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# A different mixture of vinegarand oil was 30 percent vinegar.

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Joined: 24 Mar 2019
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A different mixture of vinegarand oil was 30 percent vinegar.  [#permalink]

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24 Mar 2019, 12:07
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25% (medium)

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78% (02:26) correct 22% (02:22) wrong based on 58 sessions

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A different mixture of vinegar and oil was 30 percent vinegar. After 10 ounces of oil was added to the mixture, the resulting new mixture was only 18 percent vinegar. How many ounces were in the original mixture?

A. 15
B. 16
C. 17
D. 18
E. 19
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A different mixture of vinegarand oil was 30 percent vinegar.  [#permalink]

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Updated on: 28 Mar 2019, 11:42
1
the mixture has 30% vinegar, the oil has 0% vinegar, and we want to mix them to reach the average of 18%
so by using alligation method
the ratio of mixing of the mixture to the oil is 3:2
and if the oil is 10 ounces, so the mixture to be added to it should be 15 ounces

A
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Originally posted by MahmoudFawzy on 24 Mar 2019, 17:07.
Last edited by MahmoudFawzy on 28 Mar 2019, 11:42, edited 3 times in total.
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Re: A different mixture of vinegarand oil was 30 percent vinegar.  [#permalink]

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28 Mar 2019, 07:24
rkgp90 wrote:
A different mixture of vinegar and oil was 30 percent vinegar. After 10 ounces of oil was added to the mixture, the resulting new mixture was only 18 percent vinegar. How many ounces were in the original mixture?

A. 15
B. 16
C. 17
D. 18
E. 19

Let the quantity initial solution be Q and the quantity of vinegar in the solution 0.3Q
0.3Q/(Q+10) = 0.18 --------here 10 is the additional 10 ounces of oil that have been added to the solution and 0.18 is the proportion of vinegar in the new solution
solving the above equation, Q = 15
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Re: A different mixture of vinegarand oil was 30 percent vinegar.  [#permalink]

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28 Mar 2019, 10:59
Let's assume that the amount of mixture is x ounces.
Given that 30% of the mixture is vinegar, i.e. amount of vinegar in original mixture = (30/100)x = (3/10)x. Therefore, the amount of oil = (7/10)x
Now 10 ounces of oil was added to the mixture, the resulting new mixture was only 18% vinegar
Amount of Vinegar = (3/10)x
Amount of Oil = (7/10)x + 10
Therefore, (3/10)x/(x+10) = 18/100
3x/(x+10) = 9/5
x/(x+10) = 3/5
5x = 3x + 30
2x = 30
x = 15

Hence, the correct answer is Option A. 15

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Re: A different mixture of vinegarand oil was 30 percent vinegar.  [#permalink]

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28 Mar 2019, 16:48
rkgp90 wrote:
A different mixture of vinegar and oil was 30 percent vinegar. After 10 ounces of oil was added to the mixture, the resulting new mixture was only 18 percent vinegar. How many ounces were in the original mixture?

A. 15
B. 16
C. 17
D. 18
E. 19

let x=ounces in original mixture
.7x+10=.82(x+10)
x=15
A
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Re: A different mixture of vinegarand oil was 30 percent vinegar.  [#permalink]

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31 Mar 2019, 11:45
1
rkgp90 wrote:
A different mixture of vinegar and oil was 30 percent vinegar. After 10 ounces of oil was added to the mixture, the resulting new mixture was only 18 percent vinegar. How many ounces were in the original mixture?

A. 15
B. 16
C. 17
D. 18
E. 19

We can let n = the total number of ounces of the mixture.

Thus, 0.3n = amount of vinegar and 0.7n = amount of oil.

We are given that after 10 ounces of oil was added to the mixture, the new mixture was 18 percent vinegar. So we can create the following equation to determine n:

0.3n/(n + 10) = 18/100

0.3n x 100 = 18(n + 10)

30n = 18n + 180

12n = 180

n = 15

So originally there were 15 ounces in the mixture.

Alternate Solution:

We start with x ounces of 30% vinegar, and we add 10 ounces of 0% vinegar, to get (x + 10) ounces of 18% vinegar. We can create the equation:

0.30x + (0)(10) = (x + 10)0.18

0.30x = 0.18x + 1.8

0.12x = 1.8

12x = 180

x = 15

The initial amount of mixture was 15 ounces.

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Re: A different mixture of vinegarand oil was 30 percent vinegar.   [#permalink] 31 Mar 2019, 11:45
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