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A drink contains only 7oz. of soda and 3 oz. of juice. A

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A drink contains only 7oz. of soda and 3 oz. of juice. A [#permalink] New post 02 Jan 2010, 08:01
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A drink contains only 7oz. of soda and 3 oz. of juice. A second drink that contains only soda and juice is poured into the first. How many oz. of liquid were poured in?

1. After the second drink is poured in, 40% of the entire beverage is juice.
2. 30% of the second drink was soda.
Manager
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Kudos [?]: 5 [0], given: 3

Re: Mixture Problem [#permalink] New post 02 Jan 2010, 13:49
Can some one solve the problem and explain why c or any other choice is the answer?
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Re: Mixture Problem [#permalink] New post 02 Jan 2010, 16:36
IMO C...

Explanation:

Original drink quantity = 7+3 = 10 oz.

Let x quanitity of second drink be added which has J% of juice in it.

Statement 1: After the second drink is poured in, 40% of the entire beverage is juice.

0.4 (10+X) = 3 + X*\frac{J}{100}which is not sufficient as J and X are unknown.

Statement 2: 30% of the second drink was soda
Given only J% of the second drink as 70% but no % of the new mixture. Hence not sufficient.

Combining 1 & 2, we get the answer as now we have J = 70 in Statement 1..

Hence C....

Please share the OA...

Cheers!
JT
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Re: Mixture Problem   [#permalink] 02 Jan 2010, 16:36
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