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 ±1 or ±2⇒ (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,a,b,c,d,e,f,1⇒ where a,b,c,...,f are the unknown amounts raised.
Also a+b+c+d+e+f=21−2=19.
We can observe that 19∕6 is slightly greater than 3⇒ The average amount raised should be around 3 .
If a=3 and f=3⇒b+c+d+e=13 (not possible) as the minimum case would be (4,5,6,4)⇒ Not possible.
If a=3 and f=2⇒b+c+d+e=14 (not possible) as the minimum case would be (4,5,4,3)⇒ Not possible.
⇒a=2 and f=2⇒b+c+d+e=15 the minimum case is (3,4,5,3) or (3,5,4,3) which gives a sum of 15 .
So, the possible cases for A are:
Consider B:
The patterns looks as follows:
1,a,b,1
If a=2, b has to be equal to 3 to satisfy (1)
if a=3, b has to be equal to 2 to satisfy (1)
⇒ The possible cases for B are:Consider C:
The pattern looks as follows:
1,...,1
Let us assume there are 2 gaps between ⇒+b=7 (Not possible) as maximum case would be 1, 3, 2, 1
Let us assume there are 3 gaps between ⇒a+b+c=7, the minimum case possible is 1,2,3,2,1⇒ Satisfies.
Now, if there are 4 gaps ⇒a+b+c+d=7⇒ The average value is 7∕4 which is less than 2⇒ Not possible.
⇒ The possible cases for C are:

C	2013	2014	2015	2016	2017
	1	2	3	2	1

Consider D:
The pattern looks as follows:
1,a,b,c,1
⇒a+b+c=8
When a=2 and c=2⇒b=4⇒2,4,2⇒ Satisfies.
When a=2 and c=3, b should be 3 (Not satisfying (1))
When a=3 and c=3, b should be 2 (Not satisfying (1))
⇒ The possible cases for 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 ⇒+b+c=11, the maximum case is 3,5,3⇒ Satisfies.
Now, assume 4 gaps
⇒a+b+c+d=11, the minimum case is 2,3,4,2 or 2,4,3,2 which satisfies ( 1 ) and 2+3+4+2=11.
⇒ The possible cases for E are:In summary, the possible cases for all 5 companies is:Maximum money raised in 2013 is 5+3+1+4+4=17. ⇒