# CAT 2018 LRDI Slot 2 | Previous Year CAT Paper

**Answer the following question based on the information given below.**

An agency entrusted to accredit colleges looks at four parameters: faculty quality (F), reputation (R), placement quality (P), and infrastructure (I). The four parameters are used to arrive at an overall score, which the agency uses to give an accreditation to the colleges. In each parameter, there are five possible letter grades given, each carrying certain points: A

(50 points), B (40 points), C (30 points), D (20 points), and F (0 points). The overall score for a college is the weighted sum of the points scored in the four parameters. The weights of the parameters are 0.1, 0.2, 0.3 and 0.4 in some order, but the order is not disclosed. Accreditation is awarded based on the following scheme:

Eight colleges apply for accreditation, and receive the following grades in the four parameters (F, R, P, and I):

It is further known that in terms of overall scores:

- High Q is better than Best Ed;
- Best Ed is better than Cosmopolitan; and
- Education Aid is better than A-one.

**1. CAT 2018 LRDI Slot 2 | DI - Tables & Graphs**

What is the weight of the faculty quality parameter?

- A.
0.1

- B.
0.2

- C.
0.3

- D.
0.4

Answer: Option A

**Explanation** :

From point 2, Best Ed is better than Cosmopolitan. The grades of the two colleges on F and P are the same. If Best Ed is better than Cosmopolitan, the weightage assigned to R must be higher than that assigned for I.

From point 3, Education Aid is better than A-one. The grades of the two colleges on F and R are the same. If Education Aid is better than A-one, the weightage assigned to I must be higher than that assigned for P.

Therefore we have, R > I > P.

From point 1, High Q is better than Best Ed. High Q has worse grades than Best Ed on F and R and has same grade on P and a better grade only on I. We have already figured out that R > I > P. If a better grade only on I reverses the effect of worse grades on F and R, I must be assigned higher weightage than that assigned for F.

Therefore, we have the following two possibilities:

R > I > P > F or R > I > F > P.

That means R = 0.4, I = 0.3, P = 0.2 and F= 0.1 or R = 0.4, I = 0.3, F = 0.2 and P = 0.1 are the two possibilities.

Using the two possibilities, let us calculate the scores of High Q and Best Ed.

**Possibility 1: R= 0.4, I = 0.3, P =0.2, F = 0.1**

**Possibility 2: R= 0.4, I = 0.3, F =0.2, P = 0.1**

Since High Q is better than Best Ed, possibility 1 is valid and possibility 2 is ruled out.

Therefore, we have R = 0.4, I = 0.3, P = 0.2 and F = 0.1

Now all the questions can be answered.

Hence, option (a).

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**2. CAT 2018 LRDI Slot 2 | DI - Tables & Graphs**

How many colleges receive the accreditation of AAA?

Answer: 3

**Explanation** :

The weight assigned to parameter R is highest and is at 0.4.

A college that receives 50 points on all the four parameters scores 50 out of 50.

A college that receives 40 on parameter R loses 4 points on R and can score maximum 46.

Further, a college that receives 30 or below on parameter R cannot score 45.

Therefore we need to consider only three colleges, namely A-one, Education Aid and Fancy.

A-one lost 10 × 0.3 = 3 points on I.

Education Aid lost 10 × 0.2 = 2points on P.

Fancy lost 10 × 0.2 + 10 × 0.3 = 5points on P and I.

Therefore, these three colleges receive the accreditation AAA.

Answer: 3

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**3. CAT 2018 LRDI Slot 2 | DI - Tables & Graphs**

What is the highest overall score among the eight colleges?

Answer: 48

**Explanation** :

The weight assigned to parameter R is highest and is at 0.4.

A college that receives 50 points on all the four parameters scores 50 out of 50.

A college that receives 40 on parameter R loses 4 points on R and can score maximum 46.

Further, a college that receives 30 or below on parameter R cannot score 45.

Therefore we need to consider only three colleges, namely A-one, Education Aid and Fancy.

A-one lost 10 × 0.3 = 3 points on I.

Education Aid lost 10 × 0.2 = 2points on P.

Fancy lost 10 × 0.2 + 10 × 0.3 = 5points on P and I.

it can be seen that the highest overall score among the eight colleges is 50 - 2 = 48. Therefore the required answer is 48.

Answer: 48

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**4. CAT 2018 LRDI Slot 2 | DI - Tables & Graphs**

How many colleges have overall scores between 31 and 40, both inclusive?

- A.
0

- B.
1

- C.
2

- D.
3

Answer: Option A

**Explanation** :

We have already figured out that three colleges have overall score of 45 or above. Therefore we need to check for only the remaining five colleges.

The college that earned 0 on R (Global) lost 50 × 0.4 = 20 points on R. Therefore it can score maximum 30 points. We need not calculate the overall score of that college further.

There are three colleges that scored 20 on R (Cosmopolitan, Dominance and High Q).

They lost 30 × 0.4 = 12 points on R.

However, Cosmopolitan and Dominance colleges lost 30 × 0.3 = 9 more points on I. Therefore those colleges cannot score more than 30 points.

We have already figured out that High Q and Best Ed had an overall scores of 27 and 26 respectively.

