find a day without reference day Detailed Explanation And More Example

MOST IMPORTANT verbal reasoning - 4 EXERCISES

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The following question based on calendar topic of verbal reasoning

Questions : What day of the week was on 15th August 1949?

(a) Monday

(b) Tuesday

(c) Thursday

(d) Saturday

The correct answers to the above question in:

Answer: (a)

15th August 1949 means,

1948 complete year + First 7 months of the year 1949 + 15 days of August

Number of odd days in 1600 yrs = 0

Number of odd days in 300 yrs = 1

Number of odd days in 48 yr (36 non - leap years + 12 leap years)

= 36 × 1 + 12 × 2

= 60 = 7 × 8 + 4 = 4

odd days From 1st January 1949 to 15th August 1949

Number of odd days in 1949,

January 3
February 0
March 3
April 2
May 3
June 2
July 3
August (15 ÷ 7) = 1

Total number of odd days in 1949 = 3 + 0 + 3 + 2 + 3 + 2 + 3 + 1 = 17

= 7 × 2 + 3 = 3 odd day

Total odd days = 1 + 4 + 3 = 8

= 1 odd days

Since, 1 is the code for Monday.

Therefore, the required day was Monday.

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Read more finding day without ref day Based Verbal Reasoning Questions and Answers

Question : 1

On which day of the week does 18th September 1991 fall?

a) Wednesday

b) Tuesday

c) Friday

d) Saturday

Answer: (a)

18th September 1991 means,

1990 complete years + 8 months of 1991 + 18 days of September

Number of odd days in 1600 yrs = 0

Number of odd days in 300 yrs = 1

Number of odd days in 90 yrs (22 leap year + 68 ordinary years )

= 22 × 2 + 68 × 1

= 44 + 68 = 112

⇒ 0 odd days

Number of odd days in 1991,

January 3
February 0
March 3
April 2
May 3
June 2
July 3
August 3
September 4

= 3 + 0 + 3 + 2 +3 + 2 + 3 + 3 + 4

= 23 = 7 × 3 + 2

= 2 odd days

Total number of odd days till 18th September, 1991

= 0 + 1 + 0 + 2 = 3

So, the required day was Wednesday.

Question : 2

Ashu was born on August 19, 1992, What day of the week was the born?

a) Sunday

b) Monday

c) Tuesday

d) Wednesday

Answer: (d)

19th August 1992 means,

1991 complete years + First 7 months of 1992 + 19 days of August

Number of odd days in 1600 years = 0

Number of odd days in 300 yrs = 1

Number of odd days in 91 yrs (22 leap year + 69 non-leap years)

= 22 × 2 + 69 × 1

= 44 + 69 113

= 7 × 16 + 1 = 1 odd day

From 1st January, 1992 to 19th August, 1992

Number of odd days in 1992,

January 3
February 1
March 3
April 2
May 3
June 2
July 3
August 5

= 3 + 1 + 3 + 2 + 3 + 2 + 3 + 5

= 22 = 7 × 3 + 1

= 1 odd day

∴ Number of odd days till 19th August, 1992 = 0 + 1 + 1 + 1 = 3

So, the required day was Wednesday.

Question : 3

On which dates of April 2012 will a Sunday come?

a) 5, 12, 19, 26

b) 1, 8, 15, 22, 29

c) 3, 10, 17, 24

d) 7, 14, 21, 28

Answer: (b)

First of all, we have to find the day on 1st April, 2012 1st April 2012 means

(2011 years 3 months and 1 day)

Now, 2000 years have 0 odd days 11 years have

(2 leap years and 9 ordinary years)

= ( 2 × 2 + 9 × 1 ) odd days

= (4 + 9) odd days = 13

= 6 odd days

3 months and 1 day

January 31
February 29
March 31
April 1

= 92 days = 1 odd day

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

⇒ 0 odd day

Hence, it was Sunday on 1st April 2012. (1st Sunday).

Subsequently, Sundays of the month were on 1st, 8th, 15nd, 22nd, and 29th.

Question : 4

What was the day of the week on 30th June 1980?

a) Friday

b) Wednesday

c) Monday

d) Saturday

Answer: (c)

30th June 1980 means 1979 complete years + 6 months of 1980

Number of odd days in 1600 yrs = 0

Number of odd days in 300 yrs = 1

Number of odd days in 79 yrs = (19 leap yrs + 60 ordinary years)

= 19 × 2 + 60 × 1 = 38 + 60 = 98

⇒ o odd days

January 3
February 1
March 3
April 2
May 3
June 2

Number of odd days in 1980 = 3 + 1 + 3 + 2 + 3 + 2 = 14

⇒ 0 odd days

Total number of odd days till 30th June, 1980 = 0 + 1 + 0 + 0 = 1

So, the required day was Monday.

Question : 5

What was the day of the week on 1st April 1901?

a) Sunday

b) Monday

c) Wednesday

d) Saturday

Answer: (b)

1st April 1901 means 1900 complete years + first 3 months of 1901 + 1 day of April

Number of odd days in 1600 yrs = 0

Number of odd days in 300 yrs = 1

Number of odd days in 1901 yrs

January 3
February 0
March 3
April 1

= 3 + 0 + 3 + 1 = 7

⇒ 0 odd days

Total number of odd days till 1st April 1901 = 0 + 1 + 0 = 1

So, the required day was Monday.

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