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

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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
[Reveal] Spoiler: OA
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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.

Answer: D.

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

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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?
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New post 07 Jun 2012, 00:32
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

Answer is D

Hope it clarifies
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New post 23 Aug 2014, 02:20
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 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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New post 13 Dec 2014, 16:11
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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New post 11 Sep 2015, 07:32
when people are saying of multiplying by the inverse, do we do this because the price was the same?
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New post 02 May 2016, 08:27
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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On a certain day, orangeade was made by mixing a certain [#permalink]

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New post Updated on: 07 Dec 2016, 20:08
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
Screen Shot 2016-12-07 at 7.06.07 PM.png [ 144.29 KiB | Viewed 5375 times ]


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Originally posted by mcelroytutoring on 17 May 2016, 19:25.
Last edited by mcelroytutoring on 07 Dec 2016, 20:08, edited 5 times in total.
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Re: On a certain day, orangeade was made by mixing a certain [#permalink]

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New post 18 May 2016, 02:23
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

Answer (D)
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New post 09 Jun 2017, 20:01
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
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New post 08 Apr 2018, 05:50
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 :?
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New post 08 Apr 2018, 06:11
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....
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Re: On a certain day, orangeade was made by mixing a certain [#permalink]

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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



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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