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CAT — Connectivity

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Connectivity

15 questions

Set 1 5 questions ▶

Three countries – Pumpland (P), Xiland (X) and Cheeseland (C) – trade among themselves and with the (other countries in) Rest of World (ROW). All trade volumes are given in IC (international currency). The following terminology is used:

• Trade balance = Exports - Imports

• Total trade = Exports + Imports

• Normalized trade balance = Trade balance / Total trade, expressed in percentage terms The following information is known.

1. The normalized trade balances of P, X and C are 0%, 10%, and -20%, respectively.

2. 40% of exports of X are to P. 22% of imports of P are from X.

3. 90% of exports of C are to P; 4% are to ROW.

4. 12% of exports of ROW are to X, 40% are to P.

5. The export volumes of P, in IC, to X and C are 600 and 1200, respectively. P is the only country that exports to C.

Q1 How much is exported from C to X, in IC? TITA ▶

How much is exported from C to X, in IC?

Answer: 48

Let the total exports of P, X, C, and ROW be Ep, Ex, Ec, and Er respectively, while their total imports be Ip, Ix, Ic, and Ir respectively.

Using the normalized trade balance formula:

For P:

(Ep − Ip)/(Ep + Ip) = 0

⇒ Ep = Ip

For X:

(Ex − Ix)/(Ex + Ix) = 10%

⇒ Ex − Ix = 0.1(Ex + Ix)

⇒ 0.9Ex = 1.1Ix

⇒ Ex/Ix = 11/9

For C:

(Ec − Ic)/(Ec + Ic) = −20%

⇒ Ec − Ic = −0.2(Ec + Ic)

⇒ 1.2Ec = 0.8Ic

⇒ Ic = 1.5Ec

Imports and Exports of C

According to condition (5), P is the only country that exports to C.

Hence, C receives all its imports from P, and the import volume from P to C is 1200.

Therefore,

Ic = 1200.

Since Ic = 1.5Ec,

Ec = 800.

Thus, C's exports are distributed as follows:

90% to P ⇒ 0.9 × Ec = 720
4% to ROW ⇒ 0.04 × Ec = 32
The remaining 6% to X ⇒ C → X = 0.06 × 800 = 48

Q2 How much is exported from P to ROW, in IC? TITA ▶

How much is exported from P to ROW, in IC?

Answer: 200

Step 1:

Let the total exports of P, X, C, and ROW be Ep, Ex, Ec, and Er respectively, while their corresponding imports be Ip, Ix, Ic, and Ir.

Using the normalized trade balance equation:

For P: (Ep − Ip)/(Ep + Ip) = 0 ⇒ Ep = Ip

For X: (Ex − Ix)/(Ex + Ix) = 10% ⇒ Ex − Ix = 0.1(Ex + Ix)

⇒ 0.9Ex = 1.1Ix

⇒ Ex/Ix = 11/9

For C: (Ec − Ic)/(Ec + Ic) = −20% ⇒ Ec − Ic = −0.2(Ec + Ic)

⇒ 1.2Ec = 0.8Ic

⇒ Ic = 1.5Ec

Imports and Exports of C

As stated in condition (5), P is the only country exporting to C.

Hence, C imports only from P, and the import value is 1200.

Therefore,

Ic = 1200.

Since Ic = 1.5Ec,

Ec = 800.

Thus, C's exports are:

90% to P ⇒ 0.9Ec = 720
4% to ROW ⇒ 0.04Ec = 32
Remaining 6% to X ⇒ C → X = 0.06 × 800 = 48

Step 2:

Using condition (2):

40% of X's total exports to P (X → P) is equal to 22% of P's imports received from X.

Therefore,

0.4Ex = 0.22Ip ⇒ Ex = 11Ip/20

Since Ep = Ip, 0.4Ex = 0.22Ep ⇒ Ex = 0.55Ep

Also, 20Ix = 9Ip.

Imports and Exports of P

P imports goods from X, C, and ROW.

