Happy Birthday!

You are trying to make this as confusing as possible. Anyway

Define the following variables:

$x$ is the number of guests

$y$ is the number of kids

$z$ is the number of adults

$c$ is the number of chocolates

Now, we know that:

$x=y+z \implies$ Eq.(1)

$c=3x\implies$ Eq.(2)

$y=2z\implies$ Eq.(3)

$c=4y+10\implies$ Eq.(4)

We now have 4 linear equations with 4 variables.

Eq.(2) $=$ Eq.(4):

$3x=4y+10 \implies$ Eq.(5)

Eq.(3) $- 2 \times$ Eq.(1):

$y-2x = 2z - 2(y+z)\\ y-2x=-2y$

$2x=3y \implies$ Eq.(6)

$3 \times$ Eq.(5) $-4 \times$ Eq.(6):

$9x - 8x = 3(4y+10) - 12y\\ x=\boxed{30}$

Let the no.of kids be 'K' & the no.of adults be 'A'.

No.of guests, G = A+K

It is given that the no.of kids is twice the no.of adults. Therefore, K=2A or A=K/2

No.of birthday caps = G = A+ K

No.of chocolates = thrice the no.of birthday caps = 3G = 3 (A+K) =3(K/2 + K) -eqn (1)

4 chocolates was given to each kid. Therefore total no.of chocolates given to kids=4K

Total no.of chocolates given to adults=10

Total no.of chocolates= 4K +10

Or, 4K+10=3(K/2+K) -from (1)

8K + 20= 9K

K=20

A=K/2=10

G=A+K=30

Hence, the number of guests invited is 30.

Start with the info number of kids is double the number of adults. Let adults be $x$ and kids be $2x$ .

Total caps= $3x$ , total chocolates= $3$ ×caps= $9x$

Each child gets $4$ chocolates so $4×2x=8x$ chocolates are given to kids. Left $x$ chocolates are given to adults which is equal to number of food items which is equal to $10$ .

So $x=10$ .

Total guests $=3x=\color{#D61F06}\boxed {30}$