CAT DI LR : CAT 2018 Question Paper Slot 1

Set 1: Written Test

A company administers a written test comprising of three sections of 20 marks each – Data Interpretation (DI), Written English (WE) and General Awareness (GA), for recruitment. A composite score for a candidate (out of 80) is calculated by doubling her marks in DI and adding it to the sum of her marks in the other two sections. Candidates who score less than 70% marks in two or more sections are disqualified. From among the rest, the four with the highest composite scores are recruited. If four or less candidates qualify, all who qualify are recruited.

Ten candidates appeared for the written test. Their marks in the test are given in the table below:

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Some marks in the table are missing, but the following facts are known:
1. No two candidates had the same composite score.
2. Ajay was the unique highest scorer in WE.
3. Among the four recruited, Geeta had the lowest composite score.
4. Indu was recruited.
5. Danish, Harini, and Indu had scored the same marks the in GA.
6. Indu and Jatin both scored 100% in exactly one section and Jatin’s composite score was 10 more than Indu’s.

1. Which of the following statements MUST be true?
a. Jatin’s composite score was more than that of Danish.
b. Indu scored less than Chetna in DI.
c. Jatin scored more than Indu in GA.

Only 2
Both 2 and 3
Only 1
Both 1 and 2

EXPLANATION

Answer: Option 4

Method of solving this CAT Question on CAT DILR

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Indu and Jatin are in. The highest CS Geeta can get is 54, so Ester is also in.
Ajay could not have scored 19 in WE as his CS would then become 51 same as Danish.
So, Ajay should have scored 20 in WE resulting in an CS of 52.

Geeta should have scored a CS of more than 52 so she should have scored 19 or 20 in WE. But Ajay was the unique highest scorer in WE.
So, Geeta should have score 19 in WE.
Options 1 and 2 are both true. Option 3 is not true.
Therefore Both 1 and 2

Hence, the answer is Both 1 and 2

Choice D is the correct answer.

2. Which of the following statements MUST be FALSE?

Bala scored same as Jatin in DI
Bala’s composite score was less than that of Ester
Chetna scored more than Bala in DI
Harini’s composite score was less than that of Falak

EXPLANATION

Answer: Option A

A composite score for a candidate (out of 80) is calculated by doubling her marks in DI and adding it to the sum of her marks in the other two sections.
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Ajay was the unique highest scorer in WE.

Danish, Harini, and Indu had scored the same marks the in GA.
Indu and Jatin both scored 100% in exactly one section and Jatin’s composite score was 10 more than Indu’s.
Indu and Jatin both scored 100% in exactly one section and Jatin’s composite score was 10 more than Indu’s.
Now, if Indu had scored 100% in DI, then her CS would be 20 × 2 + 8 + x = 60 or, x should be 12.
Now, this is a problem as Indu could have less than 70% in 2 sections and so would not have been selected.
So, x = 20. CAT DI LR 2018 Slot 1
Indu and Jatin are in. The highest CS Geeta can get is 54, so Ester is also in.
Ajay could not have scored 19 in WE as his CS would then become 51 same as Danish.
So, Ajay should have scored 20 in WE resulting in an CS of 52.

Geeta should have scored a CS of more than 52 so she should have scored 19 or 20 in WE. But Ajay was the unique highest scorer in WE.
So, Geeta should have score 19 in WE.
Choice A must be false as if Bala had scored 20 in DI, his CS would have been 60, same as Indu.

Hence, the answer is Bala scored same as Jatin in DI

Choice A is the correct answer.

3. If all the candidates except Ajay and Danish had different marks in DI, and Bala’s composite score was less than Chetna’s composite score, then what is the maximum marks that Bala could have scored in DI? (TITA)

EXPLANATION

Answer: 13 marks

A composite score for a candidate (out of 80) is calculated by doubling her marks in DI and adding it to the sum of her marks in the other two sections.
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Ajay was the unique highest scorer in WE.

Danish, Harini, and Indu had scored the same marks the in GA.
Indu and Jatin both scored 100% in exactly one section and Jatin’s composite score was 10 more than Indu’s.
Indu and Jatin both scored 100% in exactly one section and Jatin’s composite score was 10 more than Indu’s.
Now, if Indu had scored 100% in DI, then her CS would be 20 × 2 + 8 + x = 60 or, x should be 12.
Now, this is a problem as Indu could have less than 70% in 2 sections and so would not have been selected.
So, x = 20. CAT DI LR 2018 Slot 1
Indu and Jatin are in. The highest CS Geeta can get is 54, so Ester is also in.
Ajay could not have scored 19 in WE as his CS would then become 51 same as Danish.
So, Ajay should have scored 20 in WE resulting in an CS of 52.

Geeta should have scored a CS of more than 52 so she should have scored 19 or 20 in WE. But Ajay was the unique highest scorer in WE.
So, Geeta should have score 19 in WE.
Let us assume Bala scored x in DI. His CS would be 2x + 20. We know 2x + 20 < 56 or, x < 18.
x = 17 gives Bala a CS of 54 which Chetna has scored.
x cannot be 16, 15 or 14 because Indu, Falak and Geetha have those scores.
x = 13 gives us a CS score of 46 which thankfully no one has.

Hence, the answer is 13 marks

4. If all the candidates scored different marks in WE then what is the maximum marks that Harini could have scored in WE? (TITA)

EXPLANATION

Answer: 14 marks

A composite score for a candidate (out of 80) is calculated by doubling her marks in DI and adding it to the sum of her marks in the other two sections.
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Ajay was the unique highest scorer in WE.