From X = 0.22Ip
From C = 720
From ROW = Ip − (0.22Ip + 720) = 0.78Ip − 720

Since exports from ROW to P constitute 40% of Er,

0.4Er = 0.78Ip − 720 ...(1)

Imports and Exports of X

Given that 12% of ROW's exports are sent to X,

ROW → X = 0.12Er.

Hence, X's imports are:

From P = 600
From C = 48
From ROW = 0.12Er

Therefore,

Ix = 648 + 0.12Er.

Since Ix = 0.45Ep, 0.45Ep = 648 + 0.12Er ...(2)

Step 3:

Using equations (1) and (2):

From (1),

Er = (0.78Ep − 720)/0.4

= 1.95Ep − 1800

Substituting this into equation (2):

0.45Ep = 648 + 0.12(1.95Ep − 1800)

⇒ 0.45Ep = 648 + 0.234Ep − 216

⇒ 0.216Ep = 432

⇒ Ep = 2000

Since Ep = Ip,

Ip = 2000

Hence, the completed trade table is:

Thus, 200 units are exported from P to ROW.

Q3 How much is exported from ROW to ROW, in IC? TITA ▶

How much is exported from ROW to ROW, in IC?

Answer: 1008

Step 1:

Let the total exports of P, X, C, and ROW be Ep, Ex, Ec, and Er respectively, while their corresponding imports be Ip, Ix, Ic, and Ir.

Using the normalized trade balance equation:

For P: (Ep − Ip)/(Ep + Ip) = 0 ⇒ Ep = Ip

For X: (Ex − Ix)/(Ex + Ix) = 10% ⇒ Ex − Ix = 0.1(Ex + Ix)

⇒ 0.9Ex = 1.1Ix

⇒ Ex/Ix = 11/9

For C: (Ec − Ic)/(Ec + Ic) = −20% ⇒ Ec − Ic = −0.2(Ec + Ic)

⇒ 1.2Ec = 0.8Ic

⇒ Ic = 1.5Ec

Imports and Exports of C

As stated in condition (5), P is the only country exporting to C.

Hence, C imports only from P, and the import value is 1200.

Therefore,

Ic = 1200.

Since Ic = 1.5Ec,

Ec = 800.

Thus, C's exports are:

90% to P ⇒ 0.9Ec = 720
4% to ROW ⇒ 0.04Ec = 32
Remaining 6% to X ⇒ C → X = 0.06 × 800 = 48

Step 2:

Using condition (2):

40% of X's total exports to P (X → P) is equal to 22% of P's imports received from X.

Therefore,

0.4Ex = 0.22Ip ⇒ Ex = 11Ip/20

Since Ep = Ip, 0.4Ex = 0.22Ep ⇒ Ex = 0.55Ep

Also, 20Ix = 9Ip.

Imports and Exports of P

P imports goods from X, C, and ROW.

From X = 0.22Ip
From C = 720
From ROW = Ip − (0.22Ip + 720) = 0.78Ip − 720

Since exports from ROW to P constitute 40% of Er,

0.4Er = 0.78Ip − 720 ...(1)

Imports and Exports of X

Given that 12% of ROW's exports are sent to X,

ROW → X = 0.12Er.

Hence, X's imports are:

From P = 600
From C = 48
From ROW = 0.12Er

Therefore,

Ix = 648 + 0.12Er.

Since Ix = 0.45Ep, 0.45Ep = 648 + 0.12Er ...(2)

Step 3:

Using equations (1) and (2):

From (1),

Er = (0.78Ep − 720)/0.4

= 1.95Ep − 1800

Substituting this into equation (2):

0.45Ep = 648 + 0.12(1.95Ep − 1800)

⇒ 0.45Ep = 648 + 0.234Ep − 216

⇒ 0.216Ep = 432

⇒ Ep = 2000

Since Ep = Ip,

Ip = 2000

Hence, the completed trade table is:

1008 IC is exported from ROW to ROW.

Q4 What is the trade balance of ROW? MCQ ▶

What is the trade balance of ROW?

