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

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prasadrg
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A drink contains only 7 oz. of soda and 3 oz. of juice. A second drink [#permalink] New post  02 Jan 2010, 08:01
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A drink contains only 7 oz. 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.
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GyMrAT
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Re: A drink contains only 7 oz. of soda and 3 oz. of juice. A second drink [#permalink] New post  03 Jun 2018, 03:23
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prasadrg wrote:
A drink contains only 7 oz. 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.


Given

Drink 1
7 oz. Soda + 3 oz. Juice = 10 oz. Drink 1

Drink 2
x oz. Soda + y oz. Juice = (x+y) oz. Drink 2

Drink 1 + Drink 2
(7+x) oz. Soda + (3+y) oz. Juice = (10+x+y) oz.

Statement 1
40 % of (10+x+y) = (3+y)

Two unknowns x & y, with only 1 equation, hence St.1 alone is Insufficient

Statement 2
x = 30% of (x+y)

Two unknowns x & y, with only 1 equation, St. 2 alone is Insufficient.

Combining St.1 & St. 2

we have two equations with two unknowns, solving we will get, the required quantity of Drink 2.

Hence St.1 & 2 together are sufficient.

Answer C.

Thanks,
GyM
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jeeteshsingh
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Re: A drink contains only 7 oz. of soda and 3 oz. of juice. A second drink [#permalink] New post  02 Jan 2010, 16:36
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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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yezz
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Re: A drink contains only 7 oz. of soda and 3 oz. of juice. A second drink [#permalink] New post  13 Nov 2016, 10:25
prasadrg wrote:
A drink contains only 7 oz. 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.


this is a weighted average problem , let W1 weight of first drink and w2 weight of 2nd drink and A2 = amount of juice in 2nd drink ( or percent) , WA is the weighted average of both drinks mix and A1 the amount of juice in 2nd drink (or percent)

w1/w2 = (A2-WA)/ (WA-A1), QUESTION ASKS FOR W2

FROM Stem we have A1 = 3/10 , W1 = 10 we need A2 and WA

FROM 1
GIVEN IS WA only .... insuff
from 2
given A2 only ... insuff

both we have A2 and WA and we can solve for w2.................C
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krish76
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Re: A drink contains only 7 oz. of soda and 3 oz. of juice. A second drink [#permalink] New post  05 Jun 2018, 17:41
took some time to figure out the weighted avg. i just learnt from bunuel's post, but i was off as soon as i thought in % terms !
I will only consider the Juice % in the soln

30% ____40% (final)______A2
10L W2

1) 10/W2=A2-40/40-30 -Two unknown NS

2) A2 = 70 (since juice is 30%) - NS

1 +2

10/W2=30/10

W2=30
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prvarm24
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Re: A drink contains only 7 oz. of soda and 3 oz. of juice. A second drink [#permalink] New post  11 Nov 2018, 10:47
Thanks GyMrAT for an easy explanation
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yashmehta27101995
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Re: A drink contains only 7 oz. of soda and 3 oz. of juice. A second drink [#permalink] New post  18 Jul 2019, 10:10
There seems to be some calculation mistake in the ans,
I get volume as 10/3 total solution added
Ans is C only
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Re: A drink contains only 7 oz. of soda and 3 oz. of juice. A second drink [#permalink]
18 Jul 2019, 10:10
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