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RRB Technician Grade I 19 Dec 2024 Shift 2 Paper

Question : 88 of 100

Marks: +1, -0

The value of

3^{3}-5^{2}+(\frac{26}{13})^{2}-8+0\times 9=

_________

Solution:

Concept:

Evaluate the expression using the BODMAS rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction).

Explanation:

Given expression:

3^{3} - 5^{2} + \left(\frac{26}{13}\right)^{2} - 8 + 0 \times 9

First, solve the powers and the bracket:

3^{3} = 27,\quad 5^{2} = 25,\quad \left(\frac{26}{13}\right)^{2} = 2^{2} = 4

Also,

0 \times 9 = 0

Substitute these values:

27 - 25 + 4 - 8 + 0

Now perform addition and subtraction from left to right:

27 - 25 = 2

2 + 4 = 6

6 - 8 = -2

-2 + 0 = -2

Answer:

-2

Hence, the correct option is D.

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