100

-200

200

Correct answer: D.

200

Step 1:

Let the total exports of P, X, C, and ROW be Ep, Ex, Ec, and Er respectively, while their corresponding imports be Ip, Ix, Ic, and Ir.

Using the normalized trade balance equation:

For P: (Ep − Ip)/(Ep + Ip) = 0 ⇒ Ep = Ip

For X: (Ex − Ix)/(Ex + Ix) = 10% ⇒ Ex − Ix = 0.1(Ex + Ix)

⇒ 0.9Ex = 1.1Ix

⇒ Ex/Ix = 11/9

For C: (Ec − Ic)/(Ec + Ic) = −20% ⇒ Ec − Ic = −0.2(Ec + Ic)

⇒ 1.2Ec = 0.8Ic

⇒ Ic = 1.5Ec

Imports and Exports of C

As stated in condition (5), P is the only country exporting to C.

Hence, C imports only from P, and the import value is 1200.

Therefore,

Ic = 1200.

Since Ic = 1.5Ec,

Ec = 800.

Thus, C's exports are:

90% to P ⇒ 0.9Ec = 720
4% to ROW ⇒ 0.04Ec = 32
Remaining 6% to X ⇒ C → X = 0.06 × 800 = 48

Step 2:

Using condition (2):

40% of X's total exports to P (X → P) is equal to 22% of P's imports received from X.

Therefore,

0.4Ex = 0.22Ip ⇒ Ex = 11Ip/20

Since Ep = Ip, 0.4Ex = 0.22Ep ⇒ Ex = 0.55Ep

Also, 20Ix = 9Ip.

Imports and Exports of P

P imports goods from X, C, and ROW.

From X = 0.22Ip
From C = 720
From ROW = Ip − (0.22Ip + 720) = 0.78Ip − 720

Since exports from ROW to P constitute 40% of Er,

0.4Er = 0.78Ip − 720 ...(1)

Imports and Exports of X

Given that 12% of ROW's exports are sent to X,

ROW → X = 0.12Er.

Hence, X's imports are:

From P = 600
From C = 48
From ROW = 0.12Er

Therefore,

Ix = 648 + 0.12Er.

Since Ix = 0.45Ep, 0.45Ep = 648 + 0.12Er ...(2)

Step 3:

Using equations (1) and (2):

From (1),

Er = (0.78Ep − 720)/0.4

= 1.95Ep − 1800

Substituting this into equation (2):

0.45Ep = 648 + 0.12(1.95Ep − 1800)

⇒ 0.45Ep = 648 + 0.234Ep − 216

⇒ 0.216Ep = 432

⇒ Ep = 2000

Since Ep = Ip,

Ip = 2000

Hence, the completed trade table is:

The trade balance of ROW = Er – Ir = 2100 – 1900 = 200 IC.

Q5 Which among the countries P, X, and C has/have the least total trade? MCQ ▶

Which among the countries P, X, and C has/have the least total trade?

Only P

Only C

Both X and C

Only X

Correct answer: C.

Both X and C

Step 1:

Let the total exports of P, X, C, and ROW be Ep, Ex, Ec, and Er respectively, while their corresponding imports be Ip, Ix, Ic, and Ir.

Using the normalized trade balance equation:

For P: (Ep − Ip)/(Ep + Ip) = 0 ⇒ Ep = Ip

For X: (Ex − Ix)/(Ex + Ix) = 10% ⇒ Ex − Ix = 0.1(Ex + Ix)

⇒ 0.9Ex = 1.1Ix

⇒ Ex/Ix = 11/9

For C: (Ec − Ic)/(Ec + Ic) = −20% ⇒ Ec − Ic = −0.2(Ec + Ic)

⇒ 1.2Ec = 0.8Ic

⇒ Ic = 1.5Ec

Imports and Exports of C

As stated in condition (5), P is the only country exporting to C.

Hence, C imports only from P, and the import value is 1200.

Therefore,

Ic = 1200.

Since Ic = 1.5Ec,

Ec = 800.

