If `0lt xlt 1000` and `[x/2]+[x/3]+[x/5]=31/30x`, where `[x]` is the greatest integer less than or equal to `x` the number of possible values of x is

Options

A. 34
B. 32
C. 33
D. None of these

Answer

C

Explanation

We have,
`[(x)/(2)]+[(x)/(3)]+[(x)/(5)]=(31)/(30)x`
`implies [(x)/(2)]+[(x)/(3)]+[(x)/(5)]=(x)/(2)+(x)/(3)+(x)/(5)`
`implies[(x)/(2)]=(x)/(2),[(x)/(3)]=(x)/(3)` and `[(x)/(5)]=(x)/(5)`
`implies (x)/(2),(x)/(3),(x)/(5)` are all integers
`implies` x is a multiple of the LCM of 2,3,5 .
`implies` x is a multiple of 30.
Also, `0 lt x lt 1000`.
`:. x=30,60.90,.......990`
or, `x=30xx1,30xx2,30xx3,.......30xx33`
Hence, there are 33 values of x.