Numismatist and coins problem?

Question

A numismatist decides to divide his coin collection between his children.

The oldest gets 1/2 of the collection, the next gets 1/4, the next gets 1/5,

and the youngest gets the remaining 49 coins. What is the total number of coins?

Answer 1

n/2 + n/4 + n/5 + 49 = n
n(1 - 1/2 - 1/4 - 1/5) = 49
n(20 - 10 - 5 - 4) = 980
n = 980

n/2 = 490
n/4 = 245
n/5 = 196
...... + 49
------------
...... 980

Answer 2

The fraction of the collection given to his 3 eldest children is
1/2 + 1/4 + 1/5
= 10/20 + 5/20 + 4/20 (put everything over a common denominator)
= 19/20.
The total collection is a whole, which as a fraction is 1 = 20/20.
So the remainder is 1 - 19/20 = 1/20.

So 49 coins represents one twentieth of the collection. Therefore the total collection consists of 49 x 20 = 980 coins.

Answer 3

let total coins be n
1/2n+1/4n+1/5n+49=n
[10+5+4]n/20+49=n(taking lcm)
19/20n+49=n

hence n/20=49
and n=980

Answer 4

980