Before we proceed to solve this, we need to understand BODMAS: B – Brackets O – OF D – Division M – Multiplication A – Addition S – Subtraction This is the order in which the operators are to be solved. We have
1
3
+(
5
8
×
4
5
)−(
2
3
÷
4
5
)+
3
4
of
2
5
We have two brackets in this case. Hence, we solve them first.
5
8
×
4
5
=
1
2
And,
2
3
÷
4
5
=
2
3
×
5
4
=
5
6
Substituting them back, we have
1
3
+
1
2
−
5
6
+
3
4
of
2
5
Next we solve that operator containing “OF”. Hence,
3
4
of
2
5
=
3
4
×
2
5
=
3
10
Substituting them back, we have
1
3
+
1
2
−
5
6
+
3
10
This is the simplified version. The LCM of the denominators is 30. Hence, we have