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UPSC CDS 1 2022 Math Paper

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Question : 27 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{4 x y + x + y}{2 x y - x - y}

5
15
25
45

Solution: 👈: Video Solution

Given:

43^{x} \times 47^{y} = (2021)^{2}, x \neq 0, y \neq 0

To find

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

Calculation:

We have

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

On factorization of 2021 we get

2021 = 43 \times 47

So, we can write the equation (i) as

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

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

On comparing we get

x = 2 \;\text{ and }\; y = 2

Now we have to find the value of

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

\Rightarrow \;\frac{4 \times 2 \times 2 + 2 + 2}{2 \times 2 \times 2 - 2 - 2} = \;\frac{20}{4} = 5

\therefore

The required value is 5 .

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