# CAT 2001 LRDI

Paper year paper questions for CAT 2001 LRDI

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

The following is a table describing garments manufactured based upon the colour and size each lay. There are four sizes: M-Medium, L-Large, XL-Extra Large and XXL-Extra Extra Large. There are three colours: Yellow, Red and White.

**1. CAT 2001 LRDI | DI - Tables & Graphs**

How many lays are used to produce Yellow-coloured fabrics?

- A.
10

- B.
11

- C.
12

- D.
14

Answer: Option D

**Explanation** :

Lay numbers 1, 3, 4, 6, 7, 8, 9, 11, 12, 15, 21, 24, 25 and 27 produce yellow coloured fabrics.

∴ Number of lays that produce yellow coloured fabrics = 14

Hence, option 4.

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**2. CAT 2001 LRDI | DI - Tables & Graphs**

How many lays are used to produce Extra-Extra Large fabrics?

- A.
15

- B.
16

- C.
17

- D.
18

Answer: Option B

**Explanation** :

Lay numbers 7, 8, 9, 10, 11, 12, 13, 14, 15, 21, 22, 23, 24, 25, 26 and 27 are used to produce XXL fabrics.

∴ Number of lays that produce XXL fabrics = 16

Hence, option 2.

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**3. CAT 2001 LRDI | DI - Tables & Graphs**

How many lays are used to produce Extra-Extra Large Yellow or Extra-Extra Large White fabrics?

- A.
8

- B.
9

- C.
10

- D.
15

Answer: Option D

**Explanation** :

From the solution to the previous question we get that, number of lays that produce XXL fabrics = 16

∵ Lay number 22 produces only XXL red fabrics.

∴ Number of lays that produce XXL yellow or XXL white fabrics = 16 − 1 = 15

Hence, option 4.

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**4. CAT 2001 LRDI | DI - Tables & Graphs**

How many varieties of fabrics, which exceed the order have been produced?

- A.
3

- B.
4

- C.
5

- D.
6

Answer: Option A

**Explanation** :

Count the numbers from the last row of the table which gives surplus.

Hence, option 1.

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

Answer these questions based on the table given below concerning the twenty busiest international airports in the world.

**5. CAT 2001 LRDI | DI - Tables & Graphs**

How many international airports of type 'A' account for more than 40 million passengers?

- A.
4

- B.
5

- C.
6

- D.
7

Answer: Option B

**Explanation** :

International airports of type A: Hartsfield, Chicago-O’Hare, Los Angeles, DFW and San Francisco account for more than 40 million passengers.

Hence, option 2.

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**6. CAT 2001 LRDI | DI - Tables & Graphs**

What percentage of top ten busiest airports is in the United States of America?

- A.
60

- B.
80

- C.
70

- D.
90

Answer: Option A

**Explanation** :

There are 6 airports of USA in top ten busiest airports. i.e. Hartsfield, Chicago-O’Hare, Los Angeles, DFW, San Francisco and Denver.

Note: Los Angeles, California is in USA.

Hence, option 1.

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**7. CAT 2001 LRDI | DI - Tables & Graphs**

Of the five busiest airports, roughly what percentage of passengers is handled by Heathrow airport?

- A.
30

- B.
40

- C.
20

- D.
50

Answer: Option C

**Explanation** :

Total passengers handled by the five busiest airports = 78 + 73 + 64 + 62 + 60 = 337

Required percentage = (62 × 100)/337 ≈ 18%

Hence, option 3.

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**8. CAT 2001 LRDI | DI - Tables & Graphs**

How many international airports not located in the USA handle more than 30 million passengers?

- A.
5

- B.
6

- C.
10

- D.
14

Answer: Option B

**Explanation** :

All the given airports handle more than 30 million passengers out of which 6 are not located in the USA. i.e. Heathrow Airport, Haneda Airport, Frankfurt Airport, Roissy-Charles de, Amsterdam and Kimpo.

Hence, option 2.

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

Figure 1 shows the amount of work distribution, in man-hours, for a software between offshore and onsite activities. Figure 2 shows the estimated and actual work effort involved in the different offshore activities in the same company during the same period. [Note: Onsite refers to work performed at the customer's premise and offshore refers to work performed at the developer's premise.]

**9. CAT 2001 LRDI | DI - Tables & Graphs**

Which of the work requires as many man-hours as that spent in coding?

- A.
Offshore, design and coding

- B.
Offshore coding

- C.
Testing

- D.
Offshore, testing and coding

Answer: Option A

**Explanation** :

Man-hours spent in coding = offshore coding + onsite coding = 410 + 100 = 510

Now, total man-hours spent in offshore design and offshore coding = 100 + 410 = 510

Hence, option 1.

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**10. CAT 2001 LRDI | DI - Tables & Graphs**

Roughly what percentage of the total work is carried out onsite?

