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# M19-33

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Math Expert
Joined: 02 Sep 2009
Posts: 55231

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16 Sep 2014, 01:07
00:00

Difficulty:

15% (low)

Question Stats:

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

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How many multiples of 3 are there between 121 and 250 ?

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

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Math Expert
Joined: 02 Sep 2009
Posts: 55231

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16 Sep 2014, 01:07
1
2
Official Solution:

How many multiples of 3 are there between 121 and 250 ?

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

The first multiple of 3 in this range is 123, the last is 249. Altogether there are $$\frac{249 - 123}{3} + 1 = 43$$ multiples of 3 between 121 and 250.

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Intern
Joined: 17 Dec 2013
Posts: 4
Concentration: Strategy, Social Entrepreneurship
GMAT 1: 540 Q43 V21
GPA: 3

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31 Aug 2017, 02:50
Confused on the part when to add 1 as the question does not mention both inclusive?
Math Expert
Joined: 02 Sep 2009
Posts: 55231

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31 Aug 2017, 02:53
anantharaman wrote:
Confused on the part when to add 1 as the question does not mention both inclusive?

Neither 121 nor 250 is a multiple of 3 so it does not make any difference.
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Joined: 04 Dec 2015
Posts: 750
Location: India
Concentration: Technology, Strategy
WE: Information Technology (Consulting)

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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 $$= (83 - 41) + 1 = 42 +1 = 43$$

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Joined: 08 Apr 2017
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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!
Math Expert
Joined: 02 Sep 2009
Posts: 55231

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07 Sep 2017, 21:46
2
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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Joined: 30 Oct 2018
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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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# M19-33

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