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

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

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