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CAT — Distribution

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Distribution

18 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 4 questions ▶

The following charts depict details of research papers written by four authors, Arman, Brajen, Chintan, and Devon. The papers were of four types, single-author, two-author, three-author, and four-author, that is, written by one, two, three, or all four of these authors, respectively. No other authors were involved in writing these papers.

The following additional facts are known.

1. Each of the authors wrote at least one of each of the four types of papers.

2. The four authors wrote different numbers of single-author papers.

3. Both Chintan and Devon wrote more three-author papers than Brajen.

4. The number of single-author and two-author papers written by Brajen were the same.

Q6 What was the total number of two-author and threeauthor papers written by Brajen? TITA ▶

What was the total number of two-author and threeauthor papers written by Brajen?

Answer: 4

Step 1:

From the first bar graph, we know the total number of titles authored by each individual. We are also given the total number of single-author, two-author, three-author, and four-author papers.

Observe that 2 four-author papers contribute 2 × 4 = 8 author counts, since all four authors are involved in each paper. The same counting principle applies to the three-author and two-author papers.

Using Condition (1), each author contributed to at least one paper of every type. Therefore, none of the entries in the table can be 0.

Since Aman has authored 5 papers in total, and 2 of them are four-author papers, the remaining 3 papers must be distributed among the other three categories. As every category must have at least one paper, Aman must have 1 single-author, 1 two-author, and 1 three-author paper.

From Condition (2), every author has a distinct number of single-author papers. Since Aman already has 1, the remaining authors must have 2, 3, and 4 single-author papers.

Step 2:

Using Condition (4), Brajen cannot have 4 single-author and two-author papers combined, because the overall total for these categories is 8. He also cannot have 3, as that would leave him with 0 three-author papers, violating the given conditions.

Hence, Brajen must have 2 single-author papers and 2 two-author papers. Since his total is 8, he must also have 2 three-author papers.

Now, applying Condition (3), both Chintan and Devon must have more than 2 three-author papers, and the total number of three-author papers contributed by all authors is 9. The only feasible allocation is 3 each for Chintan and Devon.

The deductions obtained so far are summarized in the table below.

Therefore, the total number of two-author and three-author papers written by Brajen is:

2 + 2 = 4.

Q7 Which of the following statements is/are NECESSARILY true?i. Chintan wrote exactly three … MCQ ▶

Which of the following statements is/are NECESSARILY true?

i. Chintan wrote exactly three two-author papers.

ii. Chintan wrote more single-author papers than Devon.

Neither i nor ii

Only i

Both i and ii

Only ii

Correct answer: A.

Neither i nor ii

Step 1:

From the first bar graph, we know the total number of titles authored by each individual. We are also given the total number of single-author, two-author, three-author, and four-author papers.

Using Condition (1), each author contributed to at least one paper of every type. Therefore, none of the entries in the table can be 0.

From Condition (2), every author has a distinct number of single-author papers. Since Aman already has 1, the remaining authors must have 2, 3, and 4 single-author papers.

Step 2:

Hence, Brajen must have 2 single-author papers and 2 two-author papers. Since his total is 8, he must also have 2 three-author papers.

The deductions obtained so far are summarized in the table below.

Let us check each statement:

i. The statement, Chintan wrote exactly three two-author papers may not necessarily be true.

ii. The statement, Chintan wrote more single-author papers than Devon may not necessarily be true.

Hence, neither i nor ii is definitely true.

Q8 Which of the following statements is/are NECESSARILY true?i. Arman wrote three-author pap… MCQ ▶

Which of the following statements is/are NECESSARILY true?

i. Arman wrote three-author papers only with Chintan and Devon.

ii. Brajen wrote three-author papers only with Chintan and Devon.

Neither i or ii

Both i and ii

Only ii

Only i

Correct answer: B.

Both i and ii

Step 1:

From the first bar graph, we know the total number of titles authored by each individual. We are also given the total number of single-author, two-author, three-author, and four-author papers.