Danish, Harini, and Indu had scored the same marks the in GA.
Indu and Jatin both scored 100% in exactly one section and Jatin’s composite score was 10 more than Indu’s.
Indu and Jatin both scored 100% in exactly one section and Jatin’s composite score was 10 more than Indu’s.
Now, if Indu had scored 100% in DI, then her CS would be 20 × 2 + 8 + x = 60 or, x should be 12.
Now, this is a problem as Indu could have less than 70% in 2 sections and so would not have been selected.
So, x = 20. CAT DI LR 2018 Slot 1
Indu and Jatin are in. The highest CS Geeta can get is 54, so Ester is also in.
Ajay could not have scored 19 in WE as his CS would then become 51 same as Danish.
So, Ajay should have scored 20 in WE resulting in an CS of 52.

Harini could not have scored 20, 19 or 18. If Harini had sored 17, her CS would have been 47 same as Falak. So, 17 is also ruled out.
16 and 15 are also ruled out. H_max could be 14. This gives us a CS of 44 which should be fine.
WE of 14 works.

Hence, the answer is 14 marks

Set 2 : Satellites

Q1-4. 1600 satellites were sent up by a country for several purposes. The purposes are classified as broadcasting (B), communication (C), surveillance (S), and others (O). A satellite can serve multiple purposes; however a satellite serving either B, or C, or S does not serve O.
The following facts are known about the satellites:
1.The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.
2.The number of satellites serving all three of B, C, and S is 100.
3.The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B.
4.The number of satellites serving O is the same as the number of satellites serving both C and S but not B.

1. What best can be said about the number of satellites serving C?

Must be between 400 and 800
Cannot be more than 800
Must be at least 100
Must be between 450 and 725

EXPLANATION

Answer: Option D

1600 satellites were sent up by a country for several purposes. The purposes are classified as broadcasting (B), communication (C), surveillance (S), and others (O). A satellite can serve multiple purposes; however a satellite serving either B, or C, or S does not serve O.
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Now, let us move to the equations.
We know that 2(m + 100 + 0.3y + k) = 2m + y + 100
Or, 2m + 200 + 0.6y + 2k = 2m + y + 100
100 + 2k = 0.4y
250 + 5k = y
We also know that 1.6y + 2m + 2k + 100 = 1600
1.6y + 2m + 2k = 1500
100 + 2k = 0.4y
250 + 5k = y
We also know that 1.6y + 2m + 2k + 100 = 1600
1.6y + 2m + 2k = 1500
1.6(250 + 5k) + 2m + 2k = 1500
400 + 8k + 2m + 2k = 1500
10k + 2m = 1100
5k + m = 550.
The two equations we have are
250 + 5k = y ———- (1)
5k + m = 550. ———- (2)
C has 100 + m + k + 0.3y satellites. This is equal to
100 + m + k + 0.3(250 + 5k) = 100 + m + k + 75 + 1.5k = 175 + 2.5k + m
= 175 + 550 – 2.5k = 725 – 2.5k
Think about the maximum and minimum this can take.
C has 725 – 2.5k
Think about the maximum and minimum this can take.
K can be 0, this will give us m of 550 and y = 250
This will give us a maximum value of 725.
We know that 5k + m = 550. So, maximum value of k is 110.
This gives us C_min = 725 – (2.5 × 110) = 725 – 275 = 450.
C ranges from 450 to 725.

Hence, the answer is Must be between 450 and 725

Choice D is the correct answer.

2. What is the minimum possible number of satellites serving B exclusively?

EXPLANATION

Answer: Option 3

Hence, the answer is 250

Choice C is the correct answer.

3. If at least 100 of the 1600 satellites were serving O, what can be said about the number of satellites serving S?

At least 475
No conclusion is possible based on the given information
Exactly 475
At most 475

EXPLANATION

Answer: Option D

1600 satellites were sent up by a country for several purposes. The purposes are classified as broadcasting (B), communication (C), surveillance (S), and others (O). A satellite can serve multiple purposes; however a satellite serving either B, or C, or S does not serve O.
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Now, let us move to the equations.
We know that 2(m + 100 + 0.3y + k) = 2m + y + 100
Or, 2m + 200 + 0.6y + 2k = 2m + y + 100
100 + 2k = 0.4y
250 + 5k = y
We also know that 1.6y + 2m + 2k + 100 = 1600
1.6y + 2m + 2k = 1500
100 + 2k = 0.4y
250 + 5k = y
We also know that 1.6y + 2m + 2k + 100 = 1600
1.6y + 2m + 2k = 1500
1.6(250 + 5k) + 2m + 2k = 1500
400 + 8k + 2m + 2k = 1500
10k + 2m = 1100
5k + m = 550.
The two equations we have are
250 + 5k = y ———- (1)
5k + m = 550. ———- (2)
k ≥ 100. Number serving S would be 100 + m + k + 0.3y satellites.
Number serving S = 100 + m + k + 0.3(250 + 5k) = 100 + m + k + 75 + 1.5k
Number serving S = 175 + 2.5k + m = 175 + 550 – 2.5k = 725 – 2.5k
k_min = 100. So, 2.5k_min = 250
S_max would be 725 – 250 = 475

Hence, the answer is 475

Choice D is the correct answer.

