Question

If $\mathrm{III}$ stands for $2$, $\mathrm{IIII}$ for $3$, and $\mathrm{II}$ for $1$, solve the following : $\mathrm{II}-\mathrm{IIII}+\mathrm{III}+\mathrm{IIIII}=?$

Answer 1

Solve the given pattern based on given information

It is given that,

$\mathrm{III}$ stands for $2$,

And, $\mathrm{IIII}$ stands for $3$,

And, $\mathrm{II}$ stands for $1$.

Then, $\mathrm{IIIII}$ must for $4$.

So, the value of $\mathrm{II}-\mathrm{IIII}+\mathrm{III}+\mathrm{IIIII}$ can be calculated as,

. $\mathrm{II}-\mathrm{IIII}+\mathrm{III}+\mathrm{IIIII}=1-3+2+4$

$\Rightarrow \mathrm{II}-\mathrm{IIII}+\mathrm{III}+\mathrm{IIIII}=4$

Hence, the required value of $\mathrm{II}-\mathrm{IIII}+\mathrm{III}+\mathrm{IIIII}$ is $4$.