Question

Here help me Solve please

Answer 1

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$\large\mathsf\red{Some \ Assumptions \ -}$

$\mathsf{Let,}$

$\mathsf{King's \ photo \ = \ x}$

$\mathsf{Queen's \ photo \ = \ y}$

$\mathsf{Minister's \ photo \ = \ z}$

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$\large\mathsf{A/q}$

$\mathsf{x × x \ = \ 16}$ _____(1)

$\mathsf{x × y × y \ = \ 36}$ ____(2)

$\mathsf{x × y × z \ = \ 72}$ ____(3)

$\large\mathsf\green{Solving \ (1)}$

$\mathsf{x^2 \ = \ 16}$

$\mathsf{x^2 \ = \ 4^2}$

$\mathsf\pink{\boxed{x \ = \ 4}}$ ____(4)

$\large\mathsf\green{Solving \ (2)}$

$\mathsf{x × y^2 \ = \ 36}$

$\mathsf{4 × y^2 \ = \ 36 \ [From \ (4)]}$

$\mathsf{y^2 \ = \ 9}$

$\mathsf{y^2 \ = \ 3^2}$

$\mathsf\pink{\boxed{y \ = \ 3}}$ _____(5)

$\large\mathsf\green{Solving \ (3)}$

$\mathsf{x × y × z \ = \ 72}$

$\mathsf{4 × 3 × z \ = \ 72 \ [From \ (4) \ and \ (5)]}$

$\mathsf{12 × z \ = \ 72}$

$\mathsf\pink{\boxed{z \ = \ 6}}$

$\mathsf{Now,}$

$\mathsf{x + y + z \ -}$

$\mathsf{= \ 4 + 3 + 6}$

$\mathsf{= \ 13}$

$\huge\mathsf\blue{\boxed{x + y + z \ = \ 13}}$

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Answer 2

Here is your answer...

Let the King be K, Queen be Quite and the Joker be J.

K x K=16
K^2=16
K=4

K x Q x Q=36
4xQ^2=36
Q^2=36/4
Q^2=9
Q=3

K x Q x J=72
4x3xJ=72
12xJ=72
J=72/12
J=6

K+Q+J=?
4+3+6=13

Hope it's help you.