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UPSC CDS Math Basic operation and Factorisation Questions

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Question : 23 of 77

Marks: +1, -0

If

\frac{b}{y}+\frac{z}{c}=1

\frac{c}{z}+\frac{x}{a}=1

\frac{ab+xy}{bx}

$1$
$2$
$0$
$-1$

Solution:

\frac{b}{y}+\frac{z}{c}=1

...(1)

and

\frac{c}{z}+\frac{x}{a}=1

....(2)

Taking

\frac{b}{y}+\frac{z}{c}=1

\Rightarrow \frac{z}{c}=1-\frac{b}{y}

\Rightarrow \frac{c}{z}=\frac{y}{y-b}

Putting this value in equation (2), we get

\;\;\frac{y}{y-b}+\;\;\frac{x}{a}=1

\Rightarrow \;\;\frac{ay+xy-xb}{ay-ab}=1

\Rightarrow ay+xy-xb=ay\,ab

\Rightarrow ab+xy=xb

\Rightarrow \;\;\frac{ab+xy}{bx}=1

Alternative Method: Take

b=1,\,y=2,\,z=1,\,c=2,\,x=-1,\,a=1

. Then

\;\;\frac{ab+xy}{bx}=\;\;\frac{1\times1+(-1)\times2}{1\times(-1)}=\;\;\frac{1-2}{-1}

=\frac{-1}{-1}=1

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