Practice Questions: Data Caselets - 1 | Logical Reasoning (LR) and Data Interpretation (DI) - CAT PDF Download

Directions for Questions (1 - 4):
The Enforcement Directorate (ED) of the Government of India had for a long time been keeping watch over the Education Minister of West Bengal in Kolkata regarding a scam of recruitment of Staff Selection Commission candidates.
The Director of ED, Mr Sanjay Kr Mishra (IRS), with the help of his department sleuths, had observed a very interesting trend in the activity of the Education Minister :
Once every day of a week, except Sunday, the personal assistant of the Education Minister took delivery of sealed envelopes of money from the minister personally at his office to deliver to the minister’s lady-friend at her residence. It was also known from a secret source that for two fixed days of a week he delivered a certain number of envelopes less than the previous day, and for three fixed days a week, he delivered the same number of envelopes (as mentioned in the previous line) more than the previous day. However, it was not possible to get information as to what those days were. This practice continued for exactly twenty-seven consecutive days from the first Monday of delivery. From high-definition telescopic photographs of the envelopes, it was surely concluded that each envelope held a hundred crisp two thousand rupee notes. It was also known from the secret source that every week the greatest and least number of envelopes that the personal assistant delivered was eighty and twenty respectively.
Mr Mishra took the help of a professional data interpreter to analyze the collected data and to provide the ED Department with an idea of the details and enormity of the scam. What would be the answers found out by the data-interpreter to the following questions asked to him by Mr Mishra ?

Q1:  What could be the lowest possible quantum of the total delivery of money from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister ?
(a) Rs 5.40 crores
(b) Rs 14.75 crores
(c) Rs 21.60 crores
(d) Rs 24.00 crores
Ans: (c)
Sol:
The deliveries of sealed envelopes of money from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister, were done from Monday to Saturday.

The data interpreter assumed that the number of envelopes delivered on Monday to be E.

Since it was known by Mr Sanjay Kr Mishra (IRS) from a secret source that for two fixed days of a week the delivery was a certain number of envelopes less than the previous day, and for three fixed days a week the delivery was the same number of envelopes more than the previous day, he understood that the number of different permutations of those 2+3 = 5 days from Tuesday to Saturday would be 5C2, that is 10 ways.

He also assumed that the number of envelopes less or more than the previous day be x.

Hence he tabulated the 10 probable ways in which the permutation could take place and the vis-à-vis total delivery of the week as given below. The yellow shade indicated an increase and the blue a decrease in the number of envelopes :
From the secret source, it was also known that every week the greatest and least number of envelopes that were delivered was eighty and twenty respectively. Identifying them in each case (greatest number in green and least number in pink), he got :
Now, as a result of the above the data interpreter could deduce the following :

• Case 1 : E + x = 80 and E – 2x = 20. Solving, he got x = 20 and E = 60
• Case 2 : E + x = 80 and E – x = 20. Solving, he got x = 30 and E = 50
• Case 3 : E + x = 80 and E – x = 20. Solving, he got x = 30 and E = 50
• Case 4 : E + 2x = 80 and E – x = 20. Solving, he got x = 20 and E = 40
• Case 5 : E + x = 80 and E – x = 20. Solving, he got x = 30 and E = 50
• Case 6 : E + x = 80 and E = 20. Solving, he got x = 60 and E was already 20
• Case 7 : E + 2x = 80 and E = 20. Solving, he got x = 30 and E was already 20
• Case 8 : E + 2x = 80 and E = 20. Solving, he got x = 30 and E was already 20
• Case 9 : E + 2x = 80 and E = 20. Solving, he got x = 30 and E was already 20
• Case 10 : E + 3x = 80 and E = 20. Solving, he got x = 20 and E was already 20

