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50 boxes—each containing 30 machine parts—were examined

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50 boxes—each containing 30 machine parts—were examined [#permalink] New post 20 Jan 2016, 18:42
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Question Stats:

63% (01:07) correct 36% (02:07) wrong based on 55 sessions
In a quality-control test, 50 boxes—each containing 30 machine parts—were examined for defective parts. The number of defective parts was recorded for each box, and the average (arithmetic mean) of the 50 recorded numbers of defective parts per box was 1.12. Only one error was made in recording the 50 numbers: “1” defective part in a certain box was incorrectly recorded as “10”.


Quantity A
Quantity B
The actual average number of defective parts per box
0.94


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.




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Question: 9
Page: 459
Difficulty: medium
[Reveal] Spoiler: OA

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Last edited by Carcass on 22 Jan 2016, 08:03, edited 1 time in total.
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Re: 50 boxes—each containing 30 machine parts—were examined [#permalink] New post 22 Jan 2016, 08:08
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Solution

Thinking conceptually upfront is the best way to tackle such question.

Now the average is \(\frac{sum}{number}=1.12\). The number, our measurements, is \(50\) so \(S = 50*1.12=56\)

From the stem we do know that one measurement was not 1 but 10. So we do have that \(56-9=47\)

\(\frac{Defective parts}{50 boxes}= \frac{47}{50} = 0.94\)

The answer is \(C\)
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Re: 50 boxes—each containing 30 machine parts—were examined [#permalink] New post 06 May 2016, 03:55
Other way of solving the calculation..

sum(50)/50 = 1.12,

(sum(49) + 1 + 9)/50 = 1.12

(sum(50)/50 + 9/50 = 1.12

Average of 50 defective parts = 1.12 - 9/50 = 1.12 - (18/100) = 94/ 100 = 0.94
Hence (answer = c)
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Re: 50 boxes—each containing 30 machine parts—were examined [#permalink] New post 15 Jun 2019, 04:39
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sandy wrote:
In a quality-control test, 50 boxes—each containing 30 machine parts—were examined for defective parts. The number of defective parts was recorded for each box, and the average (arithmetic mean) of the 50 recorded numbers of defective parts per box was 1.12. Only one error was made in recording the 50 numbers: “1” defective part in a certain box was incorrectly recorded as “10”.

Quantity A
Quantity B
The actual average number of defective parts per box
0.94



The number of defective parts was recorded for each box, and the average (arithmetic mean) of the 50 recorded numbers of defective parts per box was 1.12
We can write: (sum of all 50 numbers)/50 = 1.12
Multiply both sides by 50 to get: sum of all 50 numbers = 56


“1” defective part in a certain box was incorrectly recorded as “10”.
This means the sum of all 50 numbers was 9 greater than it should have been
So, to find the actual sum, we must subtract 9 from our earlier value

Actual sum = 56 - 9 = 47

So, the actual average number of defective parts per box = 47/50 = 0.94

We get:
QUANTITY A: 0.94
QUANTITY B: 0.94

Answer: C

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Re: 50 boxes—each containing 30 machine parts—were examined   [#permalink] 15 Jun 2019, 04:39
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50 boxes—each containing 30 machine parts—were examined

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