- A.
40 percent

- B.
20 percent

- C.
30 percent

- D.
50 percent

Answer: Option C

**Explanation** :

Total man-hours spent on offshore work = 100 + 410 + 290 = 800

And total man-hours spent on onsite work = 90 + 100 + 150 = 340

∴ Percentage onsite work = 340/(800 + 340) ≈ 30%

Hence, option 3.

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**11. CAT 2001 LRDI | DI - Tables & Graphs**

The total effort in man-hours spent onsite is nearest to which of the following?

- A.
The sum of the estimated and actual effort for offshore design

- B.
The estimated man-hours of offshore coding

- C.
The actual man-hours of offshore testing

- D.
Half of the man-hours of estimated offshore coding

Answer: Option C

**Explanation** :

Total man-hours spent on onsite work = 90 + 100 + 150 = 340

Sum of the estimated and actual effort for offshore design = 100 + 100 = 200

Estimated man-hours of offshore coding = 410

Actual man-hours of offshore testing = 300

Half of the man-hours of estimated offshore coding = 205

∴ Total man-hours spent on onsite work is more closer to actual man-hours of offshore testing.

Hence, option 3.

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**12. CAT 2001 LRDI | DI - Tables & Graphs**

If the total working hours were 100, which of the following tasks will account for approximately 50 hours?

- A.
Coding

- B.
Design

- C.
Offshore testing

- D.
Offshore testing plus design

Answer: Option A

**Explanation** :

Using the data from the previous question.

Total amount of work = 800 + 340 = 1140

Total coding man-hours = 410 + 100 = 510

Total design man-hours = 90 + 100 = 190

Total offshore testing man-hours = 290

Total offshore testing plus design man-hours = 290 + 190 = 480

50% of 1140 = 570

510 is more closer to 570.

Hence, option 1.

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**13. CAT 2001 LRDI | DI - Tables & Graphs**

If 50 percent of the offshore work were to be carried out onsite, with the distribution of effort between the tasks remaining the same, the proportion of testing carried out offshore would be

- A.
40 percent

- B.
30 percent

- C.
50 percent

- D.
70 percent

Answer: Option B

**Explanation** :

Man-hours required for testing offshore = 290

50% of this is now carried out onsite.

∴ Man-hours for testing offshore = 290 − 145 = 145

And man-hours for testing onsite = 150 + 145 = 295

Total man-hours of testing = 145 + 295 = 440

∴ Required percentage = (145 × 100)/440 ≈ 33%

Hence, option 2.

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**14. CAT 2001 LRDI | DI - Tables & Graphs**

If 50 percent of the offshore work were to be carried out onsite, with the distribution of effort between the tasks remaining the same, which of the following is true of all work carried out onsite?

- A.
The amount of coding done is greater than that of testing

- B.
The amount of coding done onsite is less than that of design done onsite

- C.
The amount of design carried out onsite is greater than that of testing

- D.
The amount of testing carried out offshore is greater than that of total design

Answer: Option 16

**Explanation** :

From the table, amount of coding done is greater than testing for onsite work.

Hence, option 1.

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

The following sketch shows the pipelines carrying material from one location to another. Each location has a demand for material. The demand at Vaishali is 400, at Jyotishmati is 400, at Panchal is 700, and at Vidisha is 200. Each arrow indicates the direction of material flow through the pipeline. The flow from Vaishali to Jyotishmati is 300. The quantity of material flow is such that the demands at all these locations are exactly met. The capacity of each pipeline is 1000.

**15. CAT 2001 LRDI | DI - Routes & Networks**

The quantity moved from Avanti to Vidisha is

- A.
200

- B.
800

- C.
700

- D.
1000

Answer: Option D

**Explanation** :

Demand at Panchal = 700

∴ Quantity of material flowing in the pipeline from Jyotishmati to Panchal = 700

Also, Flow from Vaishali to Jyotishmati + Flow from Vidisha to Jyotishmati = Demand at Jyotishmati + Demand at Panchal

∴ 300 + Flow from Vidisha to Jyotishmati = 400 + 700 = 1100

∴ Flow from Vidisha to Jyotishmati = 1100 − 300 = 800

Now, Flow from Avanti to Vaishali = Demand at Vaishali + Flow from Vaishali to Jyotishmati

∴ Flow from Avanti to Vaishali = 300 + 400 = 700

Also, Flow from Avanti to Vidisha = Demand at Vidisha + Flow from Vidisha to Jyotishmati

∴ Flow from Avanti to Vidisha = 200 + 800 = 1000

The diagram above shows the flow of the material in the pipelines and the demand at the respective points.

∴ Quantity of material moved from Avanti to Vidisha = 1000

Hence, option 4.

