Q. A person X wants to distribute some pens among six children A, B, C, D, E and F. Suppose A gets twice the number of pens received by B, three times that of C, four times that of D, five times that of E and six times that of F.

Explanation : 

Person X wants to distribute some pens.

Let the number of pens received by B, C, D, E, and F be represented by b, c, d, e, and f respectively. According to the problem, A gets twice the number of pens received by B, three times that of C, four times that of D, five times that of E, and six times that of F. We can write these relationships as:

A = 2b
A = 3c
A = 4d
A = 5e
A = 6f

Now, we want to find the minimum number of pens X should buy so that each child gets an even number of pens. To do this, we can find the least common multiple (LCM) of 2, 3, 4, 5, and 6. The LCM is the smallest number that is a multiple of all these numbers.

LCM(2, 3, 4, 5, 6) = 60

So, A must receive at least 60 pens to satisfy the conditions.

Using the relationships above, we can find the number of pens for each child:

A = 60
B = A/2 = 60/2 = 30
C = A/3 = 60/3 = 20
D = A/4 = 60/4 = 15
E = A/5 = 60/5 = 12
F = A/6 = 60/6 = 10

However, we need to make sure that all the children receive an even number of pens. The only odd number of pens is given to D, so we must adjust the number of pens for A to make sure D also gets an even number of pens.

If we increase A to 120 pens (the next multiple of 60), we can calculate the new numbers of pens for each child:

A = 120
B = A/2 = 120/2 = 60
C = A/3 = 120/3 = 40
D = A/4 = 120/4 = 30
E = A/5 = 120/5 = 24
F = A/6 = 120/6 = 20

Now all the children have an even number of pens. The minimum number of pens X should buy is:

Total pens = A + B + C + D + E + F
Total pens = 120 + 60 + 40 + 30 + 24 + 20
Total pens = 294

So, the correct answer is c. 294.