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CAT 2023 PYQsData Interpretation & Logical Reasoning. Free, no login required.

Logical Reasoning

past year questions, CAT 2023

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past year questions, Logical Reasoning, CAT 2023

Easy

Odsville has five firms – Alfloo, Bzygoo, Czechy, Drjbna and Elavalaki. Each of these firms was founded in some year and also closed down a few years later.

Each firm raised Rs. 1 crore in its first and last year of existence. The amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down. No firm raised the same amount of money in two consecutive years. Each annual increase and decrease was either by Rs. 1 crore or by Rs. 2 crores.

The table below provides partial information about the five firms.

FirmFirst year of existenceLast year of existenceTotal amount raised (Rs. crores)
Alfloo2009201621
Bzygoo20122015
Czechy20139
Drjbna2011201510
Elavalaki201013

If Elavalaki raised Rs. 3 crores in 2013, then what is the smallest possible total amount of money (in Rs. crores) that could have been raised by all the companies in 2012?

Solution

✅ Correct Option: 1
Slide 1/8

Understanding the set-

  1. We are given five firms – Alfloo, Bzygoo, Czechy, Drjbna and Elavalaki.
  1. Each of these firms was founded in some year and also closed down a few years later, whose data we can find the table.
  1. Each firm raised some amount of money in it's total year of existence.
  1. Each firm raised Rs. 1 crore in its first and last year of existence
  1. The amount each firm raised every year increased until it reached a maximum, and then decreased until the firm closed down.
  1. Amount raised by a firm in each year is a distinct natural number.
  1. 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.

200920102011201220132014201520162017Tot
A1121
B11
C19
D1110
E113

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.

200920102011201220132014201520162017Tot
A11-21
B---11--
C12321----9
D--11--10
E-113

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

  1. Let us assume there are 3 gaps between

=> a + b + c = 7

the minimum case possible is 1, 2, 3, 2, 1 => Satisfies

  1. if there are 4 gaps

=> a + b + c + d = 7

=> no case will satisfy => Not possible

200920102011201220132014201520162017Tot
A11-21
B---11--
C12321----9
D--12421--10
E-113

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.

  1. 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)

Therefore only possible case is 1, 2, 4, 2, 1.

200920102011201220132014201520162017Tot
A1234/55/4321-21
B---11--
C12321----9
D--12421--10
E-113

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.

  1. 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.

  1. 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, 2, 3, 4, 5, 3, 2, 1
    • 1, 2, 3, 5, 4, 3, 2, 1
200920102011201220132014201520162017Tot
A1234/55/4321-21
B---12/33/21--
C12321----9
D--12421--10
E-113

Consider B:

The patterns looks as follows: 1, a, b, 1

  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

Therefore 2 possible cases for B -

  1. 1, 2, 3, 1

  2. 1, 3, 2, 1

200920102011201220132014201520162017Tot
A1234/55/4321-21
B---12/33/21--
C12321----9
D--12421--10
E-113

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

200920102011201220132014201520162017Tot
A1234/55/4321-21
B---12/33/21--
C12321----9
D--12421--10
E-113

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

If Elavalaki raised Rs. 3 crores in 2013, then Elavalaki raised Rs. 4 crores in 2014. Hence, the smallest possible total amount of money raised in 2012 = 4 +1 +2 + 4 = Rs. 11 crores

More from this set:

Question 1

For which firm(s) can the amounts raised by them be concluded with certainty in each year?

Question 2

What best can be concluded about the total amount of money raised in 2015?

Question 3

What is the largest possible total amount of money (in Rs. crores) that could have been raised in 20132013?

Question 5

If the total amount of money raised in 20142014 is Rs. 1212 crores, then which of the following is not possible?

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