pappueshwar wrote:At a certain food stand, the price of each apple is $40 and the price of each orange is $60. Mary selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean) price of the 10 pieces of fruit is $56. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is $52?
A. 1 B. 2 C. 3 D. 4 E. 5
pls assist
Let x represent the number of apples and 10-x the number of oranges...
[40x + 60(10-x)]/10 = 56
(40x+600-60x)/10 = 56
-2x+60 = 56
x=2 apples and therefore 8 oranges
Double checking...
We get 2*40+8*60 = 560 -> average is 56 GOOD
Now, for the second part of the problem...
Let x denote the number of apples put back
Let y denote the number of oranges put back.
They DO tell us the final average, so let's use that to come up with one formula or x and y:
52 = (560-40x-60y)/(10-x-y)
520-52x-52y = 560-40x-60y
8y-12x = 40
2y-3x = 10
y = (10+3x)/2
The question asks how many oranges need to be put back, so we can assume that there are NO apples put back, which means that x = 0. This allows us to quickly solve for y.
y = 5, so choice E
Now, let's assume that the question read: "How many pieces of fruit did she put back?" In other words, if they don't specify apples or organges. Well, it turns out that we can still narrow it down, based on the choices.
In this case, since we can't really come up with another formula for x and y now, let's work with our tighest constraint. Since we only have 2 apples to start with, there are only three posible values for x (or how many apples we can put back): 0, 1, and 2. Let's plug in these values to see what we get:
a.) x = 0
y = 5, this is a choice so we're done (but I'll do the other two for completeness)
b.) x = 1
y=6.5, which doesn't work because we can't split fruit in half
c.) x = 2
y = 8, which brings the total to 10 pieces, which is NOT a choice
Again, the answer is E