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**16. CAT 2001 LRDI | DI - Routes & Networks**

The free capacity available at the Avanti-Vaishali pipeline is

- A.
0

- B.
100

- C.
200

- D.
300

Answer: Option D

**Explanation** :

From the flow diagram, we have,

Each pipeline has a capacity of 1000 and the flow in the Avanti-Vaishali pipeline is 700.

∴ Free capacity available = 1000 − 700 = 300

Hence, option 4.

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**17. CAT 2001 LRDI | DI - Routes & Networks**

What is the free capacity available in the Avanti-Vidisha pipeline?

- A.
300

- B.
200

- C.
100

- D.
0

Answer: Option D

**Explanation** :

From the flow diagram, the Avanti-Vidisha pipeline carries 1000.

∴ The Avanti-Vidisha pipeline has no free capacity.

Hence, option 4.

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

There are six companies, 1 through 6. All of these companies use six operations, A through F. The following graph shows the distribution of efforts put in by each company in these six operations.

**18. CAT 2001 LRDI | DI - Tables & Graphs**

Suppose effort allocation is interchanged between operation B and C, then C and D, and then D and E. If companies are then ranked in ascending order of effort in E, what will be the rank of company 3?

- A.
2

- B.
3

- C.
4

- D.
5

Answer: Option B

**Explanation** :

If we will interchange the efforts allocated to various operations, then the values of B will be assigned indirectly to values of E.

∴ According to B, the rank of company 3 is third in ascending order.

Hence, option 2.

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**19. CAT 2001 LRDI | DI - Tables & Graphs**

A new technology is introduced in company 4 such that the total efforts for operations B through F get evenly distributed among these. What is the change in the percentage of effort in operation E?

- A.
Reduction of 12.3

- B.
Increase of 12.3

- C.
Reduction of 5.6

- D.
Increase of 5.6

Answer: Option A

**Explanation** :

Total of B, C, D, E and F for company 4 = 100 − 18.5 = 81.5

After introducing the new technology the value of E = 81.5/5 = 16.3

∴ The change in the value of E = 16.3 − 28.6 = −12.3

Hence, option 1.

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**20. CAT 2001 LRDI | DI - Tables & Graphs**

Suppose the companies find that they can remove operations B, C and D and redistribute the effort released equally among the remaining operations. Then, which operation will show the maximum across all companies and all operations?

- A.
Operation E in company 1

- B.
Operation E in company 4

- C.
Operation F in company 5

- D.
Operation E in company 5

Answer: Option D

**Explanation** :

The value for each of the options is calculated as given in the table.

Hence, option 4.

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

The questions are based on the pie charts given below. Chart 1 shows the distribution of twelve million tons of crude oil transported through different modes over a specific period of time. Chart 2 shows the distribution of the cost of transporting this crude oil. The total cost was Rs. 30 million.

**21. CAT 2001 LRDI | DI - Tables & Graphs**

The cost in rupees per ton of oil moved by rail and road happens to be roughly

- A.
3

- B.
1.5

- C.
4.5

- D.
8

Answer: Option B

**Explanation** :

Required Cost = $\frac{\mathrm{Total}\mathrm{cost}\mathrm{by}\mathrm{Rail}\mathrm{and}\mathrm{Road}\mathrm{in}\mathrm{millions}}{\mathrm{Total}\mathrm{Crude}\mathrm{oil}\mathrm{transported}\mathrm{by}\mathrm{Rail}\mathrm{and}\mathrm{Road}\mathrm{in}\mathrm{million}\mathrm{tons}}$

$=\frac{\left({\displaystyle \frac{12+6}{100}}\right)\times 30}{\left({\displaystyle \frac{9+22}{100}}\right)\times 12}\phantom{\rule{0ex}{0ex}}=\frac{18}{12}\times \frac{30}{31}$

≈ 1.5

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**22. CAT 2001 LRDI | DI - Tables & Graphs**

From the charts given, it appears that the cheapest mode of transport is

- A.
Road

- B.
Rail

- C.
Pipeline

- D.
Ship

Answer: Option A

**Explanation** :

Ratio of the cost of transportation of crude oil per ton by Rail, Road, Ship and Pipeline are calculated below.

$\left(\frac{12}{9}\times \frac{30}{12}\right):\left(\frac{6}{22}\times \frac{30}{12}\right):\left(\frac{10}{9}\times \frac{30}{12}\right):\left(\frac{65}{49}\times \frac{30}{12}\right)\phantom{\rule{0ex}{0ex}}\left(\frac{12}{9}\right):\left(\frac{6}{22}\right):\left(\frac{10}{9}\right):\left(\frac{65}{49}\right)$

Here the smallest ratio is $\frac{6}{22}$ which corresponds to transportation of oil by Road.

∴ Road is the cheapest mode of the transport.

