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# Two assembly line inspectors, Lauren and Steven, inspect widgets as

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Joined: 02 Sep 2009
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Two assembly line inspectors, Lauren and Steven, inspect widgets as  [#permalink]

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29 Dec 2016, 12:15
00:00

Difficulty:

55% (hard)

Question Stats:

62% (02:19) correct 38% (02:04) wrong based on 120 sessions

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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

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Joined: 13 Oct 2016
Posts: 367
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Re: Two assembly line inspectors, Lauren and Steven, inspect widgets as  [#permalink]

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29 Dec 2016, 12:44
2
multiples of 5 - 19

multiples of 3 - 32

multiples of 15 - 6

98 - 19 - 32 + 6 = 53

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Re: Two assembly line inspectors, Lauren and Steven, inspect widgets as  [#permalink]

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30 Dec 2016, 11:21
1
Bunuel 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

No of widgets inspected by Lauren is $$\frac{98}{5} = 19$$

No of widgets inspected by Steven is $$\frac{98}{3} = 32$$

No of widgets inspected by Steven & Lauren is $$\frac{98}{15} = 6$$

So, Total No of widgets inspected by Steven & Lauren is 19 + 32 - 6 = 45

So, Total No of widgets not inspected by Steven & Lauren is 98 - 45 = 53

Hence, correct answer will bw (C) 53

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Re: Two assembly line inspectors, Lauren and Steven, inspect widgets as  [#permalink]

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31 Dec 2016, 03:49
Multiples of 3 till 98 are 32

Multiples of 5 till 98 are 19

Multiples of 15 till 98 are 6

So 32+19-6 = 45

Tota= 98 - 45 = 53 is the answer
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Re: Two assembly line inspectors, Lauren and Steven, inspect widgets as  [#permalink]

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07 Aug 2017, 08:43
This question can be views at multiple set question .
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Re: Two assembly line inspectors, Lauren and Steven, inspect widgets as  [#permalink]

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07 Aug 2017, 12:38
I am always curious how this difficulty level is being defined, this was an easy one

total
number of multiplies of 15 - 6 (as every 15 will be checked by both inspectors, 15 has prime factors 3 and 5)
number of multiplies of 3 - 32
number of multiplies of 5 - 19

Then 32 + 19 - 6 = 45
98 - 45 = 53
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Re: Two assembly line inspectors, Lauren and Steven, inspect widgets as  [#permalink]

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10 Aug 2017, 10:30
1
Bunuel 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.

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Re: Two assembly line inspectors, Lauren and Steven, inspect widgets as   [#permalink] 10 Aug 2017, 10:30
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