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TG EAMCET 4 May 2025 Shift 2 Paper

Question : 17 of 160

Marks: +1, -0

\frac{1}{x}+\frac{1}{y}=\frac{1}{2025}

Solution:

\frac{1}{x}+\frac{1}{y}=\frac{1}{2025}

2025(x+y)=xy

2025x+2025y-xy=0

x(2025-y)-2025(-y+2025)=-(2025)^2

(2025-y)(2025-x)=(2025)^2

\underset{5^2\times3^4}{(2025-y)} \underset{5^2\times3^4}{(2025-x)}=5^4\times3^8

\text{Or} (x-2025)(y-2025)=5^4\times3^8

Total number of positive integral solution

=\text{(ways of distribute 4 coins among 2 beggars)}\times\text{(ways to distribute 8 coins among 2 beggars)}

={}^5C_1\times{}^9C_1=45

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