The question is:
I thought of something like:
1st box: $(7-5)*(7+5)+50=74$ Yes!
2nd box: $(21-5)*(21+5)+50 =466$ Yes!
3rd box: $(21-7)*(21+7)+50=442$ No :( That's not an option.
I have tried almost everything but I can't still have seemed to figure out the answer.
Can someone please guide me on the steps to take.
Thanks a lot!
On
Zubin's method is good, but still I will show my attempt.
For first triplet ... $5*\color{red}{12}+7*\color{red}{2}=74$
For second triplet ... $5*\color{red}{26}+21*\color{red}{16}=466$
$12-2=10=26-16$
For first triplet we used bigger number for first multiplication, for second triplet we used smaller number for first multplication, so for third triplet we will use bigger number for first multiplication.
$21*\color{red}{(x+10)}+7*\color{red}{(x)}=\text{any one of the four options}$
$21x+210+7x=\text{any one of the four options}$
$28x=\text{any one of the four options}-210$
By observation option B,C and D gives fractional answer, so choosing option A gives,
$28x=490-210=280 \rightarrow x=10$
$21*\color{red}{(20)}+7\color{red}{(10)}=490$
This seems likely:
$$7^2 + 5^2 = 74$$
$$5^2 + 21^2 = 466$$
$$21^2 + 7^2 = \boxed{490}$$