Thus, C's exports are:

90% to P ⇒ 0.9Ec = 720
4% to ROW ⇒ 0.04Ec = 32
Remaining 6% to X ⇒ C → X = 0.06 × 800 = 48

Step 2:

Using condition (2):

40% of X's total exports to P (X → P) is equal to 22% of P's imports received from X.

Therefore,

0.4Ex = 0.22Ip ⇒ Ex = 11Ip/20

Since Ep = Ip, 0.4Ex = 0.22Ep ⇒ Ex = 0.55Ep

Also, 20Ix = 9Ip.

Imports and Exports of P

P imports goods from X, C, and ROW.

From X = 0.22Ip
From C = 720
From ROW = Ip − (0.22Ip + 720) = 0.78Ip − 720

Since exports from ROW to P constitute 40% of Er,

0.4Er = 0.78Ip − 720 ...(1)

Imports and Exports of X

Given that 12% of ROW's exports are sent to X,

ROW → X = 0.12Er.

Hence, X's imports are:

From P = 600
From C = 48
From ROW = 0.12Er

Therefore,

Ix = 648 + 0.12Er.

Since Ix = 0.45Ep, 0.45Ep = 648 + 0.12Er ...(2)

Step 3:

Using equations (1) and (2):

From (1),

Er = (0.78Ep − 720)/0.4

= 1.95Ep − 1800

Substituting this into equation (2):

0.45Ep = 648 + 0.12(1.95Ep − 1800)

⇒ 0.45Ep = 648 + 0.234Ep − 216

⇒ 0.216Ep = 432

⇒ Ep = 2000

Since Ep = Ip,

Ip = 2000

Hence, the completed trade table is:

Total trade:

P = Ep + Ip = 4000

X = Ex + Ix = (1100 + 900) = 2000

C = Ec + Ic = (800 + 1200) = 2000

Hence, the least total trade = X and C.

Set 2 5 questions ▶

The figure below shows a network with three parallel roads represented by horizontal lines R-A, R-B, and R-C and another three parallel roads represented by vertical lines V1, V2, and V3. The figure also shows the distance (in km) between two adjacent intersections. Six ATMs are placed at six of the nine road intersections. Each ATM has a distinct integer cash requirement (in Rs. Lakhs), and the numbers at the end of each line in the figure indicate the total cash requirements of all ATMs placed on the corresponding road. For example, the total cash requirement of the ATM(s) placed on road R-A is Rs. 22 Lakhs.

The following additional information is known.

1. The ATMs with the minimum and maximum cash requirements of Rs. 7 Lakhs and Rs. 15 Lakhs are placed on the same road.

2. The road distance between the ATM with the second highest cash requirement and the ATM located at the intersection of R-C and V3 is 12 km.

Q6 Which of the following statements is correct? MCQ ▶

Which of the following statements is correct?

The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 8 Lakhs.

The cash requirement of the ATM placed at the (R-C, V2) intersection cannot be uniquely determined.

There is no ATM placed at the (R-C, V2) intersection.

The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.

Correct answer: D.

The ATM placed at the (R-C, V2) intersection has a cash requirement of Rs. 9 Lakhs.

Step 1:

From condition (1), the ATMs with the minimum and maximum cash requirements—Rs. 7 lakh and Rs. 15 lakh, respectively—must be located on Road R-A. Placing them anywhere else would violate the given condition.

According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.

This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.

Since the Rs. 7 lakh and Rs. 15 lakh ATMs are both on R-A, they can only occupy intersections V1 and V3. One of them must be at V2; otherwise, the minimum cash requirement condition would not be satisfied.

