twenty five minutes past $4$pm.
  2. Twenty six and $\dfrac{122}{143}$ minutes past $4$pm.
  3. Twenty seven and $\dfrac{1}{3}$ minutes past $4$pm.
  4. Twenty eight minutes past $4$pm.
  5. None of the above.
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Let the time of departure be $4:x$ and time of arrival be $5:y$.

Angle made by hour hand during departure $ = \dfrac{4}{12} \times 360 + x  \dfrac{360}{12 \times 60} = 120 + \dfrac{x}{2}$.

This angle is equal to the angle made by the minute hand on arrival which is $6y$. So,$$ 240 + x = 12y \rightarrow(1)$$Similarly, the angle made by the hour hand on arrival is equal to the angle made by the minute hand on departure, which gives $$\dfrac{5}{12} \times 360 + \dfrac{y}{2} = 6x \\ \implies 300 + y = 12x \rightarrow (2).$$ Eliminating $y$ from (1) and (2), $$240 + x = 12 (12x - 300) \\ \implies 143x =3840 \\ \implies x = 26 \dfrac{122}{143}$$Correct Answer: $B$
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Person went to his friends house and return between 4 to 6 PM
Also we can predict he went after 4:25 pm=4 hr 25 min +x min

and he came after 5:20 pm=5 hr 20 min +x min 

So, angle of 4 hr 25 min +x min = angle of  5 hr 20 min +x min

for, 4 hr 25 min +x min 

60 min  a clock rotates 360 degree

:. (x+25) min  a clock rotates 360/60 *(x+25) degree= 6(x+25)

12 hours a clock rotates 360 degree

4(25+x/60)=(265+x)/60 a clock rotates ((265+x)/60)*360/12 degree=(265+x)/2 degree

Angle between two hands of clock in 4:25 pm+x min is 6(x+25) - (265+x)/2................i

for 5 hr 20 min +x min

60 min  a clock rotates 360 degree

:. (x+20) min  a clock rotates 360/60 *(x+20) degree= 6(x+20)

 

12 hours a clock rotates 360 degree

5((20+x)/60)=(320+x)/60 a clock rotates ((320+x)/60)*360/12 degree=(320+x)/2 degree

So, Angle between two hands of clock in 5:20 pm+x min is (320+x)/2 - 6(x+20)...............ii

6(x+25) - (265+x)/2=(320+x)/2 - 6(x+20)

=>11x=10+12.5

=>x=2.04

So, time will be 4 hr 27.04 min
so answer option(c)

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The person went out between $4$ P.M. and $5$ P.M., hence hour hand would be between $4$ and $5$ on the clock. On her return (between $5$ P.M. to $6$ P.M.) she observed that position of the hour hand has been replaced by minute hand and position of the minute hand has been replaced by hour hand, so minute hand would have been between $5$ and $6$ (when he went to friends' house) on the clock.

$\underline{\textbf{Observations}}$

At $4$ P.M., hour hand would be pointing to $4$ and minute hand pointing to $12$ on the clock.

If minute hand is pointing to $5$ on the clock then it would be making $150^{\circ}$ from $12$ on the clock ($\because$ angle between every two consecutive numbers on the clock is $\frac{360^{\circ}}{12} = 30^{\circ}$) 

When minute hand would be between $5$ and $6$ on the clock, making (say) $150^{\circ}+\theta_1^{\circ}$ from $12$ on the clock, then hour hand would have moved by (say) $\theta_2^{\circ}$ from $4$ on the clock, where $\theta_1^{\circ}, \theta_2^{\circ} \le 30^{\circ}$. 

When minute hand covers $360^{\circ}$ then hour hand covers $30^{\circ}$. This implies minute hand covers $12\times$ of what hour hand covers. 

Suppose hour hand covers $x^{\circ}$, and in that much time minute hand covers $y^{\circ}$ ($x^{\circ} \rightarrow y^{\circ}$) then we can say that $12x^{\circ} = y^{\circ}$. In next section, we will use this particular observation directly.


$\underline{\textbf{Calculations}}$

When hour hand covered $\theta_{2}^{\circ}$, minute hand covered $150+\theta_1^{\circ}$. 

$\theta_2^{\circ} \rightarrow 150^{\circ} + \theta_1^{\circ}$

$$\begin{align} \implies 12\theta_2^{\circ} &= 150^{\circ} + \theta_1^{\circ} \tag{1}\end{align}$$

When hour hand replaces the position of minute hand, then hour hand need to cover $(30^{\circ}-\theta_2^{\circ}) + \theta_1^{\circ}$. When minute hand replaces the position of hour hand, then minute had need to cover $360^{\circ}-((30^{\circ}-\theta_2^{\circ}) + \theta_1^{\circ})$.

$(30^{\circ}-\theta_2^{\circ}) + \theta_1^{\circ} \rightarrow 360^{\circ}-((30^{\circ}-\theta_2^{\circ}) + \theta_1^{\circ})$ 

 $$\begin{align} \implies 12(30^{\circ}-\theta_2^{\circ}) + 12\theta_1^{\circ} &= 360^{\circ}-((30^{\circ}-\theta_2^{\circ}) + \theta_1^{\circ}) \tag{2}\end{align}$$

Solving $(1)$ and $(2)$, we get $\theta_1^{\circ} = 11.118^{\circ}; \ \theta_2^{\circ} = 13.426^{\circ}$.


We have been asked to find the time when she went out to chat. For that we still need to know exactly how many minutes past $4$ she went out. As we can observe in the diagram above that it would be $\text{MinutesPastIn}(\theta_1^{\circ}) +25$ minutes.

For minute hand, in $1^{\circ}$, $\frac{60}{360^{\circ}} = \frac{1}{6}$ min passes by. So for $\theta_1^{\circ}=11.1185^{\circ}$, $\frac{11.1185^{\circ}}{6} = 1.85$ minutes passes by. This is same as $26$ and $0.85$ minutes past $4$ P.M., or $\fbox{$26$ and $\frac{122}{143}$ minutes past $4$ P.M.}$.

$\textbf{Option (B) is correct}$

Answer:

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