4. If the number of satellites serving at least two among B, C, and S is 1200, which of the following MUST be FALSE?

All 1600 satellites serve B or C or S
The number of satellites serving C cannot be uniquely determined
The number of satellites serving B exclusively is exactly 250
The number of satellites serving B is more than 1000

EXPLANATION

Answer: Option B

1600 satellites were sent up by a country for several purposes. The purposes are classified as broadcasting (B), communication (C), surveillance (S), and others (O). A satellite can serve multiple purposes; however a satellite serving either B, or C, or S does not serve O.
CAT DI LR 2018 Slot 1

Now, let us move to the equations.
We know that 2(m + 100 + 0.3y + k) = 2m + y + 100
Or, 2m + 200 + 0.6y + 2k = 2m + y + 100
100 + 2k = 0.4y
250 + 5k = y
We also know that 1.6y + 2m + 2k + 100 = 1600
1.6y + 2m + 2k = 1500
100 + 2k = 0.4y
250 + 5k = y
We also know that 1.6y + 2m + 2k + 100 = 1600
1.6y + 2m + 2k = 1500
1.6(250 + 5k) + 2m + 2k = 1500
400 + 8k + 2m + 2k = 1500
10k + 2m = 1100
5k + m = 550.
The two equations we have are
250 + 5k = y ———- (1)
5k + m = 550. ———- (2)
At least two = 2m + k + 100 = 1200 or, 2m + k = 1100.
We know that 2m + 10k = 1100. If 2m + k were also 1100, we can say k = 0. This gives us m = 550 and y = 75
Choice B is false.
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Hence, the answer is The number of satellites serving C cannot be uniquely determined

Choice B is the correct answer.

Set 3 : N x N Square Matrix

Q1-4. You are given an n×n square matrix to be ﬁlled with numerals so that no two adjacent cells have the same numeral. Two cells are called adjacent if they touch each other horizontally, vertically or diagonally. So a cell in one of the four corners has three cells adjacent to it, and a cell in the ﬁrst or last row or column which is not in the corner has five cells adjacent to it. Any other cell has eight cells adjacent to it.

1. What is the minimum number of different numerals needed to ﬁll a 3×3 square matrix? (TITA)

EXPLANATION

Answer: 4

Let us fill out the 4 corner squares with the same numeral
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The center cell has to be different from everything else, so let us worry about it right at the end. We can accommodate cells (2,1) and (2,3) with the same number. We can also fill (1, 2) and (3,2) with the same number

The center cell has to be a different number.

Now, this is feasible. Question is, can we fill with only 3 numbers?
Note that the center cell has to be different from everything else.
So, if we had to fill the entire matrix with only 4 numbers, we would have to fill the 8 border squares with only 2 numerals.
This is impossible.

Hence, the answer is 4

2. What is the minimum number of different numerals needed to ﬁll a 5×5 square matrix? (TITA)

EXPLANATION

Answer: 4

Now, we have worked with a 3 x 3. We know that the 4 corner squares of the 3 x 3 can be the same number.
So, let us fill out a set of 3 x 3’s within this 5 x 5. We can build from there.
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Now, let us fill out squares in between the 1’s row wise and then column wise.

Now, let us fill out squares in between the 1’s row wise and then column wise.
Now, this wonderful matrix fills itself. All the missing numbers should be the 4th numeral.
We know that to fill a mere 3 x 3 we needed at least 4 numerals.
If we have filled a 5 x 5 with 4 numerals, this fills me with enormous confidence.
Now, we can kill this set.
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Hence, the answer is 4

3. Suppose you are allowed to make one mistake, that is, one pair of adjacent cells can have the same numeral. What is the minimum number of different numerals required to ﬁll a 5×5 matrix?

EXPLANATION

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We can fill the 5 x 5 with 4 numerals. None of the choices is less than 4. We know that even with 1 mistake allowed, filling with just 3 numerals is impossible. So, choice A it is.

Hence, the answer is 4

Choice A is the correct answer

4. Suppose that all the cells adjacent to any particular cell must have different numerals. What is the minimum number of different numerals needed to fill a 5×5 square matrix?

EXPLANATION

Option A

This is what we get for saying this set is easy!
Let us keep all our 1’s in place and work down from there.
Now, let us take the middle 1 and fill different numerals all around it. We will worry about the rest later on.
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Hang on, this does not work. The 4 at (2, 4) has four 1’s surrounding it. That cannot be right. Our approach of filling the 1’s is not correct.
Two boxes that have only two cells in between them cannot have the same numeral. If that happens we are in trouble. So, let us restart by keeping this in mind. Fill a few 1’s.
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Let us fill 1’s that cannot conflict in any way. The way shown in the diagram is good as it ensures that there is no conflict.

So, we can fill the 5 x 5 entirely with 9 different numerals.

Hence, the answer is 9

Choice A is the correct answer

Set 4 : LED TV Sales

Q1-4. The multi-layered pie-chart below shows the sales of LED television sets for a big retail electronics outlet during 2016 and 2017. The outer layer shows the monthly sales during this period, with each label showing the month followed by sales figure of that month. For some months, the sales figures are not given in the chart. The middle-layer shows quarter-wise aggregate sales figures (in some cases, aggregate quarter-wise sales numbers are not given next to the quarter). The innermost layer shows annual sales. It is known that the sales figures during the three months of the second quarter (April, May, June) of 2016 form an arithmetic progression, as do the three monthly sales figures in the fourth quarter (October, November, December)of that year.

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1. What is the percentage increase in sales in December 2017 as compared to the sales in December 2016?

