Mathematics - Sets and Relations Question with Solution | TestHub

Let $Y_i$ be the subset of $X$ such that $y_i=7m+i,m\in I$

${\mathrm{Y}}_0=\{7,14,\dots ..98\},\mathrm{n}\left({\mathrm{Y}}_0\right)=14$

${\mathrm{Y}}_1=\{1,8,15\dots 99\},\mathrm{n}\left({\mathrm{Y}}_1\right)=15$

${\mathrm{Y}}_2=\{2,9,16\dots ,100\},\mathrm{n}\left({\mathrm{Y}}_2\right)=15$

$Y_3=\{3,10,16\dots .94\},n{Y}_3)=14$

${\mathrm{Y}}_4=\{4,11,18\dots 95\},\mathrm{n}\left({\mathrm{Y}}_4\right)=14$

$Y_5=\{5,12,\dots .96\},n{Y}_5)=14$

$Y_6=\{6,13,\dots .97\},n\left(Y_6\right)=14$

The largest $Y$ will consist of (!) an element of $Y_0$ (ii) $Y_1$ (iii) $Y_2$ (iv) $Y_3$ or $Y_4$

⇒ The maximum possible number of elements in $\mathrm{Y}=1+15+15+14=45$.