# CAT 2006 LRDI | Previous Year CAT Paper

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

In a Class X Board examination, ten papers are distributed over five Groups – PCB, Mathematics, Social Science, Vernacular and English. Each of the ten papers is evaluated out of 100. The final score of a student is calculated in the following manner. First, the Group Scores are obtained by averaging marks in the papers within the Group. The final score is the simple average of the Group Scores. The data for the top ten students are presented below. (Dipan’s score in English Paper II has been intentionally removed in the table).

**Note: **B or G against the name of a student respectively indicates whether the student is a boy or a girl.

**1. CAT 2006 LRDI | DI - Tables & Graphs**

How much did Dipan get in English Paper II?

- A.
94

- B.
96.5

- C.
97

- D.
98

- E.
99

Answer: Option C

**Explanation** :

Dipan’s Group Scores are as follows:

PCB group = 98 × $\frac{3}{3}$ = 98

Mathematics Group = 95

Social Science group = $\frac{96+95}{2}$ = 95.5

Vernacular Group = $\frac{96+94}{2}$ = 95

English Group = $\frac{96+x}{2}=48+\frac{x}{2}$

Dipan’s final score = 96

∴ Sum of Dipan’s Group Scores = 96 × 5 = 480

∴ 98 + 95 + 95.5 + 95 + 48 + $\frac{x}{2}$ = 480

$\therefore \frac{x}{2}$ = 48.5

∴ x = 97

Dipan scored 97 marks in English Paper II.

Hence, option (c).

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**2. CAT 2006 LRDI | DI - Tables & Graphs**

Students who obtained Group Scores of at least 95 in every group are eligible to apply for a prize. Among those who are eligible, the student obtaining the highest Group Score in Social Science Group is awarded this prize. The prize was awarded to:

- A.
Shreya

- B.
Ram

- C.
Ayesha

- D.
Dipan

- E.
no one from the top ten

Answer: Option D

**Explanation** :

From the table we can observe that only Dipan is eligible to apply for the prize. So Dipan gets the prize.

Hence, option (d).

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**3. CAT 2006 LRDI | DI - Tables & Graphs**

Among the top ten students, how many boys scored at least 95 in at least one paper from each of the groups?

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option A

**Explanation** :

Dipan was the only boy to score at least 95 in at least one paper from each of the groups.

Hence, option (a).

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**4. CAT 2006 LRDI | DI - Tables & Graphs**

Each of the ten students was allowed to improve his/her score in exactly one paper of choice with the objective of maximizing his/her final score. Everyone scored 100 in the paper in which he or she chose to improve. After that, the topper among the ten students was:

- A.
Ram

- B.
Agni

- C.
Pritam

- D.
Ayesha

- E.
Dipan

Answer: Option E

**Explanation** :

In order to maximize scores, each student would choose to improve his/her score in the paper which would affect the group score the most.

Consider the options.

Ram chooses Vernacular Paper I or II.

His original group score in Vernacular group = 94

His new score would change by $\frac{\left({\displaystyle \frac{(94+100)}{2}}-94\right)}{5}$ = 0.6

His new score = 96.1 + 0.6 = 96.7

Agni chooses Vernacular Paper I.

His original group score in Vernacular group = 87.5

His new score would change by $\frac{\left({\displaystyle \frac{(93+100)}{2}}-87.5\right)}{5}$ = 1.8

His new score = 94.3 + 1.8 = 96.1

Pritam chooses History.

His original group score in Social Science group = 89

His new score would change by $\frac{\left({\displaystyle \frac{(95+100)}{2}}-94\right)}{5}$ = 1.7

His new score = 93.9 + 1.7 = 95.6

Ayesha chooses Geography.

Her original group score in Social Science group = 94

His new score would change by $\frac{\left({\displaystyle \frac{(95+100)}{2}}-94\right)}{5}$ = 0.7

Her new score = 96.2 + 0.7 = 96.9

Dipan chooses Mathematics.

His original group score in Mathematics group = 95

His new score would change by $\frac{(100-95)}{5}$ = 1

His new score = 96 + 1 = 97, which is the highest among the five options.

