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![If 0 < x < 1000, and [x/2] + [x/3] + [x/5] = 31x/30 where [x] denotes](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "If 0 < x < 1000, and [x/2] + [x/3] + [x/5] = 31x/30 where [x] denotes"){kind=icon}
11 Mar 2020, 02:33
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B
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Difficulty:
55%
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Question Stats:
62% (02:22) correct
38%
(02:16)
wrong
based on 290
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If \(0 < x < 1000\), and \([\frac{x}{2}] + [\frac{x}{3}] + [\frac{x}{5}] = \frac{31x}{30}\) where [x] denotes the greatest integer less than or equal to x, then the number of possible values of x will be
A. 30
B. 33
C. 43
D. 44
E. 45
Are You Up For the Challenge: 700 Level Questions
A. 30
B. 33
C. 43
D. 44
E. 45
Are You Up For the Challenge: 700 Level Questions
11 Mar 2020, 02:48
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If \(0<x<1000\), and \([\frac{x}{2}]+[\frac{x}{3}]+[\frac{x}{5}]=\frac{31x}{30}\) where [x] denotes the greatest integer less than or equal to x, then the number of possible values of x will be
The value of \([\frac{x}{2}]+[\frac{x}{3}]+[\frac{x}{5}]\) will be always INTEGER.
--> The value of \(\frac{31x}{30}\) must be integer too.
In order the value of \(\frac{31x}{30}\) to be integer, \(x\) must be \(30\), \(60\), \(90\)..... \(990\)
\(x_1= 30\)
\(x_n = 990\)
--> \(n = \frac{(990-30)}{30} +1 = 33\)
Answer (B)
The value of \([\frac{x}{2}]+[\frac{x}{3}]+[\frac{x}{5}]\) will be always INTEGER.
--> The value of \(\frac{31x}{30}\) must be integer too.
In order the value of \(\frac{31x}{30}\) to be integer, \(x\) must be \(30\), \(60\), \(90\)..... \(990\)
\(x_1= 30\)
\(x_n = 990\)
--> \(n = \frac{(990-30)}{30} +1 = 33\)
Answer (B)
General Discussion
Updated on: 11 Mar 2020, 02:52
Originally posted by GMATinsight on 11 Mar 2020, 02:45.
Last edited by GMATinsight on 11 Mar 2020, 02:52, edited 1 time in total.
Last edited by GMATinsight on 11 Mar 2020, 02:52, edited 1 time in total.
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Bunuel
\([\frac{x}{2}] + [\frac{x}{3}] + [\frac{x}{5}] = \frac{31x}{30}\) = Integer
I.e. x Must be a multiple of 30 for 31x/30 to be an Integer
at x = 30, \([\frac{30}{2}] + [\frac{30}{3}] + [\frac{30}{5}] = 15+10+6 = 31 =\frac{31x}{30}\)
Since all options suggest that total possible values of x must be 30 or more
where total values of x as per range 0 < x < 1000 = 1000/30 = 33
Let's try with biggest value of x
at x = 990, \([\frac{990}{2}] + [\frac{990}{3}] + [\frac{990}{5}] = 495+330+198 = 1023 = \frac{31x}{30}\)
I.e. Total possible values of x must be 33 hence
ANswer: Option B
All values of x from 30 to 990 satisfy the given relation