Therefore no college scored between 31 and 40 points.

Hence, option (a).

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**Answer the following question based on the information given below.**

Each of the 23 boxes in the picture below represents a product manufactured by one of the following three companies: Alfa, Bravo and Charlie. The area of a box is proportional to the revenue from the corresponding product, while its centre represents the Product popularity and Market potential scores of the product (out of 20). The shadings of some of the boxes have got erased.

The companies classified their products into four categories based on a combination of scores (out of 20) on the two parameters – Product popularity and Market potential as given below:

The following facts are known:

- Alfa and Bravo had the same number of products in the Blockbuster category.
- Charlie had more products than Bravo but fewer products than Alfa in the No-hope category.
- Each company had an equal number of products in the Promising category.
- Charlie did not have any product in the Doubtful category, while Alfa had one product more than Bravo in this category.
- Bravo had a higher revenue than Alfa from products in the Doubtful category.
- Charlie had a higher revenue than Bravo from products in the Blockbuster category.
- Bravo and Charlie had the same revenue from products in the No-hope category.
- Alfa and Charlie had the same total revenue considering all products.

**5. CAT 2018 LRDI Slot 2 | DI - Tables & Graphs**

Considering all companies' products, which product category had the highest revenue?

- A.
Promising

- B.
Doubtful

- C.
Blockbuster

- D.
No-hope

Answer: Option C

**Explanation** :

Using the information given in the diagram, we get the following:

Note: The numbers in the table represent the area of the boxes in square units.

From statement 1, Alfa and Bravo have two products each in Blockbuster category.

From statement 6, Charlie had higher revenue than Bravo from products in Blockbuster category. Therefore the product with revenue 9 belongs to Charlie and the product with revenue 4 belongs to Bravo.

From statement 2, Alfa, Charlie and Bravo had 3, 2 and 1 products respectively in No-hope category.

From statement 7, Bravo and Charlie had same revenue from products in No-hope category. Therefore Charlie had revenue (1 + 3 = 4) and the other product with revenue 1 belongs to Alfa.

From statement 4, there were 4 products of Alfa and 3 products of Bravo in the Doubtful category.

From statement 5, the product with revenue 9 belonged to Bravo and the products with revenue 1 and 6 belonged to Alfa in the Doubtful category.

So far we have the following:

From statement 3, each company had one product each in the Promising category.

From statement 8, the product with revenue 2 belonged to Alfa, the product with revenue 9 belonged to Charlie and hence the product with revenue 3 belonged to Bravo.

Now we have the following:

Now all the questions can be answered.

Hence, option (c).

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**6. CAT 2018 LRDI Slot 2 | DI - Tables & Graphs**

Which of the following is the correct sequence of numbers of products Bravo had in No-hope, Doubtful, Promising and Blockbuster categories respectively?

- A.
2, 3, 1, 2

- B.
1, 3, 1, 3

- C.
1, 3, 1, 2

- D.
3, 3, 1, 2

Answer: Option C

**Explanation** :

Using the information given in the diagram, we get the following:

Note: The numbers in the table represent the area of the boxes in square units.

From statement 1, Alfa and Bravo have two products each in Blockbuster category.

From statement 6, Charlie had higher revenue than Bravo from products in Blockbuster category. Therefore the product with revenue 9 belongs to Charlie and the product with revenue 4 belongs to Bravo.

From statement 2, Alfa, Charlie and Bravo had 3, 2 and 1 products respectively in No-hope category.

From statement 7, Bravo and Charlie had same revenue from products in No-hope category. Therefore Charlie had revenue (1 + 3 = 4) and the other product with revenue 1 belongs to Alfa.

From statement 4, there were 4 products of Alfa and 3 products of Bravo in the Doubtful category.

From statement 5, the product with revenue 9 belonged to Bravo and the products with revenue 1 and 6 belonged to Alfa in the Doubtful category.

So far we have the following:

From statement 3, each company had one product each in the Promising category.

From statement 8, the product with revenue 2 belonged to Alfa, the product with revenue 9 belonged to Charlie and hence the product with revenue 3 belonged to Bravo.

Now we have the following:

Number of products that Bravo had in the required categories is 1, 3, 1 and 2 respectively.

Hence, option (c).

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**7. CAT 2018 LRDI Slot 2 | DI - Tables & Graphs**

Which of the following statements is NOT correct?

- A.
Alfa's revenue from Blockbuster products was the same as Charlie's revenue from Promising products

- B.
Bravo and Charlie had the same revenues from No-hope products

- C.
The total revenue from No-hope products was less than the total revenue from Doubtful products

- D.
Bravo's revenue from Blockbuster products was greater than Alfa's revenue from Doubtful Products

Answer: Option D

**Explanation** :

Using the information given in the diagram, we get the following:

Note: The numbers in the table represent the area of the boxes in square units.

From statement 1, Alfa and Bravo have two products each in Blockbuster category.

From statement 6, Charlie had higher revenue than Bravo from products in Blockbuster category. Therefore the product with revenue 9 belongs to Charlie and the product with revenue 4 belongs to Bravo.

From statement 2, Alfa, Charlie and Bravo had 3, 2 and 1 products respectively in No-hope category.

