If X = a/2^2 + b/2^3 + c/2^5, does X = 5/16 ?

Is x = 5/16 or 10/32

\(X = \frac{a}{2^2} + \frac{b}{2^3} + \frac{c}{2^5}\)
—> \(X = \frac{(8a + 4b + c)}{32}\)

(1) a, b, and c are each either 0 or 1.

When a, b, c takes either 0 or 1, we can see the numerator can never take the value 10

Possible values of
(a, b, c) = (0, 0, 0) —> X = 0/32
(a, b, c) = (0, 0, 1) —> X = 1/32
(a, b, c) = (0, 1, 0) —> X = 4/32
(a, b, c) = (1, 0, 0) —> X = 8/32
(a, b, c) = (0, 1, 1) —> X = 5/32
(a, b, c) = (1, 1, 0) —> X = 12/32
(a, b, c) = (1, 0, 1) —> X = 9/32
(a, b, c) = (1, 1, 1) —> X = 13/32

—> A definite no. X can never be 10/32
—> Sufficient

(2) The greatest possible value of X is 13/32
—> 8a + 4b + c = 13

We can adjust values of a, b & c to get the greatest value as 13 and one of the values as 10

Case 1: As per statement (1) maximum value is 13/32 and X is never equal to 10/32

Case 2: Let the Maximum values of a, b and c can be 1
—> Maximum value of X = (8+4+1)/32 = 13/32
We can also assume one possible set of values, a = 1, b = 1/2, c = 0
—> X = (8*1 + 4*1/2 + 0)/32 = 10/32

So, we get 2 outcomes —> Insufficient

IMO Option A

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