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RRB Technician Grade I 19 Dec 2024 Shift 2 Paper

Question : 55 of 100

Marks: +1, -0

The value of

1^{3}-8^{2}+(\frac{36}{4})^{2}-0+2\times 5=

______

Solution:

Concept:

Use the BODMAS rule to evaluate the expression step by step.

Explanation:

First, solve the bracket:

\frac{36}{4} = 9

The expression becomes

1^{3} - 8^{2} + 9^{2} - 0 + 2 \times 5

Now evaluate the powers (orders):

1^{3} = 1

8^{2} = 64

9^{2} = 81

The expression becomes

1 - 64 + 81 - 0 + 2 \times 5

Next, do the multiplication:

2 \times 5 = 10

Now the expression is

1 - 64 + 81 - 0 + 10

Add the positive terms:

1 + 81 + 10 = 92

Subtract

64

from

92

92 - 64 = 28

Answer:

The value of the expression is

28

So the correct option is A. 28.

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