22.22%
28.57%
50.00%
38.46%

EXPLANATION

Option 2

Let us fill the simple numbers first.
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Now, let us get to the two Arithmetic Progressions. It is known that the sales figures during the three months of the second quarter (April, May, June) of 2016 form an arithmetic progression, as do the three monthly sales figures in the fourth quarter (October, November, December)of that year.
Three terms are in AP, the middle term is the average of the three terms. So, May 2016 number is 50. And November 2016 should be 120.
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Three terms are in AP, the middle term is the average of the three terms. So, May 2016 number is 50. And November 2016 should be 120.
Now, we can get the June and December numbers as well.
Dec 2016 to Dec 2017 is a leap from 140 to 180.
Growth rate = 40/140 × 100 = 2/7 in percentage terms.
Or, 28.57%

Hence, the answer is 28.57%

Choice B is the correct answer.

2. In which quarter of 2017 was the percentage increase in sales from the same quarter of 2016 the highest?

EXPLANATION

Answer: Option 3

Let us fill the simple numbers first.

Now, let us get to the two Arithmetic Progressions. It is known that the sales figures during the three months of the second quarter (April, May, June) of 2016 form an arithmetic progression, as do the three monthly sales figures in the fourth quarter (October, November, December)of that year.
Three terms are in AP, the middle term is the average of the three terms. So, May 2016 number is 50. And November 2016 should be 120.

Three terms are in AP, the middle term is the average of the three terms. So, May 2016 number is 50. And November 2016 should be 120.
Now, we can get the June and December numbers as well.

Therefore in 2017, Q1 saw the highest percentage increase in sales when compared to it’s 2016 counterpart.

Hence, the answer is Q1

Choice C is the correct answer.

3. During which quarter was the percentage decrease in sales from the previous quarter’s sales the highest?

Q4 of 2017
Q1 of 2017
Q2 of 2017
Q2 of 2016

EXPLANATION

Option 3

Let us fill the simple numbers first.

Three terms are in AP, the middle term is the average of the three terms. So, May 2016 number is 50. And November 2016 should be 120.
Now, we can get the June and December numbers as well.

Hence, the answer is Q2 of 2017

Choice C is the correct answer.

4. During which month was the percentage increase in sales from the previous month’s sales the highest?

March of 2016
October of 2016
October of 2017
March of 2017

EXPLANATION

Option 3

Let us fill the simple numbers first.

Three terms are in AP, the middle term is the average of the three terms. So, May 2016 number is 50. And November 2016 should be 120.
Now, we can get the June and December numbers as well.

Hence, the answer is October of 2017

Choice C is the correct answer.

Set 5 : ATM

Q1-4. An ATM dispenses exactly Rs. 5000 per withdrawal using 100, 200 and 500 rupee notes. The ATM requires every customer to give her preference for one of the three denominations of notes. It then dispenses notes such that the number of notes of the customer’s preferred denomination exceeds the total number of notes of other denominations dispensed to her.

1. In how many diﬀerent ways can the ATM serve a customer who gives 500 rupee notes as her preference? (TITA)

EXPLANATION

7 ways

Wonderful question. Let us first outline the overall variables.
500x + 200y + 100z = 5000. Or, 5x + 2y + z = 50.
If the preferred denomination is 500, we know that x > (y + z).
If the preferred denomination is 200, we know that y > (x + z).
If the preferred denomination is 100, we know that z > (y + x).
After this, we are off to the questions.

There should be more 500 rupee notes than the sum of the other two. Since there is no condition that says that we should have at least one note of every denomination, we could have all notes to be of the Rs. 500 denomination.
After this, we go for the number of Rs. 500 notes to be 9. This gives us Rs. 4500. Now, we need to get the final Rs. 500 with Rs. 200 and Rs. 100 notes.
We can have 5 Rs. 100 notes, or 3 Rs. 100 notes and one Rs. 200 note, or 1 Rs. 100 note and 2 Rs. 200 notes.

After this, we go for the number of Rs. 500 notes to be 8. This gives us Rs. 4000. Now, we need to get the final Rs. 1000 with Rs. 200 and Rs. 100 notes.
We can have 5 Rs. 200 notes; or 4 Rs. 200 notes and two Rs. 100 notes; or 3 Rs. 200 notes and 4 Rs. 100 notes.
If we go for 2 Rs. 200 notes we would need 6 Rs. 100 notes making it a total of 8 non Rs. 500 notes. This is not possible.
So, with 8 Rs. 500 notes only 3 options are possible.

Let us see if we can get any combination with 7 Rs. 500 notes. This amounts to Rs. 3500. We need to get Rs. 1500 from other notes.
Even if we went for least number of non-Rs 500 notes – we would need 7 Rs. 200 notes and one Rs. 100 note.
This is not possible as we would have more non-Rs. 500 notes than Rs. 500 notes.
So, we have 7 different ways.

2. If the ATM could serve only 10 customers with a stock of fifty 500 rupee notes and a sufficient number of notes of other denominations, what is the maximum number of customers among these 10 who could have given 500 rupee notes as their preferences? (TITA)

EXPLANATION

We want to have the maximum number of customers, so each customer would need to take the minimum possible number of RS. 500 notes.
We already know that the minimum number of Rs. 500 notes that a customer (with a preference for Rs. 500 notes) can take is 8.
So, we can have deliver to a maximum of 6 customers. We do not have enough Rs. 500 notes for the 7th customer.
Answer = 6.

3. What is the maximum number of customers that the ATM can serve with a stock of fifty 500 rupee notes and a sufficient number of notes of other denominations, if all the customers are to be served with at most 20 notes per withdrawal?

