Aptitude - Calendar - Discussion
Discussion Forum : Calendar - General Questions (Q.No. 3)
3.
What was the day of the week on 17th June, 1998?
Answer: Option
Explanation:
17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)
Odd days in 1600 years = 0
Odd days in 300 years = (5 x 3) 
1
97 years has 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.
Jan. Feb. March April May June (31 + 28 + 31 + 30 + 31 + 17) = 168 days
168 days = 24 weeks = 0 odd day.
Total number of odd days = (0 + 1 + 2 + 0) = 3.
Given day is Wednesday.
Discussion:
165 comments Page 5 of 17.
Steffy said:
1 decade ago
To Nitin:
Here we have 97 yrs
97/4=24 leap yrs
97-24=73 ordinary yrs
Here we have 97 yrs
97/4=24 leap yrs
97-24=73 ordinary yrs
Ekta said:
1 decade ago
Thank you Arya. I can understand properly your solution.
Sri Harsha. P said:
1 decade ago
There is another way...
Remember the code FOR ALL PROBLEMS ABOUT CALENDER
MONTH : J F M A M J J A S O N D
CODE . : 0 3 3 6 1 4 6 2 5 0 3 5
and
CODE FOR THE YEAR
1700 -- 1799 = 4
1800 -- 1899 = 2
1900 -- 1999 = 0
2000 -- 2099 = 6
2100 -- 2199 = 4
AT FINAL EX: FOR 17 JUNE 1998
take the last two numbers in the year for ex. In 1998 take 98 and divide them by 4 and take only the quotient and leave the reminder
ex: 4)98(24
8
----
18
16
----
2
leave that reminder 2 and take only quotient 24.
add the following
month code for JUNE 4
year code for 1998 0
left quotient while
divided by 4 24
finally add the last
two numbers in year
1998 and date17(98+17) 115
------------
143
divide the amount by 7 i.e., 143/7
here take only the reminder and left the quotient.
ex: 7)143(2
14
----
03
take the value of reminder 3.
WEEK S M T W T F S
CODE 0 1 2 3 4 5 6
IN THE ABOVE PROBLEM ANSWER IS 3 THEN DAY IS W-Wednesday
ANOTHER EXAMPLE: 15 AUGUST 1947
ADD
month code 2
year code 0
quotient divided by 4 11
(47+15) 62
----
75
75/7 take only reminder 5
Answer is 5-F-Friday.
Remember the code FOR ALL PROBLEMS ABOUT CALENDER
MONTH : J F M A M J J A S O N D
CODE . : 0 3 3 6 1 4 6 2 5 0 3 5
and
CODE FOR THE YEAR
1700 -- 1799 = 4
1800 -- 1899 = 2
1900 -- 1999 = 0
2000 -- 2099 = 6
2100 -- 2199 = 4
AT FINAL EX: FOR 17 JUNE 1998
take the last two numbers in the year for ex. In 1998 take 98 and divide them by 4 and take only the quotient and leave the reminder
ex: 4)98(24
8
----
18
16
----
2
leave that reminder 2 and take only quotient 24.
add the following
month code for JUNE 4
year code for 1998 0
left quotient while
divided by 4 24
finally add the last
two numbers in year
1998 and date17(98+17) 115
------------
143
divide the amount by 7 i.e., 143/7
here take only the reminder and left the quotient.
ex: 7)143(2
14
----
03
take the value of reminder 3.
WEEK S M T W T F S
CODE 0 1 2 3 4 5 6
IN THE ABOVE PROBLEM ANSWER IS 3 THEN DAY IS W-Wednesday
ANOTHER EXAMPLE: 15 AUGUST 1947
ADD
month code 2
year code 0
quotient divided by 4 11
(47+15) 62
----
75
75/7 take only reminder 5
Answer is 5-F-Friday.
Janvi said:
1 decade ago
How you got ordinary days? could you explain?
HDX said:
1 decade ago
Here is complete solution, the year is 1998
16+3=19
See from 0 A.D
1600yr+300yr =1900yr
We are calling the odd days from the 0 A.D. So 1600 yr has 0 odd days
But the remaing 300 year has 15 odd days and rest you can do.
16+3=19
See from 0 A.D
1600yr+300yr =1900yr
We are calling the odd days from the 0 A.D. So 1600 yr has 0 odd days
But the remaing 300 year has 15 odd days and rest you can do.
Arun said:
1 decade ago
How to calculate odd days for 300 years? I don't understand this only.
Rishi said:
1 decade ago
Ishan , Arya and Raviteja
Thanks to all u for ur such a explanation
Thanks to all u for ur such a explanation
Praveen kumar said:
1 decade ago
How we gat an idea to take 1600 years and 300 years in above problem.
Muthaiyan said:
1 decade ago
Step 1:[1997+1.1.1998to17.6.1998]
step2: 1997=1600+300+97=> 1600 is 0 odd days, 300 is 1 odd day
step3: calculate 97
step4:97/4= 24,=> [24-97=73]
step5:24leap year +73 ordinary year=>(24*2)+(73*1)=127
step6:127/7=17,=>17week+2 odd days
step7:1998 is an ordinary year therefore [31+28+31+30+31+17=168]
ste8:168/7=24weeks+0 odd days
step9:add all odd days we get 3 is an wednesday
step2: 1997=1600+300+97=> 1600 is 0 odd days, 300 is 1 odd day
step3: calculate 97
step4:97/4= 24,=> [24-97=73]
step5:24leap year +73 ordinary year=>(24*2)+(73*1)=127
step6:127/7=17,=>17week+2 odd days
step7:1998 is an ordinary year therefore [31+28+31+30+31+17=168]
ste8:168/7=24weeks+0 odd days
step9:add all odd days we get 3 is an wednesday
Ravihans said:
1 decade ago
Good explanation rajesh.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers