If 6 @ 4 @ 7 = 101 and 2 @ 5 @ 11 = 150, then what is the value of A in A @ 8 @ 9 = 289?
The pattern followed is that sum of squares of the number is written on the right.
Eg = $$6^2+4^2+7^2=36+16+49=101$$
and $$2^2+5^2+11^2=4+25+121=150$$
Similarly, $$A^2+8^2+9^2=289$$
=> $$A^2+64+81=289$$
=> $$A^2=289-145=144$$
=> $$A=\sqrt{144}=12$$
=> Ans - (C)
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