Question

If (FAST) = 798,(LAST) = 906,Then , (BUSY)=??
​all explaining need

Answer 1

$\sf{\bold{\blue{\underline{\underline{Given}}}}}$

• FAST=798
• LAST=906⠀⠀⠀⠀

$\sf{\bold{\red{\underline{\underline{To\:Find}}}}}$

• BUSY=????⠀⠀

$\sf{\bold{\purple{\underline{\underline{Solution}}}}}$

Here,

FAST=798

• F is the 6th letter in English alphabet.
• A is the 1st letter in English alphabet.
• S is the 19th letter in English alphabet.
• T is the 20th letter in English alphabet. ⠀⠀⠀⠀

so,

:$\mapsto$F+A+S+T=798

:$\longrightarrow$ $\bold{6^2+1^2+19^2+20^2   }$ 

:$\longrightarrow$ $\tt{36+1+361+400   }$      

:$\longrightarrow$ $\sf{ 798  }$✔✔

Again,

LAST=906

• L is the 12th letter in English alphabet.
• A is the 1st letter in English alphabet.
• S is the 19th letter in English alphabet.
• T is the 20th letter in English alphabet.

So,

:$\mapsto$L+A+S+T=906

:$\rightsquigarrow$ $\bold{12^2+1^2+19^2+20^2   }$

:$\rightsquigarrow$ $\tt{144+1+361+400   }$

:$\rightsquigarrow$ $\sf{ 908  }$ ✔✔

Finally,

BUSY=?

• B is the 2nd letter in English alphabet.
• U is the 21th letter in English alphabet.
• S is the 19th letter in English alphabet.
• Y is the 25th letter in English alphabet.

Therefore,

:$\mapsto$  B+U+S+Y=

:$\hookrightarrow$ $\bold{2^2+21^2+19^2+25^2  }$      

:$\hookrightarrow$ $\tt{4+441+361+625   }$      

:$\hookrightarrow$ $\sf{1431 }$      

Final answer:

$\therefore$we ,get that $\underline {\colorbox{aqua}{\sf{BUSY=1431}}   }$