mrunal reasoning_ calendar related mcqs, how to solve quickly

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8/8/2014 Mrunal Reasoning: Calendar related MCQs, How to solve quickly?

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[Reasoning] Calendar Questions: Finding day or date, concepts, shortcuts

explained

1. Introduction

2. Concept: 7 Day cycle

1. SSC-CGL 2000 Question on 7-Day cycle

3. Concept: Day Gain Day loss

1. What is Leap year?

2. SSC Investigator 2010 (Anniversary)

3. SSC CGL 2011 (Anniversary)

4. CAT-2001 Question on calendar5. Question 1988 to 1989

6. Speed Techniquetip#1

7. Speed Technique tip#2

4. Concept: Date Without reference Day

1. Table#1: The odd days

2. Tablet #2: Number-Day

3. What day of week was on15th ugust,1947?

4. Speed increase tip#1:

5. Find the day on 10th May, 1857?6. Speed Technique #3

5. Mock questions

Introduction

Articleprepared with help of Mr. Deepak Singh.

There are two main types of questions from Calendar

when youre given a reference day when youre given no

reference Day

What day of week was it on 5th November, 1989

if it wasMonday on 4th April, 1988 ?

a. Monday

b. Tuesday

c. Saturday

d. Sunday

What wasthe day on 15th

august 1947?

a. Sunday

b. Monday

c. Tuesday

d. Friday

Here question tells us that 4thApril 88 was

Monday, so we have a reference day.

Here question doesnt give

us any reference day.

can be solved with just two concepts
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1. 7 day cycle

2. day gain-Day loss concept

need to just mugup 2 table,

and few steps.

Although calendar question is not asked every year in every exam. But If and

when calendar question is asked, just follow the given procedure and youll get

the accurate answer= one mark is guaranteed=in that sense, cost: benefit is

great.

Concept: 7 Day cycle

Name of the day will repeat after seven

days.

1stAugust 2013= Thursday. Therefore 1st

August + 7 =8thAugust also has to beThursday.

Then what day will be 10thAugust 2013?

1stAugust =Thursday => 1+7=8 August is

also Thursday

And 8 August + 2 =10 August will be

Thursday + 2 days =>Saturday.

lets try a simple question from SSC

SSC-CGL 2000 Question on 7-Day cycle

If 9thof the month falls on the day preceding Sunday, then on what day will 1st

of the month fall?

a. Friday

b. Saturday

c. Sunday

d. Monday

As per the question 9thof the Month=Saturday. (day preceding Sunday)

Day name repeats after 7 days.

Therefore 9 minus 7=2ndof the given month is also Saturday.

Then 1stof the given month= Saturday Minus one day = Friday, Ans A

Concept: Day Gain Day loss


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Tuesday on 30thSeptember 2003 Ans A.

Now, How about a CAT level question:

CAT-2001 Question on calendar

Dec 9, 2001 is Sunday then what was the day on Dec 9, 1971?

A. Thursday

B. Wednesday

C. Saturday

D. Sunday

Same day gain, Day loss

1971 to 2001=how many jumps?

2001-1971=Total 30 years jump

Out of those 30 years, how many leap years?

72, 76,.,00 (multiples of of 4 and 2000 is also leap year because It is multiple

of 400)

but no need to manually count leap years.

if you observe

18 x 4 =(19)72

25x 4 =(20)00

so from 18 to 25 = total 8 leap years. (plz note: 25-18=7 years, but weve to

include both years as welltherefore.. in such counting, itll be 25-18+1=8 leap

years)

back to the start: 30 years jump and out of them 8 were leap years.

Meaning 22 normal years + 8 leap years = total 30 years

22 normal years (+1 day gain) 22 x 1 =22 days

8 leap years (+2 days gain) 8 x 2= 16 days

total gain 22+16=38 days

38/7=(75)+3 remainder

Meaning whatever was the day on Dec 9, 1971, itll move to +3 days on dec 92001

Reverse is also true: Whatever was the day on Dec 2001, itll be three days less

on Dec 9 1971:


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Since Dec 2001 was Sunday therefore,

Sunday Minus 3 days= (just count in your head): SaturdayFridayThursday.

Final answer is A: Thursday

Question 1988 to 1989

What day of week was it on 5th November,1989 if it was Monday on 4th April,

1988 ?