Hence, option 1.

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**23. CAT 2001 LRDI | DI - Tables & Graphs**

If the costs per ton of transport by ship, air and road are represented by P, Q and R respectively, which of the following is true?

- A.
R > Q > P

- B.
P > R > Q

- C.
P > Q > R

- D.
R > P > Q

Answer: Option C

**Explanation** :

Road is the cheapest mode of transport of crude oil.

∴ Only option 3 satisfies this condition.

Hence, option 3.

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**24. CAT 2001 LRDI | LR - Selection & Distribution**

At a village mela, the following six nautankis (plays) are scheduled as shown in the table below. You wish to see all the six nautankis. Further, you wish to ensure that you get a lunch break from 12:30 p.m. to 1:30 p.m. In which of the following way can you do this?

- A.
Sati-Savitri is viewed first; Sundar Kand is viewed third and Jhansi ki Rani is viewed last.

- B.
Sati-Savitri is viewed last; Veer Abhimanyu is viewed third and Reshma aur Shera is viewed first.

- C.
Sati-Savitri is viewed first; Sundar Kand is viewed third and Joru ka Ghulam is viewed fourth.

- D.
Veer-Abhimanyu is viewed third; Reshma aur Shera is viewed fourth and Jhansi ki Rani is viewed fifth.

Answer: Option C

**Explanation** :

The possible sequence using the given conditions is given below.

1. Sati-Savitri 9:00 a.m. to 10:00 a.m.

2. Veer Abhimanyu 10:00 a.m. to 11:00 a.m.

3. Sunder Kand 11:00 a.m. to 11:30 a.m.

4. Joru ka Ghulam 11:30 a.m. to 12:30 p.m.

5. Lunch 12:30 p.m. to 1:30 p.m.

6. Jhansi ki Rani 1:30 p.m. to 2:00 p.m.

7. Reshma aur Shera 2:00 p.m. to 3:00 p.m.

Hence, option 3.

Alternatively,

Reshma aur Shera cannot be viewed at 9:30 a.m. as the play viewing should start at 9.00 a.m. in order to complete all the plays. It cannot be viewed at 12 noon as it is a 1 hour play and there has to be a lunch break between 12:30 p.m. and 1:30 p.m.

∴ Reshma aur Shera has to be viewed last.

Only option 3 satisfies this condition.

Hence, option 3.

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**25. CAT 2001 LRDI | LR - Selection & Distribution**

Mrs. Ranga has three children and has difficulty remembering their ages and the months of their birth. The clues below may help her remember.

A. The boy, who was born in June, is 7 years old.

B. One of the children is 4 years old, but it is not Anshuman.

C. Vaibhav is older than Suprita.

D. One of the children was born in September, but it was not Vaibhav.

E. Suprita's birthday is in April.

F. The youngest child is only 2 years old.

Based on the above clues, which one of the following statements is true?

- A.
Vaibhav is the oldest, followed by Anshuman who was born in September, and the youngest is Suprita who was born in April.

- B.
Anshuman is the oldest being born in June, followed by Suprita who is 4 years old, and the youngest is Vaibhav who is 2 years old.

- C.
Vaibhav is the oldest being 7 years old, followed by Suprita who was born in April, and the youngest is Anshuman who was born in September.

- D.
Suprita is the oldest who was born in April, followed by Vaibhav who was born in June, and Anshuman who was born in September.

Answer: Option C

**Explanation** :

Using the given conditions, we can figure out the following.

Only possibility is that Vaibhav is born in the month of June and Anshuman is born in the month of September.

∵ Anshuman is not 4 years old; Suprita has to be 4 years old.

∴ Anshuman is 2 years old and Vaibhav is 7 years old.

∴ The modified table will look like this.

Hence, option 3.

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**26. CAT 2001 LRDI | LR - Selection & Distribution**

The Banerjees, the Sharmas, and the Pattabhiramans each have a tradition of eating Sunday lunch as a family. Each family serves a special meal at a certain time of day. Each family has a particular set of chinaware used only for this meal. Use the clues below to answer the following question.

A. The Sharma family eats at noon.

B. The family that serves fried brinjal uses blue chinaware.

C. The Banerjee family eats at 2 o'clock.

DThe family that serves sambar does not use red chinaware.

E. The family that eats at 1 o'clock serves fried brinjal.

F. The Pattabhiraman family does not use white chinaware.

G. The family that eats last likes makkai-ki-roti.

Which one of the following statements is true?

- A.
The Banerjees eat makkai-ki-roti at 2 o'clock, the Sharmas eat fried brinjal at 12 o'clock and the Pattabhiramans eat sambar from red chinaware.

- B.
The Sharmas eat sambar served in white chinaware, the Pattabhiramans eat fried brinjal at 1 o'clock, and the Banerjees eat makkai-ki-roti served in blue chinaware.