Hence, option (e).

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**5. CAT 2006 LRDI | DI - Tables & Graphs**

Had Joseph, Agni, Pritam and Tirna each obtained Group Score of 100 in the Social Science Group, then their standing in decreasing order of final score would be:

- A.
Pritam, Joseph, Tirna, Agni

- B.
Joseph, Tirna, Agni, Pritam

- C.
Pritam, Agni, Tirna, Joseph

- D.
Joseph, Tirna, Pritam, Agni

- E.
Pritam, Tirna, Agni, Joseph

Answer: Option A

**Explanation** :

Group scores of Joseph, Agni, Pritam and Tirna in Social Science Group are 95.5, 95.5, 89 and 89.5 respectively.

Their final scores are 95, 94.3, 93.9, 93.7 respectively.

If their group scores in social science change to hundred their final scores will be affected by 4.5/5, 4.5/5, 11/5 and 10.5/5 respectively.

Their new final scores would be 95.9, 95.2, 96.1 and 95.8 respectively.

Their standing in decreasing order of final score would be Pritam, Joseph, Tirna, Agni.

Hence, option (a).

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

Mathematicians are assigned a number called Erdös number, (named after the famous mathematician, Paul Erdös). Only Paul Erdös himself has an Erdös number of zero. Any mathematician who has written a research paper with Erdös has an Erdös number of 1. For other mathematicians, the calculation of his/her Erdös number is illustrated below:

Suppose that a mathematician X has co-authored papers with several other mathematicians. From among them, mathematician Y has the smallest Erdös number. Let the Erdös number of Y be y. Then X has an Erdös number of y + 1. Hence any mathematician with no co-authorship chain connected to Erdös has an Erdös number of infinity.

- In a seven day long mini-conference organized in memory of Paul Erdös, a close group of eight mathematicians, call them A, B, C, D, E, F, G and H, discussed some research problems. At the beginning of the conference, A was the only participant who had an infinite Erdös number. Nobody had an Erdös number less than that of F.
- On the third day of the conference F co-authored a paper jointly with A and C. This reduced the average Erdös number of the group of eight mathematicians to 3. The Erdös numbers of B, D, E, G and H remained unchanged with the writing of this paper. Further, no other co-authorship among any three members would have reduced the average Erdös number of the group of eight to as low as 3.
- At the end of the third day, five members of this group had identical Erdös numbers while the other three had Erdös numbers distinct from each other.
- On the fifth day, E co-authored a paper with F which reduced the group‘s average Erdös number by 0.5. The Erdös numbers of the remaining six were unchanged with the writing of this paper.
- No other paper was written during the conference.

**6. CAT 2006 LRDI | LR - Mathematical Reasoning**

The person having the largest Erdös number at the end of the conference must have had Erdös number (at that time):

- A.
5

- B.
7

- C.
9

- D.
14

- E.
15

Answer: Option B

**Explanation** :

Let F and E have Erdös numbers f and e respectively at the beginning of the conference.

On the third day, A’s and C’s Erdös numbers become (f + 1)

The sum of Erdös numbers changed to 8 × 3 = 24

At the end of the third day, five members had identical Erdös numbers while the other three had distinct ones.

On the fifth day, E’s Erdös numbers became f + 1 and this reduced the group’s average by 0.5. This means that E’s Erdös numbers was not f + 1 on the third day.

Thus we have,

At the end of the third day, 5(f + 1) + f + e + y = 24

Hence 6f + 5 + e + y = 24

Hence 6f + e + y = 19

At the end of the fifth day,

6(f + 1) + f + y = 2.5 × 8 = 20

Hence 7f + y = 14

Among the eight mathematicians, F has the smallest Erdös number.

Let f = 2

∴ y = 0

However, only Paul Erdös himself has an Erdös number of 0. So f cannot be equal to 2. Any other value greater than 2, would render y as a negative number, which is also not possible.

So, f = 1

∴ y = 7

∴ e = 6

Now, we can solve all the questions.

From the above explanation, the largest Erdös number at the end of the conference would be 7.

Hence, option (b).

