IB Recruitment 2017 – Apply Online for 1300 ACIO-II/ Exe Posts

IB Recruitment 2017 – Apply Online for 1300 ACIO-II/ Exe Posts: Ministry of Home Affairs, Government of India, Intelligence Bureau (IB) has advertised a notification for the recruitment of 1300 Assistant Central Intelligence Officer (GradeII/ Executive) i.e. ACIO-II (Exe) vacancies on direct recruitment basis. Eligible candidates can apply online from 12-08-2017 to 02-09-2017 till 23:59 hours. Other details like age limit, educational qualification, selection process, application fee & how to apply are given below…

IB Vacancy Details:
Total No of Posts: 1300
Name of the Posts: Assistant Central Intelligence Officer Grade-II/ Executive Exam 2017
1. UR: 951 Posts
2. OBC: 184 Posts
3. SC: 109 Posts
4. ST: 56 Posts

Age Limit: Candidates age limit should be between 18 to 27 years. Upper age relaxation of 05 years for SC/ ST, 03 years for OBC & up to 40 years for departmental candidates, 35 years for general candidates & others is applicable as per instructions or orders issued by the Central Government from time to time.

Educational Qualification: Candidates should possess Graduation or equivalent from a recognized University.

Selection Process: Candidates will be selected based on Tier-I, Tier-II Examinaion and interview.

Examination Fee: Applicants required to pay Rs.100/- Only male candidates belonging to General and OBC  category. All SC/ST and female candidates are exempted from payment of examination fee.

Offline Mode: Applicants required to pay fee through offline, have to take a printout of Challan form and deposit the fee in cash in any branch of State Bank of India after a gap of one working day. A counterfoil of this Challan would be given back to the candidate by the bank, indicating the transaction ID, which may be retained by the candidates.

Online Mode: Applicants required to pay fee through online by using Internet Banking, ATMCum-Debit Card and Credit Card of State Bank of India or any other bank can pay the fees through ‘SBI Collect’ option. The candidates would be required to provide their registration number during the process of online payment.

How to Apply: Eligible candidates can apply online through website www.mha.nic.in from 12-08-2017 to 02-09-2017 till 23:59 hours.

Instructions to Apply online:
1.Before applying online applicants should have valid e-mail ID, scanned copies of photograph & signature
2. Log on to website www.mha.nic.in
3. Go to “Whats New” & Select the desired post
4. Read the Notificaiton & Click on “Proceed…”
5. Tick the box present & Click on “Click here to Proceed…”
6. Fill all the details carefully & submit the form
7. Take print out of online application form for future use.

Important Dates:

Starting Date for Apply Online	12-08-2017
Last Date to Apply Online	02-09-2017 till 23:59 hours
Last Date for depositing fee	05-09-2017 (till closing of banking hours)
Admit Card can be Download in	September to October, 2017.
Date for Tier-I Written Examination	15-10-2017
Date for Tier-II Written Examination	07-01-2018

For more details like pay scale, examination centers & other information click on the link given below….

Click here for Recruitment Advt

Click here to Apply Online 

Read more: IB Recruitment 2017 – Apply Online for 1300 ACIO-II/ Exe Posts

Oriental Insurance Company Ltd Recruitment 2017

Oriental Insurance Company Ltd Recruitment 2017 – Apply Online for 300 Administrative Officer Posts: Oriental Insurance Company Limited, New Delhi has given an employment notification for the recruitment of 300 Administrative Officer (Scale-I) vacancies. Eligible candidates may apply online from 18-08-2017 to 15-09-2017. Other details like age, educational qualification, selection process, application fee & how to apply are mentioned below…

Oriental Insurance Company Ltd Vacancy Details:
Total No of Posts: 300
Name of the Post: Administrative Officer 
Name of the Disciplines:
1. Accounts: 20 posts
2. Actuaries: 02 posts
3. Engineers (Automobile): 15 posts
4. Legal: 30 posts
5. Medical Officer: 10 Posts
6. Generalist: 223 Posts

Age Limit: Candidates age should be 21 to 30 years as on 31-07-2017. Age relaxation is applicable by 05 years for SC/ ST, 03 years for OBC (NCL), 10 years for PWD candidates. for more details refer the notification.