From statement 7, Bravo and Charlie had same revenue from products in No-hope category. Therefore Charlie had revenue (1 + 3 = 4) and the other product with revenue 1 belongs to Alfa.

From statement 4, there were 4 products of Alfa and 3 products of Bravo in the Doubtful category.

From statement 5, the product with revenue 9 belonged to Bravo and the products with revenue 1 and 6 belonged to Alfa in the Doubtful category.

So far we have the following:

From statement 3, each company had one product each in the Promising category.

From statement 8, the product with revenue 2 belonged to Alfa, the product with revenue 9 belonged to Charlie and hence the product with revenue 3 belonged to Bravo.

Now we have the following:

Statement 4 is false while all the other statements are correct.

Hence, option (d).

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**8. CAT 2018 LRDI Slot 2 | DI - Tables & Graphs**

If the smallest box on the grid is equivalent to revenue of Rs.1 crore, then what approximately was the total revenue of Bravo in Rs. crore?

- A.
24

- B.
30

- C.
34

- D.
40

Answer: Option C

**Explanation** :

Using the information given in the diagram, we get the following:

Note: The numbers in the table represent the area of the boxes in square units.

From statement 1, Alfa and Bravo have two products each in Blockbuster category.

From statement 2, Alfa, Charlie and Bravo had 3, 2 and 1 products respectively in No-hope category.

From statement 4, there were 4 products of Alfa and 3 products of Bravo in the Doubtful category.

So far we have the following:

From statement 3, each company had one product each in the Promising category.

Now we have the following:

Bravo’s total revenue = Rs. 34 crore.

Hence, option (c).

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**Answer the following question based on the information given below.**

There are only four brands of entry level smartphones called Azra, Bysi, Cxqi, and Dipq in a country. Details about their market share, unit selling price, and profitability (defined as the profit as a percentage of the revenue) for the year 2016 are given in the table below:

In 2017, sales volume of entry level smartphones grew by 40% as compared to that in 2016. Cxqi offered a 40% discount on its unit selling price in 2017, which resulted in a 15% increase in its market share. Each of the other three brands lost 5% market share. However, the profitability of Cxqi came down to half of its value in 2016. The unit selling prices of the other three brands and their profitability values remained the same in 2017 asthey were in 2016.

**9. CAT 2018 LRDI Slot 2 | DI - Tables & Graphs**

The brand that had the highest revenue in 2016 is:

- A.
Azra

- B.
Bysi

- C.
Cxqi

- D.
Dipq

Answer: Option A

**Explanation** :

Suppose the total sales of the four models was 100 in 2016. Accordingly, we can fill the following table:

Hence, option (a).

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**10. CAT 2018 LRDI Slot 2 | DI - Tables & Graphs**

The brand that had the highest profit in 2016 is:

- A.
Azra

- B.
Bysi

- C.
Cxqi

- D.
Dipq

Answer: Option C

**Explanation** :

Suppose the total sales of the four models was 100 in 2016. Accordingly, we can fill the following table:

Cxqi had the highest profit in 2016.

Hence, option (c).

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**11. CAT 2018 LRDI Slot 2 | DI - Tables & Graphs**

The brand that had the highest profit in 2017 is:

- A.
Azra

- B.
Bysi

- C.
Cxqi

- D.
Dipq

Answer: Option B

**Explanation** :

Suppose the total sales of the four models was 100 in 2016. Accordingly, we can fill the following table:

Bysi had the highest profit in 2017.

Hence, option (b).

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**12. CAT 2018 LRDI Slot 2 | DI - Tables & Graphs**

The complete list of brands whose profits went up in 2017 from 2016 is:

- A.
Azra, Bysi, Cxqi

- B.
Bysi, Cxqi, Dipq

- C.
Cxqi, Azra, Dipq

- D.
Azra, Bysi, Dipq

Answer: Option D

**Explanation** :

The profit of all brands, except Cxqi, went up in 2017 from 2016.

Hence, option (d).

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**Answer the following question based on the information given below.**

Fun Sports (FS) provides training in three sports – Gilli-danda (G), Kho-Kho (K), and Ludo (L). Currently it has an enrollment of 39 students each of whom is enrolled in at least one of the three sports. The following details are known:

- The number of students enrolled only in L is double the number of students enrolled in all the three sports.
- There are a total of 17 students enrolled in G.
- The number of students enrolled only in G is one less than the number of students enrolled only in L.
- The number of students enrolled only in K is equal to the number of students who are enrolled in both K and L.
- The maximum student enrollment is in L.
- Ten students enrolled in G are also enrolled in at least one more sport.

**13. CAT 2018 LRDI Slot 2 | LR - Venn Diagram**

What is the minimum number of students enrolled in both G and L but not in K?

Answer: 4

**Explanation** :

Since 10 students who play G enrolled in at least one other sport, the number of students who enrolled only in G = 17 - 10 = 7. Therefore x = 7.

Also from statement 1, the number of students = who enrolled in

all the three sports $\frac{7+1}{2}$ = 4

As 17 students enrolled in G, the number of students who did not enrol in G = 39 - 17 = 22.

