### Reasoning Aptitude

## CALENDAR MCQs

### Total Questions : 60

**Page 1 of 6 pages**

**Answer is Option C. ->**Thursday

Answer: (c) 4th Saturday = 22nd day 3rd Saturday = 22 - 7 = 15th day ∴ 13th day - Saturday - 2 = Thursday

**Answer is Option C. ->**Tuesday

Answer: (c) Here, 27th March 1995 was Monday. Now, for calculating the total number of odd days. first, we calculate the total number of days till 1 November 1994. ∴ Number of days in March, 1995 = 27 Number of days in February 1995 = 28 Number of days in January 1995 = 31 Number of days in December 1994 = 31 Number of days in November 1994 = 29 Total number of days = 146 ∴ Number of odd days = $146/7$ = 20$6/7$ So, 6 odd days ∴ On November 1, 1994 = Monday - 6 = Tuesday

**Answer is Option B. ->**Tuesday

Answer: (b) Tomorrow = Friday 3 days after tomorrow = 15th June = Friday + 3 odd days = Monday Days from 15th to 30th June = 15 15 ÷ 7 = 7)15(2 =>remainder 1 odd day 30th June = Monday + 1 odd day = Tuesday

**Answer is Option D. ->**Sunday

Answer: (d) Since Total days from 15th August, 2011 to 17 September, 2011 = 33 33 ÷ 7 => 7)33(4 = remainder 5 odd days ∴ Required day = Tuesday + 5 odd days = Sunday

**Answer is Option B. ->**Monday

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.

**Answer is Option A. ->**Monday

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.

**Answer is Option 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.

**Answer is Option C. ->**Monday

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.

**Answer is Option A. ->**Wednesday

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.

**Answer is Option B. ->**1, 8, 15, 22, 29

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.

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