Educational Qualification: Candidates should have M.Com. or MBA (Finance) or Chartered Accountants (ICAI) or Cost and Management Accountant (The Institute of Cost Accountants of India) earlier known as Cost and Wok Accountants (ICWAI) for post-1, Graduation & passed 4 Actuarial Papers from Institute of Actuaries of India (IAI) or Institute and Faculty of Actuary, UK (IFoA) for post-2, Graduation/PG Degree in engineering (4 or 5 years) with automobile engineering as a subject for post-3, Graduation in Law for post-4, MBBS for post-5, Graduation in any stream for post-6 with 55% for SC/ ST and 60% for Others from a recognized University.

Selection Process: Candidates will be selected based on interview.

Application Fee: Candidates have to pay Rs.600/- (Application Fee + Intimation Charges) for Others, Rs.100/- (Intimation Charges Only) for SC/ ST/ PWD candidates.

How to Apply: Eligible candidates can apply online through the website www.orientallnsurance.org.in from 18-08-2017 to 15-09-2017.

Important Dates:

Starting Date for Online Registration & Payment of Fee	18-08-2017
Last Date for Online Registration & Payment of Fee	15-09-2017
Date of Online Examination for Phase I	22-10-2017 (Tentative)
Date of Online Examination for Phase II	18-11-2017 (Tentative)

For more details like emoluments & other information click on the links given below…

Clcik here for Recruitment Advt

Read more: Oriental Insurance Company Ltd Recruitment 2017 – Apply Online for 300 Administrative Officer Posts

UIIC Recruitment 2017 – Apply Online for 696 Assistant Posts

UIIC Recruitment 2017 – Apply Online for 696 Assistant Posts: United India Insurance Co. Ltd (UIIC) Limited has announced notification for the recruitment of 696 Assistant vacancies. Eligible candidates can apply online from 14-08-2017 to 28-08-2017. Other details like age limit, educational qualification, selection process, application fee & how to apply are given below…

UIIC Vacancy Details:
Total No.of Posts: 696
Name of the Post: Assistant
Name of the Category:
1. UR: 414 posts
2. SC: 110 posts
3. ST: 50 posts
4. OBC: 122 posts

Age Limit: Candidates age should be between 18-28 years as on 30-06-2017. Age relaxation 05 years is admissible to SC/ ST, 03 years for OBC & for complete details refer the notification.

Educational Qualification: Candidates must be Graduate from a recognized University & Knowledge of Reading, Writing and Speaking of Regional language.

Selection Process: Candidates will be selected on the basis of Preliminary Exam & Main Exam.

Application Fee: Candidates should pay Rs. 500/- (Rs. 100/- for SC/ ST/ Persons with Disability (PWD), Payment can be made by using Debit Cards (RuPay/ Visa/ Master Card/ Maestro), Credit Cards, Internet Banking, IMPS, Cash Cards/ Mobile Wallets.

How to Apply: Eligible candidates may apply online through the website www.uiic.co.in from 14-08-2017 to 28-08-2017.

Instructions to Apply Online:
1. Before applying online candidates have valid e mail id & contact number.
2. Candidates should scan their photograph & signature.
3. Candidates log on the website www.uiic.co.in
4. Click on Recruitment——> Apply Online
5. Click on “Click here for New Registration ” and enter Name, Contact details and Email-id.
6. Fill all the details and submit the form.
7. Take a print out of online application for future use.

Important Dates:

Starting Date for Online Registration	14-08-2017
Last Date for Online Registration & Payment of Fee	28-08-2017
Tentative Date of Tier-I Online Preliminary Exam	22-09-2017
Tentative Date of Tier-II Online Main Exam	23-10-2017
Last Date for Printing candidates application	12-09-2017
Download of Call letters for Online (Preliminary & Main) Examination (Tentative)	10 days prior to the date of each exam

For more details like scale of pay, emoluments, qualification, selection process & other information click on the link given below…

Click here for Recruitment Advt

Click here for Online Registration

Read more: UIIC Recruitment 2017 – Apply Online for 696 Assistant Posts 

Problems on Calendar

Problems on Calendar - Solved Examples

21. If the first day of a year (other than leap year) was Friday, then which was the last day of that year?
A. Saturday	B. Friday
C. Tuesday	D. Monday

answer with explanation

Answer: Option B
Explanation:
Given that first day of a normal year was Friday

Odd days of the mentioned year = 1 (Since it is an ordinary year)

Hence First day of the next year = (Friday + 1 Odd day) = Saturday
Therefore, last day of the mentioned year = Friday

