Hello <Kuro>

Following is my worked solution:

Let the number of children be 1u { since the chocolates(C) and sweets(S) were given to the same number of children }

Number of C given out: 4u

Number of S given out: 3u

Let the initial number of C and S be 1p and 3p respectively.

<C>

1p - 4u = 6 -- (1)

<S>

3p - 3u = 180 -- (2)

Using <simultaneous> method to solve these 2 equations, we eliminate <p> { since we need to know the number of Children }

(1)x3:

3p - 12u = 18 -- (3)

Comparing (3),(2), we take ...

(2)-(3):

-3u - (-12u) = 180 - 18

12u - 3u --> 162

9u --> 162

1u --> 18

(a)

Number of Children = 1u --> **18**

(b)

Number of S bought = 3p

in (2):

3p = 180 + 3u --> 180 + (3 × 18)

= 180 + 54

= **234**

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Trust this helps.

Do let me know again if there's further clarification .

Cheers,

Edward