For two years having same calendar must have two common conditions: both having same length in terms of number of days and first day of the week.
The year 1991 has 365 days,,that is 1 odd day , year 1992 has 366 days , i.e. 2 odd days , year 1993 has 365 days i.e. 1 odd day. the years 1994,1995,1996 have 1 odd day each.
the sum of odd days so calculated from year 1990 to 1996 .
(1+2+1+1+1+1)=7 odd days. 5
hence, the year 1997 will have same calendar as that of year 1990.
Let us assume the present ages of Mohan and Sohan be 3x years and 4x years; then,
4 Four years ago the ages was 3x - 4 and 4x - 4 years.
According to question, 4 Four years ago the ratio was 5 : 7
( 3x - 4 ) / ( 4x - 4 ) = 5/7
? ( 3x - 4 ) 7 = 5 ( 4x - 4 )
? 21x - 28 = 20x - 20
? 21x - 20x = 28 - 20
? x = 8
Hence, their present ages, 3x = 3 x 8 = 24 years; 4x = 4 x 8 = 32 years.
So their present ages 24 and 32 years.
When Ravi was born, his mother's age was 26 yr and his elder brother was 3 yr elder to him.
? Mother's age when brother was born
= 26 - 3 = 23 yr
Ravi's father was 28 yr of age when his sister was born and his sister was 4 yr of age when his brother was born
? Age of father when brother was born
= 28 + 4 = 32 yr
Let ages of man and his son are 8k yr and k yr, respectively .
According to the question,
(8k + k)/2 = 27
? k = (27 x 2)/9 = 6
Hence, age of son after 6 yr = 6 + 6 = 12 yr
Let Arnav's age = k yr
Then, Palash's age = 3k yr
Acoording to the question, 3k + 7 = 2(k + 7)
? 3k + 7 = 2k + 14
? k = 7
? Age of Arnav after 14 yr = 7 + 14 = 21 yr
and Palash's present age = 21 yr
Hence, Palash's age is one time of Arnav's age .
4 yr ago, let Ram's age = k yr and Shyam's age = 3k/4 yr
Now, Ram's present age = (k + 4) yr
and Shyam's present age = (3k/4 + 4) yr
According to the question, 5/6(k + 4 + 4) = (3k/4 + 4 + 4)
? 4 (5k + 40) = 6(3k + 32)
? 20k + 160 = 18k + 192
? 2k = 32
? k = 16
Hence, present age of Shyam = (3/4) x 16 + 4 = 16 yr
Each day of the week is repeated after 7 days
? After 63 days, it would be Friday
So, After 62 days, it would be Thursday
The year 1984 being a leap year, it has 2 odd days. So, the day on 2nd July, 1985 is two days beyond the day on 2nd July, 1984. But, 2nd July 1985 was Wednesday.
? 2nd July, 1984 was Monday
No of days = Jan. + Feb. + March
30 + 28 + 15 = 73 days
Starting with 1988, we go on counting thye number of odd days till the sum is divisible by 7
Year 1988 odd days = 2
Year 1989 odd days = 1
Year 1990 odd days = 1
Year 1991 odd days = 1
Year 1992 odd days = 2
So total odd days = 7;
? Calendar for 1993 is the same as that of 1988.
A leap year has (52 week + 2 days).
So, the number of odd days in a leap year is 2
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