Using Condition (1), each author contributed to at least one paper of every type. Therefore, none of the entries in the table can be 0.

From Condition (2), every author has a distinct number of single-author papers. Since Aman already has 1, the remaining authors must have 2, 3, and 4 single-author papers.

Step 2:

Hence, Brajen must have 2 single-author papers and 2 two-author papers. Since his total is 8, he must also have 2 three-author papers.

The deductions obtained so far are summarized in the table below.

There are 3 three author papers and both Chintan and Devon wrote 3 three author papers whereas Aman wrote 1 and Brajen wrote 2. So the only possible combination will be {(Chintan, Devon, Aman), (Chintan, Devon, Brajen), (Chintan, Devon, Brajen)}. Hence, the statement, Arman wrote three-author papers only with Chintan and Devon, is true. Brajen wrote three-author papers only with Chintan and Devon is also true. Hence, both (i) and (ii) are true.

Q9 If Devon wrote more than one two-author papers, then how many two-author papers did Chint… TITA ▶

If Devon wrote more than one two-author papers, then how many two-author papers did Chintan write?

Answer: 3

Step 1:

From the first bar graph, we know the total number of titles authored by each individual. We are also given the total number of single-author, two-author, three-author, and four-author papers.

Using Condition (1), each author contributed to at least one paper of every type. Therefore, none of the entries in the table can be 0.

From Condition (2), every author has a distinct number of single-author papers. Since Aman already has 1, the remaining authors must have 2, 3, and 4 single-author papers.

Step 2:

Hence, Brajen must have 2 single-author papers and 2 two-author papers. Since his total is 8, he must also have 2 three-author papers.

The deductions obtained so far are summarized in the table below.

If Devon wrote more than one two-author papers, then the number of two-author papers written by Chintan is 3.

Set 3 5 questions ▶

The two most populous cities and the non-urban region (NUR) of each of three states, Whimshire, Fogglia, and Humbleset, are assigned Pollution Measures (PMs). These nine PMs are all distinct multiples of 10, ranging from 10 to 90. The six cities in increasing order of their There is only one pair of an NUR and a city (considering all cities and all NURs) where the PM of the NUR is greater than that of the city. That NUR and the city both belong to Humbleset.

The PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.

PMs are: Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, Zingaloo.

The Pollution Index (PI) of a state is a weighted average of the PMs of its NUR and cities, with a weight of 50% for the NUR, and 25% each for its two cities.

There is only one pair of an NUR and a city (considering all cities and all NURs) where the PM of the NUR is greater than that of the city. That NUR and the city both belong to Humbleset.

The PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.

Q10 What is the PI of Whimshire? TITA ▶

What is the PI of Whimshire?

Answer: 45

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

We are also told that there is only one city–NUR pair for which the NUR has a higher PI than the city. This is possible only if Blusterburg has a PI of 30, while the corresponding NUR has a PI of 40. Any other assignment would violate the given condition. Moreover, both of these belong to Humbleset.

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Next, for the PI values of all the states to remain integers, the cities with weighted PIs 12.5 and 17.5 must belong to the same state. Their combined weighted PI is 30, while the remaining pair contributes 15 + 20 = 35.

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

Hence, the PI of Whimshire is 45.

Q11 What is the PI of Fogglia? TITA ▶

What is the PI of Fogglia?

Answer: 35

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

The PI of Fogglia = 35

Q12 What is the PI of Humbleset? TITA ▶

What is the PI of Humbleset?

Answer: 50

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

The PI of Humbleset = 50

Q13 Which pair of cities definitely belong to the same state? MCQ ▶

Which pair of cities definitely belong to the same state?

Splutterville, Quackford

Mumpypore, Zingaloo

Noodleton, Quackford

Blusterburg, Mumpypore

Correct answer: C.

Noodleton, Quackford

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

The pair of cities that definitely belong to the same states Noodleton and Quackford.

