Calendars and Clocks for CUET General Aptitude

AnswerB · Saturday

1. Only the remainder after removing whole weeks matters: $45 = 6 \times 7 + 3$.

2. 6 whole weeks bring you back to Wednesday; 3 more days: Thursday, Friday, Saturday.

AnswerA · Wednesday

1. $100 = 14 \times 7 + 2$, so 100 days back is the same as 2 days back.

2. Two days before Friday: Thursday, then Wednesday.

AnswerD · Wednesday

1. The day after tomorrow is today + 2, so today = Sunday − 2 days = Friday.

2. The day before yesterday is today − 2 = Wednesday.

AnswerC · 2

1. A leap year has 366 days.

2. $366 = 52 \times 7 + 2$: 52 complete weeks and 2 days left over, so 2 odd days.

AnswerB · 2

1. April has 30 days $= 4$ weeks $+ 2$ days.

2. Every weekday occurs 4 times in the 28 days; the 2 extra days (29 and 30 April) are Sunday and Monday.

3. So Sunday and Monday occur five times: 2 days of the week.

AnswerA · Tuesday

1. From the 3rd to the 24th is $24 - 3 = 21$ days.

2. 21 is exactly 3 weeks, so the 24th is also a Tuesday.

AnswerC · Thursday

1. Days left in August after the 15th: $31 - 15 = 16$.

2. Add all 30 days of September and 2 days of October: $16 + 30 + 2 = 48$ days.

3. $48 = 6 \times 7 + 6$, so move 6 days on from Friday: Thursday.

AnswerB · Thursday

1. Complete years before $1950 = 1949$. The first 1600 years give 0 odd days; the next 300 years give 1.

2. The remaining 49 years (1901–1949) have 12 leap years and 37 ordinary years: $12 \times 2 + 37 = 61$ odd days, and $61 = 8 \times 7 + 5$ gives 5.

3. 26 days of January 1950 give $26 = 3 \times 7 + 5$, that is 5 odd days.

4. Total $1 + 5 + 5 = 11$, and $11 - 7 = 4$. With 0 = Sunday, 4 = Thursday.

AnswerD · Sunday

1. Between the two dates lies 29 February 2032, because 2032 is a leap year.

2. So the gap is 366 days, which is 52 weeks and 2 days.

3. Friday moved on by 2 days is Sunday.

AnswerC · Thursday

1. The day before yesterday was Monday − 2 = Saturday.

2. $61 = 8 \times 7 + 5$, so move 5 days on from Saturday.

3. Sunday, Monday, Tuesday, Wednesday, Thursday: the answer is Thursday.

AnswerC · 21

1. The first Thursday comes 3 days after the first Monday: the 7th.

2. The second Thursday is the 14th and the third is $7 + 14 = 21$st.

AnswerB · 3

1. Leap years in 1901–1953: 1904, 1908, …, 1952, which is 13 years. The other 40 are ordinary.

2. Odd days $= 13 \times 2 + 40 \times 1 = 66$.

3. $66 = 9 \times 7 + 3$, so 3 odd days.

AnswerA · Saturday

1. $75 = 10 \times 7 + 5$, so 75 days back is the same weekday as 5 days back.

2. 5 days before Thursday: Wednesday, Tuesday, Monday, Sunday, Saturday.

AnswerB · Saturday

1. The years 2001 to 2100 must be counted. Leap years among them: 2004, 2008, …, 2096, which is 24 years. 2100 is NOT leap (a century year not divisible by 400).

2. Odd days $= 24 \times 2 + 76 \times 1 = 124$.

3. $124 = 17 \times 7 + 5$, so 5 odd days.

4. Monday moved on by 5 days is Saturday.

AnswerC · Friday

1. Complete years = 1946: 1600 years give 0 odd days and 300 years give 1.

2. 46 years (1901–1946) have 11 leap years and 35 ordinary ones: $22 + 35 = 57$, and $57 = 8 \times 7 + 1$ gives 1.

3. January to July 1947: $3 + 0 + 3 + 2 + 3 + 2 + 3 = 16$ odd days, which is 2. Then 15 days of August give 1.

4. Total $1 + 1 + 2 + 1 = 5$. With 0 = Sunday, 5 = Friday.

AnswerB · Friday and Saturday

1. 366 days = 52 weeks + 2 days, so every weekday occurs 52 times, and the 2 extra days occur once more.

