Exercise : 3


1. What will be the day of the week on 1st January, 2010 ?
(a) Friday
(b) Saturday
(c) Sunday
(d) Monday
(e) None of these
Ans.c

2000 years have 2 odd days.

Year 2001 2002 2003 2004 2005 2006 2007 2008 2009
Odd days 1 1 1 2 1 1 1 2 1

= 11 odd days = 4 odd days.

1st January, 2010 has 1 odd day. Total number of odd days = (2 + 4 + 1) = 7 = 0.

∴ 1st January, 2010 will be Sunday.

2. The calendar for the year 2005 is the same as for the year :
(a) 2010
(b) 2011
(c) 2012
(d) 2013
(e) None of these
Ans.c

Count the number of days from 2005 onwards to get 0 odd day.

Year 2005 2006 2007 2008 2010 2011
Odd days 1 1 1 2 1 1

= 7 or 0 odd day.

∴ Calender for the year 2005 is the same as that for the year 2012.

3. If 09/12/2001 happens to be Sunday, then 09/12/1971 would have been at
(a) Wednesday
(b) Tuesday
(c) Saturday
(d) Thursday
(e) None of these
Ans.d

09/12/2001—— Sunday

No. of days between 9/ 12/ 71 & 9 / 12/ 2001

we know every year has 1 odd days

we know leap year has 2 odd days

Here, No. of normal years = 22

And no. of leap years = 8

So odd days = 22 + 16 = 38 i.e 3 odd days (remainder when 38 is divided by 7, i.e. 3)

Hence it was a Thursday

4. What was the day of the week on 15th August, 1947 ?
(a) Wednesday
(b) Tuesday
(c) Friday
(d) Thursday
(e) None of these
Ans.c

15th August, 1947 = (1946 years + Period from 1st Jan.,1947 to 15th Aug., 1947)

Counting of odd days :

1600 years have 0 odd day. 300 years have 1 odd day.

47 years = (11 leap years + 36 ordinary years)

= [(11 × 2) + (36 × 1)]odd days

= 58 odd days

⇒ 2 odd days.

Jan. Feb. March April May June July Aug.
31 28 31 30 31 30 31 15

= 227 days = (32 weeks + 3 days) = 3 odd days.

Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.

Hence, the required day was ‘Friday’.

5. The last day of a century cannot be :
(a) Monday
(b) Wednesday
(c) Friday
(d) Tuesday
(e) None of these
Ans.d

100 years contain 5 odd days. So, last day of 1st century is ‘Friday’

200 years contain (5 × 2) = 10 odd days = 3 odd days.

So, last day of 2nd century is ‘Wednesday’.

300 years contain (5 × 3) = 15 odd days = 1 odd day.

∴ Last day of 3rd century is ‘Monday’.

400 years contain 0 odd day.

∴ Last day of 4th century is ‘Sunday’

Since the order is continually kept in successive cycles,

we see that the last day of a century cannot be Tuesday,Thursday or Saturday.

6. The reflex angle between the hands of a clock at 10:25 is?
(a) 180°
(b) 19212°
(c) 195°
(d) 19712°
(e) None of these
Ans.d

Angle traced by hour hand in 12512 hrs.

\(\left ( \frac{360}{12} \times \frac{125}{12} \right )^{\circ} = 312\tfrac{1}{2}^{\circ}\)

Angle traced by minute hand in 25 min.

= \(\left ( \frac{360}{12} \times 25 \right )^{\circ} = 150^{\circ}\)

∴ Reflex angle = 360 - \(\left (312\tfrac{1}{2} - 150 \right )^{\circ} = 312\tfrac{1}{2}^{\circ}\)

7. A clock gains 5 minutes. in 24 hours. It was set right at 10 a.m. on Monday. What will be the true time when the clock indicates 10:30 a.m.on the next Sunday ?
(a) 10 a.m.
(b) 11 a.m.
(c) 25 minutes past 10 a.m.
(d) 5 minutes to 11 a.m.
(e) None of these
Ans.a

Time between 10 a.m. on Monday to 10:30 a.m. on Sunday = 14412

2412 hours of incorrect clock = 24 hours of correct time.

∴ 14412 hours of incorrect clock = x hours of correct time.

