0.36 = 36/100 = 9/25.

Sum of numerator and denominator is 9 + 25 = 34

Let ten's digit be x. Then, unit's digit = 2x + 1.

[10x + (2x + 1)] - [{10 (2x + 1) + x} - {10x + (2x + 1)}] = 1

<=> (12x + 1) - (9x + 9) = 1 <=> 3x = 9, x =  3.

So, ten's digit = 3 and unit's digit = 7. Hence, original number = 37.

Speed = `[142/6]` m/sec

= `[23.6 ** 18/5]` km/hr 

= 84.9 km/hr

Here Number of odd days in 1600 years = 0

Number of odd days in 300 years from 1600 to 1900 = 5*3 = 2 week + 1 odd day= 1 odd day

Number of odd days in 84 years= 21 leap year + 63 days = 21*2 + 63*1 = 105 days = 0 odd days

Number of odd days in 20 may = 31 days of Jan. + 28 days of feb + 31 days of mar. + 30 days in april + 20 days in may = 140 days = 0 odd day

So total number of odd days = 0+1+0+0=1 = Monday

Let speed of train be x km/hr and of car be y km/hr.

Then, `120/x + 480/y = 8`

=> `1/x + 4/y = 1/5`    --- (i)

And, `200/x + 400/y = 25/3`

=> `1/x + 2/y = 1/24`    --- (ii)

Solving (i) and (ii), we get x = 60 and y = 80.

Ratio of speeds = 60: 80 = 3:4

Consider Ayisha's present age = x
Then her far's age = 6x

Given that Ayisha 's far's age will be twice age of Shankar's age after 10 years
=> Shankar's age after 10 years = `1/2`(6x + 10) = 3x + 5

Also given that Shankar's eight birthdays was celebrated two years before
So his current age is 8 + 2 = 10 years old
=> Shankar's age after 10 years = 10 + 10 = 20
=> 3x + 5 = 20
=> x = 5

Therefore, Ayisha 's present age = 5 years

This can be illustrated with an example.

Ex: 1896 is a leap year.The next leap year comes in 1904 (1900 is not a leap year).

To eliminate this error, Gregorian calendar stipulates that a year that is evenly divisible by 100 (for example, 1900) is a leap year only if it is also evenly divisible by 400. This is because y are evenly divisible by 100 but not by 400.

Remainder of `((67^99)/7)`==R==>`((63+4)^99)/7`

63 is divisible by 7 for any power, so required remainder will depend on power of 4.

Require remainder:
`(4^99)/7==R==>(4^(96+3))/7`
`4^3/7==> 64/7==> (63+1)/7==R==>1`

Let speed in still water be x and speed of stream be y

(Speed in still water) : (Speed of stream)=x:y

Distance is same, n time taken downstream is t, and time taken to travel upstream is 3t.

d=d

`[(x+y) / t] = [(x-y) / (3t)]`

 x+y= 3x-3y

2x = 4y

`x/y = 4/2 = 2/1`

x:y = 2:1

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005.

So it has 1 odd day only.

The day on 6th March, 2005 will be 1 day beyond day on 6th March, 2004.

Given that, 6th March, 2005 is Monday.

6th March, 2004 is Sunday (1 day before to 6th March, 2005).