Question

If $\frac{(a^2{b}^{3}c{)}^4.(abc{)}^3}{(a^{3}b{c}^{-1}{)}^2}=\sqrt{a^m{b}^n{c}^p}$ then $m+n+p$ has the value equal to

• A. $24$
• B. $54$
• C. $27$
• D. $34$

Answer 1

The correct option is B $54$

Given,

$\frac{(a^2{b}^{3}c{)}^4.(abc{)}^3}{(a^{3}b{c}^{-1}{)}^2}=\sqrt{a^m{b}^n{c}^p}$

$⟹\frac{a^8.b^{12}.c^4.a^3.b^3.c^3}{a^6.b^2.c^{-2}}=\sqrt{a^m{b}^n{c}^p}$

$⟹a^5{b}^{13}{c}^9=\sqrt{a^m{b}^n{c}^p}$

Squaring on both sides

$⟹a^{10}.b^{26}.c^{18}=a^m.b^n.c^p$

On comparing we get,

$m=10,n=26,p=18$

Hence the value of $m+n+p=10+26+18=54$