Hence, he could exactly find out the individual number of envelopes that were delivered from the Education Minister’s office to the residence of his lady-friend by the personal assistant of the minister, for each day of each of the 10 probable cases. Also the total number of envelopes delivered in a week of each of the 10 probable cases. They were as below :
From the deductions made by the data interpreter, it could be seen that in Case 2, Case 7 and Case 8, the total number of envelopes delivered from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister was 270 each, which was the lowest number among all the ten cases.
From photographs of the envelopes, it was known that each envelope held hundred two thousand rupee notes, that is each envelope contained (100*2000) = Rs 2,00,000.
Hence, 270 envelopes would contain 200000*270 = Rs 5,40,00,000, that is, Rs 5.40 crores
This practice had continued for exactly twenty-seven consecutive days from the first Monday of delivery, that is for four Monday to Saturday time spans.
Hence, the lowest possible quantum of the total delivery of money from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister, could be 4*5.40 = Rs 21.60 crores

Q2: If the highest amount that could be delivered from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister, was delivered twice within Monday and Saturday, with Monday and Tuesday not being the days with the lowest delivery, what was the amount of money that was delivered on Thursday ?
(a) Rs 1.20 crores
(b) Rs 1.00 crores
(c) Rs 0.40 crores
(d) Cannot be determined
Ans: (c)
Sol:
The data-interpreter had already deducted that : 
From the deductions made by the data interpreter, it could be seen that the highest amount that could be delivered, that is money in 80 envelopes, was delivered twice within Monday and Saturday in Case 3, Case 5 and Case 9.
It could also be seen that within the above three cases, Monday and Tuesday not being the days with the lowest amount that could be delivered, that is money in 20 envelopes, was Case 5 only.
In Case 5, as per the deductions by the data interpreter, 20 envelopes were delivered on Thursday.
From photographs of the envelopes, it was known that each envelope held hundred two thousand rupee notes, that is each envelope contained (100*2000) = Rs 2,00,000.
So, 20 envelopes would contain 200000*20 = Rs 40,00,000, that is, Rs 0.40 crores
Hence, if the highest amount that could be delivered from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister, was delivered twice within Monday and Saturday, with Monday and Tuesday not being the days with the lowest delivery, the amount of money that was delivered on Thursday = Rs 0.40 crores

Q3: What could be the average daily delivery of money in the six days from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister, if on Monday and Saturday the lowest and highest number respectively of envelopes were delivered ?
(a) Rs 6.60 crores
(b) Rs 6.00 crores
(c) Rs 5.40 crores
(d) Rs 1.00 crore
Ans: (d)
Sol:
The data-interpreter had already deducted that :
It is only in Case 6 of the deductions by the data interpreter that on Monday and Saturday the lowest and highest number respectively of envelopes were delivered from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister.
In Case 6, the total number of envelopes delivered in the six days from Monday to Saturday was 300.
Hence, average envelopes delivered per day in the six days = 300/6 = 50 envelopes
From photographs of the envelopes, it was known that each envelope held hundred two thousand rupee notes, that is each envelope contained (100*2000) = Rs 2,00,000.
So, 50 envelopes would contain 200000*50 = Rs 1,00,00,000, that is, Rs 1.00 crore
Hence, the average daily delivery of money in the six days from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister, if on Monday and Saturday the lowest and highest number respectively of envelopes were delivered = 1.00 crore

Q4:  If all the probable scenarios of deliveries from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister, were to be considered, what could be the average amount of money that was delivered on Fridays ?
(a) Rs 1.10 crores
(b) Rs 2.50 crores
(c) Rs 4.50 crores
(d) Rs 5.40 crores
Ans: (a)
Sol:
The data-interpreter had already deducted that :
From the deductions made by the data interpreter, it could be seen that there are ten probable scenarios as tabulated by him.
The total number of envelopes that could be delivered on Fridays, if all the ten probable scenarios of deliveries from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister, were to be considered
= (60+50+50+80+50+20+80+20+80+60)
= 550
So, the average number of envelopes that could be delivered on Fridays, if all the ten probable scenarios of deliveries were to be considered = 550/10 = 55
From photographs of the envelopes, it was known that each envelope held hundred two thousand rupee notes, that is each envelope contained (100*2000) = Rs 2,00,000.
So, 55 envelopes would contain 200000*55 = Rs 1,10,00,000, that is, Rs 1.10 crores
Hence, the average amount of money that could be delivered on Fridays, if all the probable scenarios of deliveries from the Education Minister’s office to the residence of his lady-friend, by the personal assistant of the minister, were to be considered = Rs 1.10 crores

Q2:  Directions for Questions (5 - 8):
4 colleges A, B, C and D participated in 3 surveys BS, KIRF and GT. All the 4 colleges managed to secure a rank less than 7 in all the 3 surveys. Further, it is known that no 2 colleges were tied for the same position in any of the surveys.

