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M19-33

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M19-33  [#permalink]

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New post 16 Sep 2014, 01:07
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

75% (00:38) correct 25% (01:01) wrong based on 206 sessions

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Re M19-33  [#permalink]

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New post 16 Sep 2014, 01:07
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GMAT 1: 540 Q43 V21
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Re: M19-33  [#permalink]

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New post 31 Aug 2017, 02:50
Confused on the part when to add 1 as the question does not mention both inclusive?
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New post 31 Aug 2017, 02:53
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Re: M19-33  [#permalink]

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New post 06 Sep 2017, 08:06
Bunuel wrote:
How many multiples of 3 are there between 121 and 250 ?

A. 42
B. 43
C. 44
D. 45
E. 46

First multiple of \(3\) between \(121\) and \(250\) is \(= 123\)

\(\frac{123}{3} = 41\)

Last multiple of \(3\) between \(121\) and \(250\) is \(= 249\)

\(\frac{249}{3} = 83\)

Therefore Total number of multiples of \(3\) between \(121\) and \(250\) is \(= (83 - 41) + 1 = 42 +1 = 43\)

Answer (B)...
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Re: M19-33  [#permalink]

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New post 07 Sep 2017, 20:47
Will this work as well?

250-121=129 then take 129/3 = 43

That's how I got my answer. Maybe I just got lucky!
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Re: M19-33  [#permalink]

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New post 07 Sep 2017, 21:46
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catooey wrote:
Will this work as well?

250-121=129 then take 129/3 = 43

That's how I got my answer. Maybe I just got lucky!


If it were: "How many multiples of 3 are there between 123 and 249, inclusive ?" the answer would still be 43, but with your approach you'd get (249-123)/3 = 42, so this is not error prone approach.


\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

Check this: http://gmatclub.com/forum/totally-basic ... 20multiple
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Re: M19-33  [#permalink]

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New post 29 Nov 2018, 22:44
Sol-
multiple of 3 after 121 = 123
multiple of 3 before = 249

a+(n-1)d = 249
123 + (n-1) 3 = 249
n = 43 (B)
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Re: M19-33   [#permalink] 29 Nov 2018, 22:44
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