2. The extra days are the first two days of the year: 1 January (Friday) and 2 January (Saturday).

3. So Friday and Saturday occur 53 times each.

AnswerD · December

1. Two months begin on the same weekday when the days between their first days form whole weeks.

2. From 1 September to 1 December: $30 + 31 + 30 = 91$ days $= 13$ weeks exactly.

3. Check the others: to 1 October is 30 days (2 odd days), to 1 November 61 days (5 odd days); June begins 92 days before September (1 odd day).

AnswerD · 2400

1. A century year is a leap year only if it is divisible by 400.

2. $1900 \div 400$, $2100 \div 400$ and $2200 \div 400$ are not whole numbers; $2400 = 6 \times 400$, so 2400 is a leap year.

AnswerB · 1800

1. 2012 is divisible by 4 and is not a century year, so it is a leap year.

2. 1600 and 2000 are divisible by 400, so they are leap years.

3. 1800 is a century year not divisible by 400, so it is not a leap year.

AnswerB · 24

1. Multiples of 4 from 2004 to 2100: $(2100 - 2004) \div 4 + 1 = 25$ years.

2. 2100 is a century year not divisible by 400, so it is not leap: $25 - 1 = 24$.

AnswerD · 2060

1. A leap year's calendar returns after 28 years, as long as no non-leap century year (like 2100) falls in between.

2. $2032 + 28 = 2060$, which is also a leap year starting on the same day (Thursday).

AnswerA · 2

1. The century years in the range are 1700, 1800, 1900, 2000, 2100, 2200, 2300 and 2400.

2. Only those divisible by 400 are leap: 2000 and 2400. That is 2.

AnswerA · 0

1. Each block of 100 years has 5 odd days, but the 400th year is a leap year, adding 1 odd day.

2. Total $= 4 \times 5 + 1 = 21$, and 21 is a multiple of 7, so 0 odd days.

AnswerC · 2034

1. The later year must start on the same day as 2023 (a Sunday) and must also be an ordinary year.

2. Add odd days year by year: 2023 gives 1, 2024 gives 2, 2025–2027 give 1 each, 2028 gives 2, 2029–2031 give 1 each, 2032 gives 2, 2033 gives 1. Running total after 2033: $1+2+1+1+1+2+1+1+1+2+1 = 14$, a multiple of 7.

3. So 1 January 2034 is a Sunday, and 2034 is ordinary: the calendar repeats in 2034.

AnswerC · 2030

1. 2019 is ordinary and 2019 is 3 more than a multiple of 4 (2016). For such a year the calendar repeats after 11 years.

2. Check by odd days: 2019–2029 contain 3 leap years (2020, 2024, 2028), so odd days $= 11 + 3 = 14$, a multiple of 7.

3. 2030 is ordinary and starts on a Tuesday like 2019, so the answer is 2030.

AnswerA · 2035

1. 2029 is the year just after a leap year (2028). Such a calendar repeats after 6 years.

2. Check: 2029–2034 contain one leap year (2032), so odd days $= 6 + 1 = 7$, a multiple of 7.

3. 2035 is ordinary and starts on Monday, like 2029.

AnswerC · Friday

1. From 1 January to 31 December of an ordinary year is 364 days after the first day.

2. 364 is exactly 52 weeks, so 31 December falls on the same weekday as 1 January: Friday.

AnswerB · 2029

1. Odd days carried from each year: 2024 is leap (2), 2025 (1), 2026 (1), 2027 (1), 2028 is leap (2).

2. Running total: 2, 3, 4, 5, 7. It reaches 7 after 2028.

3. So 1 January 2029 is a Monday.

AnswerD · 2028

1. The later year must be a leap year and must begin on a Saturday.

2. From 2000 to 2027 there are 28 years with 7 leap years (2000, 2004, …, 2024): odd days $= 28 + 7 = 35 = 5 \times 7$.

3. 2028 is leap and begins on Saturday. No non-leap century year lies in between, so the 28-year rule holds.

AnswerC · Tuesday

1. 100 years have 5 odd days, 200 years have 3, 300 years have 1 and 400 years have 0.

2. So the last day of each century, counted from day 0 = Sunday, is day 5, 3, 1 or 0: Friday, Wednesday, Monday or Sunday.