Only A had a better rank than D in the survey conducted by KIRF.
B managed to get a better rank than C in 2 of the 3 surveys.
D was ranked second in the survey conducted by KIRF. It is the only survey in which D managed to bag a rank within the top 3.
GT deemed C to be a better college than A.
The average rank secured by D was 4.
D had a better rank in the survey conducted by BS than the one conducted by GT.
No college had the same rank in 2 surveys.
D was ranked the worst among the 4 colleges in 2 of the 3 surveys.
A managed a position within the top 3 in all 3 surveys.

Q1: If B was ranked third in the survey conducted by BS and the average ranks of B and C were not equal, then the rank of B in the survey conducted by KIRF is
(a) 3
(b) 4
(c) 5
(d) 6
Ans: (b)
Sol:
No college had the same rank in 2 surveys.
A managed a position within the top 3 in all the surveys.
Only A had a better rank than D in all 3 surveys.
Average rank of D is 4.
=> Sum of the ranks secured by D = 12.
We know that D secured the second rank in the survey by KIRF.
=> Sum of ranks secured by D in the BS and GT surveys = 10.
10 can be represented as (5,5) or (6,4).
We know that no college secured the same rank in 2 surveys.
Therefore, the ranks secured by D must be 6 and 4.
D had a better rank in the survey conducted by BS than the one conducted by GT.
We know that A managed a position within the top 3 in all 3 surveys.
Therefore, A must have secured second and third positions in the other 2 surveys. Also, the average rank secured by A must be 2.
Let us solve each question separately now. The points that one must bear in mind while filling the table are
(I) B managed to get a better rank than C in 2 of the 3 surveys.
(II) GT deemed C to be a better college than A.
B was ranked third in the survey conducted by BS. Also, we know that the average ranks of B and C were not equal.
Now, A must have secured second rank and C must have secured first rank in the survey conducted by BS.
In the other 2 surveys, B must have a better rank than C. A must have secured third rank in the survey conducted by GT.
C must have secured second rank (since GT deems C to be a better college than C) and B must have secured first rank.
Since B has secured third rank in BS rankings, it must have secured the fourth or fifth rank in the KIRF rankings.
Had B secured fifth rank, C must have secured the sixth rank (Since 2 of the 3 surveys place B above C).
In this case, the average rank secured by both B and C will be equal (3).
Therefore, we can eliminate this possibility.
B must have secured 4th rank and C must have secured sixth rank.
Therefore, option B is the right answer.

Q2: If the average rank secured by C is 3 and B is 2 in the 3 surveys, what is the rank secured by C in the survey conducted by KIRF?
(a) 3
(b) 4
(c) 5
(d) 6
Ans: (c)
Sol: 
We know that no college secured the same rank in 2 surveys.
Therefore, the ranks secured by D must be 6 and 4.
D had a better rank in the survey conducted by BS than the one conducted by GT.
We know that A managed a position within the top 3 in all 3 surveys.
Therefore, A must have secured second and third positions in the other 2 surveys. Also, the average rank secured by A must be 2.

Let us solve each question separately now. The points that one must bear in mind while filling the table are
(I) B managed to get a better rank than C in 2 of the 3 surveys.
(II) GT deemed C to be a better college than A.
Average rank secured by B is 2 and C is 3.
Therefore, sum of the ranks secured by B and C must be 6 and 9 respectively.
Sum of ranks secured by B is 6. Therefore, the ranks must be 1,2 and 3.
B must have secured third rank in the survey conducted by KIRF (Since the first 2 ranks are already taken up).
C had a better rank than A in the survey conducted by GT.
Had C secured first rank in the survey conducted by BS, then C must have secured second rank in the survey conducted by GT.
A must have secured third rank in the survey conducted by GT and second rank in the survey conducted by BS. However, B could not have secured third rank in the survey conducted by BS since A has secured third rank. Therefore, this case can be eliminated.
Had C secured first rank in the survey conducted by GT, B must have secured first rank in the survey conducted by BS. A must have secured second rank in the survey conducted by BS and C must have secured the third rank. The rankings will be as follows:

As we can see, C must have secured fifth rank in the survey conducted by KIRF.
Therefore, option C is the right answer.