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**7. CAT 2006 LRDI | LR - Mathematical Reasoning**

How many participants in the conference did not change their Erdös number during the conference?

- A.
2

- B.
3

- C.
4

- D.
5

- E.
Cannot be determined

Answer: Option D

**Explanation** :

Let F and E have Erdös numbers f and e respectively at the beginning of the conference.

On the third day, A’s and C’s Erdös numbers become (f + 1)

The sum of Erdös numbers changed to 8 × 3 = 24

At the end of the third day, five members had identical Erdös numbers while the other three had distinct ones.

On the fifth day, E’s Erdös numbers became f + 1 and this reduced the group’s average by 0.5. This means that E’s Erdös numbers was not f + 1 on the third day.

Thus we have,

At the end of the third day, 5(f + 1) + f + e + y = 24

Hence 6f + 5 + e + y = 24

Hence 6f + e + y = 19

At the end of the fifth day,

6(f + 1) + f + y = 2.5 × 8 = 20

Hence 7f + y = 14

Among the eight mathematicians, F has the smallest Erdös number.

Let f = 2

∴ y = 0

However, only Paul Erdös himself has an Erdös number of 0. So f cannot be equal to 2. Any other value greater than 2, would render y as a negative number, which is also not possible.

So, f = 1

∴ y = 7

∴ e = 6

Now, we can solve all the questions.

From the above explanation, the Erdös numbers of B, D, G, H and F did not change during the conference.

Hence, option (d).

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**8. CAT 2006 LRDI | LR - Mathematical Reasoning**

The Erdös number of C at the end of the conference was:

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option B

**Explanation** :

Let F and E have Erdös numbers f and e respectively at the beginning of the conference.

On the third day, A’s and C’s Erdös numbers become (f + 1)

The sum of Erdös numbers changed to 8 × 3 = 24

At the end of the third day, five members had identical Erdös numbers while the other three had distinct ones.

On the fifth day, E’s Erdös numbers became f + 1 and this reduced the group’s average by 0.5. This means that E’s Erdös numbers was not f + 1 on the third day.

Thus we have,

At the end of the third day, 5(f + 1) + f + e + y = 24

Hence 6f + 5 + e + y = 24

Hence 6f + e + y = 19

At the end of the fifth day,

6(f + 1) + f + y = 2.5 × 8 = 20

Hence 7f + y = 14

Among the eight mathematicians, F has the smallest Erdös number.

Let f = 2

∴ y = 0

However, only Paul Erdös himself has an Erdös number of 0. So f cannot be equal to 2. Any other value greater than 2, would render y as a negative number, which is also not possible.

So, f = 1

∴ y = 7

∴ e = 6

Now, we can solve all the questions.

From the above explanation, C’s Erdös number was f + 1 = 2 on the third day and thereafter.

Hence, option (b).

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**9. CAT 2006 LRDI | LR - Mathematical Reasoning**

The Erdös number of E at the beginning of the conference was:

- A.
2

- B.
5

- C.
6

- D.
7

- E.
8

Answer: Option C

**Explanation** :

Let F and E have Erdös numbers f and e respectively at the beginning of the conference.

On the third day, A’s and C’s Erdös numbers become (f + 1)

The sum of Erdös numbers changed to 8 × 3 = 24

At the end of the third day, 5(f + 1) + f + e + y = 24

Hence 6f + 5 + e + y = 24

Hence 6f + e + y = 19

At the end of the fifth day,

6(f + 1) + f + y = 2.5 × 8 = 20

Hence 7f + y = 14

Among the eight mathematicians, F has the smallest Erdös number.

Let f = 2

∴ y = 0

So, f = 1

∴ y = 7

∴ e = 6

Now, we can solve all the questions.

From the above explanation, E’s Erdös number was 6.

Hence, option (c).

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**10. CAT 2006 LRDI | LR - Mathematical Reasoning**

How many participants had the same Erdös number at the beginning of the conference?

- A.
2

- B.
3

- C.
4

- D.
5

- E.
Cannort be determined

Answer: Option B

**Explanation** :

Since 5 participants had identical Erdös numbers at the end of day three and two of these were A and C whose Erdös numbers had changed on the same day, three had the same Erdös numbers at the beginning of the conference.

