MCQ
Which of the following is not equal to $1$?
- A$\frac{2^{3}\times3^{3}}{4\times18}$
- B$\big[(-2)^{3}\times(-2)^{4}\big]\div(27)^{7}$
- C$\frac{2^{0}\times5^{3}}{5\times25}$
- ✓$\frac{2^{4}}{(7^{0}+3^{0})}$
Let option $(a)$ $\frac{2^{3}\times3^{3}}{4\times18}=\frac{2\times2\times2\times3\times3}{4\times18}$
$=\frac{4\times18}{4\times18}=1$
For option $(b)$, $[(-2)^{3}\times(-2)^{4}]\div(-2)^{7}=\frac{(-2)^{3}\times(-2)^{}4}{(-2)^{7}}$
$=\frac{(-2)^{3+4}}{(-2)^{7}}$ $\big[\because\text{a}^{\text{m}}\times\text{a}^{\text{n}}=\text{a}^{\text{m+n}}\big]$
$=\frac{(-2)^{7}}{(-2)^{7}}=1$ $\big[\because\frac{\text{a}^{\text{m}}}{\text{a}^{n}}=\text{a}^{\text{m-n}}\big]$
For option $(c)$, $=\frac{3^{0}\times5^{3}}{5\times25}=\frac{1\times5\times5\times5}{5\times25}$
$=\frac{5\times25}{5\times25}=1$ $\big[\because\text{a}^{0}=1\big]$
For option $(d)$, $\frac{2^{4}}{(7^{0}+3^{0})^{3}}=\frac{2^{4}}{(1+1)^{3}}$ $\big[\because\text{a}^{0}=1\big]$
$=\frac{2^{4}}{2^{3}}=2^{4-3}$ $\big[\because\frac{\text{a}^{m}}{\text{a}^{\text{n}}}=\text{a}^{\text{m-n}},\text{m>n}\big]$
$=-2^{1}=2$
Hence, option $(d)$ is not equal to $1$.
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