This leads to the following two possibilities:

Case 1:

R-A, V1: Rs. 15 lakh
R-A, V3: Rs. 7 lakh

In this arrangement, all ATMs have distinct cash requirements:

R-A, V1 = 15 lakh
R-A, V3 = 7 lakh
R-B, V2 = 12 lakh
R-B, V3 = 8 lakh
R-C, V2 = 9 lakh
R-C, V3 = 11 lakh

Case 2:

R-A, V1: Rs. 7 lakh
R-A, V3: Rs. 15 lakh

The corresponding cash requirements are:

R-A, V1 = 7 lakh
R-A, V3 = 15 lakh
R-B, V1 = 8 lakh
R-B, V2 = 12 lakh
R-C, V2 = 9 lakh
R-C, V3 = 11 lakh

Let us examine each of the given statements:

The ATM at the intersection of R-C and V2 has a cash requirement of Rs. 8 lakh. False
The cash requirement of the ATM at the intersection of R-C and V2 cannot be determined uniquely. False
There is no ATM at the intersection of R-C and V2. False
The ATM at the intersection of R-C and V2 has a cash requirement of Rs. 9 lakh. True

Hence, only Statement 4 is correct.

Q7 How many ATMs have cash requirements of Rs. 10 Lakhs or more? TITA ▶

How many ATMs have cash requirements of Rs. 10 Lakhs or more?

Answer: 3

Step 1:

According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.

This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.

This leads to the following two possibilities:

Case 1:

R-A, V1: Rs. 15 lakh
R-A, V3: Rs. 7 lakh

In this arrangement, all ATMs have distinct cash requirements:

R-A, V1 = 15 lakh
R-A, V3 = 7 lakh
R-B, V2 = 12 lakh
R-B, V3 = 8 lakh
R-C, V2 = 9 lakh
R-C, V3 = 11 lakh

Case 2:

R-A, V1: Rs. 7 lakh
R-A, V3: Rs. 15 lakh

The corresponding cash requirements are:

R-A, V1 = 7 lakh
R-A, V3 = 15 lakh
R-B, V1 = 8 lakh
R-B, V2 = 12 lakh
R-C, V2 = 9 lakh
R-C, V3 = 11 lakh

There are three ATMs that have cash requirements of Rs. 10 Lakhs or more.

Q8 What best can be said about the road distance (in km) between the ATMs having the second … MCQ ▶

What best can be said about the road distance (in km) between the ATMs having the second highest and the second lowest cash requirements?

4 km

5 km

7 km

Either 4 km or 7 km

Correct answer: D.

Either 4 km or 7 km

Step 1:

According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.

This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.

This leads to the following two possibilities:

Case 1:

R-A, V1: Rs. 15 lakh
R-A, V3: Rs. 7 lakh

In this arrangement, all ATMs have distinct cash requirements:

R-A, V1 = 15 lakh
R-A, V3 = 7 lakh
R-B, V2 = 12 lakh
R-B, V3 = 8 lakh
R-C, V2 = 9 lakh
R-C, V3 = 11 lakh

Case 2:

R-A, V1: Rs. 7 lakh
R-A, V3: Rs. 15 lakh

The corresponding cash requirements are:

R-A, V1 = 7 lakh
R-A, V3 = 15 lakh
R-B, V1 = 8 lakh
R-B, V2 = 12 lakh
R-C, V2 = 9 lakh
R-C, V3 = 11 lakh

The second-highest and second-lowest cash requirements are Rs. 12 lakh and Rs. 8 lakh, respectively.

In Case 1, the distance between these two ATMs is 7 km.
In Case 2, the corresponding distance is 4 km.

Hence, the required distance cannot be determined uniquely. It can be either 4 km or 7 km.

Q9 Which of the following two statements is/are DEFINITELY true?Statement A: Each of R-A, R-… MCQ ▶

Which of the following two statements is/are DEFINITELY true?

Statement A: Each of R-A, R-B, and R-C has two ATMs.

Statement B: Each of V1, V2, and V3 has two ATMs.

Both Statement A and Statement B

Only Statement B

Only Statement A

Neither Statement A nor Statement B

Correct answer: C.

Only Statement A

Step 1:

According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.

This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.