Hence, option (b).

Workspace:

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

Two traders, Chetan and Michael, were involved in the buying and selling of MCS shares over five trading days. At the beginning of the first day, the MCS share was priced at Rs. 100, while at the end of the fifth day it was priced at Rs. 110. At the end of each day, the MCS share price either went up by Rs. 10, or else, it came down by Rs. 10. Both Chetan and Michael took buying and selling decisions at the end of each trading day.

The beginning price of MCS share on a given day was the same as the ending price of the previous day.

Chetan and Michael started with the same number of shares and amount of cash, and had enough of both. Below are some additional facts about how Chetan and Michael traded over the five trading days.

- Each day if the price went up, Chetan sold 10 shares of MCS at the closing price. On the other hand, each day if the price went down, he bought 10 shares at the closing price.
- If on any day, the closing price was above Rs. 110, then Michael sold 10 shares of MCS, while if it was below Rs. 90, he bought 10 shares, all at the closing price.

**11. CAT 2006 LRDI | LR - Mathematical Reasoning**

If Chetan sold 10 shares of MCS on three consecutive days, while Michael sold 10 shares only once during the five days, what was the price of MCS at the end of day 3?

- A.
Rs. 90

- B.
Rs. 100

- C.
Rs. 110

- D.
Rs. 120

- E.
Rs. 130

Answer: Option C

**Explanation** :

Let the price of the share rise on x days and fall on y days. As the price increases by Rs. 10 in the five days, we have:

x + y = 5 and 10x – 10y = 10

Solving the above two equations simultaneously, we have

x = 3 and y = 2

The price of the share goes up on 3 days and falls on 2 days.

The three days on which the price rises can be selected in ^{5}C_{3} = 10 ways

The following are the 10 cases:

Consider Case 5:

Chetan sells on Days 1, 2 and 4 and buys on days 3 and 5.

Change in the number of shares he has = –30 + 20 = –10

Change in his cash = 10 × (110 + 120 + 120) – 10 × (110 + 110) = Rs. 1300

Michael sells on days 2 and 4, but never buys as the share price does not go below Rs. 90.

Change in the number of shares he has = –20

Change in his cash = 10 × (120 + 120) = Rs. 2400

The other cases are evaluated in a similar manner and the data is tabulated as shown above.

Chetan sold on three consecutive days: Cases 1, 2 and 3.

Michael sold only once: Case 3.

∴ The price of the share at the end of day 3 = Rs. 110

Hence, option (c).

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**12. CAT 2006 LRDI | LR - Mathematical Reasoning**

If Michael ended up with Rs. 100 less cash than Chetan at the end of day 5, what was the difference in the number of shares possessed by Michael and Chetan (at the end of day 5)?

- A.
Michael had 10 less shares than Chetan.

- B.
Michael had10 more shares than Chetan.

- C.
Chetan had 10 more shares than Michael.

- D.
Chetan had 20 more shares than Michael.

- E.
Both had the same number of shares.

Answer: Option E

**Explanation** :

Let the price of the share rise on x days and fall on y days. As the price increases by Rs. 10 in the five days, we have:

x + y = 5 and 10x – 10y = 10

Solving the above two equations simultaneously, we have

x = 3 and y = 2

The price of the share goes up on 3 days and falls on 2 days.

The three days on which the price rises can be selected in ^{5}C_{3} = 10 ways

The following are the 10 cases:

Consider Case 5:

Chetan sells on Days 1, 2 and 4 and buys on days 3 and 5.

Change in the number of shares he has = –30 + 20 = –10

Change in his cash = 10 × (110 + 120 + 120) – 10 × (110 + 110) = Rs. 1300

Michael sells on days 2 and 4, but never buys as the share price does not go below Rs. 90.

Change in the number of shares he has = –20

Change in his cash = 10 × (120 + 120) = Rs. 2400

The other cases are evaluated in a similar manner and the data is tabulated as shown above.

Chetan sold on three consecutive days: Cases 1, 2 and 3.

Michael sold only once: Case 3.

