Understanding the set- 1) We are given five firms – Alfloo, Bzygoo, Czechy, Drjbna and Elavalaki. 2) Each of these firms was founded in some year and also closed down a few years later, whose data we can find the table. 3) Each firm raised some amount of money in it's total year of existence. 4) Each firm raised Rs. 1 crore in its first and last year of existence 5) The amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down. 6) Amount raised by a firm in each year is a distinct natural number. 7) Each annual increase and decrease was either by Rs. 1 crore or by Rs. 2 crores. Now let's try to form a table where we can fill the information given and asked in the question. —--

	2009	2010	2011	2012	2013	2015	2016	Tot
A	1						1	21
B				1		1
C					1			9
D			1			1		10
E		1						13

In this table we will try to fill the amount raised by each firm in it's year of existence. Also we know each firm raised 1cr in first and last year of existence which we can fill from the question. —--

	2009	2010	2011	2012	2013	2014	2015	2016	2017	Tot
A	1							1	-	21
B	-	-	-	1			1	-	-
C	1	2	3	2	1	-	-	-	-	9
D	-	-	1				1	-	-	10
E	-	1								13

Let's first start with C, The pattern looks as follows: 1, ..., 1 1) Let us assume there are 2 gaps between => a + b = 7 (Not possible) as maximum case would be 1, 3, 2, 1 2) Let us assume there are 3 gaps between => a + b + c = 7 the minimum case possible is 1, 2, 3, 2, 1 => Satisfies 3) if there are 4 gaps => a + b + c + d = 7 => no case will satisfy => Not possible —--

	2009	2010	2011	2012	2013	2014	2015	2016	2017	Tot
A	1							1	-	21
B	-	-	-	1			1	-	-
C	1	2	3	2	1	-	-	-	-	9
D	-	-	1	2	4	2	1	-	-	10
E	-	1								13

Consider D: The pattern looks as follows: 1, a, b, c, 1 => a + b + c = 8 1) When a = 2 and c = 2 => b = 4 => 2, 4, 2 => Satisfies. 2) When a = 2 and c = 3, b should be 3 (Not satisfying) 3) When a = 3 and c = 3, b should be 2 (Not satisfying) Therefore only possible case is 1, 2, 4, 2, 1. —--

	2009	2010	2011	2012	2013	2014	2015	2016	2017	Tot
A	1	2	3	4/5	5/4	3	2	1	-	21
B	-	-	-	1			1	-	-
C	1	2	3	2	1	-	-	-	-	9
D	-	-	1	2	4	2	1	-	-	10
E	-	1								13

Consider A: It raised money for 8 years => The raising pattern looks like follows: 1, a, b, c, d, e, f, 1 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. 1) 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. 2) 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. 3) if 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. Therefore there are 2 possible case for A 1) - 1, 2, 3, 4, 5, 3, 2, 1 2) - 1, 2, 3, 5, 4, 3, 2, 1 —--

	2009	2010	2011	2012	2013	2014	2015	2016	2017	Tot
A	1	2	3	4/5	5/4	3	2	1	-	21
B	-	-	-	1	2/3	3/2	1	-	-
C	1	2	3	2	1	-	-	-	-	9
D	-	-	1	2	4	2	1	-	-	10
E	-	1								13

Consider B: The patterns looks as follows: 1, a, b, 1 1) If a = 2, b has to be equal to 3 to satisfy 2) if a = 3 b has to be equal to 2 to satisfy Therefore 2 possible cases for B - 1) 1, 2, 3, 1 2) 1, 3, 2, 1 —--

	2009	2010	2011	2012	2013	2014	2015	2016	2017	Tot
A	1	2	3	4/5	5/4	3	2	1	-	21
B	-	-	-	1	2/3	3/2	1	-	-
C	1	2	3	2	1	-	-	-	-	9
D	-	-	1	2	4	2	1	-	-	10
E	-	1								13

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 => a + 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 and 2 + 3 + 4 + 2 = 11. The possible cases for E are: 1) 1, 3, 5, 3, 1 2) 1, 2, 3, 4, 2, 1 3) 1, 2, 4, 3, 2, 1

CAT (DILR) PYQs

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