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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!

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2 Answers 2

up vote 10 down vote accepted

This seems likely:

  1. $$7^2 + 5^2 = 74$$

  2. $$5^2 + 21^2 = 466$$

  3. $$21^2 + 7^2 = \boxed{490}$$

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Wow! I don't know why I didn't think of it before!! Thanks :) –  Gaurang Tandon Mar 29 at 9:48
    
@GaurangTandon This is how I found it: I guessed it's increasing beacuse it's $(5, 21)$ is way bigger than $(7,5)$, so the only option would be $490$, that's suspicious, because $7^2=49$. So I guessed sum of $2$ squares (the lucky coincidence that $21=3\cdot7$ helped). –  user2345215 Mar 29 at 12:53
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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,

$28x=490-210=280 \rightarrow x=10$

$21*\color{red}{(20)}+7\color{red}{(10)}=490$

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It's good to have many options, thanks! –  Gaurang Tandon Mar 29 at 15:23
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