CAT 2018 Question Paper Shift 1 Solved Paper

It is given that the satellites serving either B, C or S do not serve O.
From (1), let the number of satellites serving B, C and S be 2K, K, K respectively.
Let the number of satellites exclusively serving B be x.
From (3), the number of satellites exclusively serving C and exclusively serving S will each be 0.3x
From (4), the number of satellites serving $\mathit{O}$ is same as the number of satellites serving only $\mathit{C}$ and $\mathit{S}$. Let that number be $\mathit{y}$.
Since the number of satellites serving $\mathit{C}$ is same as the number of satellites serving $\mathit{S}$, we can say that
(number of satellites serving only $\mathit{B}$ and $\mathit{C})+0.3\mathit{x}+100+\mathit{y}=($ number of satellites serving only $\mathit{B}$ and $\mathit{S})+$ $0.3\mathit{x}+100+\mathit{y}$
Let the number of satellites serving only $\mathit{B}$ and $\mathit{C}=$ the number of satellites serving only $\mathit{B}$ and $\mathit{S}$ $=\mathit{Z}$
Therefore, the venn diagram will be as followsGiven that there are a total of 1600 satellites
$1.6\mathit{x}+2\mathit{y}+2\mathit{z}=1500$............(1)
Also $\mathit{K}=0.3\mathit{x}+\mathit{z}+\mathit{y}+100$
Satellites serving $\mathit{B}=2\mathit{K}=\mathit{x}+2\mathit{z}+100$
$0.4\mathit{x}=2\mathit{y}+100$
$\mathit{x}=5\mathit{y}+250$....(2)
Substituting ( 2 ) in (1), we will get
$1.6\left(5\mathit{y}+250\right)+2\mathit{y}+2\mathit{z}=1500$
$10\mathit{y}+2\mathit{z}=1100$
$\mathit{Z}=550-5\mathit{y}$..........(3)
From $2,$ the number of satellites serving B exclusively is $\mathit{x}=5\mathit{y}+250$ This is minimum when y is minimum.
Minimum value of $\mathit{y}=0$
The minimum number of satellites serving $\mathit{B}$ exclusively $=5\times 0+250=250$.