enough of probabilities let us have some mix and alligations

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Miguel is mixing up a salad dressing. Regardless of the number of servings, the recipe requires that 5/8 of the finished dressing mix be olive oil, 1/4 vinegar, and the remainder an even mixture of salt, pepper and sugar. If Miguel accidentally doubles the vinegar and forgets the sugar altogether, what proportion of the botched dressing will be olive oil?
15/29
5/8
5/16
1/2
13/27
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by Stuart@KaplanGMAT » Fri Mar 26, 2010 9:27 am
thephoenix wrote:Miguel is mixing up a salad dressing. Regardless of the number of servings, the recipe requires that 5/8 of the finished dressing mix be olive oil, 1/4 vinegar, and the remainder an even mixture of salt, pepper and sugar. If Miguel accidentally doubles the vinegar and forgets the sugar altogether, what proportion of the botched dressing will be olive oil?
15/29
5/8
5/16
1/2
13/27
On fraction questions for which we're not provided with a starting amount, picking a starting amount makes the problem more concrete.

To keep the math simple, we generally pick the smallest common denominator of all of the fractions involved. In this case we have 5/8, 1/4 and 1/24 (there's 1/8 remaining after the oil and vinegar, and we're dividing that 1/8 into 3 parts), so let's let the total volume of dressing be 24 (ounces, ml, litres, gallons, swimming pools - pick the unit of your choice).

Now let's see how much of each item we have.

Without messing up, Miguel would have:

(5/8)24 = 15 oil
(1/4)24 = 6 vinegar
(1/24)24 = 1 salt
(1/24)24 = 1 pepper
(1/24)24 = 1 sugar

However, he doubled the vinegar, so he has 12 vinegar instead; he left out the sugar, so we can ignore that bit.

Now the question: what proportion of the final mixture is oil?

15/(15+12+1+1+0) = 15/29... choose (A).
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