Question

Simplify the given problem: $\dfrac{3}{2}\div \left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\div \left( \dfrac{3}{4}-\dfrac{1}{2} \right) \right\}$?

Answer 1

Hint: We start solving the problem by recalling the definition of the BODMAS rule to solve this rule. We then solve the operations between the numbers present in the inside most brackets. We then substitute this result in the question and do all the operations present inside the brackets following BODMAS rule. We then do the remaining operations using this result to get our final answer.

Complete step-by-step answer:
According to the problem, we need to find the value of $\dfrac{3}{2}\div \left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\div \left( \dfrac{3}{4}-\dfrac{1}{2} \right) \right\}$ ---(1).
We solve this problem by following the BODMAS (Brackets of Division, Multiplication, Addition and Subtraction) rule. We first solve the operations present inside the brackets in the following order.
From the given $\dfrac{3}{2}\div \left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\div \left( \dfrac{3}{4}-\dfrac{1}{2} \right) \right\}$, we can see that $\left( \dfrac{3}{4}-\dfrac{1}{2} \right)$ is present inside the brackets which is to be solved first.
So, we get $\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{3-2}{4}$.
$\Rightarrow \dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}$. We substitute this result in equation (1).
So, the problem in equation changes as $\dfrac{3}{2}\div \left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\div \dfrac{1}{4} \right\}$ ---(2).
We can see that $\left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\div \dfrac{1}{4} \right\}$ are present inside the having different operations. So, we need to solve this first.
We have division and addition operations present inside the bracket. According to the BODMAS (Brackets of Division, Multiplication, Addition and Subtraction) rule, we need to do the division operation present inside the brackets.
So, we have $\left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\div \dfrac{1}{4} \right\}=\left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{\dfrac{1}{6}}{\dfrac{1}{4}} \right\}$.
$\Rightarrow \left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\div \dfrac{1}{4} \right\}=\left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1\times 4}{6\times 1} \right\}$.
$\Rightarrow \left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\div \dfrac{1}{4} \right\}=\left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3} \right\}$.
Now, we perform additional operations left inside the brackets.
$\Rightarrow \left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\div \dfrac{1}{4} \right\}=\left\{ \dfrac{1}{2}+1 \right\}$.
$\Rightarrow \left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\div \dfrac{1}{4} \right\}=\left\{ \dfrac{1+2}{2} \right\}$.
$\Rightarrow \left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\div \dfrac{1}{4} \right\}=\left\{ \dfrac{3}{2} \right\}$. We substitute this result in equation (2).
So, the problem in equation changes as $\dfrac{3}{2}\div \dfrac{3}{2}$. So, we perform the division operation remaining here.
$\Rightarrow \dfrac{3}{2}\div \dfrac{3}{2}=1$.
So, we have found the value of $\dfrac{3}{2}\div \left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\div \left( \dfrac{3}{4}-\dfrac{1}{2} \right) \right\}$ as 1.
∴ The value of $\dfrac{3}{2}\div \left\{ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\div \left( \dfrac{3}{4}-\dfrac{1}{2} \right) \right\}$ is 1.

Note: We should not do the operations randomly between the numbers given without following the BODMAS rule. We should not make calculation mistakes while solving this problem. We should do the calculations step by step in order to avoid calculation mistakes and confusions. We should make sure that we are doing the division between rational numbers properly. Similarly, we can expect problems which may have more operations.