Question

How do you solve $9-\dfrac{3}{\dfrac{1}{3}}+1$ ?

Answer 1

Hint:
In the above question, we were asked to simplify $9-\dfrac{3}{\dfrac{1}{3}}+1$ , as you can see that we have to add and subtract the whole number with the fraction. For which we have to solve the fractional part first and then we will add 1 and finally we will subtract that from 9. So let’s solve this problem.

Complete step by step solution:
In this question, we will first solve the fractional part which is $\dfrac{3}{\dfrac{1}{3}}$ . This fraction can also be written as $3\div \dfrac{1}{3}$
Now, turn $\dfrac{1}{3}$ upside down and multiply
 $\therefore 3\times \dfrac{3}{1}$ = 9
Putting 9 in the place of $\dfrac{3}{\dfrac{1}{3}}$ in our problem, we will get
= 9 – 9 + 1
= 1

Therefore, $9-\dfrac{3}{\dfrac{1}{3}}+1$ = 1

Additional Information:
In this solution, we got an improper fraction which is $\dfrac{9}{1}$. But there are other forms of fractions as well like faction, unlike fraction, mixed fraction, equivalent fraction, and a proper fraction.

Note:
In the above solution, we have first solved $\dfrac{3}{\dfrac{1}{3}}$. It can also be solved in another way. Suppose we have to solve $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}$. We know that $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a\times d}{b\times c}$. Our problem was $\dfrac{3}{\dfrac{1}{3}}$, here the numerator 3 has no denominator which means that it has 1 as the denominator. So it becomes $\dfrac{\dfrac{3}{1}}{\dfrac{1}{3}}$ which is $\dfrac{3\times 3}{1\times 1}$. Therefore, we concluded that $\dfrac{3}{\dfrac{1}{3}}$ is equal to 9.