Two assembly line inspectors, Lauren and Steven, inspect...

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Two assembly line inspectors, Lauren and Steven, inspect widgets as they come off the assembly line. If Lauren inspects every fifth widget, starting with the fifth , and Steven inspects every third, starting with the third, how many of the 98 widgets produced in the first hour of operation are not inspected by either inspector?

A) 91
B) 59
C) 53
D) 47
E) 45

The OA is C.

I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.

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by OWN » Sun Dec 10, 2017 3:18 pm
Hi Luandato,

Before diving into the math for this equation, let's just think about what's going on. We have these two inspectors on the same assembly line. One inspects every 3rd widget and the other every 5th. If we want to know how many were not inspected, then we need to figure out the total that were inspected and subtract that from 98. To find the total number of inspected widgets, we must add how many widgets were inspected by each inspector AND subtract any that were inspected by both. Note that the wording of the question was crucial for us to make the distinction. If, for example, the question asked for the total number of inspected widgets, we would not subtract the number from 98. Alternatively, if the question asked for the total number of inspections that hour, we would not subtract duplicates.

Let's get to work:
Lauren inspects every 5th widget, so 98/5 = 19r3, so we have 19 from Lauren
Steven inspects every 3rd widget, so 98/3 = 32r2, so we have 32 from Steven
Now, 3 and 5 are luckily both prime numbers, so any widgets that were inspected by both inspectors would have to be multiples of 15 on the assembly line.
So the widgets inspected by both would be 98/15 = 6r8, so we have 6.
So the total number of inspected widgets is 19+32-6 = 45

So the total that were not inspected is 98-45 = 53.

OA C.

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by [email protected] » Tue Dec 12, 2017 1:56 pm
Hi LUANDATO,

We're told that two inspectors, Lauren and Steven, inspect widgets as they come off the assembly line. Lauren inspects every 5th widget, starting with the fifth , and Steven inspects every 3rd widget, starting with the third. 98 widgets are produced. We're asked for the number that are NOT inspected by either.

The answer choices are 'spread out' enough that we can answer this question with a bit of arithmetic and some logic.

To start, since Lauren inspects 1/3 of the widgets, see can quickly calculate the number of widgets she inspects - if there were 99 widgets, she'd inspect 33 of them; since there are 98 widgets, she inspects 32 of them.

Since Steven inspects 1/5 of the widgets, we can also quickly calculate the number of widgets he inspects - if there were 100 widgets, he'd inspect 20 of them; since there are 98 widgets, he inspects 19 of them.

32 + 19 = 51.... so at first glance, you might assume that 98-51 = 47 of the widgets were not inspected. HOWEVER, there is some 'overlap' with the widgets. For example, the 15th widget would be Lauren's fifth widget inspected AND Steven's third widget inspected - but it's the SAME widget, so we're not allowed to count it 'twice.' Since there are several widgets that have been counted twice in this way (it's actually 6 widgets, but you don't need to calculate that here), the actual number of widgets counted must be a bit LESS than 51, so the number of widgets NOT counted must be a bit GREATER than 47. There's only one answer that logically matches...

Final Answer: C

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by Scott@TargetTestPrep » Wed Dec 20, 2017 7:48 am
LUANDATO wrote:Two assembly line inspectors, Lauren and Steven, inspect widgets as they come off the assembly line. If Lauren inspects every fifth widget, starting with the fifth , and Steven inspects every third, starting with the third, how many of the 98 widgets produced in the first hour of operation are not inspected by either inspector?

A) 91
B) 59
C) 53
D) 47
E) 45
We need to determine how many multiples of 5, 3, and 15 there are from 1 to 98 inclusive.

Number of multiples of 5:

(95 - 5)/5 + 1 = 19

Number of multiples of 3:

(96 - 3)/3 + 1 = 32

Number of multiples of 15:

(90 - 15)/15 + 1 = 6

We add the number of multiples of 5 and of 3 to get 19 + 32 = 51, but we must subtract the number of multiples of 15 from this total because those 6 numbers represent the double-counted multiples of 3 and 5. Thus, there are 19 + 32 - 6 = 45 multiples of 3 and/or 5 from 1 to 98, which is also the number of widgets inspected.

So, 98 - 45 = 53 widgets were not inspected.

Answer: C

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