\(\frac{144\tfrac{1}{2} \times 24}{24\tfrac{1}{2}} = 144 \; hours \; i.e,\)

The true time is 10 a.m. on Sunday.

8. At what angle the hands of a clock are inclined at 15 minutes past 5 ?
(a) 7212°
(b) 64°
(c) 5812°
(d) 6712°
(e) None of these
Ans.d

At 15 minutes past 5, the minute hand is at 3 and hour hand slightly advanced from 5. Angle between their 3rd and 5th position.

Angle through which hour hand shifts in 15 minutes is

(15 &times 12)° = 712°

∴ Required angle = (60 + 712) = 6712°

9. Find the day of the week on 16th July, 1776.
(a) Tuesday
(b) Wednesday
(c) Monday
(d) Thursday
(e) None of these
Ans.a

16th July, 1776 mean (1775 years + 6 months + 16 days)

Now, 1600 years have 0 odd days.

100 years have 5 odd days

75 years contain 18 leap years and 57 ordinary years and therefore (36 + 57) or 93 or 2 odd days.

∴ 1775 years given 0 + 5 + 2 = 7 and so 0 odd days.

Also number of days from 1st Jan. 1776 to 16th July,1776

Jan.  Feb.  March    April   May   June    July
31   + 29    + 31     +   30   +  31   +  30   +  16

= 198 days = 28 weeks + 2 days = 2 odd days

∴ Total number of odd days = 0 + 2 = 2.

Hence the day on 16th July, 1776 was 'Tuesday'.

10. On January 12, 1980, it was Saturday. The day of the week on January 12, 1979 was –
(a) Saturday
(b) Friday
(c) Sunday
(d) Thursday
(e) None of these
Ans.b

The year 1979 being an ordinary year, it has 1 odd day.

So, the day on 12th January 1980 is one day beyond on the day on 12th January, 1979.

But, January 12, 1980 being Saturday.

∴ January 12, 1979 was Friday.

11. The year next to 1991 having the same calendar as that of 1990 is –
(a) 1998
(b) 2001
(c) 2002
(d) 2003
(e) None of these
Ans.c

We go on counting the odd days from 1991 onward still the sum is divisible by 7. The number of such days are 14 upto the year 2001. So, the calender for 1991 will be repeated in the year 2002.

12. A clock is set right at 5 a.m. The clock loses 16 min. in 24 hours. What will be the true time when the clock indicates 10 p.m.on the 4th day ?
(a) 11 p.m.
(b) 10 p.m.
(c) 9 p.m.
(d) 8 p.m.
(e) None of these
Ans.a

Time from 5 a.m. on a day to 10 p.m.on 4th day is 89 hours.

Now, 23 hrs. 44 min. of this clock are the same as 24 hours of the correct clock.

i.e., 36515 hrs. of this clock = 24 hrs. of correct clock.

∴ 89 hrs. of this clock = \(\left(\frac{24 \times 15}{356} \times 89 \right)\) hrs. of correct

clock = 90 hrs of correct clock.

So, the correct time is 11 p.m.

13. Find the exact time between 7 am and 8 am when the two hands of a watch meet ?
(a) 7 hrs 35 min
(b) 7 hrs 36.99 min
(c) 7 hrs 38.18 min
(d) 7 hrs 42.6 min
(e) None of these
Ans.c

55 min spaces are gained in 60 min

⇒ 35 min spaces will be gained in 38.18 min.

⇒ Answer = 7 hrs + 38.18 min.

14. A watch which gains 5 seconds in 3 minutes was set right at 7 a.m. In the afternoon of the same day, when the watch indicated quarter past 4 O’clock, the true time is –
(a) 4 p.m.
(b) 59712 minutes past 3
(c) 58711 minutes past 3
(d) 2311 minutes past 4
(e) None of these
Ans.a

Time from 7 a.m. to quarter pas 4

= 9 hours 15 min. = 555 min.

Now, 3712 min. of this watch = 3 min. of the correct watch.

555 min.of this watch = \(\left(\frac{3 \times 12}{37} \times 555 \right)\) min.

= \(\left(\frac{3 \times 12}{37} \times \frac{555}{60} \right)\) hrs. = 9 hrs. of the correct watch.

Correct time is 9 hours after 7 a.m. i.e., 4 p.m.