Michael ends up with Rs. 100 less cash than Chetan in cases 3, 6 and 8. In each of these cases, both of them hold the same number of shares at the end of day 5.

Hence, option (e).

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**13. CAT 2006 LRDI | LR - Mathematical Reasoning**

If Chetan ended up with Rs. 1300 more cash than Michael at the end of day 5, what was the price of MCS share at the end of day 4?

- A.
Rs. 90

- B.
Rs. 100

- C.
Rs. 110

- D.
Rs. 120

- E.
Not uniquely determinable

Answer: Option B

**Explanation** :

Let the price of the share rise on x days and fall on y days. As the price increases by Rs. 10 in the five days, we have:

x + y = 5 and 10x – 10y = 10

Solving the above two equations simultaneously, we have

x = 3 and y = 2

The price of the share goes up on 3 days and falls on 2 days.

The three days on which the price rises can be selected in ^{5}C_{3} = 10 ways

The following are the 10 cases:

Consider Case 5:

Chetan sells on Days 1, 2 and 4 and buys on days 3 and 5.

Change in the number of shares he has = –30 + 20 = –10

Change in his cash = 10 × (110 + 120 + 120) – 10 × (110 + 110) = Rs. 1300

Michael sells on days 2 and 4, but never buys as the share price does not go below Rs. 90.

Change in the number of shares he has = –20

Change in his cash = 10 × (120 + 120) = Rs. 2400

The other cases are evaluated in a similar manner and the data is tabulated as shown above.

Chetan sold on three consecutive days: Cases 1, 2 and 3.

Michael sold only once: Case 3.

This information corresponds to cases 4, 7, 9 and 10 from the solution table. The price at the end of day 4 in each of these cases is Rs. 100.

Hence, option (b).

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**14. CAT 2006 LRDI | LR - Mathematical Reasoning**

What could have been the maximum possible increase in combined cash balance of Chetan and Michael at the end of the fifth day?

- A.
Rs. 3700

- B.
Rs. 4000

- C.
Rs. 4700

- D.
Rs. 5000

- E.
Rs. 6000

Answer: Option D

**Explanation** :

x + y = 5 and 10x – 10y = 10

Solving the above two equations simultaneously, we have

x = 3 and y = 2

The price of the share goes up on 3 days and falls on 2 days.

The three days on which the price rises can be selected in ^{5}C_{3} = 10 ways

The following are the 10 cases:

Consider Case 5:

Chetan sells on Days 1, 2 and 4 and buys on days 3 and 5.

Change in the number of shares he has = –30 + 20 = –10

Change in his cash = 10 × (110 + 120 + 120) – 10 × (110 + 110) = Rs. 1300

Michael sells on days 2 and 4, but never buys as the share price does not go below Rs. 90.

Change in the number of shares he has = –20

Change in his cash = 10 × (120 + 120) = Rs. 2400

The other cases are evaluated in a similar manner and the data is tabulated as shown above.

Chetan sold on three consecutive days: Cases 1, 2 and 3.

Michael sold only once: Case 3.

The maximum increase in combined cash balance of Chetan and Michael = 1300 + 3700 = Rs. 5000 (case 1 from the table)

Hence, option (d).

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**15. CAT 2006 LRDI | LR - Mathematical Reasoning**

If Michael ended up with 20 more shares than Chetan at the end of day 5, what was the price of the share at the end of day 3?

- A.
Rs. 90

- B.
Rs. 100

- C.
Rs. 110

- D.
Rs. 120

- E.
Rs. 130

Answer: Option A

**Explanation** :

x + y = 5 and 10x – 10y = 10

Solving the above two equations simultaneously, we have

x = 3 and y = 2

The price of the share goes up on 3 days and falls on 2 days.

The three days on which the price rises can be selected in ^{5}C_{3} = 10 ways

The following are the 10 cases:

Consider Case 5:

Chetan sells on Days 1, 2 and 4 and buys on days 3 and 5.

Change in the number of shares he has = –30 + 20 = –10

Change in his cash = 10 × (110 + 120 + 120) – 10 × (110 + 110) = Rs. 1300

Michael sells on days 2 and 4, but never buys as the share price does not go below Rs. 90.

