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Re: Glass A and glass B both contain a mixture of alcohol and water. The [#permalink]
Ratio of alcohol and water in A = 1/3
Ratio of alcohol and water in B = 1/9

Since each contains 40 ounces, the more appropriate representation of ratio is
A=10/30 (10 ounces alcohol, 30 ounces water)
B=4/36 (4 ounces alcohol, 36 ounces water)

Now, 8 ounces (2+6) are taken from A and poured in B.
New ratios: A=8/24;B=6/42

Again, 16 ounces (2+14) are taken from B and poured in A.
New ratios: A=10/38;B=4/28.

Therefore, the new ratio of alcohol and water in A = 10/38=5/19.

Answer is 5/19.Option(D).
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Re: Glass A and glass B both contain a mixture of alcohol and water. The [#permalink]
total qty ; 40 each in A & B
so Alcohol in A ; 1/4*40 ; 10 and water ; 30
& alcohol in B ; 1/10 * 40 ; 4 and water ; 36
now given that 8 ounces is transferred from A to B
i.e 8 *1/4 ; 2 parts Alcohol and 6 parts water
which makes solution B ; 6 parts alcohol and 42 parts water ; total 48
6/42 ; 1:7 ;
now 16 ounces is added to A solution i.e 8 parts alcohol + 24 parts water
16 ounces comprises of 16*1/8 ; 2 parts alchol and 2*7 ; 14 parts water
which will make solution A; 8+2 ; 10 and 24+14 ; 38
10:38 ; 5:19
OPTION C

Glass A and glass B both contain a mixture of alcohol and water. The ratio of alcohol to water in glass A is 1 to 3. The ratio of alcohol to water in glass B is 1 to 9. 8 ounces of the mixture in glass A is taken and mixed into glass B. Then, 16 ounces of the mixture in glass B is taken and mixed into glass A. If both glasses originally contained 40 ounces of mixture, what is the final ratio of alcohol to water in glass A?

A. 1:7
B. 2:7
C. 1:3
D. 5:19
E. 10:27
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Glass A and glass B both contain a mixture of alcohol and water. The [#permalink]
Original composition of A= Water-10 Alcohol-30 [ratio=1:3]

Original composition of B= Water-4 Alcohol-36[ratio=1:9]

First shift (A to B)-

2 units of water and 6 units of alcohol will be moved from A to B.

New composition of B- 6 units of water and 42 units of alcohol [new ratio-1:7]


Second shift (B to A)-

2 units of water and 14 units of Alcohol [total 16 units in ratio of 1:7]

New composition of A= 10-2+2= 8 units of water and 30-6+14=38 units of alcohol.
New ratio= 10:38 or 5:19

Correct answer is D
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Re: Glass A and glass B both contain a mixture of alcohol and water. The [#permalink]
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Step 1:
Mixture A: 10 ounces alcohol, 30 ounces water
Mixture B: 4 ounces alcohol, 36 ounces water

Step 2:
Since removal of mixture will cause a decline in the amounts of alcohol and water in their respective ratios in the mixture
Mixture A: 8 ounces alcohol, 24 ounces water
Mixture B: 6 ounces alcohol, 42 ounces water

Step 3:
Since removal of mixture will cause a decline in the amounts of alcohol and water in their respective ratios in the mixture
Mixture A: 10 ounces alcohol, 38 ounces water
Mixture B: 4 ounces alcohol, 28 ounces water

Final Ratio: 10:38 = 5:19
Answer: D
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Re: Glass A and glass B both contain a mixture of alcohol and water. The [#permalink]
1
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The ratio of alcohol to water in glass A is 1 to 3
The ratio of alcohol to water in glass B is 1 to 9
Both glasses originally contained 40 ounces of mixture

A: 10 - 30 = 40
B: 4 - 36 = 40
8 ounces of the mixture in glass A is taken and mixed into glass B
A: 8 - 24 = 32
B: 6 - 42 = 48
16 ounces of the mixture in glass B is taken and mixed into glass A
A: 10 - 38 = 48 -> 5A to 19W = final ratio of alcohol to water in glass A
B: 4 - 28 = 32

Answer -> D
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Re: Glass A and glass B both contain a mixture of alcohol and water. The [#permalink]
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Top Contributor
Initial Amounts in A

Alcohol = \(\frac{1}{4}\) * 40 = 10 ounces

Water = \(\frac{3}{4}\) * 40 = 30 ounces

Initial Amounts in B

Alcohol= \(\frac{1}{10}\) * 40 = 4 ounces

Water = \(\frac{9}{10}\) * 40 = 36 ounces


Round 1:

8 ounces of liquid from A is transferred to B.

Alcohol in 8 ounces = \(\frac{1}{4}\) * 8 = 2 ounces

Water in 8 ounces = \(\frac{3}{4}\) * 8 = 6 ounces water.

Remaining liquid in A = 10 - 2 = 8 ounces of Alcohol and 30 - 6 = 24 ounces of Water. (Ratio is 1 :3)

Quantities in B after transfer = 4 + 2 = 6 ounces of alcohol and 36 + 6 = 42 ounces of water. (Ratio = 1 :7)


Round 2:

16 ounces of the 1:7 ratio liquid from B is transferred to A

Amount of Alcohol in 16 ounces = \(\frac{1}{8}\) * 16 = 2 ounces

Water in 16 ounces = \(\frac{7}{8}\) * 16 = 14 ounces


Amounts of liquid in A after the transfer:

Alcohol = 8 + 2 = 10 ounces

Water = 24 + 14 = 38 ounces

Therefore the ratio of Alcohol : Water in A = 10 : 38 = 5 : 19


Option D

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Re: Glass A and glass B both contain a mixture of alcohol and water. The [#permalink]
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Glass A and glass B both contain a mixture of alcohol and water. The ratio of alcohol to water in glass A is 1 to 3. The ratio of alcohol to water in glass B is 1 to 9. 8 ounces of the mixture in glass A is taken and mixed into glass B. Then, 16 ounces of the mixture in glass B is taken and mixed into glass A. If both glasses originally contained 40 ounces of mixture, what is the final ratio of alcohol to water in glass A?

A. 1:7
B. 2:7
C. 1:3
D. 5:19
E. 10:27

A glass:
1:3 -> 8 ounces
2:6

B glass:
1:9 -> 40 ounces
4:36 + 2:6
6:42
1:7 -> 48 ounces

2:14 -> 16 ounces
into glass A

1:3 -> 32 ounces
8:24 -> 32 ounces
10:38 ->48 ounces
5:19

Ans D
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Re: Glass A and glass B both contain a mixture of alcohol and water. The [#permalink]
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Re: Glass A and glass B both contain a mixture of alcohol and water. The [#permalink]
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