Question 1

Clock angle change — What is the percentage change in the (smaller) angle between the minute and hour hands from 2:00 PM to 6:00 PM?

Accepted Answer

Introduction / Context: We compare the smaller inter-hand angle at two specified times and compute the percentage increase. Given Data / Assumptions: At 2:00, hour angle = 60°, minute angle = 0° → smaller angle = 60°. At 6:00, hour angle = 180°, minute angle = 0° → smaller angle = 180°. Concept / Approach: Percentage change = ((new − old)/old) * 100. Step-by-Step Solution: Old (2:00) = 60°.New (6:00) = 180°.Increase = 180 − 60 = 120°.Percentage change = (120/60)*100 = 200%. Verification / Alternative check: Graphing the smaller angle across hours shows it attains 180° at 6:00, confirming the calculation. Why Other Options Are Wrong: 100%/250% do not match the exact computation; “None” is wrong because 200% is valid. Common Pitfalls: Using the reflex angle at 6:00 (also 180° here, but conceptually some problems request the smaller angle—identical here). Final Answer: 200 (percent).

Question 2

Clock angle — What is the (smaller) angle between the minute and hour hands at exactly 2:45?

Accepted Answer

Introduction / Context: We compute the smaller angle between the hands at 2:45 using standard formulas. Given Data / Assumptions: At 2:45 → h = 2, m = 45. Hour angle = 30h + 0.5m; minute angle = 6m. Concept / Approach: Angle difference = |(30h + 0.5m) − 6m|; report the smaller angle. Step-by-Step Solution: Hour angle = 30*2 + 0.5*45 = 60 + 22.5 = 82.5.Minute angle = 6*45 = 270.Difference = |82.5 − 270| = 187.5 degrees.Smaller angle = 360 − 187.5 = 172.5 degrees. Verification / Alternative check: Using the compact formula |30h − 5.5m| also yields |60 − 247.5| = 187.5 → smaller 172.5. Why Other Options Are Wrong: 180°, 190°, 175° are nearby round values but not exact; correct is 172.5°. Common Pitfalls: Reporting the reflex angle (187.5°) instead of the smaller one, or rounding to 180°. Final Answer: 172.5°, which is not listed; hence “None of these.”

Question 3

Clock geometry — Between 9 o’clock and 10 o’clock, at what exact time are the hands of a clock in the same straight line but not together (i.e., opposite at 180 degrees)?

Accepted Answer

Introduction / Context: “Straight line but not together” means the hands are opposite (180 degrees apart). We seek the time after 9 o’clock when this occurs. Given Data / Assumptions: At 9:00, hour angle = 270°, minute angle = 0°. Let t be minutes after 9:00. Hour angle = 270 + 0.5t; minute angle = 6t. Concept / Approach: Solve |6t − (270 + 0.5t)| = 180 for t in (0, 60). Step-by-Step Solution: 5.5t = 450 → t = 900/11 ≈ 81.818 (invalid for this hour).5.5t = 90 → t = 180/11 ≈ 16.364 (valid).Thus time = 9 : 180/11. Verification / Alternative check: Substituting t = 180/11 gives a 180° separation exactly. Why Other Options Are Wrong: 164/11, 154/11, 174/11 minutes do not satisfy the 180° condition. Common Pitfalls: Choosing the extraneous solution (>60 min) or rounding fractions. Final Answer: 9 : 180/11 (not listed), therefore “None of these.”

Question 4

Clock geometry — Between 8 o’clock and 9 o’clock, at what time are the hands of a clock at right angles (90 degrees)?

Accepted Answer

Introduction / Context: Right angles occur when the hand-angle difference is 90 degrees. We find solutions between 8:00 and 9:00. Given Data / Assumptions: At 8:00, hour angle = 240°. Let m be minutes after 8:00. Hour angle = 240 + 0.5m; minute angle = 6m. Concept / Approach: Set |(240 + 0.5m) − 6m| = 90 and solve for m within (0, 60). Step-by-Step Solution: |240 − 5.5m| = 90 → 5.5m = 240 ± 90 = 330 or 150.m = 330/5.5 = 60 (hits exactly 9:00, boundary) or m = 150/5.5 = 300/11 ≈ 27 3/11.Valid between 8 and 9: 8 : 300/11. Verification / Alternative check: Plugging m = 300/11 satisfies the 90° separation. Why Other Options Are Wrong: 163/11, 273/11, and 173/11 are not equal to 300/11 and do not yield the correct angle within the hour. Common Pitfalls: Accepting the boundary time m = 60 as “between 8 and 9”; most questions exclude the hour boundary. Final Answer: 8 : 300/11 (not listed), hence “None of these.”

Question 5

Clock speeds — By how many degrees does the minute hand move in the same time that the hour hand moves by 28 degrees?

Accepted Answer

Introduction / Context: This tests the constant angular speeds of clock hands. The minute hand moves 12 times as fast as the hour hand. Given Data / Assumptions: Hour hand speed = 0.5° per minute. Minute hand speed = 6° per minute. Hour hand displacement given = 28°. Concept / Approach: First find the time taken for the hour hand to move 28°, then compute the minute hand’s movement in the same time. Step-by-Step Solution: Time t = 28 / 0.5 = 56 minutes.Minute hand movement = 6 * 56 = 336 degrees. Verification / Alternative check: Using the 12:1 speed ratio: minute displacement = 12 * hour displacement = 12 * 28 = 336 degrees. Why Other Options Are Wrong: 168 = 6 * 28/2 and 196/376 stem from incorrect speed ratios or arithmetic slips. Common Pitfalls: Forgetting the factor 12 or mixing time with angle directly. Final Answer: 336 degrees.

