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Re: On a certain day, orangeade was made by mixing a certain [#permalink]
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Let the amount of orange juice be X.

Orangeade made on first day = X + X(same amount of water) = 2X

Orangeade made on second day = X + 2X (twice the amount of water) = 3X

So the total number of glasses made will be of the ratio 2:3

As total revenue is same on both days,

(2)(0.16) = (3)(Price on second day)

Price per glass on second day = 0.32/3 = 0.107
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Re: On a certain day, orangeade was made by mixing a certain [#permalink]
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I really like this one

So first you sold 1:1 water and orange, let's say a liter of each so total per glass is 2 liters
After you sold 2:1 water and orange, let's say a liter of each again so total per glass is 3 liters

Now since the revenue is the same and the ratio in the quantity increased to 3/2 then the price has to be the inverse 2/3

So 0.6 * 2.3 - 0.4

Answer is D

Hope it clarifies
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Re: On a certain day, orangeade was made by mixing a certain [#permalink]
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On fist day 0.60$ per glass and assume sold 10 glasses of orange = 6$
On next Day on adding double amount of water, might sold 20 glasses of orange , so 0.30$ x 20 = 6 $
So C
But My answer is wrong... Even i understood Above solutions, But why analysis is wrong
Pls help
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Re: On a certain day, orangeade was made by mixing a certain [#permalink]
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kanusha
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

On fist day 0.60$ per glass and assume sold 10 glasses of orange = 6$
On next Day on adding double amount of water, might sold 20 glasses of orange , so 0.30$ x 20 = 6 $
So C
But My answer is wrong... Even i understood Above solutions, But why analysis is wrong
Pls help

On the first day orangeade was made by mixing a certain amount of orange juice with an equal amount of water. So, if 10 glasses of orangeade was sold on the first day, then it was made by mixing 5 glasses of orange juice with 5 glasses of water.

On the next day, orangeade was made by mixing the same amount of orange juice with twice the amount of water, so it was made with 5 glasses of orange juice and 10 glasses of water, which makes total of 15 glasses of orangeade.

10*0.6 = 15*x --> x = 0.4.

Hope it's clear.
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macjas
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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Re: On a certain day, orangeade was made by mixing a certain [#permalink]
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macjas
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

Revenue = Number of glasses * Price per glass

On second day, twice the amount of water was added so number of glasses was 3/2 the number of glasses on first day. Revenue on both days was the same so

N * 0.6 = (3N/2) * Cost on day 2

Cost on day 2 = 0.4

Answer (D)
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Re: On a certain day, orangeade was made by mixing a certain [#permalink]
macjas
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


why is my reasoning wrong UPDATE :) I noticed one mistake, so I corrected it but still dont get how to reach final solution :)

first day: 1 litre of juice +1 litre of water hence ratio---> 1:1 (total 2 litres)

Second day: 1 litres of juice +2 litres of water hence ratio---> 1:2 ( total 3 litres )


If ratio of mixture on day one to mixture on day two is 2 to 3 and price per glace on day one is 0.60

so i assume from this ratio \(2:3\) that \(2\) is 0.60 and what to do next :? if \(2\) is 0.60 then what is \(3\) :? how to calculate :)

or where am i wrong :?
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macjas
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


why is my reasoning wrong UPDATE :) I noticed one mistake, so I corrected it but still dont get how to reach final solution :)

first day: 1 litre of juice +1 litre of water hence ratio---> 1:1 (total 2 litres)

Second day: 1 litres of juice +2 litres of water hence ratio---> 1:2 ( total 3 litres )


If ratio of mixture on day one to mixture on day two is 2 to 3 and price per glace on day one is 0.60

so i assume from this ratio \(2:3\) that \(2\) is 0.60 and what to do next :? if \(2\) is 0.60 then what is \(3\) :? how to calculate :)

or where am i wrong :?

Note that Revenue is the same on both the days. That is the information that will give you the relation between 2:3 and 0.60.

Revenue = Unit Price * Volume sold

UnitPrice1 * VolumeSold1 = UnitPrice2 * VolumeSold2

VolumeSold1 / VolumeSold2 = UnitPrice2 / UnitPrice1

2/3 = UnitPrice2 / 0.60

UnitPrice2 = 0.40
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On a certain day, orangeade was made by mixing a certain [#permalink]
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macjas
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

Plug in values in terms of GLASSES.

On a certain day, orangeade was made by mixing a certain amount of orange juice with an equal amount of water.
The orangeade was sold at $0.60 per glass on the first day.

Let the total orangeade = (1 glass juice) + (1 glass water) = 2 glasses.
Since each glass sells for 60 cents, the revenue = 2*60 = 120 cents.

On the next day, orangeade was made by mixing the same amount of orange juice with twice the amount of water.
Orangeade = (1 glass juice) + (2 glasses water) = 3 glasses.

The revenue from selling the orangeade was the same for both days.
What was the price per glass on the second day?

Since the revenue for the 3 glasses on the second day remains 120 cents, the price per glass \(= \frac{120}{3} = 40\) cents.

­­

Originally posted by GMATGuruNY on 14 Aug 2018, 06:03.
Last edited by GMATGuruNY on 19 Feb 2024, 04:37, edited 1 time in total.
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Re: On a certain day, orangeade was made by mixing a certain [#permalink]
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Video solution from Quant Reasoning starts at 0:27
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Re: On a certain day, orangeade was made by mixing a certain [#permalink]
First Post!
Saw this question on the 2024-2025 GMAT online question bank. None of the answers for this question seem to make sense to me. Hoping this will click with someone

Formula First:
Selling Price = Profit - Cost
Since Profit isn't mentioned, we can deduce the correlation between selling price and cost
Selling price = - Cost

Note: Also good to know:
Revenue = Profit - Total Cost
Selling Price * Total number of Units = Revenue
Cost * Total number of Units = Total Cost

First day one cup cost: 1/2 => 50% of orange juice
Selling Price = 0.6
Cost = C

Second day one cup cost: 1/3 => 33% of orange juice => 33%/50% amount of orange juice relative to day 1 = 1/2/1/3 = 2/3 C (where C is day 1 cost)
Selling Price = X (trying to solve for this)
Cost = 2/3C

Putting the two formulas together
Day 1 => 0.6 = - C
Day 2 => X = -2/3C

Find the ratio: (day 1 / day 2)
Day 1 / Day 2 = 0.6/X = -C/(-2/3C)
=> 0.6/X = 3/2
=> 0.6 = 3/2 * X
=> 0.6 / 3/2 = X
=> 0.6 * 2/3 = X
=> X = 0.4
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Re: On a certain day, orangeade was made by mixing a certain [#permalink]
­We can solve this question by following this strategy:
For example,
We sold 200 units of orangeade on day 1 @ 0.6/ unit which comes to 120 (revenue)
Revenue is same for both the days and hence we sold 300 units @ x/unit which comes to 120(revenue is same for both the days)
Hence, 300*x=120
x=0.40

Hope this helps.
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