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Solution X is 10 percent alcohol by volume, and solution Y is 30 perce

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Solution X is 10 percent alcohol by volume, and solution Y is 30 perce  [#permalink]

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New post 23 Mar 2015, 06:45
5
10
00:00
A
B
C
D
E

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  15% (low)

Question Stats:

76% (01:15) correct 24% (01:38) wrong based on 350 sessions

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Solution X is 10 percent alcohol by volume, and solution Y is 30 percent alcohol by volume. How many milliliters of solution Y must be added to 200 milliliters of solution X to create a solution that is 25 percent alcohol by volume?

A. 250/3
B. 500/3
C. 400
D. 480
E. 600

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X is 10 percent alcohol by volume, and solution Y is 30 perce  [#permalink]

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New post 25 Mar 2015, 02:16
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Answer = E = 600

Using allegation method as shown in figure below:

Attachment:
alle.png
alle.png [ 3.49 KiB | Viewed 17350 times ]


To obtain a 25% alcohol mixture, X & Y are to be mixed in ratio 5:15 = 1:3

X is already 200ml, so Y should be 200*3 = 600
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Solution X is 10 percent alcohol by volume, and solution Y is 30 perce  [#permalink]

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New post Updated on: 23 Mar 2015, 07:32
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200 mL of solution X (at 10% alcohol by volume) are mixed with \(y\) mL of solution Y (at 30% by volume). We need to solve for \(y\) to find out how many mL would make this solution 25% by volume.

\(\frac{200(.1) + y(.3)}{200 + y} = .25\)

\(20 + .3y = 50 + .25y\)

\(.05y = 30\)

\(y = \frac{30}{.05} = \frac{300}{.5} = 600.\)

Answer: E

Further: A quick check shows that 600 mL of y means 180 mL alcohol, 200 mL of x means 20 mL of alcohol, and together that's \(\frac{180 + 20}{200 + 600} = \frac{200}{800} = 25%.\)

Originally posted by eaze on 23 Mar 2015, 07:31.
Last edited by eaze on 23 Mar 2015, 07:32, edited 1 time in total.
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Solution X is 10 percent alcohol by volume, and solution Y is 30 perce  [#permalink]

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New post 23 Mar 2015, 07:31
4
Bunuel wrote:
Solution X is 10 percent alcohol by volume, and solution Y is 30 percent alcohol by volume. How many milliliters of solution Y must be added to 200 milliliters of solution X to create a solution that is 25 percent alcohol by volume?

A. 250/3
B. 500/3
C. 400
D. 480
E. 600

Kudos for a correct solution.


\(10% 200 + 30% Y = 25% (200 + Y )\)
Solving for Y = 600
Answer E .

another approach is to use weighted average trick.

we know that X is 10% , Y is 30% and W.Avg = 25% . what does this mean with respect to W.Avg technique ?
W.Avg is 1 portion away from Y and 3 portion away from X so for every 1 portion of X we will have to add 3 portions of Y.
If X = 200 then Y = 600 .
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Re: Solution X is 10 percent alcohol by volume, and solution Y is 30 perce  [#permalink]

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New post 24 Mar 2015, 22:48
1
Hi All,

This is a 'Weighted Average' question, so it can be solved in a number of different ways. For most Test Takers, setting up the Weighted Average formula and doing the necessary algebra is a straight-forward way to get to the solution, but there ARE alternatives. Since the question asks for the number of milliliters of Solution Y that are needed, and the answer choices are NUMBERS, we can TEST THE ANSWERS.

From the prompt, we are told to mix 200 milliliters of a 10% alcohol solution with an unknown number of milliliters of a 30% alcohol solution to get a 25% total alcohol solution.

IF....we mixed 200 milliliters with another 200 milliliters, then we'd end up with a 20% mixture (since the average of 10% and 30% is 20%). That result would be TOO LOW, so we clearly need a lot MORE milliliters of the 30% solution. From this, we can eliminate Answers A and B from consideration.

Let's TEST Answer D....

IF...we have 480 milliliters of Solution Y, we have....

Does [.1(200) + .3(480)]/680 = .25?