Q3: If the average rank secured by C is the same as the average rank secured by B and the rank secured by A in GT is better than the one it secured in BS, then the rank secured by B in GT survey is
(a) 3 or 5
(b) 5 or 6
(c) 4 or 5
(d) 1 or 3
Ans: (a)
Sol:
We know that no college secured the same rank in 2 surveys.
Therefore, the ranks secured by D must be 6 and 4.
D had a better rank in the survey conducted by BS than the one conducted by GT.
We know that A managed a position within the top 3 in all 3 surveys.
Therefore, A must have secured second and third positions in the other 2 surveys. Also, the average rank secured by A must be 2.
Let us solve each question separately now. The points that one must bear in mind while filling the table are
(I) B managed to get a better rank than C in 2 of the 3 surveys.
(II) GT deemed C to be a better college than A.
Average rank secured by B and C is the same.
Average rank secured by B and C cannot be 2 since in KIRF one of the 2 must have secured a rank greater than 3.
Average rank cannot be 6 as well.
Average rank cannot be 5 since both of them have a rank less than 4 in the BS survey.
Let us consider the possibility that the average rank is 4.
Therefore, sum of ranks must be 12.
Also, we know that rank secured by A in GT is better than the one it secured in the survey by BS.
Therefore, A must have secured second rank in GT and third rank in BS.
In the BS survey, B and C must have secured first and second ranks, respectively.
Since C is ranked better than A by GT, C must have secured the first rank in the GT survey.
Now, C must have secured a rank of 8 in the KIRF ranking for the average rank to be 4. As we can see, this case is impossible.
Therefore, the average rank of both B and C must be 3.
C must have secured first rank in the GT survey and second rank in the BS survey.
For the average rank to be 6, C must have secured sixth rank in the KIRF survey.
B must have secured first rank in the BS survey.
Now, the sum of the ranks secured by B in the other 2 surveys must be 9-1 = 8.
8 can be represented as (4,4), (5,3) or (6,2).
We can eliminate the case (4,4) since no college secured the same ranking in 2 colleges.
We can eliminate (6,2) too since no 2 colleges secured the same ranking in a survey.
Therefore, the final ranking will be as follows:
Rank secured by B in the GT survey will be 3 or 5.
Therefore, option A is the right answer.

Q4: If the average rank secured by B is one more than the average rank secured by A, then the rank secured by C in the survey conducted by BS is
(a) 1
(b) 2
(c) 3
(d) Can be more than one of the above
Ans: (d)
Sol:
We know that no college secured the same rank in 2 surveys.
Therefore, the ranks secured by D must be 6 and 4.
D had a better rank in the survey conducted by BS than the one conducted by GT.
We know that A managed a position within the top 3 in all 3 surveys.
Therefore, A must have secured second and third positions in the other 2 surveys. Also, the average rank secured by A must be 2.
Let us solve each question separately now. The points that one must bear in mind while filling the table are
(I) B managed to get a better rank than C in 2 of the 3 surveys.
(II) GT deemed C to be a better college than A.
Average rank secured by B = Average rank secured by A + 1.
Therefore, the average rank secured by B must be 3.
Sum of the ranks secured by B = 9.
Now, had A secured second rank in the GT survey and third rank in the BS survey, C must have secured the first rank in the GT survey and second rank in BS survey.
B must have secured the first rank in the BS survey.
Now, B must have secured either fifth or third rank in the KIRF and GT surveys.
In this case, the rank secured by C is two.
Let us consider the other possibility.
Had A secured third rank in the GT survey, A would have secured second rank in the BS survey.
C must have secured the second or first rank in the GT survey.
Had C secured first rank in the GT survey, it would have secured third rank in the BS survey.
B would have secured first rank in the BS survey.
Had C secured second rank in the GT survey, C could have secured first or third rank in the BS survey. 
As we can see, C could have secured any of the top 3 ranks and hence, option D is the right answer.