EXPLANATION

Option 3

To start with, we want to have the maximum number of customers who can be serviced with the available Rs. 500 notes. In other words, we need to figure out the minimum number of Rs. 500 notes with which one customer can be serviced.
In other words, can we give a customer only one Rs. 500 note and meet all conditions.

1 Rs. 500 note gives us Rs. 500. The remaining Rs. 4500 needs to delivered with Rs. 200 and Rs. 100 notes.
The best case scenario of 22 Rs. 200 and one Rs. 100 note, we have a total of 24 notes which is not possible.
With 2 Rs. 500 notes, we would have RS. 1000. 20 Rs. 200 notes make it 22 in total. Not possible again.
With 3 Rs. 500 notes, we would have RS. 1500. 17 Rs. 200 notes and one Rs. 100 note make it 21 in total. Not possible again.
With 4 Rs. 500 notes and 15 Rs. 200 notes, our conditions are met. So, the best case scenario involves 4 Rs. 500 notes.
So, with each customer being given 4 notes, we can service a maximum of 12 notes.

4. What is the number of 500 rupee notes required to serve 50 customers with 500 rupee notes as their preferences and another 50 customers with 100 rupee notes as their preferences, if the total number of notes to be dispensed is the smallest possible?

750
800
1400
900

EXPLANATION

Option 4

The number of notes dispensed should be the smallest possible.
50 customers have given the preference as Rs. 500 notes. For these if we need to have the least number of notes, we should go for the option of having 10 Rs. 500 notes and nothing else.
What about when the preference is Rs. 100 note? We could have all as Rs. 100 notes, but that would demand 50 notes which is a large number.
We could have, say, 45 Rs. 100 notes and one Rs. 500 note. This is also large.
Let us take a leap and say, we get to have 7 Rs. 500 notes, we would need to get RS. 1500 from other notes.
We could have 11 Rs. 100 notes and 2 Rs. 200 notes but this would still result in 20 notes.
The best scenario would have only Rs. 500 and Rs. 100 notes.

We could have 8 Rs. 500 notes and 10 Rs. 100 notes, resulting in 18 notes totally.
For servicing 50 customers with this combination, we would require 50 × 8 = 400 Rs. 500 notes.
So, in total, we need 500 + 400 = 900 customers.

Set 6 : Petrol Pumps

Q1-4. Fuel contamination levels at each of 20 petrol pumps P1, P2, …, P20 were recorded as either high, medium, or low.
1. Contamination levels at three pumps among P1 – P5 were recorded as high.
2. P6 was the only pump among P1 – P10 where the contamination level was recorded as low.
3. P7 and P8 were the only two consecutively numbered pumps where the same levels of contamination were recorded.
4. High contamination levels were not recorded at any of the pumps P16 – P20.
5. The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.

1.Which of the following MUST be true?

The contamination level at P12 was recorded as high.
The contamination level at P20 was recorded as medium.
The contamination level at P10 was recorded as high.
The contamination level at P13 was recorded as low.

EXPLANATION

Option 3

We know that three of P1 to P5 are recorded as High. This means the other two should be M. So, let us fill this data in.
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P7 and P8 are both same, so both could be H or both M. Let us try both. We put P7 and P8 in, P9 and P10 get filled automatically.
Let us fill in the totals as well.

The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.
So, of the number of low contaminations were x, then the number of high contaminations would be 2x.
H + L = 3x {3, 6, 9, 12, 15, or 18} or, M should be from {2, 5, 8, 11, 14, 17}
L + H is at least 5 so M cannot be 14 or 17. M cannot be 2 either. So, we can have M to be 5, 8 or 11.
So, we can have H, M, L to be {10, 5, 5} or {8, 8, 4} or {6, 11, 3}
There is no H in the P16 to P20 range, so total H cannot be 10. So, H, M, L being {10, 5, 5} can be ruled out.
So, we can have H, M, L to be {8, 8, 4} or {6, 11, 3}.
We cannot have more than 5 M’s in the range from P16 to P20.
So, we now know that {6, 11, 3} is also ruled out. So H, M, L has to be {8, 8, 4}.
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We know that in the first 10 pipes we could have has H, M, L could have been {6, 3, 1} or {4, 5, 1}
So, H, M, L for the last 10 should be {2, 5, 3} or {4, 3, 3}. Since there is no H in the first 5 pipes, we cannot have 4 H in the P16 to P20 range.
So, we need to have only {6, 3, 1} in the first range.
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The last 10 pipes should have H, M, L to be {2, 5, 3}.
High contamination levels were not recorded at any of the pumps P16 – P20. So, the last 5 should be either MLMLM or LMLML.
If the last 5 were LMLML, then we would not have one more L for P11 to P15. So, P11 to P15 should be MHMHM. Now, let us fine-tune this.
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We know that P11 cannot be H otherwise P10- and P11 would be H.
So, we need to accommodate 2 H’s in the P12 to P15 range.
We could have them at P12 and P14, P12 and P15 and P13 and P15.

If P15 were not H, it has to be L. So, we can fill the first column completely.
P12 need not have been high. Choice A need not be true.
P20 could have been L, so Choice B is also out.
P10 was high. This is indeed true.
P13 need not have been low.
Choice C is true.

2. What best can be said about the number of pumps at which the contamination levels were recorded as medium?

Exactly 8
At most 9
At least 8
More than 4

EXPLANATION

Option 1

We know that three of P1 to P5 are recorded as High. This means the other two should be M. So, let us fill this data in.
CAT DI LR 2018 Slot 1
P7 and P8 are both same, so both could be H or both M. Let us try both. We put P7 and P8 in, P9 and P10 get filled automatically.
Let us fill in the totals as well.

