CAT 2023 Slot 2 Solved Paper

In this set, we are told that the amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down and no firm raised the same amount of money in two consecutive years.
The increase or decrease can be $\pm 1$ or $\pm 2\Rightarrow$ (1)
We are also told that each firm raised Rs. 1 crore in its first and last year of existence
Consider A:
It raised money for 8 years
⇒The raising pattern looks like follows:
$1,\mathit{a},\mathit{b},\mathit{c},\mathit{d},\mathit{e},\mathit{f},1\Rightarrow$ where $\mathit{a},\mathit{b},\mathit{c},...,\mathit{f}$ are the unknown amounts raised.
Also $\mathit{a}+\mathit{b}+\mathit{c}+\mathit{d}+\mathit{e}+\mathit{f}=21-2=19$.
We can observe that $19∕6$ is slightly greater than $3\Rightarrow$ The average amount raised should be around 3 .
If $\mathit{a}=3$ and $\mathit{f}=3\Rightarrow \mathit{b}+\mathit{c}+\mathit{d}+\mathit{e}=13$ (not possible) as the minimum case would be $\left(4,5,6,4\right)\Rightarrow$ Not possible.
If $\mathit{a}=3$ and $\mathit{f}=2\Rightarrow \mathit{b}+\mathit{c}+\mathit{d}+\mathit{e}=14$ (not possible) as the minimum case would be $\left(4,5,4,3\right)\Rightarrow$ Not possible.
$\Rightarrow \mathit{a}=2$ and $\mathit{f}=2\Rightarrow \mathit{b}+\mathit{c}+\mathit{d}+\mathit{e}=15$ the minimum case is $\left(3,4,5,3\right)$ or $\left(3,5,4,3\right)$ which gives a sum of 15 .
So, the possible cases for $\mathrm{A}$ are:
Consider B:
The patterns looks as follows:
$1,\mathit{a},\mathit{b},1$
If $\mathit{a}=2$, $\mathit{b}$ has to be equal to 3 to satisfy (1)
if $\mathit{a}=3$, $\mathit{b}$ has to be equal to 2 to satisfy (1)
$\Rightarrow$ The possible cases for $\mathit{B}$ are:Consider C:
The pattern looks as follows:
$1,...,1$
Let us assume there are 2 gaps between $\Rightarrow +\mathit{b}=7$ (Not possible) as maximum case would be 1, 3, 2, 1
Let us assume there are 3 gaps between $\Rightarrow \mathit{a}+\mathit{b}+\mathit{c}=7$, the minimum case possible is $1,2,3,2,1\Rightarrow$ Satisfies.
Now, if there are 4 gaps $\Rightarrow \mathit{a}+\mathit{b}+\mathit{c}+\mathit{d}=7\Rightarrow$ The average value is $7∕4$ which is less than $2\Rightarrow$ Not possible.
$\Rightarrow$ The possible cases for $\mathrm{C}$ are:

$\mathrm{C}$	2013	2014	2015	2016	2017
	1	2	3	2	1

Consider D:
The pattern looks as follows:
$1,\mathit{a},\mathit{b},\mathit{c},1$
$\Rightarrow \mathit{a}+\mathit{b}+\mathit{c}=8$
When $\mathit{a}=2$ and $\mathit{c}=2\Rightarrow \mathit{b}=4\Rightarrow 2,4,2\Rightarrow$ Satisfies.
When $\mathrm{a}=2$ and $\mathrm{c}=3$, $\mathrm{b}$ should be 3 (Not satisfying (1))
When $\mathrm{a}=3$ and $\mathrm{c}=3$, b should be 2 (Not satisfying (1))
$\Rightarrow$ The possible cases for $\mathit{D}$ are:

D	2011	2012	2013	2014	2015
	1	2	4	2	1

Consider E:
The pattern looks as follows:
$1,.....,1$
For 1 or 2 gaps, we can't get a sum of 11 .
Assume 3 gaps $\Rightarrow +\mathit{b}+\mathit{c}=11$, the maximum case is $3,5,3\Rightarrow$ Satisfies.
Now, assume 4 gaps
$\Rightarrow \mathit{a}+\mathit{b}+\mathit{c}+\mathit{d}=11$, the minimum case is $2,3,4,2$ or $2,4,3,2$ which satisfies ( 1 ) and $2+3+4+2=11$.
$\Rightarrow$ The possible cases for $\mathit{E}$ are:In summary, the possible cases for all 5 companies is:Maximum money raised in 2013 is $5+3+1+4+4=17.$ ⇒