On a certain day, orangeade was made by mixing a certain

Lets say on 1st day we can make 2 glasses of mixture with 1:1 ratio, There fore cost of 2 glasses of juice = 2 X .60
On the second day we will have 3 glasses of juice since we are mixing in a 1:2 ratio. Cost = 3 X P (P=cost per glass)

Since revenue is same we can equate both cost, ie 2X .60 = 3 X P => P = .40
Hence Answer D

Lets say two glasses (one glass juice + one glass water) were sold on the first day at 0.6$ per glass.
Revenue on first day = 1.2$
So, revenue on second day = 1.2$
Three glass (one glass juice + 2 glasses water) were sold on second day.
So, price per glass = 1.2/3 = 0.40$
Answer is D

Since solution by picking numbers is already given by many, hence to add to it the algebric solution is given below:

Assume volume of Orange and water added on first day is each V.
Hence total orangeade on day 1=V+V = 2V

On day 2, volume of water is doubled,
Hence total orangeade on day 2 = V + 2V = 3V

Assume, volume of 1 glass = G

For given question, if reveneue on Day 1 and Day 2 are same that would mean:

2 V * P1/ G = 3V *P2/G,
where P1 and P2 are price per glass on Day1 and Day2 respectively.

=> P2 = 2/3 *P1
We have P1 = 0.60 hence, P2 = 2/3*0.60 = 0.40

Ans D

Let total orangeade on first day be 2x units.
On second day,orangeade made will be 3x units.

R1=2x*.60=1.20x
R2=3x*T(say)

R1=R2
T=1.20x/3x=.40 per glass

Ans. D

First day:
a + a= 2a
Total revenue= 0.60*2a

Second day:
a+2a= 3a
Let revenue = x


0.60*2a= x*3a
x= 0.40

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 as a ratio using a variable multiplier:

W : OJ = x : x

Thus, amount of orangeade = amount of water + amount of OJ = x + x = 2x.

We are next given that orangeade is made on Day 2 by mixing the SAME AMOUNT of orange juice with TWICE THE AMOUNT of water. Again, we can set this information up as a ratio using a variable multiplier:

W : OJ = 2x : x

Thus, amount of orangeade = 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)

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)(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

Check out our video solution to this problem here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#-soluti ... olving_208

Hi All,

This question is really wordy – and gives us a lot of information to keep track of - but it gives us that information in a logical order, so we just have to be diligent about taking notes and we can get everyone on our note-pad in one read-through.

We’re told:
1) On the 1st day, Orangeade is made by mixing a certain amount of orange juice with an EQUAL amount of water.
2) On the 2nd day, Orangeade was made by mixing the SAME amount of orange juice with TWICE the amount of water.
3) ALL of the Orangeade was sold on both days AND the revenue was the SAME on both days.
4) On the 1st day, Orangeade was $0.60 per glass

We’re asked how much the price was per glass on the 2nd day. This question can be solved with some basic Arithmetic and TESTing VALUES.

IF… on the 1st day, we mix 1 glass of orange juice with 1 glass of water, then we have 2 glasses of Orangeade… which we then sell for (2 glasses)($0.60 glass) = $1.20

On the 2nd day, we mix 1 glass of orange juice with 2 glasses of water, which gives us 3 glasses of Orangeade. Since the revenue on both days was the SAME, we ended up selling…. ($1.20)/(3 glasses) = $0.40 per glass.

Final Answer:

GMAT Assassins aren’t born, they’re made,
Rich

The mention of the price per glass of orangeade, that too, at the end of the question, can put test takers on a wild goose chase. It can make them take unnecessary variables and complicate matters, wasting time and energy in the process.

But, if you give it a thought, you will understand that the assumption on which this question works is that the glasses used in both the cases are the same, except that more glasses are needed on day two.
If that’s the case, clearly, the number of glasses will also increase at the same rate as the increase in the total volume of orangeade.

The ratio of the volumes of orangeade prepared on the two days, is 2:3. The volume has increased by 50% from day 1 to day 2. Therefore, the number of glasses will also increase by 50%.

For example, if we had 2 litres of orangeade on day 1 and hence 3 litres of orangeade on day 2; if the capacity of each glass were to be 100ml, 20 glasses would be sold on day 1 while 30 glasses would be sold on day 2.

Now, Revenue = Number of glasses * Price per glass
To keep the revenue constant, increase in the number of glasses should correspond with a decrease in the price per glass.

Since number of glasses on day 2 = \(\frac{3}{2}\) (number of glasses on day 1),

Price per glass on day 2 = \(\frac{2}{3}\) (price per glass on day 1)

Substituting the value of price per glass on day 1, Price per glass on day 2 = 0.40

The correct answer option is D.

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