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Re: There are 40 marbles in a jar. If of the marbles are blue [#permalink]
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achirarulz wrote:
Are we suppose to write answer as 7/20 or 14/40?

In GRE you are allowed to use both the fractions unless it is mentioned to reduce to lowest terms in which case it is \(\frac{7}{20}\)
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Re: There are 40 marbles in a jar. If of the marbles are blue [#permalink]
thank you :-D

sandy wrote:
achirarulz wrote:
Are we suppose to write answer as 7/20 or 14/40?

In GRE you are allowed to use both the fractions unless it is mentioned to reduce to lowest terms in which case it is \(\frac{7}{20}\)
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Re: There are 40 marbles in a jar. If of the marbles are blue [#permalink]
There are 40 marbles in a jar. If 1/5 of the marbles are blue, 1/4 of the remaining marbles are red, and 10 marbles are green. If a marble is selected at random, then what is the probability that the marble will not be blue, red, or green?

Blue marbles: 1/5*40 = 8
Red Marbles: 1/4 of remaining marbles = 1/4 (40-8) = 1/4*32 = 8
Green Marbles = 10

Red + Blue + Green = 8 + 8 + 10 = 26
Probability of not getting RBG marbles = 1 - (26/40) = 14/40 = 7/20
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Re: There are 40 marbles in a jar. If of the marbles are blue [#permalink]
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sandy wrote:
There are 40 marbles in a jar. If \(\frac{1}{5}\) of the marbles are blue, \(\frac{1}{4}\) of the remaining marbles are red, and 10 marbles are green. If a marble is selected at random, then what is the probability that the marble will not be blue, red, or green?

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\(\frac{14}{40}\)

There are 40 marbles in a jar, and 1/5 of the marbles are blue
1/5 of 40 = 8. So, there are 8 blue marbles.

1/4 of the remaining marbles are red
We have already accounted for 8 marbles, which means there are 32 marbles remaining.
So, the number of red marbles = 1/4 of 32 = 8

Also given: 10 marbles are green.

8 + 8 + 10 = 26

40 - 26 = 14, which means 14 of the 40 marbles are neither blue, red, or green

So, P(selected marble will not be blue, red, or green) = 14/40 = 7/20

Answer: 7/20 (or any equivalent fraction)
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