A clock is set at 10 a.m. The clock loses 16 minute in 24 hours. What will be the true time when the clock indicates 3 a.m. on 4th day?

From Sunday noon to the following Sunday at 2 p.m. there are 7 days 2 hours or 170 hours. The watch gains 2 + 4⁴/₅ min in 170 hrs. Therefor, the watch gains 2 min in 2 x 170 hrs i.e., 50 hours 6⁴/₅. Now 50 hours Sunday noon = 2 p.m on Tuesday.

44

In 12 hours both hands are in straight line (either coincides or in opposite direction) 22 times. Hence in 24 hours both hands are in straight line 44 times.

22

The minute hand to coincide with the hour hand it should trace at first 5 minute spaces and then the hands of the clocks to be opposite to each other minute hand should trace 30 minute spaces i.e. totally it should gain 5 + 30 = 35 minute spaces to be opposite to that of hour hand.
We know that, Minute hand gains 55 minute spaces over hour hand in 1 hour. Therefore, Minute hand gain 40 minute spaces over hand in 35 x (60/55) = 38(2/11).
Hence the hand of the clock will minutes be opposite to each at 38 (2/11) past 1'O clock. Therefore, Correct option is B.

Time from 12 p.m. on Sunday to 2 p.m. on the following Sunday = 7 days 2 hours.
= 24 x 7 + 2 = 170 hours.
The watch gain = (2 + 4 x 4/5) min = 34/5 minute in 170 hrs.
Since, 34/5 min are gained in 170 hrs.
2 min are gained in (170 x 5/34 x 2) hrs = 50 hours i.e., 2 days 2 hrs. after 12 p.m. on Sunday i.e., it will be correct at 2 p.m. on Tuesday.

When we divide 79 by 7 we will get remainder 2 so we have 2 odd days, so required day must be 2 days back from today (i.e Sunday) and that day should be Friday.

Consider from 2^nd June 2010 to 2^nd June 2013 we have total 2 non leap year and one leap year so number of odd days are 1 + 1 + 2 = 4 so 2^nd June 2010 must be 4 days back from Sunday and that day is Wednesday.
From Zeller's Formula:
f = k + [13 x m - 1/ 5 ] + D + [D/4] + [C/4] - 2 x C.
In this case k = 2 (since 2^nd June)
Month m = 4 (As march = 1, April = 2, May = 3, June = 4 )
D is the last two digit of year here D = 10 (As year is 2010)
C is 1 st two digit of century here C = 20 (As year is 2010)
f = 2 + [13 x 4 - 1 / 5] + 10 + [10/ 4] + [ 20/4] - 2 x 20.
f = 2 + [51/5] + 10 +[2.5] + [5] - 40.
f = 2 + 10 + 10 + 2 + 5 - 40 = -11
This - ve value of f can be made positive by adding multiple of 7
So f = - 11 + 14 = 3
When divided by 7 we will get remainder 3, hence number of odd days is 3,
So 2^nd June 2010 is 3 days more than Monday, i.e Wednesday.

From Zeller's Formula:
f = k + [13 x m - 1 / 5] + D + [D/4] + [C/4] - 2 x C.
In this case k = 15 (since 15^th August)
Month m = 6 (As march = 1, April = 2, May = 3, August = 6)
D is the last two digit of year here D = 47 (As year is 1947)
C is the 1^st two digit of century here C = 19 (As year is 1947)
f = 15 + [13 x 6 - 1 / 5] + 47 + [47/4] + [19/4] - 2 x 19.
f = 15 + [77/5] + 47 + [11.75] + [4.75] - 38.
f = 15 + 15 + 47 + 11 + 4 - 38 = 54.
When divided by 7 we will get remainder 5, hence number of odd days is 3,
A remainder of 0 corresponds to Sunday, 1 means Monday,
So 15^th August 1947 is 5 days more than Sunday, i.e Friday.

From Zeller's Formula
f = k + [13 x m - 1 / 5] + D + [D/4] + [C/4] - 2 x C.
In this case k = 18 (since 18^th October)
Month m = 8 (As march = 1, April = 2, May = 3, October = 8)
D is the last two digit of year here D = 50 (As year is 2050)
C is the 1^st two digit of century here C = 20 (As year is 1950)
f = 18 + [13 x 8 - 1 / 5] + 50 + [50/4] + [20/4] - 2 x 20.
f = 18 + [103/5] + 50 + [12.5] + [5] - 40.
f = 18 + 20 + 50 + 12 + 5 - 40 = 65.
When divided by 7 we will get remainder 2, hence number of odd days is 2,
So 18^th October 2050 is 2 days more than Sunday, i.e Tuesday.