1. Monday

2. Tuesday

3. Saturday

4. Sunday

Recall our Day gain Day loss principle

For Non-Leap year, When we proceed forward by one year, then 1 day is gained

and vice-versa.

If 4th April, 1988 = Monday, then 4th April, 1989 = Tuesday (Because 1989 is a

non-leap year)

Remaining days until 5thNov.89

Month DaysApril (4 to 30) 26

May 31

June 30

July 31

Aug 31

Sep 30

Oct 31

Nov 5

total 215

=Tuesday + 215

Divide 215 with 7 to find the remainder

215=(30*7)+5 Hence five is the remainder

Back to the problem

= Tuesday + [5 days]

What is the 5thday after Tuesday? Count on your fingertip: Wed, Thurs., Fri,

Sat., Sunday


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=Sunday (Final answer D)

Speed Technique tip#1

in above problem, if you dont want to waste time in adding the days like

26+31+30. then use following approach

30 days=(7*4)+2=> 2 remainder

31 days=(7*4)+3=> 3 remainder.

Month Days Remainder with 7

April (4 to 30) 2 6 5 (because 26=7*3+5)

May 31 3

June 30 2

July 31 3

Aug 31 3Sep 30 2

Oct 31 3

Nov 5 5

total dont need 26

Divide this by 7 and find remainder: 26=(7*3)+5 so remainder is 5

=Tuesday + 5

= Tuesday + 5 days

=Sunday

Speed Technique tip#2

in speed tech #1, if you dont want to waste time in adding the remainders of

seven: 5+3+2. then do following

pickup remainders that add upto 7 and cancel them.

for example 5+2=7. So Ill cancel each such pair in the table. Observe

Month Days Remainder with 7after cancelling numbers that add upto

7 (e.g. 5+2)

April (4 to

30) 26 5

May 31 3

June 30 2July 31 3

Aug 31 3

Sep 30 2


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Oct 31 3

Nov 5 5

total dont

need 12

After cancelling the two pairs of (5+2), Im left with four 3s= 12 (because 4 x

3=12)

=Tuesday + 12

now divide 12 with 7 find remainder: 12=7*1 +5

= Tuesday + 5 days

=Sunday

Concept: Date Without reference Day

Example: What was the day on 15 thAugust 1947?

To solve this type of questions, youve to first mug up these two tables:

Table#1: The odd days

100 5

200 3

300 1

400, 800, 1200, 1600 etcmultiples of 400 0

^ok but what is the use of above table? It tells us the number of odd days in

that given year. I dont want to bore you with the theory so just mug up those

values.

Tablet #2: Number-Day0, 7 Sunday

1 Monday

2 Tuesday

3 Wednesday

4 Thursday

5 Friday

6 Saturday

Just remember that 1 to 6 is Monday to Saturday and 0 or 7=Sunday.

^ok but what is the use of above table? just hang on for a few more paragraphs

and youll know! Now lets try a sum


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What day of week was on 15th ugust,1947?

Step 1:

Subtract the given year by 1.

1947 -1 = 1946

Step 2:

Break into relevant years as in table #1.

100 5

200 3

300 1

400, 800, 1200, 1600 etc (multiples of 400) 0

Therefore,

1946 = 1600 + 300 + ( 46 )

Step 3:

Now write corresponding values from the table: 1600=0 and 300=1

1946=>0+1+(46)

Step 4:

Now for the number in bracket (46), divide it by 4 and add quotient in same

line.

in this case 46=(11x4)+2 therefore, 11 is quotient and 2 is remainder. Were

concerned with number and quotient here

1946=0+1+(46+11)

Step5:

Now add all these numbers and divide by 7

0+1+(46+11)

=58

And when you divide 58 by 7, you get 58= (7*8)+2. Therefore remainder is 2.

we got the number 2 = well call this our 31Dec Number

Observe second table


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0, 7 Sunday

1 Monday

2 Tuesday

3 Wednesday

4 Thursday

5 Friday

6 Saturday

From this table we can see that 2=>Tuesday

It means 31stDecember 1946 was Tuesday. Now we apply our 7day cycle

concept to find out the day on 15thAugust 1947 using the following formula

Given day= (our 31Dec Number 2)+ *how many days till we reach 15 th

August?*

Step 6

from 1stJan 1947 till we reach 15thAugust 1947 days in given month

Jan 31

feb (not leap year) 28

march 31

april 30

may 31

june 30july 31

august (upto 15thAug) 15

total 227 days

back to the bold part

(Our 31Dec Number)+ *how many days till we reach 15thAugust?*

=2 + 227

=229

Step 7

Divide this number (229) by 7 and whatever remainder you get= that is our day

from table #2

229/7=(32*7)+5

so 5 is the remainder and as per table#2

0, 7 Sunday

1 Monday


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2 Tuesday

3 Wednesday

4 Thursday

5 Friday

6 Saturday

5=>Friday.

