Bunuel wrote:
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 five more heavier pumpkins than lighter pumpkins.
(2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.
Kudos for a correct solution.
800score Official Solution: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.
Writing an equation for the average weight of the pumpkins is (10x + ry) / (x + y) = 12. This equation has three unknowns.
Solve for r to get:
r = 12 + (2x/y)
Statement (1) defines the relationship between x and y as y = x + 5 or y = x – 5. Plug in the formula for r to get
r = 12 + (2(y – 5)/y)
r = 14 – 10/y
Plug in couple values of y (greater than 5) to see that they yield different results for r.
Statement (1) is not sufficient.
Statement (2) gives the relationship between y and r as r = y + 3. This isn't enough information as you can plug it into the original equation and see that too many variables remain. Plug in couple values for y to see that we get different possible valuee of r and x.
Using the given with the equations from (1) and (2), the system is
(10x + ry) / (x + y) = 12
y = x + 5
r = y + 3
Rewrite r = y + 3 as r – 3 = y.
Substitute r – 3 = y into y = x + 5 and get r – 3 = x + 5 so r = x + 8.
This gives an equation for y in terms of x, and an equation for r in terms of x.
Substitute these equations into the original equation.
(10x + ry) / (x + y) = 12. Substitute y = x + 5 and r = x + 8.
(10x + (x + 8)(x + 5)) / (x + (x + 5)) = 12
10x + (x² + 13x + 40) = 12(2x + 5)
x² + 23x + 40 = 24x + 60
x² – x – 20 = 0
(x – 5)(x + 4) = 0
x = 5 or x = -4 = 0
So there are 5 of the smaller pumpkins.
Using y = x + 5, y = 5 + 5 = 10, so there are 10 of the larger pumpkins.
And using r = y + 3, r = 13, the larger pumpkins weigh 13 pounds.