Comprehension
On each of the three consecutive days Monday through Wednesday, exactly two employees are to staff a company's information booth. The three available employees — Feng, Gómez, and Hull— will staff the booth in accordance with the following conditions:
Gómez and Hull must each staff the booth on at least one of the days, but Feng must staff it on at least two the days.
The booth cannot be staffed by the same two employees on any two consecutive days.
If Hull staffs the booth on Monday, Gómez must be the other employee staffing the booth on Monday.
Gómez and Hull must each staff the booth on at least one of the days, but Feng must staff it on at least two the days.
The booth cannot be staffed by the same two employees on any two consecutive days.
If Hull staffs the booth on Monday, Gómez must be the other employee staffing the booth on Monday.
Question: 1
Which of the following can be the schedule of employees staffing the booth on the three days?
Monday | Tuesday | Wednesday
Which of the following can be the schedule of employees staffing the booth on the three days?
Monday | Tuesday | Wednesday
Monday | Tuesday | Wednesday
Show Hint
When checking a full schedule, start with the most restrictive rules. Here, checking for consecutive repeats (Rule 2) or the Monday conditional (Rule 3) can often eliminate options faster than counting frequencies.
Updated On: Oct 4, 2025
- Feng,Gómez | Feng,Gómez | Feng,Hull
- Feng,Gómez | Feng,Hull | Gómez,Hull
- Feng,Hull | Feng,Gómez | Gómez,Hull
- Gómez,Hull | Feng,Gómez | Gómez,Hull
- Gómez,Hull | Feng,Gómez | Feng,Hull
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
This question asks for a valid, complete schedule. We need to check each option against all the rules.
Step 2: Key Formula or Approach:
Go through each option and test it against the Frequency, Consecutive Pairs, and Conditional rules. The first one that passes all tests is the answer.
Step 3: Detailed Explanation:
Let's check each option:
- (A) {F,G | {F,G} | {F,H}:} This violates Rule 2, as the same pair ({F,G}) staffs the booth on two consecutive days (Monday and Tuesday).
- (B) {F,G | {F,H} | {G,H}:} - \textit{Frequency:} F works twice, G works twice, H works twice. This satisfies F\(\geq\)2, G\(\geq\)1, H\(\geq\)1. (OK)
- \textit{Consecutive Pairs:} {F,G} \(\neq\) {F,H}. {F,H} \(\neq\) {G,H}. (OK)
- \textit{Conditional:} H does not work on Monday, so the rule does not apply. (OK)
This schedule is valid.
- (C) {F,H | {F,G} | {G,H}:} This violates Rule 3. H works on Monday, so G must also work on Monday. The Monday pair must be {G,H}, but here it is {F,H}.
- (D) {G,H | {F,G} | {G,H}:} Let's check the frequency count. F works once. This violates Rule 1 (F\(\geq\)2).
- (E) {G,H | {F,G} | {F,H}:} This schedule is also a valid possibility (F=2, G=2, H=2; Mon is GH; no consecutive repeats). In a standard test, there should only be one correct answer. Re-checking (B), there are no violations. Both B and E appear valid based on the rules as stated, suggesting a potential ambiguity in the question, but (B) is a confirmed valid schedule.
Step 4: Final Answer:
Option (B) represents a schedule that satisfies all the given conditions.
This question asks for a valid, complete schedule. We need to check each option against all the rules.
Step 2: Key Formula or Approach:
Go through each option and test it against the Frequency, Consecutive Pairs, and Conditional rules. The first one that passes all tests is the answer.
Step 3: Detailed Explanation:
Let's check each option:
- (A) {F,G | {F,G} | {F,H}:} This violates Rule 2, as the same pair ({F,G}) staffs the booth on two consecutive days (Monday and Tuesday).
- (B) {F,G | {F,H} | {G,H}:} - \textit{Frequency:} F works twice, G works twice, H works twice. This satisfies F\(\geq\)2, G\(\geq\)1, H\(\geq\)1. (OK)
- \textit{Consecutive Pairs:} {F,G} \(\neq\) {F,H}. {F,H} \(\neq\) {G,H}. (OK)
- \textit{Conditional:} H does not work on Monday, so the rule does not apply. (OK)
This schedule is valid.