Therefore, x + 1 + z + y + z = 22. We know that x = 7 and y = 4. Therefore we have the following: 8 + 4 + 2z = 22 or z = 5.

From statement 5, 6-w>wor w=0 or 1 or 2.

Now first two questions can be answered.

The required answer is 6 - 2 = 4.

Answer: 4

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**14. CAT 2018 LRDI Slot 2 | LR - Venn Diagram**

If the numbers of students enrolled in K and L are in the ratio 19:22, then what is the number of students enrolled in L?

- A.
17

- B.
19

- C.
22

- D.
18

Answer: Option C

**Explanation** :

If the ratio of the numbers of students enrolled in K and L are in the ratio 19:22,

$\frac{18+w}{23-w}=\frac{19}{22}$

Therefore w = 1.

Therefore total enrollment in L

= 23 − 1 = 22.

Hence, option (c).

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**15. CAT 2018 LRDI Slot 2 | LR - Venn Diagram**

Due to academic pressure, students who were enrolled in all three sports were asked to withdraw from one of the three sports. After the withdrawal, the number of students enrolled in G was six less than the number of students enrolled in L, while the number of students enrolled in K went down by one. After the withdrawal, how many students were enrolled in both G and K?

Answer: 2

**Explanation** :

Out of 4 students who are enrolled in all the three, suppose ‘a’ students dropped out of L and ‘b’ students dropped out of K. Therefore the number of students who dropped out of G = 4 - a - b.

Therefore we have the following:

If the number of students enrolled in K reduced by 1 that means out of the 4 students who had enrolled in all the three, one student dropped out of K i.e. b = 1.

Now, if the number of students enrolled in G was 6 less than the number of students enrolled in L, we have the following:

(7 + w + a + 6 − w + b) + 6

= 6 − w + b + 9 − a − b + 8

∴19 + a + b = 23 − w − a

∴2a + b + w = 4

Since b = 1, the only solution for the equation 2a + b + w = 4 is a = 1, b = 1 and w = 1.

Now both the questions can be answered.

The required number of students = w + a = 2.

Answer: 2

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**16. CAT 2018 LRDI Slot 2 | LR - Venn Diagram**

Due to academic pressure, students who were enrolled in all three sports were asked to withdraw from one of the three sports. After the withdrawal, the number of students enrolled in G was six less than the number of students enrolled in L, while the number of students enrolled in K went down by one. After the withdrawal, how many students were enrolled in both G and L?

- A.
7

- B.
5

- C.
8

- D.
6

Answer: Option D

**Explanation** :

The required number of students

= 6 − w + b = 6 − 1 + 1 = 6.

Hence, option (d).

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**Answer the following question based on the information given below.**

According to a coding scheme the sentence Peacock is designated as the national bird of India is coded as

5688999 35 1135556678 56 458 13666689 1334 79 13366

This coding scheme has the following rules:

- The scheme is case-insensitive (does not distinguish between upper case and lower case letters).
- Each letter has a unique code which is a single digit from among 1,2,3, …, 9.
- The digit 9 codes two letters, and every other digit codes three letters.
- The code for a word is constructed by arranging the digits corresponding to its letters in a non-decreasing sequence.

**17. CAT 2018 LRDI Slot 2 | LR - Coding Decoding**

What best can be concluded about the code for the letter L?

- A.
1

- B.
6

- C.
8

- D.
1 or 8

Answer: Option A

**Explanation** :

is ≡ 35 and as ≡ 56 ⇒ s ≡ 5

∴ a ≡ 6 and i ≡ 3

Letters ‘i’ and ‘d’ are common in words ‘bird’ and ‘india’. Numbers ‘1’ and ‘3’ are common in there codes. We know that i ≡ 3. Therefore, d ≡ 1. Also, br = 34

India ≡ 13366, d ≡ 1 a ≡ 6 and i ≡ 3 ⇒ n = 6

of ≡ 79 means o = 7 or 9

As peacock ≡ 5688999, code for ‘o’ must be ‘9’. Therefore, ‘f ≡ 7’.

national ≡ 13666689, a ≡ 6, i ≡ 3 , o ≡ 9 and n = 6 ⇒ tl ≡ 19

the ≡ 458 and tl ≡ 18 ⇒ t ≡ 8 and hence, l ≡ 1

Consider designated ≡ 1135556678.

As d ≡ 1 a ≡ 6, i ≡ 3, n ≡ 6, s ≡ 5, t ≡ 8; eeg ≡ 557. Therefore, e ≡ 5 and g ≡ 7

Now in peacock ≡ 5688999 we know codes for letters e, a, and o.

Therefore, pcck ≡ 8899 i.e., c = 8 or 9

If c ≡ 8, 9 codes both ‘p’ and ‘k’. As ‘9’ codes two letters and one of them in ‘o’, it can not be code for both ‘p’ and ‘k’. Hence, ‘c’ must be coded as 9. And 8 must be the code for both ‘p’ and ‘k’.

Thus, we have

br = 34 and

1 ≡ d, l

3 ≡ i

5 ≡ s, e

6 ≡ a, n

7 ≡ f, g

8 ≡ t, p, k

9 ≡ o, c

Now both the questions can be answered.

