The number of defects in the first five cars to come through

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swerve: Legendary Member; Posts: 2499; Joined: Sun Oct 29, 2017 2:04 pm; Followed by:6 members

The number of defects in the first five cars to come through

by swerve » Fri Jun 29, 2018 11:55 am

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The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the sixth car through the production line has either 3, 7, or 12 defects, for which of these values does the mean number of defects per car for the first six cars equal the median?

I. 3
II. 7
III. 12

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

The OA is D.

Please, can anyone explain this PS question? I tried to solve it but I can't get the correct answer. I need help. Thanks.

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Source: Beat The GMAT — Problem Solving | Fri Jun 29, 2018 11:55 am

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[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pm; Location: Palo Alto, CA; Thanked: 2867 times; Followed by:511 members; GMAT Score:800

by [email protected] » Sat Jun 30, 2018 1:56 pm

Hi swerve,

We're told that the number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively and that the sixth car through the production line has either 3, 7, or 12 defects. We're asked which of the following three values does the MEAN number of defects per car for the first six cars equal the MEDIAN. This question comes down to organizing the data (to find the Median) and using the Average Formula.

I. 3 defects

IF... the 6th car has 3 defects, then the six numbers would be: 3, 4, 6, 7, 9 and 10
The Median would be (6+7)/2 = 6.5
The Average would be (3+4+6+7+9+10)/6 = 39/6 = 6.5
The Average and the Median ARE equal.
Eliminate Answers B and C.

II. 7 defects

IF... the 6th car has 7 defects, then the six numbers would be: 4, 6, 7, 7, 9 and 10
The Median would be (7+7)/2 = 7
The Average would be (4+6+7+7+9+10)/6 = 43/6 = 7 1/6
The Average and the Median are NOT equal.
Eliminate Answer E

III. 12 defects

IF... the 6th car has 12 defects, then the six numbers would be: 4, 6, 7, 9, 10 and 12
The Median would be (7+9)/2 = 8
The Average would be (4+6+7+9+10+12)/6 = 48/6 = 8
The Average and the Median ARE equal.
Eliminate Answer A.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Mon Jul 23, 2018 5:35 pm

swerve wrote:The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the sixth car through the production line has either 3, 7, or 12 defects, for which of these values does the mean number of defects per car for the first six cars equal the median?

I. 3
II. 7
III. 12

A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III

Let's check each number in the given Roman numerals.

I. 3

The average is (3 + 4 + 6 + 7 + 9 + 10)/6 = 6.5.

The median is (6 + 7)/2 = 13/2 = 6.5.

We see that I is correct.

II. 7

The average is (4 + 6 + 7 + 7 + 9 + 10)/6 â‰ˆ 7.17.

The median is (7 + 7)/2 = 14/2 = 7.

We see that II is not correct.

III. 12

The average is (4 + 6 + 7 + 9 + 10 + 12)/6 = 8.

The median is (7 + 9)/2 = 16/2 = 8.

We see that III is correct.

Answer: D

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