Question

The value of $\dfrac{16}{5}-2\dfrac{1}{10}+\dfrac{8}{15}$ is
$\begin{align}
  & a)1\dfrac{17}{30} \\
 & b)1\dfrac{19}{30} \\
 & c)1\dfrac{23}{30} \\
 & d)\text{ none of these} \\
\end{align}$

Answer 1

Hint: Now we are given with the expression $\dfrac{16}{5}-2\dfrac{1}{10}+\dfrac{8}{15}$ . First we will convert $2\dfrac{1}{10}$ into a proper fraction as we know $a\dfrac{p}{q}=\dfrac{a\times q+p}{q}$ . Now we will take LCM of the denominators and convert them in a way such that all the denominators are the same. Now since the denominators are the same we can add and subtract three fractions and hence find the result. Now we will again convert the obtained fraction into a mixed fraction.

Complete step by step answer:
Now let us first consider the term $2\dfrac{1}{10}$ .
Now we know that $2\dfrac{1}{10}$ means $\dfrac{2\times 10+1}{10}$
Hence we get,
$\begin{align}
  & 2\dfrac{1}{10}=\dfrac{20+1}{10} \\
 & \Rightarrow 2\dfrac{1}{10}=\dfrac{21}{10}..............\left( 1 \right) \\
\end{align}$
Now consider the given equation $\dfrac{16}{5}-2\dfrac{1}{10}+\dfrac{8}{15}$
Substituting the value of $2\dfrac{1}{10}$ from equation (1) we get,
$\dfrac{16}{5}-2\dfrac{1}{10}+\dfrac{8}{15}=\dfrac{16}{5}-\dfrac{21}{10}+\dfrac{8}{15}$
Now we know to add or subtract fractions the denominator must be the same.
But here we have 5, 10 and 15 as denominators which are all different.
Now we will make the denominators the same by taking LCM.
Now LCM of 5 = 5 × 1, 10 = 5 × 2 and 15 = 5 × 3.
Hence the LCM will be 5 × 3 × 2 = 30.
Now we will convert each fraction such that its denominator is 30.
To do so we will multiply the numerators and denominators of $\dfrac{16}{5},\dfrac{21}{10}$ and $\dfrac{8}{15}$ with 6, 3 and 2 respectively. Hence we get,
$\begin{align}
  & \dfrac{16}{5}-2\dfrac{1}{10}+\dfrac{8}{15}=\dfrac{16\times 6}{5\times 6}-\dfrac{21\times 3}{10\times 3}+\dfrac{8\times 2}{15\times 2} \\
 & \Rightarrow \dfrac{16}{5}-2\dfrac{1}{10}+\dfrac{8}{15}=\dfrac{96}{30}-\dfrac{63}{30}+\dfrac{16}{30} \\
\end{align}$
Now since the denominators are the same we can use addition of fraction to simplify.
We know that $\dfrac{a}{b}-\dfrac{c}{b}=\dfrac{a-c}{b}$ and $\dfrac{a}{b}+\dfrac{c}{b}=\dfrac{a+c}{b}$
$\dfrac{16}{5}-2\dfrac{1}{10}+\dfrac{8}{15}=\dfrac{96-63+16}{30}$
$\begin{align}
  & \dfrac{16}{5}-2\dfrac{1}{10}+\dfrac{8}{15}=\dfrac{49}{30} \\
 & \Rightarrow \dfrac{16}{5}-2\dfrac{1}{10}+\dfrac{8}{15}=\dfrac{30+19}{30} \\
 & \Rightarrow \dfrac{16}{5}-2\dfrac{1}{10}+\dfrac{8}{15}=\dfrac{30\times 1+19}{30} \\
 & \Rightarrow \dfrac{16}{5}-2\dfrac{1}{10}+\dfrac{8}{15}=1\dfrac{19}{30} \\
\end{align}$
Hence we have the value of given expression is $1\dfrac{19}{30}$

So, the correct answer is “Option b”.

Note: Now note that $2\dfrac{1}{10}=2+\dfrac{1}{10}$ and not $2\dfrac{1}{10}\ne 2\times \dfrac{1}{10}$ . Hence note that we cannot cancel 10 with 2. Writing $2\dfrac{1}{10}=\dfrac{1}{5}$ . Also note that while adding or subtracting fractions we want the denominators to be the same. We cannot add or subtract fractions with different denominators.