The code for the letter L = 1.

Hence, option (a).

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**18. CAT 2018 LRDI Slot 2 | LR - Coding Decoding**

What best can be concluded about the code for the letter B?

- A.
1

- B.
3

- C.
3 or 4

- D.
1 or 3 or 4

Answer: Option C

**Explanation** :

is ≡ 35 and as ≡ 56 ⇒ s ≡ 5

∴ a ≡ 6 and i ≡ 3

Letters ‘i’ and ‘d’ are common in words ‘bird’ and ‘india’. Numbers ‘1’ and ‘3’ are common in there codes. We know that i ≡ 3. Therefore, d ≡ 1. Also, br = 34

India ≡ 13366, d ≡ 1 a ≡ 6 and i ≡ 3 ⇒ n = 6

of ≡ 79 means o = 7 or 9

As peacock ≡ 5688999, code for ‘o’ must be ‘9’. Therefore, ‘f ≡ 7’.

national ≡ 13666689, a ≡ 6, i ≡ 3 , o ≡ 9 and n = 6 ⇒ tl ≡ 19

the ≡ 458 and tl ≡ 18 ⇒ t ≡ 8 and hence, l ≡ 1

Consider designated ≡ 1135556678.

As d ≡ 1 a ≡ 6, i ≡ 3, n ≡ 6, s ≡ 5, t ≡ 8; eeg ≡ 557. Therefore, e ≡ 5 and g ≡ 7

Now in peacock ≡ 5688999 we know codes for letters e, a, and o.

Therefore, pcck ≡ 8899 i.e., c = 8 or 9

If c ≡ 8, 9 codes both ‘p’ and ‘k’. As ‘9’ codes two letters and one of them in ‘o’, it can not be code for both ‘p’ and ‘k’. Hence, ‘c’ must be coded as 9. And 8 must be the code for both ‘p’ and ‘k’.

Thus, we have

br = 34 and

1 ≡ d, l

3 ≡ i

5 ≡ s, e

6 ≡ a, n

7 ≡ f, g

8 ≡ t, p, k

9 ≡ o, c

The code for the letter B = 3 or 4.

Hence, option (c).

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**19. CAT 2018 LRDI Slot 2 | LR - Coding Decoding**

For how many digits can the complete list of letters associated with that digit be identified?

- A.
0

- B.
1

- C.
2

- D.
3

Answer: Option C

**Explanation** :

is ≡ 35 and as ≡ 56 ⇒ s ≡ 5

∴ a ≡ 6 and i ≡ 3

Letters ‘i’ and ‘d’ are common in words ‘bird’ and ‘india’. Numbers ‘1’ and ‘3’ are common in there codes. We know that i ≡ 3. Therefore, d ≡ 1. Also, br = 34

India ≡ 13366, d ≡ 1 a ≡ 6 and i ≡ 3 ⇒ n = 6

of ≡ 79 means o = 7 or 9

As peacock ≡ 5688999, code for ‘o’ must be ‘9’. Therefore, ‘f ≡ 7’.

national ≡ 13666689, a ≡ 6, i ≡ 3 , o ≡ 9 and n = 6 ⇒ tl ≡ 19

the ≡ 458 and tl ≡ 18 ⇒ t ≡ 8 and hence, l ≡ 1

Consider designated ≡ 1135556678.

As d ≡ 1 a ≡ 6, i ≡ 3, n ≡ 6, s ≡ 5, t ≡ 8; eeg ≡ 557. Therefore, e ≡ 5 and g ≡ 7

Now in peacock ≡ 5688999 we know codes for letters e, a, and o.

Therefore, pcck ≡ 8899 i.e., c = 8 or 9

If c ≡ 8, 9 codes both ‘p’ and ‘k’. As ‘9’ codes two letters and one of them in ‘o’, it can not be code for both ‘p’ and ‘k’. Hence, ‘c’ must be coded as 9. And 8 must be the code for both ‘p’ and ‘k’.

Thus, we have

br = 34 and

1 ≡ d, l

3 ≡ i

5 ≡ s, e

6 ≡ a, n

7 ≡ f, g

8 ≡ t, p, k

9 ≡ o, c

Only for 8 and 9, the complete list of letters associated is identified.

Hence, option (c).

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**20. CAT 2018 LRDI Slot 2 | LR - Coding Decoding**

Which set of letters CANNOT be coded with the same digit?

- A.
X,Y,Z

- B.
S,U,V

- C.
I,B,M

- D.
S,E,Z

Answer: Option B

**Explanation** :

is ≡ 35 and as ≡ 56 ⇒ s ≡ 5

∴ a ≡ 6 and i ≡ 3

India ≡ 13366, d ≡ 1 a ≡ 6 and i ≡ 3 ⇒ n = 6

of ≡ 79 means o = 7 or 9

As peacock ≡ 5688999, code for ‘o’ must be ‘9’. Therefore, ‘f ≡ 7’.

national ≡ 13666689, a ≡ 6, i ≡ 3 , o ≡ 9 and n = 6 ⇒ tl ≡ 19

the ≡ 458 and tl ≡ 18 ⇒ t ≡ 8 and hence, l ≡ 1

Consider designated ≡ 1135556678.