This leads to the following two possibilities:

Case 1:

R-A, V1: Rs. 15 lakh
R-A, V3: Rs. 7 lakh

In this arrangement, all ATMs have distinct cash requirements:

R-A, V1 = 15 lakh
R-A, V3 = 7 lakh
R-B, V2 = 12 lakh
R-B, V3 = 8 lakh
R-C, V2 = 9 lakh
R-C, V3 = 11 lakh

Case 2:

R-A, V1: Rs. 7 lakh
R-A, V3: Rs. 15 lakh

The corresponding cash requirements are:

R-A, V1 = 7 lakh
R-A, V3 = 15 lakh
R-B, V1 = 8 lakh
R-B, V2 = 12 lakh
R-C, V2 = 9 lakh
R-C, V3 = 11 lakh

Let us evaluate each statement individually:

Statement A: Each of R-A, R-B, and R-C contains exactly two ATMs.

This condition is satisfied in both possible cases. Hence, Statement A is true.

Statement B: Each of V1, V2, and V3 contains exactly two ATMs.

This condition is not satisfied in Case 1. Hence, Statement B is false.

Therefore, only Statement A is true.

Q10 What is the number of ATMs whose locations and cash requirements can both be uniquely det… TITA ▶

What is the number of ATMs whose locations and cash requirements can both be uniquely determined?

Answer: 3

Step 1:

According to condition (2), the road distance between the ATM with the second-highest cash requirement and the ATM at the intersection of R-C and V3 is 12 km.

This implies that the ATM with the second-highest cash requirement must be placed at the intersection of R-B and V2.

This leads to the following two possibilities:

Case 1:

R-A, V1: Rs. 15 lakh
R-A, V3: Rs. 7 lakh

In this arrangement, all ATMs have distinct cash requirements:

R-A, V1 = 15 lakh
R-A, V3 = 7 lakh
R-B, V2 = 12 lakh
R-B, V3 = 8 lakh
R-C, V2 = 9 lakh
R-C, V3 = 11 lakh

Case 2:

R-A, V1: Rs. 7 lakh
R-A, V3: Rs. 15 lakh

The corresponding cash requirements are:

R-A, V1 = 7 lakh
R-A, V3 = 15 lakh
R-B, V1 = 8 lakh
R-B, V2 = 12 lakh
R-C, V2 = 9 lakh
R-C, V3 = 11 lakh

From the two possible arrangements, the positions of the ATMs with cash requirements of Rs. 12 lakh, Rs. 11 lakh, and Rs. 9 lakh remain unchanged in both cases.

However, the positions of the ATMs with cash requirements of Rs. 7 lakh, Rs. 8 lakh, and Rs. 15 lakh vary between the two arrangements.

Therefore, the positions of 3 ATMs cannot be determined uniquely.

Hence, the correct answer is 3.

Set 3 5 questions ▶

A train travels from Station A to Station E, passing through stations B, C, and D, in that order. The train has a seating capacity of 200. A ticket may be booked from any station to any other station ahead on the route, but not to any earlier station.

A ticket from one station to another reserves one seat on every intermediate segment of the route. For example, a ticket from B to E reserves a seat in the intermediate segments B – C, C – D, and D – E.

The occupancy factor for a segment is the total number of seats reserved in the segment as a percentage of the seating capacity. The total number of seats reserved for any segment cannot exceed 200.

The following information is known.

1. Segment C – D had an occupancy factor of 95%.

Only segment B – C had a higher occupancy factor.

2. Exactly 40 tickets were booked from B to C and 30 tickets were booked from B to E.

3. Among the seats reserved on segment D – E,

exactly four-sevenths were from stations before C.

4. The number of tickets booked from A to C was equal to that booked from A to E, and it was higher than that from B to E.

5. No tickets were booked from A to B, from B to D and from D to E.

6. The number of tickets booked for any segment was a multiple of 10.

Q11 What was the occupancy factor for segment D – E? MCQ ▶

What was the occupancy factor for segment D – E?

84%

35%

70%

77%

Correct answer: C.

70%

Step 1:

From Condition (2), 40 tickets were booked from B to C, while 30 tickets were booked from B to E.

