Question 1

If 6th March, 2005 is Sunday, what was the day of the week on 6 th March, 2002?

Accepted Answer

Number of odd days between 6th march 2002 to 6th march 2003 is 1 Number of odd days between 6th march 2003 to 6th march 2004 is 2 Number of odd days between 6th march 2004 to 6th march 2005 is 1 Hence total Number of odd days between 6th march 2002 to 6th march 2005 is 1 + 2 + 1 = 4 odd days so 6th march 2002 is 4 days before Sunday i.e. Wednesday.

Question 2

Find the angle between minute hand hour hand at 11 : 50 AM

Accepted Answer

At 11 : 50 A.M. from the formula here P = 11 and Q = 50 so required angle is 11 x 50/2 - 30 x 11 = 275 - 330 = -55 or ignoring the negative sign the required angle is 55°

Question 3

At what time between 2 PM and 3 PM the angle between minute hand and hour hand is 100°.

Accepted Answer

Same as previous question Option A is correct.

Question 4

Find the angle between minute hand and hour hand at 2 : 18 AM

Accepted Answer

39°

Question 5

Find the percentage change in angle between minute hand hour hand from 2 PM to 6 PM.

Accepted Answer

At 2 PM angle between minute hand and hour hand is 60° and that at 6 PM is 180° hence percentage change is 200%.

Question 6

Find the angle between minute hand and hour hand at 2 : 45

Accepted Answer

From the formula the required angle is 187.5°

Question 7

At what time between 9 and 10 o'clock will the hands of a clock be in the same straight line but not together?

Accepted Answer

9 :164/11

Question 8

At what time between 8 and 9 o'clock will the hands of a clock be at right angle?

Accepted Answer

Consider exactly at 8 : 00, minute hand is at position '12' and hour hand is at '8' hence angle between these two hands at 8 : 00 is 240°. Case (i) when minute hand is behind Hour hand then required time is (240 - 90)/(11/2) = 150/(11/2) = 300/11 = 27 3/11 minute Case (ii) when minute hand is ahead of four hand then required time is (240 + 90)/(11/2) = 330/(11/2) = 330 x 2/11 = 60 minute i.e., at 9 : 00 PM

Question 9

By how many degree does the minute hand move in the same time, in which the hour hand move by 28?

Accepted Answer

28 x 2 x 6 = 336 degree

Question 10

At what time between 1'O clock and 2'O clock the hands of the clock are opposite to each other.

Accepted Answer

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.

Question 11

How many times in a day (24 Hrs) are the hands of a clock coincide?

Accepted Answer

22

Question 12

How many times in a day (24 Hrs) are the hands of a clock are in a straight line?

Accepted Answer

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.

Question 13

How many times in a day (24 Hrs) are the hands of a clock are in a perpendicular to each other?

Accepted Answer

44

Question 14

A watch, which gains uniformly, is 2 min, slow at noon on Saturday, and is 4 min 48 seconds fast at 3 p.m on the following Sunday when was it correct?

Accepted Answer

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.

Question 15

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?

Accepted Answer

Time from 10 a.m on a day to 3 a.m on 4th day = 24 x 3 + 17 = 89 hours. Now 23 hrs 44 min. of this clock = 24 hours of correct clock. 89 hrs of faulty clock = (24 x 15/356 x 89) hrs = 90 hrs. So, the correct time is 11 p.m

Question 16

A watch which gains uniformly is 2 minutes slow at noon on Sunday and is 4 min. 48 sec fast at 2 p.m. on the following Sunday. When it has shown the correct time?

Accepted Answer

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.

Question 17

Today is Sunday what day of the week was 79 days back.

Accepted Answer

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.

Question 18

If 2 nd June 2013 is Sunday then which day was on 2 nd June 2010?

Accepted Answer

Consider from 2nd June 2010 to 2nd June 2013 we have total 2 non leap year and one leap year so number of odd days are 1 + 1 + 2 = 4 so 2nd 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 2nd 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 2nd June 2010 is 3 days more than Monday, i.e Wednesday.

Question 19

Which of the following day of the week was 15 th august 1947?

Accepted Answer

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 15th 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 1st 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 15th August 1947 is 5 days more than Sunday, i.e Friday.

Question 20

Which of the following day could be the 18 th October 2050?

Accepted Answer

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 18th 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 1st 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 18th October 2050 is 2 days more than Sunday, i.e Tuesday.

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