The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.
So, of the number of low contaminations were x, then the number of high contaminations would be 2x.
H + L = 3x {3, 6, 9, 12, 15, or 18} or, M should be from {2, 5, 8, 11, 14, 17}
L + H is at least 5 so M cannot be 14 or 17. M cannot be 2 either. So, we can have M to be 5, 8 or 11.
So, we can have H, M, L to be {10, 5, 5} or {8, 8, 4} or {6, 11, 3}
There is no H in the P16 to P20 range, so total H cannot be 10. So, H, M, L being {10, 5, 5} can be ruled out.
So, we can have H, M, L to be {8, 8, 4} or {6, 11, 3}.
We cannot have more than 5 M’s in the range from P16 to P20.
So, we now know that {6, 11, 3} is also ruled out. So H, M, L has to be {8, 8, 4}.
CAT DI LR 2018 Slot 1
We know that in the first 10 pipes we could have has H, M, L could have been {6, 3, 1} or {4, 5, 1}
So, H, M, L for the last 10 should be {2, 5, 3} or {4, 3, 3}. Since there is no H in the first 5 pipes, we cannot have 4 H in the P16 to P20 range.
So, we need to have only {6, 3, 1} in the first range.
CAT DI LR 2018 Slot 1
The last 10 pipes should have H, M, L to be {2, 5, 3}.
High contamination levels were not recorded at any of the pumps P16 – P20. So, the last 5 should be either MLMLM or LMLML.
If the last 5 were LMLML, then we would not have one more L for P11 to P15. So, P11 to P15 should be MHMHM. Now, let us fine-tune this.
CAT DI LR 2018 Slot 1
We know that P11 cannot be H otherwise P10- and P11 would be H.
So, we need to accommodate 2 H’s in the P12 to P15 range.
We could have them at P12 and P14, P12 and P15 and P13 and P15.

If P15 were not H, it has to be L. So, we can fill the first column completely.
The number of pumps at which the contamination levels were recorded as medium is exactly 8(from the table).

3. If the contamination level at P11 was recorded as low, then which of the following MUST be true?

The contamination level at P18 was recorded as low.
The contamination level at P15 was recorded as medium.
The contamination level at P14 was recorded as medium.
The contamination level at P12 was recorded as high.

EXPLANATION

Option 3

We know that three of P1 to P5 are recorded as High. This means the other two should be M. So, let us fill this data in.
CAT DI LR 2018 Slot 1
P7 and P8 are both same, so both could be H or both M. Let us try both. We put P7 and P8 in, P9 and P10 get filled automatically.
Let us fill in the totals as well.

The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.
So, of the number of low contaminations were x, then the number of high contaminations would be 2x.
H + L = 3x {3, 6, 9, 12, 15, or 18} or, M should be from {2, 5, 8, 11, 14, 17}
L + H is at least 5 so M cannot be 14 or 17. M cannot be 2 either. So, we can have M to be 5, 8 or 11.
So, we can have H, M, L to be {10, 5, 5} or {8, 8, 4} or {6, 11, 3}
There is no H in the P16 to P20 range, so total H cannot be 10. So, H, M, L being {10, 5, 5} can be ruled out.
So, we can have H, M, L to be {8, 8, 4} or {6, 11, 3}.
We cannot have more than 5 M’s in the range from P16 to P20.
So, we now know that {6, 11, 3} is also ruled out. So H, M, L has to be {8, 8, 4}.
CAT DI LR 2018 Slot 1
We know that in the first 10 pipes we could have has H, M, L could have been {6, 3, 1} or {4, 5, 1}
So, H, M, L for the last 10 should be {2, 5, 3} or {4, 3, 3}. Since there is no H in the first 5 pipes, we cannot have 4 H in the P16 to P20 range.
So, we need to have only {6, 3, 1} in the first range.
CAT DI LR 2018 Slot 1
The last 10 pipes should have H, M, L to be {2, 5, 3}.
High contamination levels were not recorded at any of the pumps P16 – P20. So, the last 5 should be either MLMLM or LMLML.
If the last 5 were LMLML, then we would not have one more L for P11 to P15. So, P11 to P15 should be MHMHM. Now, let us fine-tune this.
CAT DI LR 2018 Slot 1
We know that P11 cannot be H otherwise P10 and P11 would be H.
So, we need to accommodate 2 H’s in the P12 to P15 range.
We could have them at P12 and P14, P12 and P15 and P13 and P15.

If P15 were not H, it has to be L. So, we can fill the first column completely.

P11 was L, then two possibilities get ruled out. P11 should have been L.
The remaining 2 should have been M’s. But even in this, one possibility has M in P13 and P14 – this is not possible.
So, we have only one combination.
CAT DI LR 2018 Slot 1
Choice A , B and D are not true.
Choice C is true.

4. If contamination level at P15 was recorded as medium, then which of the following MUST be FALSE?

Contamination levels at P11 and P16 were recorded as the same.
Contamination levels at P10 and P14 were recorded as the same.
Contamination level at P14 was recorded to be higher than that at P15.
Contamination levels at P13 and P17 were recorded as the same.

EXPLANATION

Option 1

We know that three of P1 to P5 are recorded as High. This means the other two should be M. So, let us fill this data in.
CAT DI LR 2018 Slot 1
P7 and P8 are both same, so both could be H or both M. Let us try both. We put P7 and P8 in, P9 and P10 get filled automatically.
Let us fill in the totals as well.

