A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?
(1) There are six more heavier pumpkins than lighter pumpkins.
(2) There are four times fewer lighter pumpkins than heavier ones.
A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
D. Either statement BY ITSELF is sufficient to answer the question.
E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.
(B) The original statement says the average weight of a pumpkin is 12 pounds. Since x pumpkins weigh 10 pounds each, the y pumpkins must be heavier. If the y pumpkins weighed less than 12, then the average weight of all the pumpkins would be less than 12 as well, because all the pumpkins would simply weigh less than 12.
The equation for the average weight of the pumpkins is (10x + ry) / (x + y) = 12. Solve for r.
10x + ry = 12x + 12y
r = (2x + 12y) / y
r = 12 + 2x/y
Statement (1) tells us y = x + 6 . Plug it in the formula for r:
r = 12 + 2x/(x + 6)
Different values of x yield different values of r. For example the two following situations are possible: x = 2, y = 8, r = 12.5 and x = 6, y = 12, r = 13. Thus statement (1) is not sufficient.
Statement (2) tells us x = y/4 . Rewrite it as y = 4x and plug into the formula for r.
r = 12 + 2x/4x
r = 12.5
The value of r is definite, so statement (2) is sufficient.
The correct answer is B.
Your answer is incorrect!
From (1) => 10X + RY = 12X + 12Y => (R – 12)Y = 2X, since X, Y, R are positive, thus
=> R – 12 > 0 => R > 12 => R = 10 + 6 = 16
Therefore, (1) by itself is sufficient. => A is the right answer.