Solution? (Question:380826.0)

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Anuj Kumar

· started a discussion

· 1 Months ago

Solution? (Question:380826.0)

Where is detailed solution?

Question:

If 6 @ 4 @ 7 = 101 and 2 @ 5 @ 11 = 150, then what is the value of A in A @ 8 @ 9 = 289?

Options:

Solution:

$\begin{matrix} 6 @ 4 @ 7 = (6)^{2} + (4)^{2} + (7)^{2} = 36 + 16 + 49 = 101 \\ 2 @ 5 @ 11 = (2)^{2} + (5)^{2} + (11)^{2} = 4 + 25 + 121 = 150 \end{matrix}$

Similarly,

$\begin{matrix} A @ 8 @ 9 = 289 \\ = > A^{2} + (8)^{2} + (9)^{2} = 289 \\ = > A^{2} + 64 + 81 = 289 \\ = > A^{2} + 145 = 289 \\ = > A^{2} + 145 = 289 \\ A^{2} = 289 - 145 \\ A^{2} = 144 \\ A = \sqrt[2]{144} = 12 \end{matrix}$

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Solution? (Question:380826.0)

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