The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.
So, of the number of low contaminations were x, then the number of high contaminations would be 2x.
H + L = 3x {3, 6, 9, 12, 15, or 18} or, M should be from {2, 5, 8, 11, 14, 17}
L + H is at least 5 so M cannot be 14 or 17. M cannot be 2 either. So, we can have M to be 5, 8 or 11.
So, we can have H, M, L to be {10, 5, 5} or {8, 8, 4} or {6, 11, 3}
There is no H in the P16 to P20 range, so total H cannot be 10. So, H, M, L being {10, 5, 5} can be ruled out.
So, we can have H, M, L to be {8, 8, 4} or {6, 11, 3}.
We cannot have more than 5 M’s in the range from P16 to P20.
So, we now know that {6, 11, 3} is also ruled out. So H, M, L has to be {8, 8, 4}.
CAT DI LR 2018 Slot 1
We know that in the first 10 pipes we could have has H, M, L could have been {6, 3, 1} or {4, 5, 1}
So, H, M, L for the last 10 should be {2, 5, 3} or {4, 3, 3}. Since there is no H in the first 5 pipes, we cannot have 4 H in the P16 to P20 range.
So, we need to have only {6, 3, 1} in the first range.
CAT DI LR 2018 Slot 1
The last 10 pipes should have H, M, L to be {2, 5, 3}.
High contamination levels were not recorded at any of the pumps P16 – P20. So, the last 5 should be either MLMLM or LMLML.
If the last 5 were LMLML, then we would not have one more L for P11 to P15. So, P11 to P15 should be MHMHM. Now, let us fine-tune this.
CAT DI LR 2018 Slot 1
We know that P11 cannot be H otherwise P10 and P11 would be H.
So, we need to accommodate 2 H’s in the P12 to P15 range.
We could have them at P12 and P14, P12 and P15 and P13 and P15.

If P15 were not H, it has to be L. So, we can fill the first column completely.

P15 was recorded as maximum. This eliminates 3 possibilities.
CAT DI LR 2018 Slot 1
From the table we can see that, except choice A everything is true and therefore our answer is Choice A

Set 7 : Management Electives

Q1-4. Adriana, Bandita, Chitra, and Daisy are four female students, and Amit, Barun, Chetan, and Deb are four male students. Each of them studies in one of three institutes – X, Y, and Z. Each student majors in one subject among Marketing, Operations, and Finance, and minors in a different one among these three subjects. The following facts are known about the eight students:

1. Three students are from X, three are from Y, and the remaining two students, both female, are from Z.
2. Both the male students from Y minor in Finance, while the female student from Y majors in Operations.
3. Only one male student majors in Operations, while three female students minor in Marketing.
4. One female and two male students major in Finance.
5. Adriana and Deb are from the same institute. Daisy and Amit are from the same institute.
6. Barun is from Y and majors in Operations. Chetan is from X and majors in Finance.
7. Daisy minors in Operations.

1. Who are the students from the institute Z?

Chitra and Daisy
Adriana and Daisy
Bandita and Chitra
Adriana and Bandita

EXPLANATION

Option 3

Let us call the females A, B, C and D; and the males AM, BR, C10 and DB.
We know that there are 2 males and one female each in X and Y and two females in Z.
CAT DI LR 2018 Slot 1

Female in Institute Y cannot be Daisy.
So, Daisy should be from institute X. So, Amit should also be from institute X.

A and DB should be from Y. So, B and C should be from Z.

We get the minors or A, B and C.

AM majors in Fin. Only one male student majors in Ops. Two male students major in Fin, so the last male student, DB should have majored in Marketing.
Now, with this let us move on to the questions.
From the above table we can see that Bandita and Chitra are the students from institute Z.

2. Which subject does Deb minor in?

Operations
Marketing
Finance
Cannot be determined uniquely from the given information

EXPLANATION

Option 3

3. Which subject does Amit major in?

Operations
Marketing
Finance
Cannot be determined uniquely from the given information

EXPLANATION

Option 3

4. If Chitra majors in Finance, which subject does Bandita major in?

Finance
Operations
Cannot be determined uniquely from the given information
Marketing

EXPLANATION

Option 2

Let us call the females A, B, C and D; and the males AM, BR, C10 and DB.
We know that there are 2 males and one female each in X and Y and two females in Z.
CAT DI LR 2018 Slot 1

Female in Institute Y cannot be Daisy.
So, Daisy should be from institute X. So, Amit should also be from institute X.

A and DB should be from Y. So, B and C should be from Z.

We get the minors or A, B and C.

AM majors in Fin. Only one male student majors in Ops. Two male students major in Fin, so the last male student, DB should have majored in Marketing.
Now, with this let us move on to the questions.
CAT DI LR 2018 Slot 1
From the above table we can see that if Chitra majors in Finance then Bandita major in Operations.

Set 8 : Committees

Q1-4. Twenty four people are part of three committees which are to look at research, teaching, and administration respectively. No two committees have any member in common. No two committees are of the same size. Each committee has three types of people: bureaucrats, educationalists, and politicians, with at least one from each of the three types in each committee. The following facts are also known about the committees:

1. The numbers of bureaucrats in the research and teaching committees are equal, while the number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee.
2. The number of educationalists in the teaching committee is less than the number of educationalists in the research committee. The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees.
3. 60% of the politicians are in the administration committee, and 20% are in the teaching committee.