- (C) {F,H | {F,G} | {G,H}:} This violates Rule 3. H works on Monday, so G must also work on Monday. The Monday pair must be {G,H}, but here it is {F,H}.
- (D) {G,H | {F,G} | {G,H}:} Let's check the frequency count. F works once. This violates Rule 1 (F\(\geq\)2).
- (E) {G,H | {F,G} | {F,H}:} This schedule is also a valid possibility (F=2, G=2, H=2; Mon is GH; no consecutive repeats). In a standard test, there should only be one correct answer. Re-checking (B), there are no violations. Both B and E appear valid based on the rules as stated, suggesting a potential ambiguity in the question, but (B) is a confirmed valid schedule.
Step 4: Final Answer:
Option (B) represents a schedule that satisfies all the given conditions.
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Question: 2
If Gómez staffs the booth on Monday and Tuesday, which of the following must be true?
If Gómez staffs the booth on Monday and Tuesday, which of the following must be true?
Show Hint
For "must be true" questions, always look to the most restrictive rules. The frequency requirement (F\(\geq\)2) is a powerful constraint that often determines the answer.
Updated On: Oct 4, 2025
- Feng staffs the booth on Monday.
- Feng staffs the booth on Tuesday.
- Feng staffs the booth on Wednesday.
- Hull staffs the booth on Monday.
- Hull staffs the booth on Tuesday.
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
We are given a new condition and must determine what necessarily follows from it.
Step 2: Key Formula or Approach:
Start with the new condition and the fixed rules, and deduce the consequences. The frequency rule for Feng (F\(\geq\)2) will be critical.
Step 3: Detailed Explanation:
1. Gómez (G) works on Monday and Tuesday. This uses up two of the six total employee slots.
2. The pairs on Monday and Tuesday cannot be the same (Rule 2). So, G works with two different people on those days. The pairs must be {G,F} and {G,H} in some order.
3. Hull (H) must work at least once, and Feng (F) must work at least twice (Rule 1).
4. In any scenario where G works on Monday and Tuesday, the other slots on those days must be filled by F and H. For example, M:{G,F} and T:{G,H}. In this case, F has worked once and H has worked once.
5. Now consider Wednesday. Feng (F) must work at least one more time to satisfy the F\(\geq\)2 rule. Therefore, F must be one of the two employees on Wednesday.
6. The two possible sets of pairs for M/T are ({G,F}, {G,H}) or ({G,H}, {G,F}). In both cases, F has worked only once by the end of Tuesday. To meet the requirement of working at least two days, F absolutely must work on Wednesday.
Let's check the options:
- (A) Feng staffs on Monday: Possible (if M={G,F}), but not a must (if M={G,H}).
- (B) Feng staffs on Tuesday: Possible (if T={G,F}), but not a must (if T={G,H}).
- (C) Feng staffs on Wednesday: As deduced above, this is necessary to fulfill F's \(\geq\)2 day requirement. This must be true.
- (D) Hull staffs on Monday: Possible, but not a must.
- (E) Hull staffs on Tuesday: Possible, but not a must.
Step 4: Final Answer:
Given that G works Monday and Tuesday with different partners (F and H), F has only worked one day. To meet the minimum requirement of two days, F must work on Wednesday.
We are given a new condition and must determine what necessarily follows from it.
Step 2: Key Formula or Approach:
Start with the new condition and the fixed rules, and deduce the consequences. The frequency rule for Feng (F\(\geq\)2) will be critical.
Step 3: Detailed Explanation:
1. Gómez (G) works on Monday and Tuesday. This uses up two of the six total employee slots.
2. The pairs on Monday and Tuesday cannot be the same (Rule 2). So, G works with two different people on those days. The pairs must be {G,F} and {G,H} in some order.
3. Hull (H) must work at least once, and Feng (F) must work at least twice (Rule 1).
4. In any scenario where G works on Monday and Tuesday, the other slots on those days must be filled by F and H. For example, M:{G,F} and T:{G,H}. In this case, F has worked once and H has worked once.
5. Now consider Wednesday. Feng (F) must work at least one more time to satisfy the F\(\geq\)2 rule. Therefore, F must be one of the two employees on Wednesday.