As d ≡ 1 a ≡ 6, i ≡ 3, n ≡ 6, s ≡ 5, t ≡ 8; eeg ≡ 557. Therefore, e ≡ 5 and g ≡ 7

Now in peacock ≡ 5688999 we know codes for letters e, a, and o.

Therefore, pcck ≡ 8899 i.e., c = 8 or 9

Thus, we have

br = 34 and

1 ≡ d, l

3 ≡ i

5 ≡ s, e

6 ≡ a, n

7 ≡ f, g

8 ≡ t, p, k

9 ≡ o, c

(X, Y, Z) can be coded with the same digit. (I, B, M) can have code as ‘3’. (S, E, Z) can be coded with number ‘5’. As ‘S’ and ‘E’ are coded 5, only one more letter has code 5. Thus, (S, U, V) cannot be coded with the same digit.

Hence, option (b).

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**Answer the following question based on the information given below.**

The base exchange rate of a currency X with respect to a currency Y is the number of units of currency Y which is equivalent in value to one unit of currency X. Currency exchange outlets buy currency at buying exchange rates that are lower than base exchange rates, and sell currency at selling exchange rates that are higher than base exchange rates.

A currency exchange outlet uses the local currency L to buy and sell three international currencies A, B, and C, but does not exchange one international currency directly withanother. The base exchange rates of A, B and C with respect to L are in the ratio 100:120:1. The buying exchange rates of each of A, B, and C with respect to L are 5% below the corresponding base exchange rates, and their selling exchange rates are 10% above their corresponding base exchange rates.

The following facts are known about the outlet on a particular day:

- The amount of L used by the outlet to buy C equals the amount of L it received by selling C.
- The amounts of L used by the outlet to buy A and B are in the ratio 5 : 3.
- The amounts of L the outlet received from the sales of A and B are in the ratio 5 : 9.
- The outlet received 88000 units of L by selling A during the day.
- The outlet started the day with some amount of L, 2500 units of A, 4800 units of B, and 48000 units of C.
- The outlet ended the day with some amount of L, 3300 units of A, 4800 units of B, and 51000 units of C.

**21. CAT 2018 LRDI Slot 2 | LR - Mathematical Reasoning**

How many units of currency A did the outlet buy on that day?

Answer: 1200

**Explanation** :

Suppose the base exchange rates of A, B and C w.r.t L are 100m, 120m and m respectively.

Therefore we have the following:

Given: The outlet received 88000 units of L by selling A. Therefore the number of units of A sold

$=\frac{88000}{110m}=\frac{800}{m}$

From point 3, the number of units of L received by selling B = $\frac{9}{5}$ × 88000 = 158400

Therefore the number of units of B sold = $\frac{158400}{132m}=\frac{1200}{m}$

From points 5 and 6, the number of units of A bought = 800 + $\frac{800}{m}$

Therefore, the number of units of L paid to buy A

= 95m $\left(800+\frac{800}{m}\right)$ = 76000 (1 + m)

From point 2, the number of units of L paid to buy B =

$\frac{3}{5}$ × 76000 (1 + m) = 45600 (1 + m)

Therefore the number of units of B bought = $\frac{45600(1+m)}{114m}=400+\frac{400}{m}$

From points 5 and 6,

$\frac{1200}{m}=400+\frac{400}{m}.$ Solving for m, m = 2

Therefore we have the following

If the number of units of C sold = x, from points 5 and 6, the number of units of C bought = x + 3000.

Therefore, we get, 1.9(x+3000)=2.2x or solving for x, x = 19000.

Using the value of m, we get the following table for the number of units of A, B and C bought and sold.

Answer: 1200

Workspace:

**22. CAT 2018 LRDI Slot 2 | LR - Mathematical Reasoning**

How many units of currency C did the outlet sell on that day?

- A.
3000

- B.
6000

- C.
19000

- D.
22000

Answer: Option C

**Explanation** :

Suppose the base exchange rates of A, B and C w.r.t L are 100m, 120m and m respectively.

Therefore we have the following:

Given: The outlet received 88000 units of L by selling A. Therefore the number of units of A sold

$=\frac{88000}{110m}=\frac{800}{m}$

From point 3, the number of units of L received by selling B = $\frac{9}{5}$ × 88000 = 158400

Therefore the number of units of B sold = $\frac{158400}{132m}=\frac{1200}{m}$

From points 5 and 6, the number of units of A bought = 800 + $\frac{800}{m}$

Therefore, the number of units of L paid to buy A

= 95m $\left(800+\frac{800}{m}\right)$ = 76000 (1 + m)

From point 2, the number of units of L paid to buy B =

$\frac{3}{5}$ × 76000 (1 + m) = 45600 (1 + m)

Therefore the number of units of B bought = $\frac{45600(1+m)}{114m}=400+\frac{400}{m}$

From points 5 and 6,

$\frac{1200}{m}=400+\frac{400}{m}.$ Solving for m, m = 2

Therefore we have the following

If the number of units of C sold = x, from points 5 and 6, the number of units of C bought = x + 3000.

Therefore, we get, 1.9(x+3000)=2.2x or solving for x, x = 19000.

