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

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08 Oct 2012, 02:01
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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) \$015
(B) \$0.20
(C) \$0.30
(D) \$0.40
(E) \$0.45

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08 Oct 2012, 02:02
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SOLUTION

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) \$015
(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.

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08 Oct 2012, 02:45
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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
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Last edited by SOURH7WK on 08 Oct 2012, 03:33, edited 1 time in total.

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08 Oct 2012, 03:07
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It is better to take smart numbers rather than getting confused in ratio's.
So lets say on Day1 = 10 liters was sold ( Orange :Water, 5:5 lt) [1 glass = 1liter]
So Revenue on Day1 = 10 * 0.6

Ratio on 2nd day ( Orange :Water, 5:10 lt) 15 lt = 15 glassess
Day1 revenue = Day2 revenue
10 * 0.6 = 15 *x
x = 0.4
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08 Oct 2012, 03:10
fameatop wrote:
It is better to take smart numbers rather than getting confused in ratio's.
So lets say on Day1 = 10 liters was sold ( Orange :Water, 5:5 lt) [1 glass = 1liter]
So Revenue on Day1 = 10 * 0.6

Ratio on 2nd day ( Orange :Water, 5:10 lt) 15 lt = 15 glassess
Day1 revenue = Day2 revenue
10 * 0.6 = 15 *x
x = 0.4

+1 D
nice explaination..!!

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08 Oct 2012, 03:41
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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\$
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08 Oct 2012, 09:58
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What a big misktake i was actually doing ..i was just assuming ..that if its .60 on first day..then .30 will be on second because of addding twwice water..

after luking at above solution ..i change took D too :D

well

10 +10=20 *.60=12

10 is orange juice and 10 litre water...

10+20=30..

10 is orange...20 litter is water ...as mentioned twice the amount..

12=30*x

12/30=4/10=.40=D=ans=:)
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09 Oct 2012, 23:48
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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

Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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) \$015
(B) \$0.20
(C) \$0.30
(D) \$0.40
(E) \$0.45

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Question: 60
Page: 160
Difficulty: 600

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10 Oct 2012, 19:34
reached D by methodof substitution..
aftern a basic algebra..
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11 Oct 2012, 13:42
SOLUTION

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) \$015
(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.

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10 Jan 2014, 01:18
For the first day, we are given: X*(OJ) + X*(W) = Orangeade ----> For each glass, X accounts for 2 units

For the second day we are given X*(OJ) + 2X*(W) = Orangeade -----> For each glass, X accounts for 3 units

If, for 2 X, the "per unit" price is 0.3 (0.6/2), then for 3 X, the "per unit" price is 0.2 (0.6/3) ---> subtract the difference from the original: 0.6 - 0.2 = 0.4

The reason we subtract the difference is because we are told that the revenue is the same for both days.

Thus, the difference in per-glass revenue is equal to subtracting 0.2 from each sale because our 3/2 units of X in day two compared to day one dilutes our "per unit of X" revenue. So, when we know the magnitute of dilution, we know the cost of a glass in day 2.

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10 Jan 2014, 02:04
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Let total orangeade on first day be 2x units.

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

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

Ans. D

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06 Jun 2014, 13:44
never mind my question

Last edited by ilpadre123 on 11 Jun 2014, 04:55, edited 1 time in total.

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11 Jun 2014, 04:20
On the 2nd the amount sold was greater 1/3 than on the 1st day --> So the price should be 1/3 lower then 0,60
0,60x2/3 = 40
or one paid 0,60 for 1 liter mix on the 1st day -> the Volume got 1/3 greater - 1,5X = 60 ; X = 40

Update 2015
It's a proportion here R = Glases*Price if we raise the # of glases by 3/2 our price should change by 2/3 to keep the revenue unchanged (in balance or equal to the revenue of the 1st day) --> 0,60*2/3=0,40
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02 May 2016, 14:28
Simple question here, hoping this helps anyone!

OJ + W = D1
OJ + 2W = D2

1 part OJ + 1 part W = 2 parts * .6 = 1.20

1 part OJ + 2 part W = 3 parts * x = 1.20
3 parts * x = 1.20
1.20 / 3 = x = .4

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03 May 2016, 20:34
assume they sell 100 glasses on day 1 at 60¢ per glass, for total revenue of \$60
on day 2 they sell (3/2)(100)=150 glasses, for same total revenue of \$60
\$60/150 glasses=40¢ per glass

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20 Apr 2017, 15:49
good one... thanks.... i by mistake counted half and marked it 0.3.............. idiot...... thanks for a great question and a clearly expressed explaination by all ... thanks again

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08 Sep 2017, 06:27
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
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11 Sep 2017, 15:53
Bunuel 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) \$015
(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 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.

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