DRDO Multi tasking staff Model paper 2

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

10

30

+

15

30

−

25

30

+

9

30

=

9

30

=

3

10