Using Condition (5), there were no bookings on the routes A to B, B to D, and D to E.

From Condition (4), the number of tickets booked from A to C is equal to the number of tickets booked from A to E, and this common value is greater than 30.

Now, applying Condition (3), exactly four-sevenths of the passengers travelling on segment D–E had boarded before Station C, i.e., from Stations A and B.

Since the total occupancy on segment D–E must be divisible by both 7 and 10, the only possible values are 70 and 140.

An occupancy of 70 is not possible because A to E + B to E > 60. Therefore, the occupancy on segment D–E must be 140.

Hence,

A to E + B to E = (4/7) × 140 = 80

Since B to E = 30,

A to E = 80 − 30 = 50

Therefore,

C to E = 140 − (50 + 30) = 60

Step 2:

From Conditions (1) and (6), the occupancy on segment C–D is

0.95 × 200 = 190

Also, the occupancy on segment B–C is 200.

For segment B–C,

AC + AD + AE + BC + BD + BE = 200

Substituting the known values,

50 + AD + 50 + 40 + 0 + 30 = 200

AD = 30

Similarly, for segment C–D,

AD + AE + BD + BE + CD + CE = 190

Substituting the known values,

30 + 50 + 0 + 30 + CD + 60 = 190

CD = 20

The completed booking table is shown below.

The occupancy on segment D–E is:

A to E + B to E + C to E + D to E

= 50 + 30 + 60 + 0 = 140

Hence, the occupancy factor for segment D–E is

(140/200) × 100 = 70%.

Q12 How many tickets were booked from Station A to Station E? TITA ▶

How many tickets were booked from Station A to Station E?

Answer: 50

Step 1:

From Condition (2), 40 tickets were booked from B to C, while 30 tickets were booked from B to E.

Using Condition (5), there were no bookings on the routes A to B, B to D, and D to E.

From Condition (4), the number of tickets booked from A to C is equal to the number of tickets booked from A to E, and this common value is greater than 30.

Now, applying Condition (3), exactly four-sevenths of the passengers travelling on segment D–E had boarded before Station C, i.e., from Stations A and B.

Since the total occupancy on segment D–E must be divisible by both 7 and 10, the only possible values are 70 and 140.

An occupancy of 70 is not possible because A to E + B to E > 60. Therefore, the occupancy on segment D–E must be 140.

Hence,

A to E + B to E = (4/7) × 140 = 80

Since B to E = 30,

A to E = 80 − 30 = 50

Therefore,

C to E = 140 − (50 + 30) = 60

Step 2:

From Conditions (1) and (6), the occupancy on segment C–D is

0.95 × 200 = 190

Also, the occupancy on segment B–C is 200.

For segment B–C,

AC + AD + AE + BC + BD + BE = 200

Substituting the known values,

50 + AD + 50 + 40 + 0 + 30 = 200

AD = 30

Similarly, for segment C–D,

AD + AE + BD + BE + CD + CE = 190

Substituting the known values,

30 + 50 + 0 + 30 + CD + 60 = 190

CD = 20

The completed booking table is shown below.

The number of tickets booked from Station A to Station E was 50.

Q13 How many tickets were booked from Station C? TITA ▶

How many tickets were booked from Station C?

Answer: 80

Step 1:

From Condition (2), 40 tickets were booked from B to C, while 30 tickets were booked from B to E.

Using Condition (5), there were no bookings on the routes A to B, B to D, and D to E.

From Condition (4), the number of tickets booked from A to C is equal to the number of tickets booked from A to E, and this common value is greater than 30.

Now, applying Condition (3), exactly four-sevenths of the passengers travelling on segment D–E had boarded before Station C, i.e., from Stations A and B.

Since the total occupancy on segment D–E must be divisible by both 7 and 10, the only possible values are 70 and 140.

An occupancy of 70 is not possible because A to E + B to E > 60. Therefore, the occupancy on segment D–E must be 140.

