Hi,
Alternate but lengthy approach
Let number of cars be n , and numbers of sales persons be S
then
From Stmt 1: n= 4S+23
From Stmt 2: n= 6S+5
Combining 1 and 2 we have
From Stmt 2 :\(\frac{n}{6}\)reminder 5
or \(\frac{4S+23}{6}\)
Note that we will have a cylicity of 3 for reminders here,this is what i mean,
For s=1 n= 27 and \(\frac{27}{6}\) reminder will be 3
For s=2 n= 31 and \(\frac{31}{6}\) reminder will be 1
For s=3 n= 35 and \(\frac{35}{6}\) reminder will be 5
For s=4 n= 39 and \(\frac{39}{6}\) reminder will be 3
we get reminder 5 when n=35 or 47 or 59
Now lets check at which value does it give reminder of 23 when divided by 4.
But when \(\frac{35}{6}\) we have 5 sales person and 5 cars remaining but when \(\frac{35}{4}\) when we have 5 sales person each selling 4 cars we will have 15 cars remaining
But when \(\frac{47}{6}\) we have 7 sales person and 5 cars remaining but when \(\frac{47}{4}\) when we have 7 sales person each selling 4 cars we will have 19 cars remaining
But when \(\frac{59}{6}\) we have 9 sales person and 5 cars remaining but when \(\frac{59}{4}\) when we have 9 sales person each selling 4 cars we will have 23 cars remaining Bingo
So Answer is C
Probus
_________________
Probus
~You Just Can't beat the person who never gives up~ Babe Ruth