Question

The value of $\dfrac{3}{5} \times 1\dfrac{7}{8} \div 1\dfrac{2}{3}of\dfrac{3}{{16}} - (3\dfrac{2}{5} \div 4\dfrac{1}{2}of5\dfrac{1}{3}) \times 2\dfrac{3}{2} + \dfrac{3}{2} + \dfrac{3}{8} \div \dfrac{2}{4}$ is:-

Answer 1

Hint: In this question we will first convert the mixed form of fractions into improper form of fractions then we will apply BODMAS whose full form is mentioned below:-
B – Brackets
O – Of
D – Division
M – Multiplication
A – Addition
S – Subtraction

Complete step-by-step answer:
First of all we will convert the mixed form of fractions into improper form of fractions
$ \Rightarrow \dfrac{3}{5} \times \dfrac{{15}}{8} \div \dfrac{5}{3}of\dfrac{3}{{16}} - (\dfrac{{17}}{5} \div \dfrac{9}{2}of\dfrac{{16}}{3}) \times \dfrac{7}{2} + \dfrac{3}{2} + \dfrac{3}{8} \div \dfrac{2}{4}$
Now we will solve according to the BODMAS. As in this equation there are only one brackets so we will first solve that and then ‘of’ which is multiplication.
$ \Rightarrow \dfrac{3}{5} \times \dfrac{{15}}{8} \div \dfrac{5}{3}of\dfrac{3}{{16}} - (\dfrac{{17}}{5} \div \dfrac{9}{2} \times \dfrac{{16}}{3}) \times \dfrac{7}{2} + \dfrac{3}{2} + \dfrac{3}{8} \div \dfrac{2}{4}$
Here we had applied BODMAS inside the bracket and first converted ‘of’ into multiplication then will now do division.
$ \Rightarrow \dfrac{3}{5} \times \dfrac{{15}}{8} \div \dfrac{5}{3}of\dfrac{3}{{16}} - (\dfrac{{17}}{5} \div 24) \times \dfrac{7}{2} + \dfrac{3}{2} + \dfrac{3}{8} \div \dfrac{2}{4}$
Now we will remove the brackets and solve the division.
 $ \Rightarrow \dfrac{3}{5} \times \dfrac{{15}}{8} \div \dfrac{5}{3}of\dfrac{3}{{16}} - \dfrac{{17}}{{720}} \times \dfrac{7}{2} + \dfrac{3}{2} + \dfrac{3}{8} \div \dfrac{2}{4}$
Now we solve ‘of’ and then division.
$ \Rightarrow \dfrac{3}{5} \times \dfrac{{15}}{8} \times \dfrac{3}{5} \times \dfrac{3}{{16}} - \dfrac{{17}}{{720}} \times \dfrac{7}{2} + \dfrac{3}{2} + \dfrac{3}{8} \times \dfrac{4}{2}$
Now we will solve the multiplication sign and cancel terms from numerator and denominator.
$ \Rightarrow \dfrac{{81}}{{640}} - \dfrac{{119}}{{240}} + \dfrac{3}{2} + \dfrac{3}{4}$
Now we will solve addition and then subtraction
$ \Rightarrow \dfrac{{81}}{{640}} - \dfrac{{119}}{{240}} + \dfrac{9}{4}$
Now we will take L.C.M
$ \Rightarrow \dfrac{{81 \times 3 - 119 \times 8 + 480 \times 9}}{{1920}}$
$ \Rightarrow \dfrac{{243 - 952 + 4320}}{{1920}} = \dfrac{{3611}}{{1920}}$
We can also change this answer into mixed form of fraction $\therefore \dfrac{{3611}}{{1920}} = 1\dfrac{{1671}}{{1920}}$ so both are correct.

Note: Students may find difficulty in conversion of mixed to improper and vice versa so here below is a brief explanation of that:-
First of all improper fractions are those whose numerator is greater than denominator and proper fractions are those whose numerator is less than denominator.
Let us take one example $5\dfrac{1}{3}$ is a mixed fraction so to convert it into improper fraction multiply 5 with 3 and then add 1 to it will give $\dfrac{{16}}{3}$