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MAT Exam 2023 Paper

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Question : 108 of 200

Marks: +1, -0

The value of

\frac{1.073 \times 1.073-0.927 \times 0.927}{1.073-0.927}+\frac{\left(3^{4}\right)^{4} \times 9^{6}}{(27)^{7} \times(3)^{9}} \text { is }

2
$\frac{1}{9}$
$2 \frac{1}{9}$
$3 \frac{1}{9}$

Solution:

Concept:

Use the algebraic identity

a^2 - b^2 = (a+b)(a-b)

and the laws of exponents to simplify both parts.

Explanation:

Let

a = 1.073

and

b = 0.927

.

First part:

\frac{a^2 - b^2}{a - b} = \frac{(a+b)(a-b)}{a-b} = a+b

.

So,

a + b = 1.073 + 0.927 = 2

.

Second part:

(3^4)^4 \times 9^6 = 3^{16} \times (3^2)^6 = 3^{16} \times 3^{12} = 3^{28}

.

Denominator:

(27)^7 \times 3^9 = (3^3)^7 \times 3^9 = 3^{21} \times 3^9 = 3^{30}

.

Thus, second part

= \frac{3^{28}}{3^{30}} = 3^{-2} = \frac{1}{9}

.

Total value

= 2 + \frac{1}{9} = 2\frac{1}{9}

.

Answer:

2\frac{1}{9}

, i.e. option C.

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