22. If 1st October is Sunday, then 1st November will be
A. Saturday	B. Thursday
C. Wednesday	D. Tuesday

answer with explanation

Answer: Option C
Explanation:
Given that 1^st October is Sunday

Number of days in October = 31
31 days = 3 odd days
(As we can reduce multiples of 7 from odd days which will not change anything)

Hence 1^st November = (Sunday + 3 odd days) = Wednesday

23. Arun went for a movie nine days ago. He goes to watch movies only on Thursdays. What day of the week is today?
A. Wednesday	B. Saturday
C. Friday	D. Sunday

answer with explanation

Answer: Option B
Explanation:
Clearly it can be understood from the question that 9 days ago was a Thursday

Number of odd days in 9 days = 2 (As 9-7 = 2, reduced perfect multiple of 7 from total days)

Hence today = (Thursday + 2 odd days) = Saturday

24. 1.12.91 is the first Sunday. Which is the fourth Tuesday of December 91?
A. 20.12.91	B. 22.12.91
C. 24.12.91	D. 25.12.91

answer with explanation

Answer: Option C
Explanation:
Given that 1.12.91 is the first Sunday

Hence we can assume that 3.12.91 is the first Tuesday

If we add 7 days to 3.12.91, we will get second Tuesday
If we add 14 days to 3.12.91, we will get third Tuesday
If we add 21 days to 3.12.91, we will get fourth Tuesday
=> fourth Tuesday = (3.12.91 + 21 days) = 24.12.91

25. If the day before yesterday was Thursday, when will Sunday be?
A. Day after tomorrow	B. Tomorow
C. Two days after today	D. Today

answer with explanation

Answer: Option B
Explanation:
Day before yesterday was Thursday
=>Yesterday was a Friday
=> Today is a Saturday
=> Tomorrow is a Sunday

26. The second day of a month is Friday, What will be the last day of the next month which has 31 days?
A. Friday	B. Saturday
C. Wednesday	D. Data inadequate

answer with explanation

Answer: Option D
Explanation:
We cannot find out the answer because the number of days of the current month is not given.

27. How many days will there be from 26th January,1996 to 15th May,1996(both days included)?
A. 102	B. 103
C. 111	D. 120

answer with explanation

Answer: Option C
Explanation:
Number of days from 26-Jan-1996 to 15-May-1996 (both days included)
= 6(Jan) + 29(Feb) + 31 (Mar) + 30(Apr)+ 15(May) = 111

28. If 25th of August in a year is Thursday, the number of Mondays in that month is
A. 4	B. 5
C. 2	D. 3

answer with explanation

Answer: Option B
Explanation:
Given that 25^th August = Thursday
Hence 29^th August = Monday
So 22^nd,15^th and 8^th and 1^st of August also will be Mondays

Number of Mondays in August = 5

29. If the seventh day of a month is three days earlier than Friday, What day will it be on the nineteenth day of the month?
A. Saturday	B. Monday
C. Sunday	D. Wednesday

answer with explanation

Answer: Option C
Explanation:
Given that seventh day of a month is three days earlier than Friday
=> Seventh day is Tuesday
=> 14^th is Tuesday
=> 19^th is Sunday

30. Second Saturday and every Sunday is a holiday. How many working days will be there in a month of 30 days beginning on a Saturday?
A. 24	B. 23
C. 18	D. 21

answer with explanation

Answer: Option A
Explanation:
Mentioned month begins on a Saturday and has 30 days

Sundays = 2^nd, 9^th, 16^th, 23^rd, 30^th
=> Total Sundays = 5

Every second Saturday is holiday.
1 second Saturday in every month

Total days in the month = 30

Total working days = 30 - (5 + 1) = 2
4

Problems on Calendar

Problems on Calendar - Solved Examples

11. The last day of a century cannot be
A. Monday	B. Wednesday
C. Tuesday	D. Friday

answer with explanation

Answer: Option C
Explanation:
We know that number of odd days in 100 years = 5
Hence last day of the first century is Friday

Number of odd days in 200 years = 5 x 2 = 10 = 3 (As we can reduce multiples of 7 from odd days which will not change anything)
Hence last day of the 2^nd century is Wednesday

Number of odd days in 300 years = 5 x 3 = 15 = 1
Hence last day of the 3^rd century is Monday

We know that umber of odd days in 400 years = 0. (∵ 5 x 4 + 1 = 21 = 0)
Hence last day of the 4^th century is Sunday

Now this cycle will be repeated. Hence last day of a century will not be Tuesday or Thursday or Saturday.