Using the value of m, we get the following table for the number of units of A, B and C bought and sold.

The outlet sold 19000 units of currency C on that day.

Hence, option (c).

Workspace:

**23. CAT 2018 LRDI Slot 2 | LR - Mathematical Reasoning**

What was the base exchange rate of currency B with respect to currency L on that day?

Answer: 240

**Explanation** :

Suppose the base exchange rates of A, B and C w.r.t L are 100m, 120m and m respectively.

Therefore we have the following:

Given: The outlet received 88000 units of L by selling A. Therefore the number of units of A sold

$=\frac{88000}{110m}=\frac{800}{m}$

From point 3, the number of units of L received by selling B = $\frac{9}{5}$ × 88000 = 158400

Therefore the number of units of B sold = $\frac{158400}{132m}=\frac{1200}{m}$

From points 5 and 6, the number of units of A bought = 800 + $\frac{800}{m}$

Therefore, the number of units of L paid to buy A

= 95m $\left(800+\frac{800}{m}\right)$ = 76000 (1 + m)

From point 2, the number of units of L paid to buy B =

$\frac{3}{5}$ × 76000 (1 + m) = 45600 (1 + m)

Therefore the number of units of B bought = $\frac{45600(1+m)}{114m}=400+\frac{400}{m}$

From points 5 and 6,

$\frac{1200}{m}=400+\frac{400}{m}.$ Solving for m, m = 2

Therefore we have the following

If the number of units of C sold = x, from points 5 and 6, the number of units of C bought = x + 3000.

Therefore, we get, 1.9(x+3000)=2.2x or solving for x, x = 19000.

Using the value of m, we get the following table for the number of units of A, B and C bought and sold.

The base exchange rate of B w.r.t. L on that day was 240

Therefore the required answer is 240.

Answer: 240

Workspace:

**24. CAT 2018 LRDI Slot 2 | LR - Mathematical Reasoning**

What was the buying exchange rate of currency C with respect to currency L on that day?

- A.
0.95

- B.
1.10

- C.
1.90

- D.
2.20

Answer: Option C

**Explanation** :

Suppose the base exchange rates of A, B and C w.r.t L are 100m, 120m and m respectively.

Therefore we have the following:

Given: The outlet received 88000 units of L by selling A. Therefore the number of units of A sold

$=\frac{88000}{110m}=\frac{800}{m}$

From point 3, the number of units of L received by selling B = $\frac{9}{5}$ × 88000 = 158400

Therefore the number of units of B sold = $\frac{158400}{132m}=\frac{1200}{m}$

From points 5 and 6, the number of units of A bought = 800 + $\frac{800}{m}$

Therefore, the number of units of L paid to buy A

= 95m $\left(800+\frac{800}{m}\right)$ = 76000 (1 + m)

From point 2, the number of units of L paid to buy B =

$\frac{3}{5}$ × 76000 (1 + m) = 45600 (1 + m)

Therefore the number of units of B bought = $\frac{45600(1+m)}{114m}=400+\frac{400}{m}$

From points 5 and 6,

$\frac{1200}{m}=400+\frac{400}{m}.$ Solving for m, m = 2

Therefore we have the following

Therefore, we get, 1.9(x+3000)=2.2x or solving for x, x = 19000.

The buying exchange rate of C w.r.t. L on that day was 1.90

Hence, option (c).

Workspace:

**Answer the following question based on the information given below.**

Seven candidates, Akil, Balaram, Chitra, Divya, Erina, Fatima, and Ganeshan, were invited to interview for a position. Candidates were required to reach the venue before 8 am. Immediately upon arrival, they were sent to one of three interview rooms: 101, 102, and 103. The following venue log shows the arrival times for these candidates. Some of the names have not been recorded in the log and have been marked as ‘?’.

Additionally here are some statements from the candidates: Balaram: I was the third person to enter Room 101. Chitra: I was the last person to enter the room I was allotted to. Erina: I was the only person in the room I was allotted to. Fatima: Three people including Akil were already in the ro om that I was allotted to when entered it. Ganeshan : I was one among the two candidates allotted to Room 102.

**25. CAT 2018 LRDI Slot 2 | LR - Selection & Distribution**

What best can be said about the room to which Divya was allotted?

- A.
Definitely Room 101

- B.
Definitely Room 102

- C.
Definitely Room 103

- D.
Either Room 101 or Room 102

Answer: Option A

**Explanation** :

From Ganeshan’s statement, there were two candidates in room 102. From Erina’s statement, she was the only candidate in her room.

However, from Balram’s statement, there were at least three candidates in the room 101. Therefore, there were 4 candidates in room 101, 2 candidates in room 102 and 1 candidate in room 103.

Using the given statements, we can generate the following:

Now all the questions can be answered.

Hence, option (a).

Workspace:

**26. CAT 2018 LRDI Slot 2 | LR - Selection & Distribution**

Who else was in Room 102 when Ganeshan entered?

- A.
Akil

- B.
Divya

- C.
Chitra

- D.
No one

Answer: Option D

**Explanation** :

From Ganeshan’s statement, there were two candidates in room 102. From Erina’s statement, she was the only candidate in her room.