- C.
The Sharmas eat sambar at noon, the Pattabhiramans eat fried brinjal served in blue chinaware, and the Banerjees eat makkai-ki-roti served in red chinaware.

- D.
The Banerjees eat makkai-ki-roti served in white chinaware, the Sharmas eat fried brinjal at 12 o'clock and the Pattabhiramans eat sambar from red chinaware.

Answer: Option C

**Explanation** :

The information can be summarised as follows.

The Sharma family eats at noon and the Bannerjees family eats at 2 o’ clock.

∴ The Pattabhiramanas family must be eating at 1 o’ clock.

Also, the family that eats at 1 o'clock serves fried brinjal i.e. Pattabhiramanas family and they use blue chinaware.

The family that serves sambhar does not use red chinaware, the family that eats last likes makkai-ki-roti.

∴ The Sharma family must be eating sambhar in the white chinaware and Bannerjees family must be eating Makkai-ki-roti in the red chinaware.

Hence, option 3.

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**27. CAT 2001 LRDI | LR - Binary Logic**

While Balbir had his back turned, a dog ran into his butcher shop, snatched a piece of meat off the counter and ran out. Balbir was mad when he realised what had happened. He asked three other shopkeepers, who had seen the dog, to describe it. The shopkeepers really didn't want to help Balbir. So each of them made a statement which contained one truth and one lie.

A. Shopkeeper Number 1 said: "The dog had black hair and a long tail."

B. Shopkeeper Number 2 said: "The dog has a short tail and wore a collar."

C. Shopkeeper Number 3 said: "The dog had white hair and no collar."

Based on the above statements, which of the following could be a correct description?

- A.
The dog had white hair, short tail and no collar.

- B.
The dog had white hair, long tail and a collar.

- C.
The dog had black hair, long tail and a collar.

- D.
The dog had black hair, long tail and no collar.

Answer: Option B

**Explanation** :

According to the statement of the shopkeeper number 1, the dog had either black hair or a long tail, but definitely not both.

∴ Options 3 and 4 can be eliminated.

According to the statement of shopkeeper number 3, the dog had either white hair or wore a collar, but not both.

∴ Option 1 can be eliminated.

Hence, option 2.

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

Elle is three times older than Yogesh. Zaheer is half the age of Wahida. Yogesh is older than Zaheer.

**28. CAT 2001 LRDI | LR - Rank & Ordering**

Which of the following can be inferred?

- A.
Yogesh is older than Wahida.

- B.
Elle is older than Wahida.

- C.
Elle may be younger than Wahida.

- D.
None of the above

Answer: Option B

**Explanation** :

Let us assume that Zaheer is as old as Yogesh (which is the maximum possible age for Zaheer).

∴ Waheeda is twice older than Yogesh but Elle is three times older than Yogesh.

∴ Elle is the eldest.

Hence, option 2.

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**29. CAT 2001 LRDI | LR - Rank & Ordering**

Which of the following information will be sufficient to estimate Elle's age?

- A.
Zaheer is 10 years old.

- B.
Both Yogesh and Wahida are older than Zaheer by the same number of years.

- C.
Both 1 and 2 above

- D.
None of the above

Answer: Option C

**Explanation** :

Using the information in options 1 and 2 together, we have,

Waheeda’s age = 2 × Zaheer’s age = 2 × 10 = 20

∴ Yogesh’s age must be equal to 20 years.

∴ Elle’s age = 3 × 20 = 60

Hence, option 3.

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

A group of three or four has to be selected from seven persons. Among are two women: Fiza and Kavita, and five men: Ram, Shyam, David, Peter and Rahim. Ram would not like to be in the group if Shyam is also selected. Shyam and Rahim want to be selected together in the group. Kavita would like to be in the group only if David is also there. David, if selected, would not like Peter in the group. Ram would like to be in the group only if Peter is also there. David insists that Fiza be selected in case he is there in the group.

**30. CAT 2001 LRDI | LR - Selection & Distribution**

Which of the following is a feasible group of three?

- A.
David, Ram, Rahim

- B.
Peter, Shyam, Rahim

- C.
Kavita, David, Shyam

- D.
Fiza, David, Ram

Answer: Option B

**Explanation** :

Option 1 and 3 can be eliminated because Shyam and Rahim always go together.

Option 4 can be eliminated because Ram will be in the group if Peter is selected.

∴ Only possible option is 2.

Hence, option 2.

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**31. CAT 2001 LRDI | LR - Selection & Distribution**

Which of the following is a feasible group of four?

- A.
Ram, Peter, Fiza, Rahim

- B.
Shyam, Rahim, Kavita, David

- C.
Shyam, Rahim, Fiza, David

- D.
Fiza, David, Ram, Peter

Answer: Option C

**Explanation** :

Option 1 can be eliminated because Shyam and Rahim always go together.

