IPL Fever #2

By $1^{st}$ condition ,

$\dfrac{y}{x}=10\\ \Rightarrow y=10x \dots (1)$

By $2^{nd}$ condition ,

$\dfrac{y-3}{x-1}=10+1\\ \Rightarrow \dfrac{y-3}{x-1}=11 \\ \Rightarrow y-3=11x-11 \\ \Rightarrow 11x-y=8\\ \Rightarrow 11x-10x=8 \text{[Using (1)]} \\ \Rightarrow\large\boxed{x=8 \quad , \quad y=80 \quad ,\quad x+y=88}$

Let The overs remaining =y

Let the Runs required=x

Therefore,

$\dfrac{x}{y} = 10.... (1)$

And After 1 over, Since the run rate becomes 11 (10+1),

$\dfrac{x-3}{y-1}=11..... (2)$

Now, From 1, We have $x=10y$

Substituting, We get,

$10y-3=11(y-1)$

$10y-3=11y-11$

$y=8$

Now Since x=10y, $x=80$

Therefore, $x+y=88$

Cheers! :D