Y 3u =96
Then later 2u also 96???

In the end, all have 3u...agree? Refer to A have and ate ¼. Ate 1u left 3u.

B at first have 3u since he gave D ⅓. Give 1u left 2u (this 2u is in the end B have =96)... So, this is ⅔ of B have, is 2u).. may be I should use ⅔ .

AC Lim’s workings are correct.

At first...
A have 128
B have 72
C have 144
D have 48
Total 392.

Since C gave D ⅓, so 144 × ⅓ = 48
C left ⅔ = 144-48 =96...

This is what I would do.

A eats 1/4 of his cookies.
B collects 24 cookies.
C gives away 1/3 of his cookies to D.

Now all of them have the same number of cookies.

The easiest to start with is C.

Let the number of cookies by C be 3 units. Of course, giving away 1/3 of C’s cookies means C gives away 1 unit of cookies. Now, both C and D have 2 units of cookies each, so D must have started out with 1 unit of cookies.

This means that A must have has 2 cookies at the end. However, we soon discover that it is hard to find A’s total number of cookies because 2 units is not divisible by 3.

Let’s redefine the problem (to be multiplied by 3 at the start) such that C has 9 units of cookies at first. After C gives away 1/3 of his cookies, or in other words, 3 units of cookies, C would have 6 units of cookies.

D must have had 3 units of cookies at the start because he gained 3 units of cookies from C to have 6 units of cookies at the end.

Now, A and B must have had 6 units of cookies at the end as well.

This is after A eats 1/4 of his cookies. He must have started out with 8 units of cookies.

B must have started out with (6 units - 24) cookies.

Let’s put them together.

A has 8 units of cookies.
B has (6 units - 24) of cookies.
C has 9 units of cookies.
D has 3 units of cookies.

These total (26 units - 24) of cookies, equaling 392 cookies.

Therefore,
26 units = 416 cookies
1 unit = 16 cookies

C has 9 units of cookies, or 144 cookies at first.