Option 2 can be eliminated because David can only be selected if Fiza is selected.

Option 4 can be eliminated because David and Peter cannot be selected together.

Hence, option 3.

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**32. CAT 2001 LRDI | LR - Selection & Distribution**

Which of the following statements is true?

- A.
Kavita and Ram can be part of a group of four

- B.
A group of four can have two women

- C.
A group of four can have all four men

- D.
None of the above

Answer: Option D

**Explanation** :

Option 1 is false because for Kavita to be in the group, David has to be in the group and for Ram to be in the group Peter has to be selected. But David and Peter cannot be selected together.

Option 2 is false because along with the two women Kavita and Fiza, David has to be selected and Ram, Shyam, Peter and Rahim cannot be selected alone.

Option 3 is false because David and Peter cannot be selected together and Ram and Shyam cannot be selected together.

Hence, option 4.

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**33. CAT 2001 LRDI | LR - Venn Diagram**

On her walk through the park, Hansa collected 50 coloured leaves, all either maple or oak. She sorted them by category when she got home, and found the following:

A. The number of red oak leaves with spots is even and positive.

B. The number of red oak leaves without any spot equals the number of red maple leaves without spots.

C. All non-red oak leaves have spots, and there are five times as many of them as there are red spotted oak leaves.

D. There are no spotted maple leaves that are not red.

E. There are exactly 6 red spotted maple leaves.

F. There are exactly 22 maple leaves that are neither spotted nor red.

How many oak leaves did she collect?

- A.
22

- B.
17

- C.
25

- D.
18

Answer: Option B

**Explanation** :

Let the number of red but not spotted oak leaves be ‘x’ and number of red spotted oak leaves be ‘y’.

From the given conditions, we get the following tabulated table.

But, 6 + x + 22 + y + x + 5y = 50

∴ 28 + 2x + 6y = 50

∴ x + 3y = 11

∵ Red spotted oak leaves has to be positive and even.

∴ y = 2

∴ x = 11 − 3 × 2 = 5

∴ Number of Oak leaves = x + y + 5y = 5 + 6 × 2 = 17

Hence, option 2.

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**34. CAT 2001 LRDI | LR - Selection & Distribution**

Eight people carrying food baskets are going for a picnic on motorcycles. Their names A, B, C, D, E, F, G, and H. They have four motorcycles, M1, M2, M3 and M4 among them. They also have four food baskets O, P, Q and R of different sizes and shapes and each can be carried only on motorcycles M1, M2, M3, or M4, respectively. No more than two persons can travel on a motorcycle and no more than one basket can be carried on a motorcycle. There are two husband-wife pairs in this group of eight people and each pair will ride on a motorcycle together. C cannot travel with A or B. E cannot travel with B or F. G cannot travel with F or H or D. The husband-wife pairs must carry baskets O and P. Q is with A and P is with D. F travels on M1 and E travels on M2 motorcycles. G is with Q, and B cannot go with R. Who is travelling with H?

- A.
A

- B.
B

- C.
C

- D.
D

Answer: Option C

**Explanation** :

As, O, P, Q, R are on motorcycle M1, M2, M3, M4 respectively. Hence we get,

Now, A is with Q and G is with Q hence A is with G on motorcycle M3.

And F is on M1 and E is on M2.

P is with D hence D is with E.

As B cannot go with R hence he has to go with F on motorcycle M1.

Hence C and H go together on motorcycle M4.

Hence we get,

Hence C goes with H.

Hence, option 3.

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**35. CAT 2001 LRDI | LR - Blood Relation**

In a family gathering there are two males who are grandfathers and four males who are fathers. In the same gathering there are two females who are grandmothers and four females who are mothers. There is at least one grandson or a granddaughter present in this gathering. There are two husband-wife pairs in this group. These can either be a grandfather and a grandmother, or a father and a mother. The single grandfather (who wife is not present) has two grandsons and a son present. The single grandmother (whose husband is not present) has two granddaughters and a daughter present. A grandfather or a grandmother present with their spouses does not have any grandson or granddaughter present. What is the minimum number of people present in this gathering?

- A.
10

- B.
12

- C.
14

- D.
16

Answer: Option B

**Explanation** :

Here "-" represents the Husband-Wife pair.

∵ We want the minimum number of people, let us assume that the two grandfathers and two grandmothers have already been counted as fathers and mothers respectively.

∴ We have 4 fathers, 4 mothers, 2 grandsons and 2 granddaughters.

∴ Minimum number of people present = 4 + 4 + 2 + 2 = 12

Hence, option 2.

