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### HideShow timer Statistics 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
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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

On the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade;
On the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade;

So, the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3. Naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3.

We are told that "the revenue from selling the orangeade was the same for both days" so the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day.

Say the price of the glass of the orangeade on the second day was $x then 2*0.6=3*x --> x=$0.4.

Hope it's clear.
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Bunuel

Is it ok to plugin for this sum.

100 ltrs of the orangeade= 50 W + 50 Concentrate
150 ltrs of the orangeade-= 100W + 50 concentrate

since total revenue is the same
100*.6 = 150* x
thus x = 0.4

Can i take the above approach?
##### General Discussion
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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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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

Hope it clarifies
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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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kanusha 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

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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I also like this question. I just wanted to post another way to think about this question.

The first day was a ratio of 1:1, which gives a total of 2 parts: 1L + 1W = 2X
The second day was a ratio of 1:2, which gives a total of 3 parts: 1L + 2W = 3X

Since the revenue was the same for both says, and since there was a quantity increase of 50% (3 is an increase of 50% from 2), you can think of the answer this way:

What number when increased by 50% equals .6? That number is .4

Also, as previously mentioned, since the increase was 3/2, you can also multiply .6 by 2/3, which is the inverse of 3/2. Admittedly, this method is probably better. I just wanted to point out a different method / different way to think about the problem.
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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

1o +1w = 2j*0.60=1.20
1o + 2w= 3j*0.40=1.20
the correct option is D
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when people are saying of multiplying by the inverse, do we do this because the price was the same?
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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.

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Attached is a visual that should help. Part of the key here is realizing that a "glass" is a constant, and that you can insert numbers to "make it true".
Attachments Screen Shot 2016-12-07 at 7.06.07 PM.png [ 144.29 KiB | Viewed 9120 times ]

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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

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

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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

g=number of glasses sold day 1
x=price per glass day 2
r=revenue both days
g*.6=r
3/2*g*x=r
3/2*g*x=g*.6
x=$.40 D VP  D Joined: 09 Mar 2016 Posts: 1273 On a certain day, orangeade was made by mixing a certain [#permalink] ### Show Tags 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 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 VP  D Joined: 09 Mar 2016 Posts: 1273 Re: On a certain day, orangeade was made by mixing a certain [#permalink] ### Show Tags Bunuel wrote: 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. $0.45 On the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade; On the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade; So, the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3. Naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3. We are told that "the revenue from selling the orangeade was the same for both days" so the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day. Say the price of the glass of the orangeade on the second day was$x then 2*0.6=3*x --> x=$0.4. Answer: D. Hope it's clear. Bunuel can you explain the logic when we receive the ratio, know the price of glass 0.60 Say the price of the glass of the orangeade on the second day was$x then 2*0.6=3*x --> x=$0.4. I dont get why are you multiplying 0.60 by 2 ( 0.60 costs one glass and not two litres ) right ? if 2 were number of glasses than it would be ok.... Veritas Prep GMAT Instructor D Joined: 16 Oct 2010 Posts: 9446 Location: Pune, India Re: On a certain day, orangeade was made by mixing a certain [#permalink] ### Show Tags 1 dave13 wrote: 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 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 _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Intern  B Joined: 16 Sep 2017 Posts: 5 Location: United States GPA: 4 On a certain day, orangeade was made by mixing a certain [#permalink] ### Show Tags 1 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 LET THE TOTAL QUANTITY OF THE MIXTURE BE 'X' FIRST DAY: AMOUNT OF ORANGEADE = 1/2X SECOND DAY: AMOUNT OF ORANGEADE = 1/3X [ SINCE 2 PARTS ARE WATER AND ONE PART IS ORANGEADE ] SINCE THE AMOUNT QUANTIFIES ONLY TO THE ORANGEADE WHILE WATER IS CONSIDERED TO HAVE NO COST - FIRST DAY : (1/2) X = 0.60 THEREFORE, SECOND DAY : (1/3) X = (0.60 * 2)/3 = 0.40 [ CHOICE D ] Manager  B Joined: 16 May 2017 Posts: 51 GPA: 3.8 WE: Medicine and Health (Health Care) Re: On a certain day, orangeade was made by mixing a certain [#permalink] ### Show Tags Let the quantity in volume be "x". First case revenue=2x*0.60 (orange:water=1:1 and price per cup is 0.60) Second case revenue=3x*P (orange:water=1:2 and price per cup is "P") As they have mentioned revenue is same, 2x*0.60=3x*P P=0.40 Answer (D) Senior Manager  G Joined: 04 Aug 2010 Posts: 435 Schools: Dartmouth College Re: On a certain day, orangeade was made by mixing a certain [#permalink] ### Show Tags 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 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 $$= \frac{120}{30} = 40$$ cents.

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