Change in the number of shares he has = –20

Change in his cash = 10 × (120 + 120) = Rs. 2400

The other cases are evaluated in a similar manner and the data is tabulated as shown above.

Chetan sold on three consecutive days: Cases 1, 2 and 3.

Michael sold only once: Case 3.

This information corresponds to case 2 from the table. The price at the end of day 3 was Rs. 90.

Hence, option (a).

Workspace:

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

A significant amount of traffic flows from point S to point T in the one-way street network shown below. Points A, B, C, and D are junctions in the network, and the arrows mark the direction of traffic flow. The fuel cost in rupees for travelling along a street is indicated by the number adjacent to the arrow representing the street.

Motorists travelling from point S to point T would obviously take the route for which the total cost of travelling is the minimum. If two or more routes have the same least travel cost, then motorists are indifferent between them. Hence, the traffic gets evenly distributed among all the least cost routes.

The government can control the flow of traffic only by levying appropriate toll at each junction. For example, if a motorist takes the route S-A-T (using junction A alone), then the total cost of travel would be Rs. 14 (i.e. Rs. 9 + Rs. 5) plus the toll charged at junction A.

**16. CAT 2006 LRDI | DI - Routes & Networks**

If the government wants to ensure that all motorists travelling from S to T pay the same amount (fuel costs and toll combined) regardless of the route they choose and the street from B to C is under repairs (and hence unusable), then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:

- A.
2, 5, 3, 2

- B.
0, 5, 3, 1

- C.
1, 5, 3, 2

- D.
2, 3, 5, 1

- E.
1, 3, 5, 1

Answer: Option B

**Explanation** :

Let the toll charged at junctions A, B, C and D be a, b, c and d respectively.

Since the cost of travel including toll on routes S-A-T, S-D-T, S-B-A-T and S-D-C-T is the same,

∴ 14 + a = 13 + d = 9 + a + b = 10 + c + d

Thus, b = 5, d – a = 1, c = 3

If a = 0, d = 1, If a = 1, d = 2 and if a = 2, d = 3

Hence, both options 2 and 3 satisfy the given criteria.

Note: The question makers took care of this inconsistency while calculating scores.

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**17. CAT 2006 LRDI | DI - Routes & Networks**

If the government wants to ensure that no traffic flows on the street from D to T, while equal amount of traffic flows through junctions A and C, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:

- A.
1, 5, 3, 3

- B.
1, 4, 4, 3

- C.
1, 5, 4, 2

- D.
0, 5, 2, 3

- E.
0, 5, 2, 2

Answer: Option E

**Explanation** :

Since the cost of travel including toll on routes S-A-T, S-B-C-T, S-B-A-T and S-D-C-T is the same,

∴ 14 + a = 7 + b + c = 9 + a + b = 10 + c + d

∴ b = 5, d = 2, c – a = 2

Only option 5 satisfies these criteria.

Hence, option (e).

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**18. CAT 2006 LRDI | DI - Routes & Networks**

If the government wants to ensure that all routes from S to T get the same amount of traffic, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:

- A.
0, 5, 2, 2

- B.
0, 5, 4, 1

- C.
1, 5, 3, 3

- D.
1, 5, 3, 2

- E.
1, 5, 4, 2

Answer: Option D

**Explanation** :

Since the cost of travel including toll on all routes is the same.

∴ 14 + a = 7 + b + c = 13 + d = 9 + a + b = 10 + c + d

∴ b = 5, d = 2, c = 3 and a = 1

Hence, option (d).

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**19. CAT 2006 LRDI | DI - Routes & Networks**

If the government wants to ensure that the traffic at S gets evenly distributed along streets from S to A, from S to B, and from S to D, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:

- A.
0, 5, 4, 1

- B.
0, 5, 2, 2

- C.
1, 5, 3, 3

- D.
1, 5, 3, 2

- E.
0, 4, 3, 2

Answer: Option A

**Explanation** :

If we make the cost of travelling on all the routes equal, traffic along S-B will be twice that along S-A.

But we want traffic along S-A, S-B and S-D to be the same.

