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Telangana TET Paper 1 Exam Model Paper 2

Section: Mathematics

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Question : 120 of 150

Marks: +1, -0

Simplify:

\frac{(8.3)^{3}+(9.2)^{3}+(6.1)^{3}-3\times 8.3\times 9.2\times 6.1}{(8.3)^{2}+(9.2)^{2}+(6.1)^{2}-8.3\times 9.2-9.2\times 6.1-6.1\times 8.3}

30.2
23.6
28.7
25.5

Solution:

Concept:

This expression follows the algebraic identity:

a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

. The given fraction simplifies directly to

a + b + c

.

Explanation:

Let

a = 8.3

,

b = 9.2

, and

c = 6.1

.

Write numerator as

a^3 + b^3 + c^3 - 3abc

and denominator as

a^2 + b^2 + c^2 - ab - bc - ca

.

Using the identity, the numerator equals

(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

.

Cancel the common factor

(a^2 + b^2 + c^2 - ab - bc - ca)

from numerator and denominator.

The result is

a + b + c = 8.3 + 9.2 + 6.1 = 23.6

.

Answer:

23.6 (Option B)

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