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In a jar there are balls in different colors: blue, red, green and yel

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In a jar there are balls in different colors: blue, red, green and yel [#permalink]

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New post 13 Jul 2016, 01:01
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In a jar there are balls in different colors: blue, red, green and yellow.
The probability of drawing a blue ball is 1/8.
The probability of drawing a red ball is 1/5.
The probability of drawing a green ball is 1/10.
If a jar cannot contain more than 50 balls, how many yellow balls are in the Jar?

A. 25.
B. 24.
C. 23.
D. 20.
E. 17.
[Reveal] Spoiler: OA

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In a jar there are balls in different colors: blue, red, green and yel [#permalink]

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New post 13 Jul 2016, 02:25
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Bunuel wrote:
In a jar there are balls in different colors: blue, red, green and yellow.
The probability of drawing a blue ball is 1/8.
The probability of drawing a red ball is 1/5.
The probability of drawing a green ball is 1/10.
If a jar cannot contain more than 50 balls, how many yellow balls are in the Jar?

A. 25.
B. 24.
C. 23.
D. 20.
E. 17.


I tried this question without probability formula and used only logic.

We are given that the jar doesn't contain more than 50 balls then we can take that the number can be lesser than 50 or equal to 50.

If we assume exactly 50 numbers of balls we get different incorrect probability.

We need to get some number which is less than 50 and exactly divisible by 8,5 and 10

Lets take LCM of three balls 8,5 and 10 = 40.

The probability of drawing a blue ball is 1/8...For every 8 balls we get 1 BLUE ball then out of 40 balls we get 5 BLUE balls.
The probability of drawing a red ball is 1/5...For every 5 balls we get 1 RED ball then out of 40 balls we get 8 REB balls.
The probability of drawing a green ball is 1/10. For every 10 balls we get 1 GREEN ball then out of 40 balls we get 4 GREEN BALLS.

Then out of 40 - ( 5 + 8 + 4 ) = 23 YELLOW balls.

IMO option C is correct.

OA please...will correct if I missed anything.

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Re: In a jar there are balls in different colors: blue, red, green and yel [#permalink]

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New post 13 Jul 2016, 03:01
Bunuel wrote:
In a jar there are balls in different colors: blue, red, green and yellow.
The probability of drawing a blue ball is 1/8.
The probability of drawing a red ball is 1/5.
The probability of drawing a green ball is 1/10.
If a jar cannot contain more than 50 balls, how many yellow balls are in the Jar?

A. 25.
B. 24.
C. 23.
D. 20.
E. 17.


The probabilities for blue red and green balls are reduced fractions. Let the no. of yellow balls be x.

A number expressing the probability (p) that a specific event will occur, expressed as the ratio of the number of actual occurrences (n) to the number of possible occurrences (N).

p=\(\frac{n}{N}\)

Total no of balls in the Jar should be the Lease Common Multiple LCM of 8/ 5 & 10 i.e 40

We get \(\frac{5}{40}\) Blue + \(\frac{8}{40}\) Red + \(\frac{4}{40}\) Green + \(\frac{y}{40}\) = 1

=> Y = 23
Ans is C

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Re: In a jar there are balls in different colors: blue, red, green and yel [#permalink]

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New post 13 Jul 2016, 07:11
p(yellow)=1-p(remaining)
=1-(1/8+1/5+1/10)
=23/40
no of yellow balls = 23

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Re: In a jar there are balls in different colors: blue, red, green and yel [#permalink]

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New post 08 Nov 2017, 16:43
Bunuel wrote:
In a jar there are balls in different colors: blue, red, green and yellow.
The probability of drawing a blue ball is 1/8.
The probability of drawing a red ball is 1/5.
The probability of drawing a green ball is 1/10.
If a jar cannot contain more than 50 balls, how many yellow balls are in the Jar?

A. 25.
B. 24.
C. 23.
D. 20.
E. 17.


We see that the total number of balls must be a multiple of the LCM of 8, 5, and 10, which is 40. However, since there are no more than 50 balls in the jar, the number of balls must be 40.

Let’s determine the probability of selecting a yellow ball. Since the selected ball will be either blue, red, green, or yellow, we have:

P(blue) + P(red) + P(green) + P(yellow) = 1

1/8 + 1/5 + 1/10 + P(yellow) = 1

5/40 + 8/40 + 4/40 + P(yellow) = 1

17/40 + P(yellow) = 1

P(yellow) = 1 - 17/40 = 23/40

Since the probability of selecting a yellow ball is 23/40, we can create the following proportion to determine the number of yellow balls:

23/40 = y/40

23 = y

Alternative solution:

After we have determined that the total number of balls in the jar must be 40, we can say that the number of blue balls must be 1/8 x 40 = 5, the number of red balls must be 1/5 x 40 = 8, and the number of green balls must be 1/10 x 40 = 4. Thus, the number of yellow balls in the jar must be 40 - (5 + 8 + 4) = 40 - 17 = 23.

Answer: C
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Kudos [?]: 980 [0], given: 5

Re: In a jar there are balls in different colors: blue, red, green and yel   [#permalink] 08 Nov 2017, 16:43
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