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Re: A candy jar has 4 lime, 10 cherry, 8 orange, and x grape ca [#permalink]
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I do think the problem is not so much clear to you.

What we do know is that the sum of lime orange and cherry is 22 + x that represents the grape.

Now, you have to pick the oranges that are 8 over 22 + x.

Using the answer choices if you do have 22 + 10 or 14, the 8 / 32 or 8/36 or 8/40 the probability is ALWAYS > 20%.

If you take the fraction 8/ 22 + 18 = 8/40 then the probability is \(= 20%\) but we need only the probability \(> 20%\).

As such, only the two first answer choices are correct.

A and B are the correct answer.

Hope now is clear.

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Re: A candy jar has 4 lime, 10 cherry, 8 orange, and x grape ca [#permalink]
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4 lime, 10 cherry, 8 orange, and 18 grape candies =

10/ (4+10+8+18) = 25% which is grater than 20. So the answer is A, B, & C
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Re: A candy jar has 4 lime, 10 cherry, 8 orange, and x grape ca [#permalink]
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4L | 10C | 8OR | X.G

we are given that probability for picking an orange is greater than 20%.

Lets put it into the probability equation: number of oranges/total number of candy

8/ (4+10+8+X) > 20/100 ( simplify and cross-multiply)

80 > 44 + 2x ( divide by 2)

40 > 22 + x

=> 18 > x
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Re: A candy jar has 4 lime, 10 cherry, 8 orange, and x grape ca [#permalink]
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