Does 20 + 144 = .25(680)?

Does 164 = 170?

No it doesn't, so Answer D is NOT the answer. We likely need to add MORE of the 30% solution to get the two 'sides' to match, but we can confirm this by TESTing Answer E.

IF....we have 600 milliliters of Solution Y, we have...

Does [.1(200) + .3(600)]/800 = .25?

Does 20 + 180 = .25(800)?

Does 200 = 200?

Yes it does, so this MUST be the answer.

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Re: Solution X is 10 percent alcohol by volume, and solution Y is 30 perce  [#permalink]

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New post 25 Mar 2015, 01:07
1
10 % Alcohol of Solution X = 20
Let amount of Solution Y to be added be X

(20 + 0.3x)/(200+X) = 25/100

x = 600

Ans : E
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Re: Solution X is 10 percent alcohol by volume, and solution Y is 30 perce  [#permalink]

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New post 30 Mar 2015, 03:23
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2
Bunuel wrote:
Solution X is 10 percent alcohol by volume, and solution Y is 30 percent alcohol by volume. How many milliliters of solution Y must be added to 200 milliliters of solution X to create a solution that is 25 percent alcohol by volume?

A. 250/3
B. 500/3
C. 400
D. 480
E. 600

Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:
Attachment:
alcoholsolutionsxandy_text.PNG
alcoholsolutionsxandy_text.PNG [ 24.13 KiB | Viewed 17261 times ]

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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Solution X is 10 percent alcohol by volume, and solution Y is 30 perce  [#permalink]

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New post 30 Mar 2015, 10:32
ans 600, Using allegations method.
X : 10% : 200 Y : 30% : 'V'
Mixture: 25%
X : 5% : 200 Y : 15% : 'V'

Thus, 5/15 = 200/V => V = 200*3 = 600.
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Solution X is 10 percent alcohol by volume, and solution Y is 30 perce  [#permalink]

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New post 02 Oct 2016, 01:00
lalania1 wrote:
Solution X is 10 percent alcohol by volume, and solution Y is 30 percent alcohol by volume. How many milliliters of solution Y must be added to 200 milliliters of solution X to create a solution that is 25 percent alcohol by volume?

A) 250/3
B) 500/3
C) 400
D) 480
E) 600


Easy one.

Alcohol in 200 ml of X = 20 ml.

Alcohol in 100y ml of Y = 30y ml

Now as per question,

20 + 30y = 25% of (200 + 100y)

or y = 6.

Therefore, required quantity of solution y = 600 ml. Hence, E
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Re: Solution X is 10 percent alcohol by volume, and solution Y is 30 perce  [#permalink]

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New post 22 Oct 2017, 21:22
Bunuel wrote:
Solution X is 10 percent alcohol by volume, and solution Y is 30 percent alcohol by volume. How many milliliters of solution Y must be added to 200 milliliters of solution X to create a solution that is 25 percent alcohol by volume?

A. 250/3
B. 500/3
C. 400
D. 480
E. 600

Kudos for a correct solution.



x = 10%
y= 30%
x= 200 milliliters

.1x + .3Y = .25(x+Y)

.1(200) + .3Y = .25(200 + Y)

20 + .3Y = 50 + .25Y

0.05Y = 30

Y = 600 milliliters E
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Re: Solution X is 10 percent alcohol by volume, and solution Y is 30 perce  [#permalink]

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New post 30 Aug 2018, 20:01
Let x be the total vol.
Then .2*100+.3*(x-200)=.25x
so, x=40/.05=800 and hence 800-200=600
E is the correct choice
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Solution X is 10 percent alcohol by volume, and solution Y is 30 perce  [#permalink]

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New post 04 Sep 2018, 23:11
Using weighted average method -

Y (y ml) X (200 ml)
30 10
25 (200+y ml)
15 5
3 : 1

Therefore Y:X is in the ratio 3:1
Since X is 200 ml, Y will be 3 times 200 ml
Hence Y = 600 ml
Ans E.

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Solution X is 10 percent alcohol by volume, and solution Y is 30 perce &nbs [#permalink] 04 Sep 2018, 23:11
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