Let the two-digit number be 10a + b
a + b = 12 --- (1)
If a>b, a - b = 6
If b>a, b - a = 6
If a - b = 6, adding it to equation (1), we get
2a = 18 => a =9
so b = 12 - a = 3
Number would be 93.
if b - a = 6, adding it to the equation (1), we get
2b = 18 => b = 9
a = 12 - b = 3.
Number would be 39.
There fore, Number would be 39 or 93.

Let the number be in the form of 10a + b
Number formed by interchanging a and b = 10b + a.
a + b = 13 --- (1)
10b + a = 10a + b - 45
45 = 9a - 9b => a - b = 5 --- (2)
Adding (1) and (2), we get
2a = 18 => a = 9 and b = 4
The number is: 94.

Let the five consecutive even numbers be 2(x - 2), 2(x - 1), 2x, 2(x + 1) and 2(x + 2)
Their sum = 10x = 440
x = 44 => 2(x - 2) = 84
Second least number of the other set = 2(84) - 121 = 47
This set has its least number as 46.
Sum of the numbers of this set = 46 + 47 + 48 + 49 + 50
= 48 - 2 + 48 - 1 + 48 + 48 + 1 + 48 + 2 => 5(48) = 240

Let the numbers be x, x + 2, x + 4, x + 6, x + 8 and x + 10.
Given (x + 2) + (x + 10) = 24
=> 2x + 12 = 24 => x = 6
The fourth number = x + 6 = 6 + 6 = 12.

(f + m + d)/3 = 35
=> f + m + d = 105 --- (1)
m + 15 = f + d
Substituting f + d as m + 15 in (1), we get
2m + 15 = 105
2m = 90 => m = 45 years.

Let the present ages of Anand and Bala be 'a' and 'b' respectively.
a - 10 = 1/3 (b - 10) --- (1)
b = a + 12
Substituting b = a + 12 in first equation,
a - 10 = 1/3 (a + 2) => 3a - 30 = a + 2
=> 2a = 32 => a = 16.

Let the number be
2 + 1/2[1/3(a/5)] = a/15
=> 2 = a/30 => a = 60

Let Mudit's present age be 'm' years.
m + 18 = 3(m - 4)
=> 2m = 30 => m = 15 years.

Let the three digit numbers be 100a + 10b + c
a = b + 2
c = b - 2
a + b + c = 3b = 18 => b = 6
So a = 8 and b = 4
Hence the three digit number is: 864

Let the four consecutive even numbers be 2(x - 2), 2(x - 1), 2x, 2(x + 1)
Their sum = 8x - 4 = 292 => x = 37
Smallest number is: 2(x - 2) = 70.