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**36. CAT 2001 LRDI | LR - Mathematical Reasoning**

I have a total of Rs. 1,000. Item A costs Rs. 110, item B costs Rs. 90, item C costs Rs. 70, item D costs Rs. 40 and item E costs Rs. 45. For every item D that I purchase, I must also buy two of item B. For every item A, I must buy one of item C. For every item E, I must also buy two of item D and one of item B. For every item purchased I earn 1000 points and for every rupee not spent I earn a penalty of 1500 points. My objective is to maximise to points I earn. What is the number of items that I must purchase to maximise my points?

- A.
13

- B.
14

- C.
15

- D.
16

Answer: Option B

**Explanation** :

To maximise the points, I should spent the entire amount and should get as many items as possible.

∴ I can maximize my points by purchasing 13C’s and 1B as there is no condition on the purchase of C and B.

∴ Total cost = 13 × 70 + 90 = 910 + 90 =1000

∵ The entire amount has been spent, there is no penalty.

∴ Number of points earned = 14 × 1000 = 14000

Hence, option 2.

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**37. CAT 2001 LRDI | LR - Mathematical Reasoning**

Four friends Ashok, Bashir, Chirag and Deepak are out shopping. Ashok has less money than three times the amount that Bashir has. Chirag has more money than Bashir. Deepak has an amount equal to the difference of amounts with Bashir and Chirag. Ashok has three times the money with Deepak. They each have to buy at least one shirt, or one shawl, or one sweater, or one jacket that are priced Rs. 200, Rs. 400, Rs. 600, and Rs. 1,000 a piece, respectively. Chirag borrows Rs. 300 from Ashok and buys a jacket. Bashir buys a sweater after borrowing Rs. 100 from Ashok and is left with no money. Ashok buys three shirts. What is the costliest item the Deepak could buy with his own money?

- A.
A shirt

- B.
A shawl

- C.
A sweater

- D.
A jacket

Answer: Option B

**Explanation** :

Let Ashok, Bashir, Chirag and Deepak have amounts Rs. A, Rs. B, Rs. C and Rs. D respectively. From the data given in the question, we have following equations.

A < 3B ... (i)

C > B ... (ii)

D = C − B ... (iii)

A = 3D ... (iv)

Chirag borrows Rs. 300 from Ashok and buys a jacket worth Rs. 1000.

∴ C ≥ 700 ... (v)

Bashir buys a sweater after borrowing Rs. 100 from Ashok and left with no money.

∴ B = 500 ... (vi)

Ashok buys 3 shirts (costing Rs. 600 in total) with Rs. A − 300 − 100

∴ A ≥ 1000

From (i) and (vi), we get,

A < 1500

∴ 1000 ≤ A < 1500 ... (vii)

From equation (iv) and (vii), we get,

333 < D < 500

∴ Deepak could buy a shawl worth Rs. 400.

Hence, option 2.

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**38. CAT 2001 LRDI | LR - Selection & Distribution**

In a "keep-fit" gymnasium class there are fifteen females enrolled in a weight-loss program. They all have been grouped in any one of the five weight-groups W1, W2, W3, W4, or W5. One instructor is assigned to one weight-group only. Sonali, Shalini, Shubhra, and Shahira belong to the same weight-group. Sonali and Rupa are in one weight-group, Rupali and Renuka are also in one weight-group. Rupa, Radha, Renuka, Ruchika and Ritu belong to different weight-groups. Somya cannot be with Ritu, and Tara cannot be with Radha. Komal cannot be with Radha, Somya, or Ritu. Shahira is in W1 and Somya is in W4 with Ruchika. Sweta and Jyotika cannot be with Rupali, but are in a weight-group with total membership of four. No weight-group can have more than five or less than one member. Amita, Babita, Chandrika, Deepika, and Elina are instructors of weight-groups with membership sizes 5, 4, 3, 2 and 1, respectively. Who is the instructor of Radha?

- A.
Babita

- B.
Elina

- C.
Chandrika

- D.
Deepika

Answer: Option B

**Explanation** :

Based on the information provided, the data can be summarized as follows.

From the first two statements, we get that Sonali, Shalini, Shubhra, Shahira and Rupa are in the same weight group with a group size of 5.

It is given that Rupa, Radha, Renuka, Ruchika and Ritu belong to different weight-groups.

∴ There are 4 groups (i.e. of Radha, Renuka, Ruchika and Ritu) remaining.

∴ In the group along with Ruchika and Somya, there cannot be Radha, Renuka and Ritu, thus forming a group of 2.

Also, Komal cannot be with Radha, Ritu and Somya (Ruchika) and Rupa.

∴ She has go with Rupali and Renuka, thus forming a group of size 3.

If W1, W2, W3, W4 and W5 represent weight groups in the descending order of the group size then Sweta and Jyotika must contain Ritu and Tara as their group members as Somya and Ruchika belong to W4 thus forming a group of 4.