its better to learn this by heart which will be helpful to save time in objective type exams

12. January 1, 2008 is Tuesday. What day of the week lies on Jan 1, 2009?
A. Saturday	B. Wednesday
C. Thursday	D. Saturday

answer with explanation

Answer: Option C
Explanation:
Number of odd days in 2008 = 2 (since it is a leap year)
(we have taken the complete year 2008 because we need to find out the odd days from 01-Jan-2008 to 31-Dec-2008, that is the whole year 2008)

Given that January 1, 2008 is Tuesday.
Hence January 1, 2009 = (Tuesday + 2 odd days) = Thursday

13. If Jan 1, 2006 was a Sunday, What was the day of the week Jan 1, 2010?
A. Friday	B. Thursday
C. Tuesday	D. Saturday

answer with explanation

Answer: Option A
Explanation:
Given that Jan 1 2006 was a Sunday

Number of odd days in the period 2006-2009
= 3 normal years + 1 leap year
= 3 x 1 + 1 x 2 = 5 (note that we have taken the complete year 2006 because the period in 2006 is from 01-Jan-2006 to 31-Dec-2006, which is the whole year 2006. Then the complete years 2007, 2008 and 2009 are also involved)

Hence Jan 1 2010 = (Sunday + 5 odd days) = Friday

14. What was the day of the week on 17th June 1998?
A. Monday	B. Sunday
C. Wednesday	D. Friday

answer with explanation

Answer: Option C
Explanation:
17 Jun 1998 = (1997 years + period from 1-Jan-1998 to 17-Jun-1998)

We know that number of odd days in 400 years = 0
Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)

Number of odd days in the period 1601-1900
= Number of odd days in 300 years
= 5 x 3 = 15 = 1
(As we can reduce perfect multiples of 7 from odd days without affecting anything)

Number of odd days in the period 1901-1997
= 73 normal years + 24 leap year
= 73 x 1 + 24 x 2 = 73 + 48 = 121 = (121 - 7 x 17) = 2 odd days

Number of days from 1-Jan-1998 to 17-Jun-1998
= 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 31(may) + 17(Jun)
= 168 = 0 odd day

Total number of odd days = (0 + 1 + 2 + 0) = 3
3 odd days = Wednesday
Hence 17th June 1998 is Wednesday.

15. 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004?
A. Friday	B. Saturday
C. Wednesday	D. Sunday

answer with explanation

Answer: Option D
Explanation:
Number of days from 6th March, 2004 to 5th March 2005 = 365 days
(Though Feb 2004 has 29 days as it is a leap year, it will not come in the required period)

365 days = 1 odd day

Given that 6th March, 2005 is Monday
Hence 6th March, 2004 = (Monday - 1 odd day) = Sunday

16. What day of the week was 1 January 1901
A. Monday	B. Tuesday
C. Saturday	D. Friday

answer with explanation

Answer: Option B
Explanation:
1 Jan 1901 = (1900 years + 1^st Jan 1901)

We know that number of odd days in 400 years = 0
Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)

Number of odd days in the period 1601-1900
= Number of odd days in 300 years
= 5 x 3 = 15 = 1
(As we can reduce perfect multiples of 7 from odd days without affecting anything)

1^st Jan 1901 = 1 odd day

Total number of odd days = (0 + 1 + 1) = 2
2 odd days = Tuesday
Hence 1 January 1901 is Tuesday.

17. What day of the week will 22 Apr 2222 be?
A. Monday	B. Tuesday
C. Sunday	D. Thursday

answer with explanation

Answer: Option A
Explanation:
22 Apr 2222 = (2221 years + period from 1-Jan-2222 to 22-Apr-2222)

We know that number of odd days in 400 years = 0
Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)

Number of odd days in the period 2001-2200
= Number of odd days in 200 years
= 5 x 2 = 10 = 3
(As we can reduce perfect multiples of 7 from odd days without affecting anything)

Number of odd days in the period 2201-2221
= 16 normal years + 5 leap years
= 16 x 1 + 5 x 2 = 16 + 10 = 26 = 5 odd days

Number of days from 1-Jan-2222 to 22 Apr 2222
= 31 (Jan) + 28 (Feb) + 31 (Mar) + 22(Apr) = 112
112 days = 0 odd day

Total number of odd days = (0 + 3 + 5 + 0) = 8 = 1 odd day
1 odd days = Monday
Hence 22 Apr 2222 is Monday.