3. Tuesday, Thursday and Saturday can never be the last day of a century.

AnswerA · 2106

1. 2100 is a century year not divisible by 400, so it is an ORDINARY year, beginning on a Friday.

2. Look for the next ordinary year beginning on Friday. Odd days: 2100 (1), 2101 (1), 2102 (1), 2103 (1), 2104 leap (2), 2105 (1). Total $= 7$.

3. So 1 January 2106 is a Friday, and 2106 is ordinary: the calendar repeats in 2106.

AnswerB · 2108

1. The 28-year rule fails here because 2100, which lies in between, is NOT a leap year.

2. Odd days from 2096: 2096 (2), 2097 (1), 2098 (1), 2099 (1), 2100 (1), 2101–2103 (1 each), 2104 (2), 2105–2107 (1 each). Total $= 2+1+1+1+1+3+2+3 = 14$.

3. 14 is a multiple of 7, and 2108 is a leap year, so 2108 repeats the calendar of 2096.

AnswerD · 2048

1. Her birthday exists only in leap years: 2024, 2028, 2032, …

2. From one 29 February to the next is $4 \times 365 + 1 = 1461$ days $= 208$ weeks $+ 5$ days, so each birthday moves 5 days on.

3. Saturday + 5 = Thursday (2024), then Tuesday (2028), Sunday (2032), Friday (2036), Wednesday (2040), Monday (2044), Saturday (2048).

4. After 7 steps of 5 days ($35 = 5$ weeks) it is Saturday again: 2048.

AnswerA · 130°

1. Angle $= |30h - 5.5m| = |30 \times 3 - 5.5 \times 40|$.

2. $= |90 - 220| = 130^\circ$. This is less than $180^\circ$, so it is the angle asked for.

AnswerB · 100°

1. Angle $= |30 \times 7 - 5.5 \times 20| = |210 - 110| = 100^\circ$.

AnswerB · 172.5°

1. $|30 \times 9 - 5.5 \times 15| = |270 - 82.5| = 187.5^\circ$.

2. That is more than $180^\circ$, so the smaller angle is $360 - 187.5 = 172.5^\circ$.

AnswerA · 105°

1. Half past two is 2:30.

2. Angle $= |30 \times 2 - 5.5 \times 30| = |60 - 165| = 105^\circ$.

AnswerC · 100°

1. 11 a.m. to 2:20 p.m. is 3 hours 20 minutes $= 200$ minutes.

2. The hour hand turns $30^\circ$ per hour, that is $0.5^\circ$ per minute: $200 \times 0.5 = 100^\circ$.

AnswerC · 44

1. In every 12 hours the minute hand gains 11 full turns on the hour hand, and in each gain the hands are at right angles twice.

2. So 22 times in 12 hours, and $2 \times 22 = 44$ times in 24 hours.

AnswerC · 162.5°

1. Angle $= |30 \times 10 - 5.5 \times 25| = |300 - 137.5| = 162.5^\circ$.

2. $162.5^\circ$ is less than $180^\circ$, so it is the angle between the hands.

AnswerD · 282°

1. Smaller angle $= |30 \times 4 - 5.5 \times 36| = |120 - 198| = 78^\circ$.

2. The reflex angle is the other side of the dial: $360 - 78 = 282^\circ$.

AnswerC · $21\frac{9}{11}$ minutes past 4

1. At 4:00 the minute hand is $120^\circ$ behind the hour hand.

2. The minute hand gains $6 - 0.5 = 5.5^\circ$ per minute on the hour hand.

3. Time to close the gap $= 120 \div 5.5 = \frac{240}{11} = 21\frac{9}{11}$ minutes, so $21\frac{9}{11}$ minutes past 4.

AnswerB · $5\frac{5}{11}$ minutes past 7

1. At 7:00 the hour hand is $210^\circ$ ahead of the minute hand. For the hands to be opposite, the gap must shrink to $180^\circ$.

2. Gap to close $= 210 - 180 = 30^\circ$, at $5.5^\circ$ per minute: $30 \div 5.5 = \frac{60}{11} = 5\frac{5}{11}$ minutes.

3. So the hands are opposite at $5\frac{5}{11}$ minutes past 7.

AnswerD · $43\frac{7}{11}$ minutes past 5

1. At 5:00 the minute hand is $150^\circ$ behind the hour hand. The second right angle comes after the minute hand has passed the hour hand and is $90^\circ$ ahead of it.

