Question

Simplify: \[{\left( {100} \right)^{\dfrac{1}{2}}} \times {\left( {0.001} \right)^{\dfrac{1}{3}}} - {\left( {0.0016} \right)^{\dfrac{1}{4}}} \times {3^0} + {\left( {\dfrac{5}{4}} \right)^{ - 1}}\]
A. \[1.6\]
B. \[2.4\]
C. 2
D. \[1.4\]

Answer 1

Hint:
Here we will firstly convert the decimal digits into the fraction form. Then we will write it in the exponent form to cancel out the exponent outside the brackets. Then we will simplify and solve the equation to get the required value.

Complete Step by step Solution:
Given equation is \[{\left( {100} \right)^{\dfrac{1}{2}}} \times {\left( {0.001} \right)^{\dfrac{1}{3}}} - {\left( {0.0016} \right)^{\dfrac{1}{4}}} \times {3^0} + {\left( {\dfrac{5}{4}} \right)^{ - 1}}\]
First, we will write the decimal numbers in the fractional form for the simplification of the equation. Therefore, we get
\[ \Rightarrow {\left( {100} \right)^{\dfrac{1}{2}}} \times {\left( {0.001} \right)^{\dfrac{1}{3}}} - {\left( {0.0016} \right)^{\dfrac{1}{4}}} \times {3^0} + {\left( {\dfrac{5}{4}} \right)^{ - 1}} = {\left( {100} \right)^{\dfrac{1}{2}}} \times {\left( {\dfrac{1}{{1000}}} \right)^{\dfrac{1}{3}}} - {\left( {\dfrac{{16}}{{10000}}} \right)^{\dfrac{1}{4}}} \times {3^0} + {\left( {\dfrac{5}{4}} \right)^{ - 1}}\]
Now we will write the numbers in terms of the square or the exponent form so as to cancel out its own exponential power outside the bracket. Therefore, we get
\[ \Rightarrow {\left( {100} \right)^{\dfrac{1}{2}}} \times {\left( {0.001} \right)^{\dfrac{1}{3}}} - {\left( {0.0016} \right)^{\dfrac{1}{4}}} \times {3^0} + {\left( {\dfrac{5}{4}} \right)^{ - 1}} = {\left( {{{10}^2}} \right)^{\dfrac{1}{2}}} \times {\left( {\dfrac{1}{{{{10}^3}}}} \right)^{\dfrac{1}{3}}} - {\left( {\dfrac{{{2^4}}}{{{{10}^4}}}} \right)^{\dfrac{1}{4}}} \times {3^0} + {\left( {\dfrac{5}{4}} \right)^{ - 1}}\]
Now simplifying the equation by cancelling out the exponent of the terms, we get
\[ \Rightarrow {\left( {100} \right)^{\dfrac{1}{2}}} \times {\left( {0.001} \right)^{\dfrac{1}{3}}} - {\left( {0.0016} \right)^{\dfrac{1}{4}}} \times {3^0} + {\left( {\dfrac{5}{4}} \right)^{ - 1}} = 10 \times \dfrac{1}{{10}} - \dfrac{2}{{10}} \times 1 + \dfrac{4}{5}\]
Multiplying the terms, we get
\[ \Rightarrow {\left( {100} \right)^{\dfrac{1}{2}}} \times {\left( {0.001} \right)^{\dfrac{1}{3}}} - {\left( {0.0016} \right)^{\dfrac{1}{4}}} \times {3^0} + {\left( {\dfrac{5}{4}} \right)^{ - 1}} = 1 - \dfrac{1}{5} + \dfrac{4}{5}\]
Converting the fraction into decimal, we get
\[ \Rightarrow {\left( {100} \right)^{\dfrac{1}{2}}} \times {\left( {0.001} \right)^{\dfrac{1}{3}}} - {\left( {0.0016} \right)^{\dfrac{1}{4}}} \times {3^0} + {\left( {\dfrac{5}{4}} \right)^{ - 1}} = 1 - 0.2 + 0.8\]
Adding the terms, we get
\[ \Rightarrow {\left( {100} \right)^{\dfrac{1}{2}}} \times {\left( {0.001} \right)^{\dfrac{1}{3}}} - {\left( {0.0016} \right)^{\dfrac{1}{4}}} \times {3^0} + {\left( {\dfrac{5}{4}} \right)^{ - 1}} = 1.6\]
Hence the value of the expression \[{\left( {100} \right)^{\dfrac{1}{2}}} \times {\left( {0.001} \right)^{\dfrac{1}{3}}} - {\left( {0.0016} \right)^{\dfrac{1}{4}}} \times {3^0} + {\left( {\dfrac{5}{4}} \right)^{ - 1}}\] is equal to \[1.6\].

So, option A is the correct option.

Additional information:
We perform different mathematical operations while solving an arithmetic expression such as addition, subtraction, etc. Addition is the operation in which two numbers are combined to get the result. Subtraction is the operation that gives us the difference between the two numbers. Multiplication is the operation in which the one number is added to itself for some particular number of times. Exponent is the defined as the number, which represents how many times a number is being multiplied to itself.

Note:
Here we have to note that while solving these types of equations we have to use the rule of BODMAS. We know that in mathematics, a mathematical expression is solved according to the BODMAS Rule. We should apply the mathematical operations in a particular order which is given by the letters of BODMAS and letters of BODMAS are defined as B-Brackets, O-Of, D-Division, M-Multiplication, A-Addition and S-Subtraction.