GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
A shipment consists of 1,800 parts, some of which are defect
[#permalink]
24 Jul 2020, 10:54
24 Jul 2020, 10:54
Expert Reply
00:00
Question Stats:
87% (01:31) correct
12% (01:07) wrong based on 31 sessions
Hide Show timer Statistics
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 1/9 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
Kudos for the right solution and explanation
(A) 90
(B) 120
(C) 180
(D) 200
(E) 900
Kudos for the right solution and explanation
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1111
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
Re: A shipment consists of 1,800 parts, some of which are defect
[#permalink]
24 Jul 2020, 22:27
24 Jul 2020, 22:27
1
Let number of defective shipments be x
=> Non defective = Total - defective = 1800 - x
Probability of Picking Defective = \(\frac{Total Defective}{ Total}\) [i.e. = \(\frac{x}{ 1800}\) ]
Probability of Picking Non-Defective = \(\frac{Total Non-Defective}{ Total}\) [i.e. = \(\frac{1800-x}{ 1800}\) ]
\(\frac{Probability Of Picking Defective}{Probability Of Picking Non-Defective}\) = \(\frac{Total Defective}{ Total }\)/ \(\frac{Total Non-Defective}{ Total}\)
= \(\frac{Total Defective}{ Total Non-Defective}\) = \(\frac{x}{ 1800-x}\) = \(\frac{1}{9}\) [ Given]
=> 9x = 1800 - x
=> 10x = 1800
=> x = 180
So, answer will be C
Hope it helps!
=> Non defective = Total - defective = 1800 - x
Probability of Picking Defective = \(\frac{Total Defective}{ Total}\) [i.e. = \(\frac{x}{ 1800}\) ]
Probability of Picking Non-Defective = \(\frac{Total Non-Defective}{ Total}\) [i.e. = \(\frac{1800-x}{ 1800}\) ]
\(\frac{Probability Of Picking Defective}{Probability Of Picking Non-Defective}\) = \(\frac{Total Defective}{ Total }\)/ \(\frac{Total Non-Defective}{ Total}\)
= \(\frac{Total Defective}{ Total Non-Defective}\) = \(\frac{x}{ 1800-x}\) = \(\frac{1}{9}\) [ Given]
=> 9x = 1800 - x
=> 10x = 1800
=> x = 180
So, answer will be C
Hope it helps!
Re: A shipment consists of 1,800 parts, some of which are defect
[#permalink]
25 Jul 2020, 00:13
25 Jul 2020, 00:13
Is my approach correct?
1/9 * (1-1/9)
1/9*8/9
8/81 = 0.098
0.098 * 1800 = 176.4 approx which is close to 180
1/9 * (1-1/9)
1/9*8/9
8/81 = 0.098
0.098 * 1800 = 176.4 approx which is close to 180
Hence the confusion remains
