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02 Dec 2021, 02:30
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Difficulty:
95%
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Question Stats:
24% (02:31) correct
76%
(02:23)
wrong
based on 141
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A quality control worker at a warehouse inspects x separate boxes, each containing units of product. For each box, she records in List A a 1 if there are no defective units in the box and a 0 if there is at least one defective unit. She also records in List B the number of defective units in each box. What is the mode of List B ?
(1) The average of List B is \(\frac{5}{2}\)
(2) The sum of List A is \(\frac{2x}{3}\)
(1) The average of List B is \(\frac{5}{2}\)
(2) The sum of List A is \(\frac{2x}{3}\)
22 Dec 2021, 16:03
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Bunuel
Breaking Down the Info:
Note List B will probably contain a lot of 0's as list B directly records the number of defective units for EACH box, and this number would be 0 if there are no defective units.
Statement 1 Alone:
The average of B alone does not tell us much. We can only infer there are \(\frac{5}{2}x\) defective units in total, which isn't that helpful. Then statement 1 alone is insufficient.
Statement 2 Alone:
This tells us \(\frac{2}{3}\) of the set is NOT defective, hence \(\frac{2}{3}\) of list B will be 0's. Thus the mode of B is 0. Statement 2 alone is sufficient.
Answer: B
03 Oct 2023, 00:48
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Statement 1: The average of List B is 5/2.
We are already given that List B records number of defective units in each box (including the boxes that have no defects) but we are concerned about the mode of the List B and hence what is the mode of # of defective units.
In statement 1, we are given the average but we don't know how many numbers are there in the set and how well they are spaced, it is difficult to comment on the mode.
So insufficient.
Statement 2: The sum of List A is 2x/3
Since list A will contain either 1 or 0, we can infer that 2x/3 will be an integer and hence x is a multiple of 3 (3k,6k etc).
So if # of terms are 3, the sum is 2 and thus 1 defective box (and not unit) with a value 0.
if # of terms are 6, the sum is 4 and thus 2 defective boxes (and not units)
But what we can infer from here is that the number of defective boxes are half of non-defective boxes. Therefore, the mode will always be ZERO.
Hence sufficient.
chetan2u - Is the reasoning correct?
We are already given that List B records number of defective units in each box (including the boxes that have no defects) but we are concerned about the mode of the List B and hence what is the mode of # of defective units.
In statement 1, we are given the average but we don't know how many numbers are there in the set and how well they are spaced, it is difficult to comment on the mode.
So insufficient.
Statement 2: The sum of List A is 2x/3
Since list A will contain either 1 or 0, we can infer that 2x/3 will be an integer and hence x is a multiple of 3 (3k,6k etc).
So if # of terms are 3, the sum is 2 and thus 1 defective box (and not unit) with a value 0.
if # of terms are 6, the sum is 4 and thus 2 defective boxes (and not units)
But what we can infer from here is that the number of defective boxes are half of non-defective boxes. Therefore, the mode will always be ZERO.
Hence sufficient.
chetan2u - Is the reasoning correct?
