CTET 2 Math and Science 21 Jan 2024 Paper

take option 1:
$\mathit{x}=3$ and $\mathit{y}=1$
means $9471\times 9\mathit{y}2$ becomes 94713912
$9+4+7+1+\mathit{x}+9+\mathit{y}+2=32+\mathit{x}+\mathit{y}=32+3+1=36$ is divisble by 9
Last three digits 912 is also divisible by 8
Hence, Divisble by 72
take option 2:
$\mathit{x}=8$ and $\mathit{y}=5$
means $9471\times 9\mathit{y}2$ becomes 94718952
$9+4+7+1+\mathit{x}+9+\mathit{y}+2=32+\mathit{x}+\mathit{y}=32+8+5=45$ is divisble by 9
Last three digits 952 is also divisible by 8
Hence, Divisble by 72
take option 3:
$\mathit{x}=4$ and $\mathit{y}=9$
means $9471\times 9\mathit{y}2$ becomes 94714992
$9+4+7+1+4+9+9+2=32+\mathit{x}+\mathit{y}=32+4+9=45$ is divisble by 9
Last three digits 992 is also divisible by 8
Hence, Divisble by 72
take option 4:
$\mathit{x}=9$ and $\mathit{y}=5$
means $9471\times 9\mathit{y}2$ becomes 94719952
$9+4+7+1+9+9+5+2=32+\mathit{x}+\mathit{y}=32+9+5=46$ is not divisble by 9
Last three digits 952 is also divisible by 8
Hence, Not Divisble by 72
Hence, The Correct option is option 4.