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UPSC CDS 1 2022 Math Paper

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Question : 13 of 100

Marks: +1, -0

If

43^{x} \times 47^{y} = (2021)^{2}

x \neq 0

y \neq 0

, then what is the value of the following?

\frac{4xy + x + y}{2xy - x - y}

5
15
25
45

Solution:

Concept:

Use prime factorization to equate exponents and then evaluate the given rational expression.

Explanation:

Given:

43^{x} \times 47^{y} = (2021)^{2}

, with

x \neq 0

,

y \neq 0

.

Factorize

2021

:

2021 = 43 \times 47

.

So,

(2021)^{2} = (43 \times 47)^{2} = 43^{2} \times 47^{2}

.

Thus,

43^{x} \times 47^{y} = 43^{2} \times 47^{2}

.

Comparing exponents of the same base gives

x = 2

and

y = 2

.

Now compute the expression

\frac{4xy + x + y}{2xy - x - y}

.

Substitute

x = 2

,

y = 2

:

Numerator:

4(2)(2) + 2 + 2 = 16 + 2 + 2 = 20

.

Denominator:

2(2)(2) - 2 - 2 = 8 - 2 - 2 = 4

.

Therefore,

\frac{20}{4} = 5

.

Answer:

5

(Option A).

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