Q14 For how many of the cities and NURs is it possible to identify their PM and the state the… TITA ▶

For how many of the cities and NURs is it possible to identify their PM and the state they belong to?

Answer: 9

Step 1:

From the given information, the nine PI values are distinct multiples of 10, namely:

10, 20, 30, 40, 50, 60, 70, 80, and 90.

The cities are arranged in increasing order of PI as follows:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

Accordingly, the remaining cities take the PI values 50, 60, 70, 80, and 90 in increasing order, while the remaining two NURs receive the PI values 10 and 20.

The partial information is summarized in the table below.

Step 2:

The PI values of all three states are distinct integers, with Humbleset having the highest PI and Fogglia the lowest.

For Humbleset to have an integer PI, its remaining city must contribute a weighted PI of 12.5, 17.5, or 22.5.

However, choosing 12.5 or 17.5 would prevent Humbleset from having the highest PI. Therefore, the remaining city in Humbleset must be Zingaloo, whose weighted PI is 22.5.

Thus,

PI(Humbleset) = 7.5 + 20 + 22.5 = 50

Since Fogglia must have the lowest PI, the only feasible allocation is:

PI(Fogglia) = 12.5 + 17.5 + 5 = 35

PI(Whimshire) = 15 + 20 + 10 = 45

The completed table is shown below.

We can identify the pIs of all cities and NURs and also identify the state they belong to.

Set 4 4 questions ▶

Six web surfers M, N, O, P, X, and Y each had 30 stars which they distributed among four bloggers A, B, C, and D. The number of stars received by A and B from the six web surfers is shown in the figure below.

The following additional facts are known regarding the number of stars received by the bloggers from the surfers.

1. The numbers of stars received by the bloggers from the surfers were all multiples of 5 (including 0).

2. The total numbers of stars received by the bloggers were the same.

3. Each blogger received a different number of stars from M.

4. Two surfers gave all their stars to a single blogger.

5. D received more stars than C from Y.

Q15 What was the total number of stars received by D? TITA ▶

What was the total number of stars received by D?

Answer: 45

Common Solution:

The available information can initially be represented as follows:

A receives: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
B receives: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
C receives: P = 0 (Total = 45)
D receives: P = 0 (Total = 45)

Since M distributed a distinct number of stars to each blogger, C and D must have received 5 and 15 stars from M, in some order.

Also, O and X are the only web surfers who assigned all their stars to a single blogger.

Since D received more stars than C from Y, D must have received 5 stars from Y, while C received 0 stars from Y.

Thus, the remaining allocations are:

C: M = 5/15, N = 0/5, O = 0/30, P = 0, X = 30/0, Y = 0
D: M = 15/5, N = 5/0, O = 30/0, P = 0, X = 0/30, Y = 5

If D were assigned 15 stars from M, D's total would exceed 45.

Therefore, D must receive 5 stars from M, and C must receive 15 stars from M.

The final allocations are:

A: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
B: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
C: M = 15, N = 0, O = 0, P = 0, X = 30, Y = 0 (Total = 45)
D: M = 5, N = 5, O = 30, P = 0, X = 0, Y = 5 (Total = 45)

Each web surfer distributed 30 stars, and every blogger received an equal total of 45 stars.

Hence, the number of stars received by D is:

= (30 × 6) / 4

= 45.

Q16 What was the number of stars received by D from Y? MCQ ▶

What was the number of stars received by D from Y?

cannot be determined

Correct answer: B.

The available information can initially be represented as follows:

A receives: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
B receives: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
C receives: P = 0 (Total = 45)
D receives: P = 0 (Total = 45)

Since M distributed a distinct number of stars to each blogger, C and D must have received 5 and 15 stars from M, in some order.

Also, O and X are the only web surfers who assigned all their stars to a single blogger.

Since D received more stars than C from Y, D must have received 5 stars from Y, while C received 0 stars from Y.

