Mathematics - Functions Question with Solution | TestHub

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

$Y_0=\{7,14,\dots ,98\}$, $n(Y_0)=14$

$Y_1=\{1,8,15,\dots ,99\}$, $n(Y_1)=15$

$Y_2=\{2,9,16,\dots ,100\}$, $n(Y_2)=15$

$Y_3=\{3,10,16,\dots ,94\}$, $n(Y_3)=14$

$Y_4=\{4,11,18,\dots ,95\}$, $n(Y_4)=14$

$Y_5=\{5,12,\dots ,96\}$, $n(Y_5)=14$

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

The largest $Y$ will consist of:

(i) an element of $Y_0$

(ii) $Y_1$

(iii) $Y_2$

(iv) $Y_3$ or $Y_4$

Therefore, the maximum possible number of elements in $Y=1+15+15+14=45$.