Hence,

A to E + B to E = (4/7) × 140 = 80

Since B to E = 30,

A to E = 80 − 30 = 50

Therefore,

C to E = 140 − (50 + 30) = 60

Step 2:

From Conditions (1) and (6), the occupancy on segment C–D is

0.95 × 200 = 190

Also, the occupancy on segment B–C is 200.

For segment B–C,

AC + AD + AE + BC + BD + BE = 200

Substituting the known values,

50 + AD + 50 + 40 + 0 + 30 = 200

AD = 30

Similarly, for segment C–D,

AD + AE + BD + BE + CD + CE = 190

Substituting the known values,

30 + 50 + 0 + 30 + CD + 60 = 190

CD = 20

The completed booking table is shown below.

The number of tickets booked from Station C was = 20 + 60 = 80.

Q14 What is the difference between the number of tickets booked to Station C and the number o… TITA ▶

What is the difference between the number of tickets booked to Station C and the number of tickets booked to Station D?

Answer: 40

Step 1:

From Condition (2), 40 tickets were booked from B to C, while 30 tickets were booked from B to E.

Using Condition (5), there were no bookings on the routes A to B, B to D, and D to E.

From Condition (4), the number of tickets booked from A to C is equal to the number of tickets booked from A to E, and this common value is greater than 30.

Now, applying Condition (3), exactly four-sevenths of the passengers travelling on segment D–E had boarded before Station C, i.e., from Stations A and B.

Since the total occupancy on segment D–E must be divisible by both 7 and 10, the only possible values are 70 and 140.

An occupancy of 70 is not possible because A to E + B to E > 60. Therefore, the occupancy on segment D–E must be 140.

Hence,

A to E + B to E = (4/7) × 140 = 80

Since B to E = 30,

A to E = 80 − 30 = 50

Therefore,

C to E = 140 − (50 + 30) = 60

Step 2:

From Conditions (1) and (6), the occupancy on segment C–D is

0.95 × 200 = 190

Also, the occupancy on segment B–C is 200.

For segment B–C,

AC + AD + AE + BC + BD + BE = 200

Substituting the known values,

50 + AD + 50 + 40 + 0 + 30 = 200

AD = 30

Similarly, for segment C–D,

AD + AE + BD + BE + CD + CE = 190

Substituting the known values,

30 + 50 + 0 + 30 + CD + 60 = 190

CD = 20

The completed booking table is shown below.

Required difference = (50 + 40) – (30 + 20) = 40.

Q15 How many tickets were booked to travel in exactly one segment? TITA ▶

How many tickets were booked to travel in exactly one segment?

Answer: 60

Step 1:

From Condition (2), 40 tickets were booked from B to C, while 30 tickets were booked from B to E.

Using Condition (5), there were no bookings on the routes A to B, B to D, and D to E.

From Condition (4), the number of tickets booked from A to C is equal to the number of tickets booked from A to E, and this common value is greater than 30.

Now, applying Condition (3), exactly four-sevenths of the passengers travelling on segment D–E had boarded before Station C, i.e., from Stations A and B.

Since the total occupancy on segment D–E must be divisible by both 7 and 10, the only possible values are 70 and 140.

An occupancy of 70 is not possible because A to E + B to E > 60. Therefore, the occupancy on segment D–E must be 140.

Hence,

A to E + B to E = (4/7) × 140 = 80

Since B to E = 30,

A to E = 80 − 30 = 50

Therefore,

C to E = 140 − (50 + 30) = 60

Step 2:

From Conditions (1) and (6), the occupancy on segment C–D is

0.95 × 200 = 190

Also, the occupancy on segment B–C is 200.

For segment B–C,

AC + AD + AE + BC + BD + BE = 200

Substituting the known values,

50 + AD + 50 + 40 + 0 + 30 = 200

AD = 30

Similarly, for segment C–D,

AD + AE + BD + BE + CD + CE = 190

Substituting the known values,

30 + 50 + 0 + 30 + CD + 60 = 190

CD = 20

The completed booking table is shown below.

The number of tickets booked to travel in exactly one segment = B to C + C to D = 40 + 20 = 60.

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