Question

The square root of $\dfrac{0.324\times 0.081\times 4.624}{1.5625\times 0.0289\times 72.9\times 64}$ is
(a) 0.024
(b) 0.204
(c) 0.24
(d) 240

Answer 1

Hint: Write the given expression by removing the decimals from all the numbers given. Write each number as a product of their prime factors and make groups of two identical factors so that it can be written as a square of that factor. Now, cancel all the common terms and simplify the expression. Finally take the square root of the simplified number obtained, to get the answer.

Complete step-by-step answer:
Let us assume a given expression: $\dfrac{0.324\times 0.081\times 4.624}{1.5625\times 0.0289\times 72.9\times 64}$ as ‘E’. Therefore,
$E=\dfrac{0.324\times 0.081\times 4.624}{1.5625\times 0.0289\times 72.9\times 64}$
Removing the decimals from all the numbers, we get,
$\begin{align}
  & E=\dfrac{\dfrac{324}{1000}\times \dfrac{81}{1000}\times \dfrac{4624}{1000}}{\dfrac{15625}{10000}\times \dfrac{289}{10000}\times \dfrac{729}{10}\times 64} \\
 & \Rightarrow E=\dfrac{324\times 81\times 4624}{15625\times 289\times 729\times 64} \\
\end{align}$
Now, writing all the terms as a product of their prime factors, we get,
$E=\dfrac{\left( 3\times 3\times 2\times 3\times 3\times 2 \right)\times \left( 3\times 3\times 3\times 3 \right)\times \left( 2\times 2\times 2\times 2\times 17\times 17 \right)}{\left( 5\times 5\times 5\times 5\times 5\times 5 \right)\times \left( 17\times 17 \right)\times \left( 3\times 3\times 3\times 3\times 3\times 3 \right)\times \left( 2\times 2\times 2\times 2\times 2\times 2 \right)}$
This expression can be written in the exponential form as:
\[E=\dfrac{\left( {{3}^{4}}\times {{2}^{2}} \right)\times \left( {{3}^{4}} \right)\times \left( {{2}^{4}}\times {{17}^{2}} \right)}{\left( {{5}^{6}} \right)\times \left( {{17}^{2}} \right)\times \left( {{3}^{6}} \right)\times \left( {{2}^{6}} \right)}\]
Using the formula: \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\], in the numerator, we get,
\[\begin{align}
  & E=\dfrac{\left( {{3}^{4+4}}\times {{2}^{2+4}} \right)\times \left( {{17}^{2}} \right)}{\left( {{5}^{6}} \right)\times \left( {{17}^{2}} \right)\times \left( {{3}^{6}} \right)\times \left( {{2}^{6}} \right)} \\
 & \Rightarrow E=\dfrac{\left( {{3}^{8}}\times {{2}^{6}} \right)\times \left( {{17}^{2}} \right)}{\left( {{5}^{6}} \right)\times \left( {{17}^{2}} \right)\times \left( {{3}^{6}} \right)\times \left( {{2}^{6}} \right)} \\
\end{align}\]
Now, cancelling the common terms and using the formula: \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\], we get,
\[\begin{align}
  & E=\dfrac{\left( {{3}^{8-6}} \right)}{\left( {{5}^{6}} \right)} \\
 & \Rightarrow E=\dfrac{\left( {{3}^{2}} \right)}{\left( {{5}^{6}} \right)} \\
 & \Rightarrow E=\dfrac{\left( {{3}^{2}} \right)}{\left( {{5}^{2\times 3}} \right)} \\
\end{align}\]
Using the formula: \[{{a}^{m\times n}}={{\left( {{a}^{m}} \right)}^{n}}\], this can be written as:
\[E=\dfrac{\left( {{3}^{2}} \right)}{{{\left( {{5}^{3}} \right)}^{2}}}\]
Now, taking the square root, we have,
\[\begin{align}
  & \sqrt{E}=\sqrt{\dfrac{\left( {{3}^{2}} \right)}{{{\left( {{5}^{3}} \right)}^{2}}}} \\
 & \Rightarrow \sqrt{E}=\dfrac{3}{{{5}^{3}}} \\
\end{align}\]
Now, multiplying the numerator and denominator by ${{2}^{3}}$, we have,
\[\sqrt{E}=\dfrac{3}{{{5}^{3}}}\times \dfrac{{{2}^{3}}}{{{2}^{3}}}\]
Using the formula: \[{{a}^{n}}\times {{b}^{n}}={{\left( a\times b \right)}^{n}}\], we have,
\[\begin{align}
  & \sqrt{E}=\dfrac{3\times 8}{{{\left( 5\times 2 \right)}^{3}}} \\
 & \Rightarrow \sqrt{E}=\dfrac{24}{{{10}^{3}}} \\
 & \Rightarrow \sqrt{E}=0.024 \\
\end{align}\]
Hence, option (a) is the correct answer.

Note: One may note that we have broken all the given numbers into its prime factors because we have to form groups of even powers of the factors so that the square root can be taken easily in the end. The basic reason behind the removal of decimal is to reduce calculation mistakes. If we will break the numbers into their prime factors without removing the decimal then it may lead to confusion and subsequently calculation mistakes. Also one more thing to note is that we have multiplied the numerator and denominator of the obtained square root by ${{2}^{3}}$. This is done because we have to convert the answer into the decimal form and if there is 10 or any power of 10 in the denominator then the calculation becomes easy.