macjas wrote:
On a certain day, orangeade was made by mixing a certain amount of orange juice with an equal amount of water. On the next day, orangeade was made by mixing the same amount of orange juice with twice the amount of water. On both days, all the orangeade that was made was sold. If the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $0.60 per glass on the first day, what was the price per glass on the second day?
A. $0.15
B. $0.20
C. $0.30
D. $0.40
E. $0.45
We are given that orangeade is made on Day 1 with an EQUAL AMOUNT of water and orange juice. We can set this information up into a ratio using a variable multiplier:
W : OJ = x : x
Thus, orangeade quantity = amount of water + amount of OJ = x + x = 2x
We are next given that orangeade on Day 2 was made by mixing the SAME AMOUNT of orange juice with TWICE THE AMOUNT of water. We can set this information up into a ratio using a variable multiplier:
W : OJ = 2x : x
Thus, orangeade quantity = amount of water + amount of OJ = 2x + x = 3x
We also know that all orangeade made was sold and that the revenue on both days was the same. We can therefore set up the following equation:
Day 1 Revenue = Day 2 Revenue
That is,
(quantity sold Day 1)(price per glass Day 1) = (quantity sold Day 2)(price per glass Day 2)
Let a be the amount of orangeade a glass can hold, then
quantity sold Day 1 = the number of glasses of orangeade sold on Day 1 = 2x/a
Similarly,
quantity sold Day 2 = the number of glasses of orangeade sold on Day 2 = 3x/a
We also know that the price per glass on day 1 = $0.6
But we don’t know the price per glass on day 2, so let’s label it as variable p.
We now have:
(2x/a)(0.6) = (3x/a)(p)
Multiplying both sides by a, we have:
(2x)(0.6) = (3x)(p)
1.2x = 3xp
1.2 = 3p
p = 0.4
Thus, each glass of orangeade was sold for $0.40 on Day 2.
Answer D.
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