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A candy jar has 4 lime, 10 cherry, 8 orange, and x grape ca

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A candy jar has 4 lime, 10 cherry, 8 orange, and x grape ca [#permalink] New post   06 Jun 2017, 06:40
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A candy jar has 4 lime, 10 cherry, 8 orange, and x grape candies. If Tom selects a candy from the jar at random and the probability that he selects an orange candy is greater than 20 percent, which of the following could be the value of x ?

Indicate all such values.

❑ 10

❑ 14

❑ 18

❑ 22

❑ 24

❑ 28
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A,B
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IlCreatore
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Re: A candy jar has 4 lime, 10 cherry, 8 orange, and x grape ca [#permalink] New post   21 Sep 2017, 06:06
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Since we know that the probability of extracting a orange candy is greater than 20%, we can rewrite this information as \(\frac{8}{22+x}>0.2\). In other words, there are eight favorable cases, the eight orange candies out of a total number of cases equal to the total number of candies. Now, we just have to substitute x with the provided answers to get that only A and B satisfies the inequality above.
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Re: A candy jar has 4 lime, 10 cherry, 8 orange, and x grape ca [#permalink] New post   26 Mar 2018, 20:08
why can't we solve the inequality and get x<28 as the answer?
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Re: A candy jar has 4 lime, 10 cherry, 8 orange, and x grape ca [#permalink] New post   27 Mar 2018, 10:04
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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] New post   02 Apr 2018, 10:47
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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] New post   26 Oct 2020, 05:48
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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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