type 3 faulty clock Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (a)

Total time from 3 O' clock Sunday afternoon to 10 O' clock Tuesdy morning = 43 h

Total increased time = 12/3 × 43 = 172s = 2 min 52s

Therefore, Time at 10 O' clock Tuesday morning = 2 min 52 s past 10

Answer: (b)

Time from Monday noon (12 pm) to 2 pm the following Monday = 7 days 2 h = 170 h

Now, the watch gains (2 + 4 4/5) min from Monday (12 pm) to 2 pm,

the following Monday. In other words, the watch gains 34/5 min in 170 h.

Therefore, it will gain 2 min in (170 × 5/34 × 2) = 50 h = 2 days 2 h

Therefore, the watch is correct after 2 days 2 h from Monday noon or at 2 pm Wednesday.

Answer: (b)

Time from 5 am of a particular day to 10 pm on the 3rd day is 89 h.

Now, the clock loses 16 min in 24 h or in other words,

we can say that 23 h 44 min of this clock is equal to 24 h of the correct clock.

$(23 + 44/60) = 356/15$ h of this clock = 24h of the correct clock

Therefore, 89 h of this clock = $(24 × 15 / 356 × 89) h$

the correct clock = 90 h of the correct clock or 89 h of this clock = 90 h of the correct clock.

Therefore, it is clear that in 90 h this clock loses 1 h and hence, the correct time is 11 pm when this clock shows 10 pm.

Answer: (c)

Time of the last train leaving the station = (18 : 00 - 2 : 30)h = 15 : 30h

But this happens 40 min before the announcement is made.

Hence, Time of making announcement = (15 : 30 + 0 : 40) = 16 : 10h

Answer: (b)

Clearly, Raveena left home 10 min before 8 : 40 am i.e., at 8 : 30 am

but it was 15 min earlier than usual,

so she usually leaves for the stop at ( 8 : 30 + 0 : 15 ) = 8  : 45 am.