2. So the minute hand must gain $150 + 90 = 240^\circ$.

3. Time $= 240 \div 5.5 = \frac{480}{11} = 43\frac{7}{11}$ minutes past 5.

AnswerC · $65\frac{5}{11}$ minutes

1. The minute hand has to gain a full $360^\circ$ on the hour hand.

2. It gains $5.5^\circ$ per minute, so time $= 360 \div 5.5 = \frac{720}{11} = 65\frac{5}{11}$ minutes.

AnswerD · 7:35

1. For a mirror image, actual time $= 11{:}60 -$ the time seen.

2. $11{:}60 - 4{:}25 = 7{:}35$.

AnswerA · 3:28

1. First find the actual time: $11{:}60 - 3{:}48 = 8{:}12$.

2. 20 minutes later the actual time is 8:32.

3. Reflection of 8:32 $= 11{:}60 - 8{:}32 = 3{:}28$.

AnswerD · $52\frac{8}{11}$ minutes past 8

1. At 8:00 the minute hand is $240^\circ$ behind the hour hand.

2. The second time, the minute hand has passed the hour hand and is $50^\circ$ ahead: it has gained $240 + 50 = 290^\circ$.

3. Time $= 290 \div 5.5 = \frac{580}{11} = 52\frac{8}{11}$ minutes past 8.

AnswerB · 7

1. Normally the hands are at right angles twice in each hour, which would suggest $4 \times 2 = 8$.

2. But 3:00 is itself a right angle, and it is shared by the hour 2–3 and the hour 3–4. The two right angles of 2 to 3 o'clock are at $27\frac{3}{11}$ minutes past 2 and at 3:00.

3. List: $21\frac{9}{11}$ and $54\frac{6}{11}$ past 1; $27\frac{3}{11}$ past 2; 3:00; $32\frac{8}{11}$ past 3; $5\frac{5}{11}$ and $38\frac{2}{11}$ past 4. That is 7.

AnswerD · 2:30 p.m.

1. From 8 a.m. to 2 p.m. is 6 hours.

2. Gain $= 6 \times 5 = 30$ minutes, so the clock shows 2:30 p.m.

AnswerA · 7:36 p.m.

1. From noon to 8 p.m. is 8 hours.

2. Loss $= 8 \times 3 = 24$ minutes, so the watch shows $8{:}00 - 24$ min $=$ 7:36 p.m.

AnswerC · 28

1. From 10 a.m. on 1 March to 10 a.m. on 15 March is $15 - 1 = 14$ days.

2. Gain $= 14 \times 2 = 28$ minutes.

AnswerB · 8

1. One day has 24 hours, so the loss $= 24 \times 20 = 480$ seconds.

2. $480 \div 60 = 8$ minutes.

AnswerD · 6:12 a.m.

1. From 6 a.m. Monday to 6 a.m. Wednesday is 48 hours.

2. Gain $= 48 \times 15 = 720$ seconds $= 12$ minutes, so it shows 6:12 a.m.

AnswerB · 4:00 p.m.

1. In every 60 true minutes this clock shows 65 minutes.

2. Since 8 a.m. it has shown 8 h 40 min $= 520$ minutes.

3. True time passed $= 520 \times \frac{60}{65} = 480$ minutes $= 8$ hours, so the correct time is 4:00 p.m.

AnswerD · 6:00 p.m.

1. For every 60 true minutes the clock shows only 56 minutes.

2. Time shown since 6 a.m. $= 11$ h 12 min $= 672$ minutes.

3. True time passed $= 672 \times \frac{60}{56} = 720$ minutes $= 12$ hours, so the correct time is 6:00 p.m.

AnswerA · 2:42 p.m.

1. First find the true time. The first clock shows 62 minutes for every 60 true minutes.

2. It has shown 6 h 12 min $= 372$ minutes, so true time passed $= 372 \times \frac{60}{62} = 360$ minutes: it is 3:00 p.m.

3. The second clock has lost $6 \times 3 = 18$ minutes, so it shows 2:42 p.m.

AnswerB · 16

1. 9 a.m. to 2:20 p.m. is 5 hours 20 minutes $= 5\frac{1}{3}$ hours.

2. Gain $= 5\frac{1}{3} \times 3 = 16$ minutes.