As routes lead to C from both B and D, we can increase the toll at C so that the cost of travelling along S-B-C-T and S-D-C-T is more than that along the other three routes.

Now, 14 + a = 9 + a + b = 13 + d

∴ a = 0, b = 5 and d =1

Also, 7 + b + c > 14 and 10 + d + c > 14

∴ c > 3

Hence, option (a).

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**20. CAT 2006 LRDI | DI - Routes & Networks**

The government wants to devise a toll policy such that the total cost to the commuters per trip is minimized. The policy should also ensure that not more than 70 per cent of the total traffic passes through junction B. The cost incurred by the commuter travelling from point S to point T under this policy will be:

- A.
Rs. 7

- B.
Rs. 9

- C.
Rs. 10

- D.
Rs. 13

- E.
Rs. 14

Answer: Option C

**Explanation** :

If toll charges at all junctions are made 0, 100% traffic will pass through S-B-C-T. This is not possible.

If toll charges at A and B are made 0, then 100% traffic will pass through S-B-A-T. This is also not possible.

If toll charges at C and D are made 0, that at B are made Rs. 3, then the traffic will get equally divided between S-D-C-T and S-B-C-T.

Thus, the cost incurred will be Rs. 10.

Hence, option (c).

Workspace:

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

K, L, M, N, P, Q, R, S, U and W are the only ten members in a department. There is a proposal to form a team from within the members of the department, subject to the following conditions:

- A team must include exactly one among P, R, and S.
- A team must include either M or Q, but not both.
- If a team includes K, then it must also include L, and vice versa.
- If a team includes one among S, U, and W, then it must also include the other two.
- L and N cannot be members of the same team.
- L and U cannot be members of the same team.
- The size of a team is defined as the number of members in the team.

**21. CAT 2006 LRDI | LR - Selection & Distribution**

What could be the size of a team that includes K?

- A.
2 or 3

- B.
2 or 4

- C.
3 or 4

- D.
Only 2

- E.
Only 4

Answer: Option E

**Explanation** :

As K is included, L is included. So, N and U cannot be included. As U is not included, S and W are not included. One out of M and Q and one out of P and R will be included.

Thus, the team will include: K, L, (M or Q) and (P or R).

Hence, option (e).

Workspace:

**22. CAT 2006 LRDI | LR - Selection & Distribution**

In how many ways a team can be constituted so that the team includes N?

- A.
2

- B.
3

- C.
4

- D.
5

- E.
6

Answer: Option E

**Explanation** :

If the team includes N, it does not include L and K.

One out of M and Q can be included and one out of P, S and R can be included.

If S is a member, so are U and W.

Thus the possible teams are:

- N, M, P
- N, M, R
- N, Q, P
- N, Q, R
- N, M, S, U, W
- N, Q, S, U, W

Hence, option (e).

Workspace:

**23. CAT 2006 LRDI | LR - Selection & Distribution**

What would be the size of the largest possible team?

- A.
8

- B.
7

- C.
6

- D.
5

- E.
Cannot be determined

Answer: Option D

**Explanation** :

If S is not included, the team can have P or R, M or Q, K and L.

If S is included, the team will have S, U, W, M or Q, N.

This is the largest possible team.

Hence, option (d).

Workspace:

**24. CAT 2006 LRDI | LR - Selection & Distribution**

Who can be a member of a team of size 5?

- A.
K

- B.
L

- C.
M

- D.
P

- E.
R

Answer: Option C

**Explanation** :

If K or L are included, N, U, S and W are excluded. One out of P and R and one out of M and Q are included. Thus the team has only 4 members.

If P or R are included, the team can have M or Q, K and L. This team also has 4 members.

A team having M can have S, U, W and N i.e., 5 members.

Hence, option (c).

Workspace:

**25. CAT 2006 LRDI | LR - Selection & Distribution**

Who cannot be a member of a team of size 3?

- A.
L

- B.
M

- C.
N

- D.
P

- E.
Q

Answer: Option A

**Explanation** :

A team sized 3 has to have M or Q and P or R. The only other member that can be selected all alone is N.

L cannot be selected as K has to be selected with him.

Hence, option (a).

Workspace:

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