let us say, the number is "a"
one less than "a" is ... well a - 1
2/5 of that will be... well, 2/5 * ( a - 1)
one more than "a" is... well a + 1
3/5 of that is.. hmmm 3/5 * (a + 1)
now, we're told that, the first expression is "less than" the second one, thus
[tex]\bf \cfrac{2}{5}(a-1)\ \textless \ \cfrac{3}{5}(a+1)[/tex]
solve for "a", or make it "x" if you wish, same thing anyway