18. Today is Thursday. The day after 59 days will be?
A. Monday	B. Tuesday
C. Saturday	D. Sunday

answer with explanation

Answer: Option D
Explanation:
59 days = 8 weeks 3 days = 3 odd days

Hence if today is Thursday, After 59 days, it will be = (Thursday + 3 odd days)
= Sunday

19. What is the year next to 1990 which will have the same calendar as that of the year 1990?
A. 1992	B. 2001
C. 1995	D. 1996

answer with explanation

Answer: Option B
Explanation:
For a year to have the same calendar with 1990 ,total odd days from 1990 should be 0.

Take the year 1992 from the given choices.
Total odd days in the period 1990-1991= 2 normal years
≡ 2 x 1 = 2 odd days

Take the year 1995 from the given choices.
Number of odd days in the period 1990-1994 = 4 normal years + 1 leap year
≡ 4 x 1 + 1 x 2 = 6 odd days

Take the year 1996 from the given choices.
Number of odd days in the period 1990-1995= 5 normal years + 1 leap year
≡ 5 x 1 + 1 x 2 = 7 odd days ≡ 0 odd days
(As we can reduce multiples of 7 from odd days which will not change anything)

Though number of odd days in the period 1990-1995 is 0, there is a catch here.
1990 is not a leap year whereas 1996 is a leap year.
Hence calendar for 1990 and 1996 will never be the same.

Take the year 2001 from the given choices.
Number of odd days in the period 1990-2000= 8 normal years + 3 leap years
≡ 8 x 1 + 3 x 2 = 14 odd days ≡ 0 odd days
Also, both 1990 and 2001 are normal years.
Hence 1990 will have the same calendar as that of 2001

20. January 1, 2004 was a Thursday, what day of the week lies on January 1 2005.
A. Saturday	B. Monday
C. Saturday	D. Tuesday

answer with explanation

Answer: Option C
Explanation:
Given that January 1, 2004 was Thursday.

Odd days in 2004 = 2 (because 2004 is a leap year)
(Also note that we have taken the complete year 2004 because we need to find out the odd days from 01-Jan-2004 to 31-Dec-2004, that is the whole year 2004)

Hence January 1, 2005 = (Thursday + 2 odd days) = Saturda
y

Problems on Calendar

Problems on Calendar - Solved Examples

1. What day of the week does May 28 2006 fall on
A. Saturday	B. Monday
C. Sunday	D. Thursday

answer with explanation

Answer: Option C
Explanation:
28th May 2006 = (2005 years + period from 1-Jan-2006 to 28-May-2006)

We know that number of odd days in 400 years = 0
Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)

Number of odd days in the period 2001-2005
= 4 normal years + 1 leap year
= 4 x 1 + 1 x 2 = 6

Days from 1-Jan-2006 to 28-May-2006 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 28(may)
= 148

148 days = 21 weeks + 1 day = 1 odd day

Total number of odd days = (0 + 6 + 1) = 7 odd days = 0 odd day
0 odd day = Sunday

Hence May 28 2006 is Sunday.

2. What will be the day of the week 15th August, 2010?
A. Thursday	B. Sunday
C. Monday	D. Saturday

answer with explanation

Answer: Option B
Explanation:
15th Aug 2010 = (2009 years + period from 1-Jan-2010 to 15-Aug-2010)

We know that number of odd days in 400 years = 0
Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)

Number of odd days in the period 2001-2009
= 7 normal years + 2 leap year
= 7 x 1 + 2 x 2 = 11 = (11 - 7x1) odd day = 4 odd day

Days from 1-Jan-2010 to 15-Aug-2010
= 31 (Jan) + 28 (Feb) + 31 (Mar) + 30 (Apr) + 31(may) + 30(Jun) + 31(Jul) + 15(Aug)
= 227

227 days = 32 weeks + 3 day = 3 odd day

Total number of odd days = (0 + 4 + 3) = 7 odd days = 0 odd day
0 odd day = Sunday
Hence 15th August, 2010 is Sunday.

