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\(\frac{[a*b]}{c}\)= a*b-c

\(\frac{[3*4]}{5}\)= 3*4-5 => 12-5 => 7.

\(\frac{[5*6]}{7}\)= 5*6-7 => 30-7 => 23.

Now \(\frac{[3*4]}{5}\)+\(\frac{[5*6]}{7}\)= 23+7 => 30.

Therefore, the OA is B.
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Bunuel
If \([\frac{a.b}{c}]\) is defined as being equal to \(ab – c\), then \([\frac{3.4}{5}] + [\frac{5.6}{7}]\) is equal to

(A) 40
(B) 30
(C) 15
(D) 11
(E) 6


\(\frac{[3.4]}{5}\) = \((3*4) - 5 = 12 - 5 = 7\)

\(\frac{[5.6]}{7}\) = \((5*6) - 7 = 30 - 7 = 23\)


So, \(23 + 7 = 30\)

The answer is \(B\)
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Bunuel
If \([\frac{a.b}{c}]\) is defined as being equal to \(ab – c\), then \([\frac{3.4}{5}] + [\frac{5.6}{7}]\) is equal to

(A) 40
(B) 30
(C) 15
(D) 11
(E) 6
Solution:

[3.4/5] + [5.6/7]

= (3 x 4 - 5) + (5 x 6 - 7)

= 7 + 23

= 30

Answer: B
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