Question

The value of $\frac{2^{m+1}\times 3^{2m-n}\times 5^{m+n}\times 6^n}{6^m\times {10}^{n+2}\times {15}^m}$ is equal to

• A. $50$
• B. $\frac{1}{25}$
• C. $\frac{1}{50}$
• D. $\frac{1}{5}$

Answer 1

Answer: (C)

$\frac{1}{50}$

Now $\frac{2^{m+1}\times 3^{2m-n}\times 5^{m+n}\times 6^n}{6^m\times {10}^{n+2}\times {15}^m}$

$=\frac{2^m{\times 2}^1\times \frac{3^{2m}}{3^n}\times 5^m\times 5^n\times {(2\times 3)}^n}{{(2\times 3)}^m{\times (2\times 5)}^{n+2}\times {(3\times 5)}^m}$

$=\frac{2^m{\times 2}^1\times 3^{2m}\times 5^m\times 5^n\times {(2\times 3)}^n}{{3^n\times (2\times 3)}^m{\times (2\times 5)}^{n+2}\times {(3\times 5)}^m}$

$=\frac{2^m{\times 2}^1\times 3^{2m}\times 5^m\times 5^n\times 2^n\times 3^n}{3^n\times 2^m\times 3^m\times 2^{n+2}\times 5^{n+2}\times 3^m\times 5^m}$

$=\frac{2\times 3^{2m}\times 5^{n-(n+2)}\times 2^{n-(n+2)}}{3^{m+m}}$

$=\frac{2\times 3^{2m}\times 5^{-2}\times 2^{-2}}{3^{2m}}$

$=2\times 5^{-2}\times 2^{-2}$

$=2\times {(5\times 2)}^{-2}$

$=\frac{2}{25\times 4}$

$=\frac{1}{50}$