That means 15thAugust 1947 was Friday.

Speed increase tip#1:

in above 15thAugust example, Back to the step6

31Dec Number+ *how many days till we reach 15thAugust?*

after that, we did following:

from 1stJan 1947 till we reach 15thAugust 1947 days in given month

Jan 31

feb (not leap year) 28

march 31

april 30

may 31

june 30july 31

august (upto 15thAug) 15

total 227 days

^as you see we had to sum of 31+28+31.=lot of time taken in doing the

addition (+).

So, it is better to just add remainders for each month with 7

30=(7*4)+2=> 2 remainder

31=(7*4)+3=> 3 remainder.

Now observe again

from 1stJan 1947till we reach 15thAugust

1947

days in given

month

remainder with

7

Jan 31 3

feb (not leap year) 28 0march 31 3

april 30 2

may 31 3


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june 30 2

july 31 3

august (upto 15thAug) 15 1 (because

14+1)

total 227 days 17

now instead of 227 we can simply write 17

Our 31Dec Number+ *how many days till we reach 15thAugust?*

=2+17

=19

Therefore 19/7 = its remainder will tell us the final day

19=(7*2) + 5=therefore remainder is 5 and as per table#2, 5 means Friday.

Find the day on 10th May, 1857?

Subtract 1 from the given year

1857-1=1856

Now breakup 1856 as per our table #1

100 5200 3

300 1

400, 800, 1200, 1600 etc (multiples of 400) 0

1856

=1600+200+(56)

write corresponding values from table#1

=0+3+(56)

Divide the bracket number by 4 and write quotient along with number

56=14*4 + 0 therefore quotient is 14

1856

=0+3+(56+14)

add these numbers and divide by 7 find remainder

=73/7


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=remainder 3

Speed Technique #3

after doing the division with 4 to find quotient step, you got following

=0+3+(56+14)

Whenever you see the multiple of 7, just scratch that number (because itll give

remainder zero anyways)

for example

=0+3+(56+14)

Here both 56 and 14 are multiples of 7 so even if you divide them by 7, youre

going to get zero as remainder.

=0+3+0+0=3 is 31Dec Number

as per table#2: 3 means Wednesday. Meaning 31stDecember 1856 was

Wednesday

now our 31Dec Number+ how many days till 10th May, 1857?

from 1stJan 1857till we reach 31th

May 1857

days in given

month remainder with 7

Jan 31 3

feb (not leap year in 1857) 28 0

march 31 3

april 30 2

may 10 3, Because 10=(7 x

1)+3

total *not needed* 11

back to our problem

our 31Dec Number+ how many days till 10th May, 1857?

=3 + 11

=14

when 14 is divided by 7 we get zero as remainder.

as per table#2

0, 7 Sunday

1 Monday


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2 Tuesday

3 Wednesday

4 Thursday

5 Friday

6 Saturday

Zero means Sunday. so final answer 10thMay 1857 was Sunday.

Extra facts:

1. The 1st day of a century must be Tuesday, Thursday, or Saturday.

2. The last day of a century cannot be Tuesday, Thursday, or Saturday.

Mock questions

1. In 2013, Gandhis birth-anniversary is on Wednesday. In which nearest

future year, will his birth-anniversary be on Monday?

2. If 29thApril 2013 is Monday then what is the day on 30thNovember 2013?

3. If June 11, 2013 is Monday, what was the day on July 11, 2000

4. What was the day on 9/11 attacks in 2001

5. What was the day on 26/11 attacks in 2008

Answers

1. 2017

2. 30/nov/13=Saturday3. 11/jul/00=Monday

4. 11/09/01=Tuesday

5. 26/11/08=Wednesday

For more articles on aptitude, visit Mrunal.org/aptitude

URL to article: http://mrunal.org/2013/08/reasoning-calendar-questions-finding-

day-or-date-concepts-shortcuts-explained.html

Posted By Mrunal On 09/08/2013 @ 22:58 In the category Aptitude
http://www.mrunal.org/aptitude

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