Thus, the remaining allocations are:

C: M = 5/15, N = 0/5, O = 0/30, P = 0, X = 30/0, Y = 0
D: M = 15/5, N = 5/0, O = 30/0, P = 0, X = 0/30, Y = 5

If D were assigned 15 stars from M, D's total would exceed 45.

Therefore, D must receive 5 stars from M, and C must receive 15 stars from M.

The final allocations are:

A: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
B: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
C: M = 15, N = 0, O = 0, P = 0, X = 30, Y = 0 (Total = 45)
D: M = 5, N = 5, O = 30, P = 0, X = 0, Y = 5 (Total = 45)

Each web surfer distributed 30 stars, and every blogger received an equal total of 45 stars.

Hence, the number of stars received by D is:

= (30 × 6) / 4

= 45.

Q17 How many surfers distributed their stars among exactly 2 bloggers? TITA ▶

How many surfers distributed their stars among exactly 2 bloggers?

Answer: 2

The available information can initially be represented as follows:

A receives: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
B receives: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
C receives: P = 0 (Total = 45)
D receives: P = 0 (Total = 45)

Since M distributed a distinct number of stars to each blogger, C and D must have received 5 and 15 stars from M, in some order.

Also, O and X are the only web surfers who assigned all their stars to a single blogger.

Since D received more stars than C from Y, D must have received 5 stars from Y, while C received 0 stars from Y.

Thus, the remaining allocations are:

C: M = 5/15, N = 0/5, O = 0/30, P = 0, X = 30/0, Y = 0
D: M = 15/5, N = 5/0, O = 30/0, P = 0, X = 0/30, Y = 5

If D were assigned 15 stars from M, D's total would exceed 45.

Therefore, D must receive 5 stars from M, and C must receive 15 stars from M.

The final allocations are:

A: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
B: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
C: M = 15, N = 0, O = 0, P = 0, X = 30, Y = 0 (Total = 45)
D: M = 5, N = 5, O = 30, P = 0, X = 0, Y = 5 (Total = 45)

Each web surfer distributed 30 stars, and every blogger received an equal total of 45 stars.

Hence, the number of stars received by D is:

= (30 × 6) / 4

= 45.

Q18 Which of the following can be determined with certainty?I. The numbers of stars received … MCQ ▶

Which of the following can be determined with certainty?

I. The numbers of stars received by C from M

II. The number of stars received by D from O

Only I

Only II

Both I and II

Neither I nor II

Correct answer: A.

Only I

The available information can initially be represented as follows:

A receives: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
B receives: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
C receives: P = 0 (Total = 45)
D receives: P = 0 (Total = 45)

Since M distributed a distinct number of stars to each blogger, C and D must have received 5 and 15 stars from M, in some order.

Also, O and X are the only web surfers who assigned all their stars to a single blogger.

Since D received more stars than C from Y, D must have received 5 stars from Y, while C received 0 stars from Y.

Thus, the remaining allocations are:

C: M = 5/15, N = 0/5, O = 0/30, P = 0, X = 30/0, Y = 0
D: M = 15/5, N = 5/0, O = 30/0, P = 0, X = 0/30, Y = 5

If D were assigned 15 stars from M, D's total would exceed 45.

Therefore, D must receive 5 stars from M, and C must receive 15 stars from M.

The final allocations are:

A: M = 10, N = 25, O = 0, P = 5, X = 0, Y = 5 (Total = 45)
B: M = 0, N = 0, O = 0, P = 25, X = 0, Y = 20 (Total = 45)
C: M = 15, N = 0, O = 0, P = 0, X = 30, Y = 0 (Total = 45)
D: M = 5, N = 5, O = 30, P = 0, X = 0, Y = 5 (Total = 45)

Each web surfer distributed 30 stars, and every blogger received an equal total of 45 stars.

Hence, the number of stars received by D is:

= (30 × 6) / 4

= 45.

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