Otherwise Sweta and Jyotika can be either with Ritu and Tara or with Somya and Ruchika, but in both the cases Radha is alone and doesnot have any other member in the group.

∴ Elina is the instructor of Radha.

Hence, option 2

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**39. CAT 2001 LRDI | LR - Selection & Distribution**

A king has unflinching loyalty from eight of his ministers M1 to M8, but he has to select only four to make a cabinet committee. He decides to choose these four such that each selected person shares a liking with at least one of the other three selected. The selected persons must also hate one of the likings of any of the other three persons selected.

A. M1 likes fishing and smoking, but hates gambling.

B. M2 likes smoking and drinking, but hates fishing.

C. M3 likes gambling, but hates smoking.

D. M4 likes mountaineering, but hates drinking.

E. M5 likes drinking, but hates smoking and mountaineering.

F. M6 likes fishing, but hates smoking and mountaineering.

G. M7 likes gambling and mountaineering, but hates fishing.

H. M8 likes smoking and gambling, but hates mountaineering.

Who are the four people selected by the king?

- A.
M1, M2, M5, M6

- B.
M3, M4, M5, M6

- C.
M4, M5, M6, M8

- D.
M1, M2, M4, M7

Answer: Option D

**Explanation** :

For option 1:

M1 hates gambling but no one in the group likes it.

∴ Option 1 cannot be the answer.

For option 2:

In this group nobody has a common liking.

∴ Option 2 cannot be the answer.

For option 3:

In this group also, nobody has a common liking.

∴ Option 3 cannot be the answer.

∴ Only option remaining is 4.

Hence, option 4.

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

A and B are two sets (e.g. A = mothers, B = women). The elements that could belong to both the sets (e.g. women who are mothers) is given by the set C = A.B. The elements which could belong to either A or B, or both, is indicated by the set D = A ∪ B. A set that does not contain any elements is known as a null set, represented by ϕ (for example, if none of the women in the set B is a mother, they C = A.B is a null set, or C = ϕ).

Let 'V' signify the set of all vertebrates; 'M' the set of all mammals; 'D' dogs; 'F' fish; 'A' Alsatian and 'P', a dog named Pluto.

**40. CAT 2001 LRDI | LR - Venn Diagram**

Given that X = M.D is such that X = D, which of the following is true?

- A.
All dogs are mammals.

- B.
Some dogs are mammals.

- C.
X = ϕ

- D.
All mammals are dogs.

Answer: Option A

**Explanation** :

X = M.D implies some dogs are mammals and X = D implies the set of dogs.

This implies all dogs are mammals.

Hence, option 1.

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**41. CAT 2001 LRDI | LR - Venn Diagram**

If Y = F. (D.V), is not a null set, it implies that

- A.
All fish are vertebrates

- B.
All dogs are vertebrates

- C.
Some fish are dogs

- D.
None of the above

Answer: Option C

**Explanation** :

Y = F.(D.V) implies some vertebrates are dogs and some vertebrate dogs are fishes.

∴ Some fishes are dogs.

Hence, option 3.

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**42. CAT 2001 LRDI | LR - Venn Diagram**

If Z = (P.D) ∪ M, then

- A.
The elements of Z consist of Pluto the dog or any other mammal.

- B.
Z implies any dog or mammal.

- C.
Z implies Pluto or any dog that is a mammal.

- D.
Z is a null set.

Answer: Option A

**Explanation** :

Z = (P.D) ∪ M

P.D means a dog called Pluto and P ∪ M means Pluto the dog and the other mammals.

Hence, option 1.

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**43. CAT 2001 LRDI | LR - Venn Diagram**

If P.A = ϕ and P ∪ A = D, then which of the following is true?

- A.
Pluto and Alsatians are dogs.

- B.
Pluto is an Alsatian.

- C.
Pluto is not an Alsatian.

- D.
D is a null set.

Answer: Option C

**Explanation** :

P. A = ϕ means that "Pluto is not an Alsatian".

Hence, option 3.

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**44. CAT 2001 LRDI | LR - Calendar**

If 09/12/2001 happens to be Sunday, then 09/12/1971 would have been a

- A.
Wednesday

- B.
Tuesday

- C.
Saturday

- D.
Thursday

Answer: Option D

**Explanation** :

A normal year has 52 weeks and 1extra day.

∴ 365 = 52 × 7 + 1

A leap year has 52 weeks and 2 extra days.

∴ 366 = 52 × 7 + 2

There are 30 extra days between 1971 and 2001.

In this period there were 8 leap years, therefore another 8 extra days.

∴ 38 = 7 × 5 + 3

∴ We are left with 3 extra days.

If 9/12/2001 is a Sunday, then 12/12/1971 was a Sunday and 9/12/1971 was a Thursday.

Hence, option 4.

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