1. Based on the given information, which of the following statements MUST be FALSE?

In the administration committee the number of bureaucrats is equal to the number of educationalists
The size of the research committee is less than the size of the teaching committee
The size of the research committee is less than the size of the administration committee
In the teaching committee the number of educationalists is equal to the number of politicians

EXPLANATION

Option 2

The number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee i.e.
The ratio of number of bureaucrats in Research to Admin is 3 : 4.
Number of bureaucrats in Research and Teaching are equal. So, Research : Teaching : Admin = 3 : 3 : 4.

The number of politicians in research, teaching and Admin are 20%, 20% and 60% or, the ratio should of number of politicians is 1 : 1 : 3.

The number of educationalists in the teaching committee is less than the number of educationalists in the research committee.
The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees.
If number of educationalists in teaching = a, and in Admin = b, then Research = a+b2�+�2. a < b.

Now, we know that Number of Educationalists doing research is the average of Educationalists in Teaching and Admin or, the number of educationalists in Teaching, Research and Admin are in AP or, the sum of the three numbers should be a multiple of 3.
Let us look at the totals.
Number of Bureaucrats is a multiple of 10 – it should be 10 or 20
Number of Educationalists is a multiple of 3 – it should be 3, 6, 9, 12,..
Number of Politicians is a multiple of 5 – it should be 5, 10, 15, 20,..
We have at least one from each of the three types from each of the committees.
So, number of politicians should be at least 5, so the number of bureaucrats should be 10.

Now, 3K + 5y = 14. 5y can be 5 or 10. It cannot be 10 as 3K cannot be 4.
So, we know that 3K should be 9 and 5K should be 5.

Number of Educationalists in Research is the average of the three numbers in educationalists. So, the number of educationalists in Research is 3.
Number of educationalists in Teaching and Admin are either 2, 4 or 1, 5.

With this, we are done with the whole grid. Now, let us move to the questions.
From the table, we can see that Choice B The size of the research committee is less than the size of the teaching committee is false.

2. What is the number of bureaucrats in the administration committee? (TITA)

EXPLANATION

Answer: 4

The number of politicians in research, teaching and Admin are 20%, 20% and 60% or, the ratio should of number of politicians is 1 : 1 : 3.

Now, 3K + 5y = 14. 5y can be 5 or 10. It cannot be 10 as 3K cannot be 4.
So, we know that 3K should be 9 and 5K should be 5.

With this, we are done with the whole grid. Now, let us move to the questions.
From the table, we can see that the number of bureaucrats in the administration committee is 4.

3. What is the number of educationalists in the research committee? (TITA)

EXPLANATION

Answer: 3

CAT DI LR 2018 Slot 1

The number of bureaucrats in the research committee is 75% of the number of bureaucrats in the administration committee i.e.
The ratio of number of bureaucrats in Research to Admin is 3 : 4.
Number of bureaucrats in Research and Teaching are equal. So, Research : Teaching : Admin = 3 : 3 : 4.
CAT DI LR 2018 Slot 1
The number of politicians in research, teaching and Admin are 20%, 20% and 60% or, the ratio should of number of politicians is 1 : 1 : 3.

The number of educationalists in the teaching committee is less than the number of educationalists in the research committee.
The number of educationalists in the research committee is the average of the numbers of educationalists in the other two committees.
If number of educationalists in teaching = a, and in Admin = b, then Research = a+b2�+�2. a < b.
CAT DI LR 2018 Slot 1
Now, we know that Number of Educationalists doing research is the average of Educationalists in Teaching and Admin or, the number of educationalists in Teaching, Research and Admin are in AP or, the sum of the three numbers should be a multiple of 3.
Let us look at the totals.
Number of Bureaucrats is a multiple of 10 – it should be 10 or 20
Number of Educationalists is a multiple of 3 – it should be 3, 6, 9, 12,..
Number of Politicians is a multiple of 5 – it should be 5, 10, 15, 20,..
We have at least one from each of the three types from each of the committees.
So, number of politicians should be at least 5, so the number of bureaucrats should be 10.
CAT DI LR 2018 Slot 1
Now, 3K + 5y = 14. 5y can be 5 or 10. It cannot be 10 as 3K cannot be 4.
So, we know that 3K should be 9 and 5K should be 5.

Number of Educationalists in Research is the average of the three numbers in educationalists. So, the number of educationalists in Research is 3.
Number of educationalists in Teaching and Admin are either 2, 4 or 1, 5.
CAT DI LR 2018 Slot 1
With this, we are done with the whole grid. Now, let us move to the questions.
From the table, we can see that the number of educationalists in the research committee 3.

4. Which of the following CANNOT be determined uniquely based on the given information?

The total number of educationalists in the three committees
The total number of bureaucrats in the three committees
The size of the teaching committee
The size of the research committee

EXPLANATION

Option 3

The number of politicians in research, teaching and Admin are 20%, 20% and 60% or, the ratio should of number of politicians is 1 : 1 : 3.

Now, 3K + 5y = 14. 5y can be 5 or 10. It cannot be 10 as 3K cannot be 4.
So, we know that 3K should be 9 and 5K should be 5.

With this, we are done with the whole grid. Now, let us move to the questions.
From the table, we can see that the size of teaching committee cannot be determined uniquely.

CAT DI LR : CAT 2018 Question Paper Slot 1

Set 1: Written Test

Set 2 : Satellites

Set 3 : N x N Square Matrix

Set 4 : LED TV Sales

Set 5 : ATM

Set 6 : Petrol Pumps

Set 7 : Management Electives

Set 8 : Committees

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