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A shipment consists of 1,800 parts, some of which are defective. If a

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A shipment consists of 1,800 parts, some of which are defective. If a  [#permalink]

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New post 20 Jul 2017, 16:32
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A shipment consists of 1,800 parts, some of which are defective. If a part is chosen from the shipment at random, the probability of it being defective is 1/19 the probability that it is not defective. How many of the parts in the shipment are defective?

(A) 90
(B) 120
(C) 180
(D) 200
(E) 900
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Re: A shipment consists of 1,800 parts, some of which are defective. If a  [#permalink]

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New post 20 Jul 2017, 19:32
Turkish wrote:
A shipment consists of 1,800 parts, some of which are defective. If a part is chosen from the shipment at random, the probability of it being defective is 1/19 the probability that it is not defective.
How many of the parts in the shipment are defective?

(A) 90
(B) 120
(C) 180
(D) 200
(E) 900



Hi...

Let the probability that it is NOT defective is x.
So the probability it is defective is\(\frac{x}{19}\)..
Do \(x+\frac{x}{19}=1\)...
\(\frac{20x}{19}=1\)..
\(x=\frac{19}{20}\)...
Therefore defective pieces= \(1800*\frac{1}{20}=90\)
A
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Re: A shipment consists of 1,800 parts, some of which are defective. If a  [#permalink]

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New post 20 Jul 2017, 21:03
A is correct as probability of defective is x/19 where x is probability of not defective.Adding both the probability overall should be 1.On solving you get x as 19/20,which is probability of not defective.Therefore probability of defective is 1/20 and since total is 1800 ,we get 90 as answer.

A is correct choice.


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A shipment consists of 1,800 parts, some of which are defective. If a  [#permalink]

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New post 21 Jul 2017, 10:48
chetan2u wrote:
Turkish wrote:
A shipment consists of 1,800 parts, some of which are defective. If a part is chosen from the shipment at random, the probability of it being defective is 1/19 the probability that it is not defective.
How many of the parts in the shipment are defective?

(A) 90
(B) 120
(C) 180
(D) 200
(E) 900



Hi...

Let the probability that it is NOT defective is x.
So the probability it is defective is\(\frac{x}{19}\)..
Do \(x+\frac{x}{19}=1\)...
\(\frac{20x}{19}=1\)..
\(x=\frac{19}{20}\)...
Therefore defective pieces= \(1800*\frac{1}{20}=90\)
A


Probability of it being defective = 1/19 (Probability of being not defective)
No of defective items = x; Probability of defective items = x/1800
No of not defective items = (1800-x); Probability of defective items = (1800-x)/1800
x/1800 = 1/19 (1800 - x)/1800
20x = 1800
x = 90
No of defective items = 90
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Re: A shipment consists of 1,800 parts, some of which are defective. If a  [#permalink]

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New post 30 Jul 2017, 17:50
Turkish wrote:
A shipment consists of 1,800 parts, some of which are defective. If a part is chosen from the shipment at random, the probability of it being defective is 1/19 the probability that it is not defective. How many of the parts in the shipment are defective?

(A) 90
(B) 120
(C) 180
(D) 200
(E) 900


We can let x = the number of parts that are defective, and thus we have 1,800 - x parts that are not defective. If a part is chosen at random, the probability that it is defective is x/1,800 and the probability that is is not defective is (1,800 - x)/1,800. Since the probability of the part being defective is 1/19 the probability that it is not defective, we can create the following equation:

x/1,800 = (1/19) * (1,800 - x)/1,800

19x/1,800 = (1,800 - x)/1,800

19x = 1,800 - x

20x = 1,800

x = 90

Answer: A
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Re: A shipment consists of 1,800 parts, some of which are defective. If a   [#permalink] 30 Jul 2017, 17:50
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