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(1)
average = (5f+7s+11e)/(f+s+e) = (10e+7s+11e)/(2e+s+e) = (21e+7s)/(3e+s) = 7

Sufficient

(2)
e=2t
average = (5f+7s+11e)/(f+s+e) = (20t+14t+22t)/8t = 56t/8t = 7

Sufficient

The correct answer is D
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Let,
number of 5lb boxes be x
number of 7lb boxes = y
number of 11lb boxes = z

Mean =(5x + 7y + 11z)/(x + y + z).......A

(1) x = 2z

Applying this in A, we get (21z + 7y)/(3z + y) which gives us 7 (sneaky!)

(1) is sufficient

(2) x : y : z = 4 : 2 : 2

Assuming the constant of this proportion as a,

x = 4a; y = 2a & z = 2a

Applying these in the equation A, we get 56a/8a which gives us 7.

Thus, (2) is sufficient too.

Answer is (D)
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Statement 1:
only ratio of 5 pounds and 11 pounds is given. Can't determine weighted average unless 7 pounds one is mentioned
Insufficient

Statement 2:
all ratios given.

Sufficient

Ans: B
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