However, from Balram’s statement, there were at least three candidates in the room 101. Therefore, there were 4 candidates in room 101, 2 candidates in room 102 and 1 candidate in room 103.

Using the given statements, we can generate the following:

Noone was there in the room when Ganeshan entered.

Hence, option (c).

Workspace:

**27. CAT 2018 LRDI Slot 2 | LR - Selection & Distribution**

When did Erina reach the venue?

- A.
7:10 am

- B.
7:15 am

- C.
7:25 am

- D.
7:45 am

Answer: Option D

**Explanation** :

From Ganeshan’s statement, there were two candidates in room 102. From Erina’s statement, she was the only candidate in her room.

However, from Balram’s statement, there were at least three candidates in the room 101. Therefore, there were 4 candidates in room 101, 2 candidates in room 102 and 1 candidate in room 103.

Using the given statements, we can generate the following:

Erina reached the venue at 7:45 am

Hence, option (c).

Workspace:

**28. CAT 2018 LRDI Slot 2 | LR - Selection & Distribution**

If Ganeshan entered the venue before Divya, when did Balaram enter the venue?

- A.
7:10 am

- B.
7:15 am

- C.
7:25 am

- D.
7:45 am

Answer: Option C

**Explanation** :

Using the given statements, we can generate the following:

If Ganeshan entered before Divya, Balaram entered at 7.25 am.

Hence, option (c).

Workspace:

**Answer the following question based on the information given below.**

Each visitor to an amusement park needs to buy a ticket. Tickets can be Platinum, Gold, or Economy. Visitors are classified as Old, Middle-aged, or Young. The following facts are known about visitors and ticket sales on particular day:

- 140 tickets were sold.
- The number of Middle-aged visitors was twice the number of Old visitors, while the number of Young visitors was twice the number of Middle-aged visitors.
- Young visitors bought 38 of the 55 Economy tickets that were sold, and they bought half the total number of Platinum tickets that were sold.
- The number of Gold tickets bought by Old visitors was equal to the number of Economy tickets bought by Old visitors.

**29. CAT 2018 LRDI Slot 2 | LR - Mathematical Reasoning**

If the number of Old visitors buying Platinum tickets was equal to the number of Middle-aged visitors buying Platinum tickets, then which among the following could be the total number of Platinum tickets sold?

- A.
32

- B.
34

- C.
38

- D.
36

Answer: Option A

**Explanation** :

From points 1 and 2, the number of tickets bought by Young people = 80, the number of tickets bought by the middle aged people = 40 and the number of tickets bought by the old people = 20.

Using points 3 & 4, we get the following

Now all the questions can be answered.

We have, 20 - 2y = x + 2y - 20. Therefore, x + 4y = 40.

Therefore, x = 40 - 4y = 4(10 - y)

or 2x = 8(10 - y). Therefore, x is a multiple of 8. Out of the given options, 32 is a possible answer.

Hence, option (a).

Workspace:

**30. CAT 2018 LRDI Slot 2 | LR - Mathematical Reasoning**

If the number of Old visitors buying Platinum tickets was equal to the number of Middleaged visitors buying Economy tickets, then the number of Old visitors buying Gold tickets was

Answer: 3

**Explanation** :

We have, 20 − 2y = 17 − y or y = 3. Therefore the number of old visitors buying Gold tickets = 3.

Therefore the required answer is 3.

Answer: 3

Workspace:

**31. CAT 2018 LRDI Slot 2 | LR - Mathematical Reasoning**

If the number of Old visitors buying Gold tickets was strictly greater than the number of Young visitors buying Gold tickets, then the number of Middle-aged visitors buying Gold tickets was

Answer: 0

**Explanation** :

We have, y > 42 − x or x + y > 42. Now, the number of middle-aged visitors buying Gold tickets

= 43 − (x + y). Since x + y > 42, the number of middle-aged visitors buying Gold tickets = 0.

Therefore the required answer is 0.

Answer: 0

Workspace:

**32. CAT 2018 LRDI Slot 2 | LR - Mathematical Reasoning**

Which of the following statements MUST be FALSE?

- A.
The numbers of Middle-aged and Young visitors buying Gold tickets were equal

- B.
The numbers of Old and Middle-aged visitors buying Economy tickets were equal

- C.
The numbers of Gold and Platinum tickets bought by Young visitors were equal

- D.
The numbers of Old and Middle-aged visitors buying Platinum tickets were equal

Answer: Option B

**Explanation** :

Option 1: The numbers of Middle-aged and Young visitors buying Gold tickets were equal i.e. 43 − x − y = 42 − x or y = 1. No condition precludes the possibility of y having a value of 1.

Option 2: The numbers of Old and Middle-aged visitors buying Economy tickets were equal i.e. y = 17 − y. This is not possible because in that case y will not be a natural number.

Option 3: The numbers of Gold and Platinum tickets bought by Young visitors were equal i.e. x = 42 − x i.e. x = 21. This is possible.

Option 4: The numbers of Old and Middle-aged visitors buying Platinum tickets were equal i.e. 20 - 2y = x + 2y - 20 or x + 4y = 40. This is possible.

Hence, option (b).

Workspace:

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