# Exercise : 3

(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.

(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.

(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

(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’.

(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.

(b) 192

^{1}⁄

_{2}°

(c) 195°

(d) 197

^{1}⁄

_{2}°

(e) None of these

**Ans.d**

Angle traced by hour hand in ^{125}⁄_{12} hrs.

Angle traced by minute hand in 25 min.

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

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

(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 = 144^{1}⁄_{2}

24^{1}⁄_{2} hours of incorrect clock = 24 hours of correct time.

∴ 144^{1}⁄_{2} 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.

^{1}⁄

_{2}°

(b) 64°

(c) 58

^{1}⁄

_{2}°

(d) 67

^{1}⁄

_{2}°

(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 × ^{1}⁄_{2})° = 7^{1}⁄_{2}°

∴ Required angle = (60 + 7^{1}⁄_{2}) = 67^{1}⁄_{2}°

(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'.

(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.

(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.

(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., ^{365}⁄_{15} 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.

(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.

(b) 59

^{7}⁄

_{12}minutes past 3

(c) 58

^{7}⁄

_{11}minutes past 3

(d) 2

^{3}⁄

_{11}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, ^{37}⁄_{12} 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.