Question

I left home form bringing milk between 7 am and 8 am. The angle between the hour-hand and the minute-hand was ${90}^o$ I returned home between 7 am and 8 am. Then also the angle between the minute-hand and hour-hand was ${90}^o$. At what time (nearest to second) did I leave and return home?

• A. 7h 18m 35s and 7h 51m 24s
• B. 7h 19m 24s and 7h 52m 14s
• C. 7h 20m 42s and 7h 53m 11s
• D. 7h 21m 49s and 7h 54m 33s

Answer 1

Answer: (D)

7h 21m 49s and 7h 54m 33s

The answer is option $D$

A minute hand covers an angle of ${360}^o$ in $60mins$, whereas an hour hand covers the same angle in $12hours$

Speed of minute hand $=\frac{{360}^o}{60}=6^o/min$

Speed of hour hand $=\frac{{360}^o}{60\times 12}={0.5}^o/min$

Relative speed of minute hand wrt hour hand$=6-0.5={5.5}^o/min$

At 7am angle between the two hands $={210}^o$

For the angle between the two hands to be ${90}^o$, the minute hand needs to cover ${120}^o({210}^o-{90}^o)$ wrt the hour hand

Time taken to cover this distance $=\frac{120}{5.5}=21\frac{9}{11}$min =$21min49sec$

When the person arrives also the angle between the two hands is ${90}^o$

From the time of starting, the minute hand needs to cover ${180}^o$ to arrive at a position that makes the angle between the two hands as ${90}^o$

Time taken to cover ${180}^o=\frac{180}{5.5}=32\frac{8}{11}$ min$=32min44secs$

$\therefore$ The time of departure $=7h21m49sec$

and Time of arrival $=7h21m49sec+32min44sec=7h54min33sec$