3. Today is Monday. After 61 days, it will be
A. Thursday	B. Sunday
C. Monday	D. Saturday

answer with explanation

Answer: Option D
Explanation:
61 days = 8 weeks 5 days = 5 odd days

Hence if today is Monday, After 61 days, it will be = (Monday + 5 odd days)
= Saturday

4. On what dates of April, 2001 did Wednesday fall?
A. 2nd, 9th, 16th, 23rd	B. 4th, 11th, 18th, 25th
C. 3rd, 10th, 17th, 24th	D. 1st, 8th, 15th, 22nd, 29th

answer with explanation

Answer: Option B
Explanation:
We need to find out the day of 01-Apr-2001

01-Apr-2001 = (2000 years + period from 1-Jan-2001 to 01-Apr-2001)

We know that number of odd days in 400 years = 0
Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)

Days from 1-Jan-2001 to 01-Apr-2001 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 1(Apr) = 91
91 days = 13 weeks = 0 odd day

Total number of odd days = (0 + 0) = 0 odd days
0 odd day = Sunday. Hence 01-Apr-2001 is Sunday.

Hence first Wednesday of Apr 2011 comes in 04^th and successive Wednesdays come in 11^th, 18^th and 25^th

5. How many days are there in x weeks x days
A. 14x	B. 8x
C. 7x2	D. 7

answer with explanation

Answer: Option B
Explanation:
$x$ weeks $x$ days $=(7\times x)+x=7x+x=8x$ days

6. The calendar for the year 2007 will be the same for the year
A. 2017	B. 2018
C. 2014	D. 2016

answer with explanation

Answer: Option B
Explanation:
For a year to have the same calendar with 2007 ,the total odd days from 2007 should be 0.

Year:	2007	2008	2009	2010	2011	2012	2013	2014	2015	2016	2017
Odd Days:	1	2	1	1	1	2	1	1	1	2	1

Take the year 2014 given in the choice.

Total odd days in the period 2007-2013 = 5 normal years + 2 leap year
= 5 x 1 + 2 x 2 = 9 odd days
= 2 odd day (As we can reduce multiples of 7 from odd days which will not change anything)

Take the year 2016 given in the choice.

Number of odd days in the period 2007-2015 = 7 normal years + 2 leap year
= 7 x 1 + 2 x 2 = 11 odd days
= 4 odd days
(Even if the odd days were 0, calendar of 2007 will not be same as the calendar of 2016 because 2007 is not a leap year whereas 2016 is a leap year. In fact, you can straight away ignore this choice due to this fact without even bothering to check the odd days)

Take the year 2017 given in the choice.

Number of odd days in the period 2007-2016 = 7 normal years + 3 leap year
= 7 x 1 + 3 x 2 = 13 odd days
= 6 odd days

Take the year 2018 given in the choice.

Number of odd days in the period 2007-2017 = 8 normal years + 3 leap year
= 8 x 1 + 3 x 2 = 14 odd days
= 0 odd day (As we can reduce multiples of 7 from odd days which will not change anything)

Also, both 2007 and 2018 are not leap years.
Since total odd days in the period 2007-2017 = 0 and both 2007 and 2018 are of same type, 2018 will have the same calendar as that of 2007

7. Which of the following is not a leap year?
A. 1200	B. 800
C. 700	D. 2000

answer with explanation

Answer: Option C
Explanation:
Remember the leap year rule (Given in the formulas)

1. Every year divisible by 4 is a leap year, if it is not a century.
2. Every 4th century is a leap year, but no other century is a leap year.

800,1200 and 2000 comes in the category of 4th century (such as 400,800,1200,1600,2000 etc).
Hence 800,1200 and 2000 are leap years

700 is not a 4th century, but it is a century. Hence it is not a leap year

8. 01-Jan-2007 was Monday. What day of the week lies on 01-Jan-2008?
A. Wednesday	B. Sunday
C. Friday	D. Tuesday

answer with explanation

Answer: Option D
Explanation:
Given that January 1, 2007 was Monday.

Odd days in 2007 = 1 (we have taken the complete year 2007 because we need to find out the odd days from 01-Jan-2007 to 31-Dec-2007, that is the whole year 2007)

Hence January 1, 2008 = (Monday + 1 Odd day) = Tuesday

9. 8th Dec 2007 was Saturday, what day of the week was it on 8th Dec, 2006?
A. Sunday	B. Tuesday
C. Friday	D. Tuesday

answer with explanation

Answer: Option C
Explanation:
Given that 8th Dec 2007 was Saturday

Number of days from 8th Dec, 2006 to 7th Dec 2007 = 365 days
365 days = 1 odd day

Hence 8th Dec 2006 was = (Saturday - 1 odd day) = Friday

10. On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004?
A. Sunday	B. Friday
C. Saturday	D. Monday

answer with explanation

Answer: Option A
Explanation:
Given that 8th Feb, 2005 was Tuesday

Number of days from 8th Feb, 2004 to 7th Feb, 2005 = 366 (Since Feb 2004 has 29 days as it is a leap year)

366 days = 2 odd days
Hence 8th Feb, 2004 = (Tuesday - 2 odd days) = Sunda
y

Calendar

Important Formulas - Calendar1. Odd Days

Number of days more than the complete weeks are called odd days in a given period.

