PLS HELP ME

Question

Four children shared a sum of money equally. Then A gave B 1/3 of his money. B then gave C 1/7 of his money.C then gave D 3/5 and $50.D had $770 in the end

1)How much did each child have at first?

2)How much money did Thomas have in the end?

Answer 1

Call x the amount of money that each child starts with.

 

A starts with x but after step1 (A gives to B), he has 2/3 x and B has 4/3 x.

 

In step 2 (B gives to C), B starts with 4/3 x and gives 1/7 of that to C.  1/7 of 4/3 x is  4/21 x.  So now B has 6/7 of what he had at the beginning of the step, which is 6/7 * 4/3 x = 24/21 x = 8/7 x.  And C has x + 4/21 x = 25/21 x.

 

In step 3, (C gives to D), C gives 3/5 of what he has to D, plus $50, which is 3/5 * 25/21 x + 50 = 5/7 x + 50.  So C is left with 25/21 x - (5/7 x + 50) = (25/21 - 5/7) x - 50 = 10/21 x - 50.  And D has x + 5/7 x + 50 = 12/7 x + 50

 

So now we have an equation:

 

12/7 x + 50 = 770

12/7 x = 720

x = 720 * 7/12 = 60 * 7 = 420

 

So each child started with $420.

 

Who's Thomas???