Question 6

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 7

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

Accepted Answer

22

Question 8

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 9

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 10

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 11

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 12

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 13

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 14

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 15

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 16

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.

Question 17

If 5 th march of a particular year is Friday then which day of the week will be on 5 th November.

Accepted Answer

Here we have to find the number of odd days between, 5th march and 5th November, Number of days in March is 26 or 5 odd days (Here we have not included 5th march) Number of days in April is 30 or 2 odd days Number of days in May is 31 or 3 odd days Number of days in June is 30 or 2 odd days Number of days in July is 31 or 3 odd days Number of days in August is 31 or 3 odd days Number of days in September is 30 or 2 odd days Number of days in October is 31 or 3 odd days Number of days in November is 5 or 5 odd days (Here 5th November is included) So total number of odd days = 5 + 2 + 3 + 2 + 3 + 3 +2 + 3 + 5 = 28 when divided by 7 gives remainder 0 hence 5th November will be same as that of 5th march.

Question 18

In March 2009, how many weekend days (Saturdays and Sundays together) are there in total?

Accepted Answer

Introduction / Context: This calendar question asks for the total number of weekend days (all Saturdays and all Sundays) in March 2009. Many candidates confuse “weekends” with “full Saturday–Sunday pairs.” Here, we count the individual weekend days. Given Data / Assumptions: Target month: March 2009. Weekend days considered: Saturdays and Sundays. March has 31 days. Concept / Approach: Find the weekday of March 1, 2009, then list the Saturdays and Sundays within the month. Every 7 days the same weekday repeats. Count occurrences within 31 days. Step-by-Step Solution: March 1, 2009 was a Sunday.Sundays in March 2009: 1, 8, 15, 22, 29 → 5 Sundays.Saturdays in March 2009: 7, 14, 21, 28 → 4 Saturdays.Total weekend days = 5 + 4 = 9. Verification / Alternative check: 31 days = 4 full weeks (28 days) + 3 extra days. Starting on Sunday, the extras are Sun, Mon, Tue, which contributes one extra Sunday beyond the 4 week cycle, confirming 5 Sundays and 4 Saturdays. Why Other Options Are Wrong: 10 assumes 5 Saturdays + 5 Sundays (not possible with a 31-day month starting on Sunday). 8 counts only 4+4 or only pairs. “None of these” is unnecessary since 9 is valid. Common Pitfalls: Interpreting “weekends” as “number of Saturday–Sunday pairs” (which would be 4). The question asks total weekend days. Final Answer: 9

Question 19

In the same calendar year, which two months definitely share the same calendar (same weekdays on the same dates)?

Accepted Answer

Introduction / Context: Two months have “the same calendar” if every date in one month falls on the same weekday as the corresponding date in the other month (e.g., the 1st is the same weekday in both, the 2nd is the same weekday in both, etc.). Given Data / Assumptions: We consider a single year (either leap or non-leap). We seek a pair that always matches within that same year. Concept / Approach: Shifts between month start weekdays equal (days in month) mod 7. The total shift from March to November sums to 21 days (mod 7 = 0) regardless of leap status because February has already passed and does not affect the March→November spacing. Step-by-Step Solution: Month length shifts (Mar→Nov): Mar(31)=+3, Apr(30)=+2, May(31)=+3, Jun(30)=+2, Jul(31)=+3, Aug(31)=+3, Sep(30)=+2, Oct(31)=+3.Total shift = 3+2+3+2+3+3+2+3 = 21 ≡ 0 (mod 7).Therefore, March and November always start on the same weekday and share the same date–weekday pattern. Verification / Alternative check: Test with sample years (leap and non-leap) and observe that March 1 and November 1 align on the same weekday. Why Other Options Are Wrong: January–August and January–October depend on whether the year is leap or not (February changes the offset), so they are not guaranteed every year. Common Pitfalls: Assuming “pairs” like Jan–Oct always match without considering February's 28/29 days. Final Answer: March and November

Question 20

On Saturday, 14 February 2009 it was Valentine’s Day. What weekday was 14 February 2008?

Accepted Answer

Introduction / Context: We relate the same month–day across consecutive years while accounting for leap years. Given Data / Assumptions: 14 Feb 2009 = Saturday. 2008 is a leap year. Concept / Approach: From a leap year date to the same date next year, the weekday advances by 2 days. Conversely, going backward from a non-leap year date to the same date in the leap year, the weekday moves back 2 days. Step-by-Step Solution: Known: 14 Feb 2009 is Saturday.Going back to 14 Feb 2008 (leap year) shifts by −2 days.Saturday − 1 = Friday, − 1 = Thursday. Verification / Alternative check: Count the total days between the two dates or apply standard day-of-week congruences; both confirm Thursday for 14 Feb 2008. Why Other Options Are Wrong: Tuesday/Sunday are off by multiple days; “None of these” is unnecessary since Thursday fits. Common Pitfalls: Forgetting that 2008 is leap, which changes the 1-day step to a 2-day step for dates on/after Feb 29 considerations. Final Answer: Thursday

Calendar Questions