2. Leap Year

A leap year has 366 days.

In a leap year, the month of February has 29 days.

Every year divisible by 4 is a leap year, if it is not a century.

Examples:
1952, 2008, 1680 etc. are leap years.
1991, 2003 etc. are not leap years

Every 4th century is a leap year and no other century is a leap year.

Examples:
400, 800, 1200 etc. are leap years.
100, 200, 1900 etc. are not leap years

3. Ordinary Year

The year which is not a leap year is an ordinary year.

An ordinary year has 365 days

4. Counting odd days and calculating day of any particular date

1 ordinary year ≡ 365 days ≡ (52 weeks + 1 day)
Hence number of odd days in 1 ordinary year= 1.

1 leap year ≡ 366 days ≡ (52 weeks + 2 days)
Hence number of odd days in 1 leap year= 2.

100 years ≡ (76 ordinary years + 24 leap years )
≡ (76 x 1 + 24 x 2) odd days
≡ 124 odd days.
≡ (17 weeks + 5 days)
≡  5 odd days.

Hence number of odd days in 100 years = 5.

Number of odd days in 200 years = (5 x 2) = 10 ≡ 3 odd days

Number of odd days in 300 years = (5 x 3) = 15 ≡ 1 odd days

Number of odd days in 400 years = (5 x 4 + 1) = 21 ≡ 0 odd days

Similarly, the number of odd days in all 4th centuries (400, 800, 1200 etc.) = 0

Mapping of the number of odd day to the day of the week

Number of odd days	Day of the week
0	Sunday
1	Monday
2	Tuesday
3	Wednesday
4	Thursday
5	Friday
6	Saturday

5. Additional Notes

Last day of a century cannot be Tuesday or Thursday or Saturday.

For the calendars of two different years to be the same, the following conditions must be satisfied.

(1)Both years must be of the same type. i.e., both years must be ordinary years or both years must be leap years.

(2)1^st January of both the years must be the same day of the week.

Problems on Boats and Streams(Set 4) - Solved Examples

Problems on Boats and Streams(Set 4) - Solved Examples

31. If a man's rate with the current is 15 km/hr and the rate of the current is 11⁄2 km/hr, then his rate against the current is
A. 12 km/hr	B. 10 km/hr
C. 10.5 km/hr	D. 12.5 km/hr

answer with explanation

Answer: Option A
Explanation:
Speed downstream = 15 km/hr
Rate of the current= 1¹⁄₂ km/hr

Speed in still water = 15 - 1¹⁄₂ = 13¹⁄₂ km/hr

Rate against the current = 13¹⁄₂ km/hr - 1¹⁄₂ = 12 km/hr

32. The speed of the boat in still water in 12 kmph. It can travel downstream through 45 kms in 3 hrs. In what time would it cover the same distance upstream?
A. 8 hours	B. 6 hours
C. 4 hours	D. 5 hours

answer with explanation

Answer: Option D
Explanation:
Speed of the boat in still water = 12 km/hr

Speed downstream $=\frac{45}{3}$ = 15 km/hr

Speed of the stream = 15-12 = 3 km/hr

Speed upstream = 12-3 = 9 km/hr

Time taken to cover 45 km upstream $=\frac{45}{9}$ = 5 hours

33. A boat goes 8 km upstream in 24 minutes. The speed of stream is 4 km/hr. The speed of boat in still water is:
A. 25 km/hr	B. 26 km/hr
C. 22 km/hr	D. 24 km/hr

answer with explanation

Answer: Option D
Explanation:
Speed upstream $=\frac{8}{\left(\frac{24}{60}\right)}$ = 20 km/hr

Speed of the stream = 4 km/hr

speed of boat in still water = (20+4) = 24 km/hr

34. The speed of a boat in still water is 25 kmph. If it can travel 10 km upstream in 1 hr, what time it would take to travel the same distance downstream?
A. 22 minutes	B. 30 minutes
C. 40 minutes	D. 15 minutes

answer with explanation

Answer: Option D
Explanation:
Speed of boat in still water = 25 km/hr

Speed upstream $=\frac{10}{1}$ = 10 km/hr

Speed of the stream = (25-10) = 15 km/hr

Speed downstream = (25+15) = 40 km/hr

Time taken to travel 10 km downstream $=\frac{10}{40}\text{hours}=\frac{10\times 60}{40}$ = 15 minutes

