Find the value of $$\frac{0.5×0.5+0.09-0.15}{0.125+0.027}$$​​

Correct option is B

Given:

$$\frac{0.5×0.5+0.09-0.15}{0.125+0.027}$$

Concept Used: 

$$\begin{array}{|c|c|} \hline \textbf{Operation preference wise} & \textbf{Symbol} \\\hline \text{Brackets} &[],{}, () \\ \hline \text{Orders, of} & (power), √ (root) , of \\ \hline \text{Division}& ÷ \\ \hline \text{Multiplication} & × \\ \hline \text{Addition} & + \\ \hline \text{Subtraction} & - \\\hline \end{array}$$​

Solution:

$$\frac{0.5×0.5+0.09-0.15}{0.125+0.027}$$​​

= $$\frac{0.25+0.09-0.15}{0.125+0.027}$$

= $$\frac{0.34-0.15}{0.152}$$

= $$\frac{0.19}{0.152}$$

= $$\frac{190}{152}$$

= $$\frac{5}{4}$$

Thus, the correct answer is (b). 

Alternate Solution: ​

​$$a^3 +b^3 = (a+b) (a^2-ab+b^2)$$ 

Using the identity; 

​$$=\frac{0.5×0.5+0.09-0.15}{0.125+0.027} \\ \ \\ =\frac{0.5×0.5+0.09-0.15}{(0.5)^3+(0.3)^3}\\ \ \\ =\frac{0.5×0.5+0.09-0.15}{(0.5+0.3)(0.5×0.5+0.09-0.15)} \\ \ \\ = \frac{1}{(0.5 +0.3)} \\ \ \\ = \frac{1}{0.8}\\ \ \\ = \frac{10}{8} = \frac{5}{4}$$​​