35. A boat can travel with a speed of 12 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream.
A. 4 hr	B. 4.25 hr
C. 4.5 hr	D. 6 hr

answer with explanation

Answer: Option B
Explanation:
speed of boat in still water = 12 km/hr

speed of the stream = 4 km/hr

Speed downstream = (12+4) = 16 km/hr

Time taken to travel 68 km downstream $=\frac{68}{16}=\frac{17}{4}$ = 4.25 hours

36. A man can row 7.5 kmph in still water and he finds that it takes him twice as long to row up as to row down the river. Find the rate of stream.
A. 10 km/hr	B. 2.5 km/hr
C. 5 km/hr	D. 7.5 km/hr

answer with explanation

Answer: Option B
Explanation:
Given that, time taken to travel upstream = 2 × time taken to travel downstream

When distance is constant, speed is inversely proportional to the time
Hence, 2 × speed upstream = speed downstream

Let speed upstream $=x$
Then speed downstream $=2x$

we have, $\frac{1}{2}(x+2x)$ = speed in still water
$\Rightarrow \frac{1}{2}\left(3x\right)=7.5\Rightarrow 3x=15\Rightarrow x=5$

i.e., speed upstream = 5 km/hr

Rate of stream $=\frac{1}{2}(2x-x)=\frac{x}{2}=\frac{5}{2}=2.5\text{km/hr}$

37. A boat moves downstream at the rate of one km in 5 minutes and upstream at the rate of 4 km an hour. What is the velocity of the current?
A. 4 km/hr	B. 2 km/hr
C. 3 km/hr	D. 1 km/hr

answer with explanation

Answer: Option A
Explanation:
Speed downstream = $\frac{1}{\left(\frac{5}{60}\right)}=12\text{km/hr}$

Speed upstream $=\frac{4}{1}\text{= 4 km/hr}$

velocity of the current $=\frac{1}{2}(12-4)=\text{4 km/hr}$

38. The speed of a boat in still water is 15 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 24 minutes is
A. 3.6 km	B. 2.4 km
C. 3.2 km	D. 7.2 km

answer with explanation

Answer: Option D
Explanation:
speed of a boat in still water = 15 km/hr
Speed of the current = 3 km/hr

Speed downstream = (15+3) = 18 km/hr

Distance travelled downstream in 24 minutes $=\frac{24}{60}\times 18=\frac{2\times 18}{5}\text{= 7.2 km}$

39. The current of a stream runs at the rate of 2 km per hr. A motor boat goes 10 km upstream and back again to the starting point in 55 min. Find the speed of the motor boat in still water?
A. 22 km/hr	B. 12 km/hr
C. 20 km/hr	D. 16 km/hr

answer with explanation

Answer: Option A
Explanation:
Let the speed of the boat in still water $=x$ km/hr
Speed of the current = 2 km/hr

Then, speed downstream $=(x+2)$ km/hr
speed upstream $=(x-2)$ km/hr

Total time taken to travel 10 km upstream and back = 55 minutes $=\frac{55}{60}$ hour = $\frac{11}{12}$hour
$\Rightarrow \frac{10}{x-2}+\frac{10}{x+2}=\frac{11}{12}120(x+2)+120(x-2)=11(x^2-4)240x=11x^2-4411x^2-240x-44=011x^2-242x+2x-44=011x(x-22)+2(x-22)=0(x-22)(11x+2)=0x=22\text{or}\frac{-2}{11}$

Since $x$ cannot be negative, $x$ = 22

i.e., speed of the boat in still water = 22 km/hr

40. The speed of a boat in still water is 8 kmph. If it can travel 1 km upstream in 1 hr, what time it would take to travel the same distance downstream?
A. 1 minute	B. 2 minutes
C. 3 minutes	D. 4 minutes

answer with explanation

Answer: Option D
Explanation:
Speed of the boat in still water = 8 km/hr

Speed upstream $=\frac{1}{1}$ = 1 km/hr

Speed of the stream = 8-1 = 7 km/hr

Speed downstream = (8+7) = 15 km/hr

Time taken to travel 1 km downstream $=\frac{1}{15}\text{hr =}\frac{1\times 60}{15}\text{= 4 minutes}$