6. The two possible sets of pairs for M/T are ({G,F}, {G,H}) or ({G,H}, {G,F}). In both cases, F has worked only once by the end of Tuesday. To meet the requirement of working at least two days, F absolutely must work on Wednesday.
Let's check the options:
- (A) Feng staffs on Monday: Possible (if M={G,F}), but not a must (if M={G,H}).
- (B) Feng staffs on Tuesday: Possible (if T={G,F}), but not a must (if T={G,H}).
- (C) Feng staffs on Wednesday: As deduced above, this is necessary to fulfill F's \(\geq\)2 day requirement. This must be true.
- (D) Hull staffs on Monday: Possible, but not a must.
- (E) Hull staffs on Tuesday: Possible, but not a must.
Step 4: Final Answer:
Given that G works Monday and Tuesday with different partners (F and H), F has only worked one day. To meet the minimum requirement of two days, F must work on Wednesday.
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Question: 3
If Hull staffs the booth on Monday and Wednesday, which of the following must be true?
If Hull staffs the booth on Monday and Wednesday, which of the following must be true?
Show Hint
Conditional rules (If... then...) are very powerful. The condition about H on Monday immediately fixes the Monday pair, which then creates a cascade of deductions, often locking the whole schedule into place.
Updated On: Oct 4, 2025
- Feng and Gómez staff the booth on Tuesday.
- Feng and Hull staff the booth on Monday.
- Feng and Hull staff the booth on Tuesday.
- Gómez and Hull staff the booth on Tuesday.
- Gómez and Hull staff the booth on Wednesday.
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Given a new condition about Hull's schedule, we must determine a necessary consequence for the overall schedule.
Step 2: Key Formula or Approach:
Apply the new condition and follow the chain of deductions from the original rules.
Step 3: Detailed Explanation:
1. Hull (H) works on Monday. According to Rule 3, this means Gómez (G) must also work on Monday. So, the Monday pair is {G,H}.
2. We now have the partial schedule: M:{G,H}, T:{?,?}, W:{H,?}.
3. Feng (F) must work at least two days (Rule 1). Since F is not on Monday, F must work on both Tuesday and Wednesday.
4. This determines one person for Tuesday (F) and the second person for Wednesday (F). The schedule is now: M:{G,H}, T:{F,?}, W:{H,F}.
5. Now we must determine the second person for Tuesday. Let's check the total days worked so far: H=2, F=2, G=1. This satisfies all frequency rules. The last open slot is Tuesday's second person. The only person available is G.
6. So, the Tuesday pair must be {F,G}.
7. The complete, unique schedule is: M:{G,H}, T:{F,G}, W:{F,H}. Let's double check it. Freq: F=2, G=2, H=2 (OK). Consec: {G,H}\(\neq\){F,G}, {F,G}\(\neq\){F,H} (OK). Conditional: H on Mon, G is there (OK). The schedule is valid.
8. Based on this fixed schedule, let's evaluate the options:
- (A) Feng and Gómez staff the booth on Tuesday. Our schedule shows T:{F,G}. This must be true.
- (B) Feng and Hull staff the booth on Monday. False, it's {G,H}.
- (C) Feng and Hull staff the booth on Tuesday. False, it's {F,G}.
- (D) Gómez and Hull staff the booth on Tuesday. False, it's {F,G}.
- (E) Gómez and Hull staff the booth on Wednesday. False, it's {F,H}.
Step 4: Final Answer:
The initial condition forces a unique schedule where Feng and Gómez must staff the booth together on Tuesday.
Given a new condition about Hull's schedule, we must determine a necessary consequence for the overall schedule.
Step 2: Key Formula or Approach:
Apply the new condition and follow the chain of deductions from the original rules.
Step 3: Detailed Explanation:
1. Hull (H) works on Monday. According to Rule 3, this means Gómez (G) must also work on Monday. So, the Monday pair is {G,H}.
2. We now have the partial schedule: M:{G,H}, T:{?,?}, W:{H,?}.
3. Feng (F) must work at least two days (Rule 1). Since F is not on Monday, F must work on both Tuesday and Wednesday.
4. This determines one person for Tuesday (F) and the second person for Wednesday (F). The schedule is now: M:{G,H}, T:{F,?}, W:{H,F}.
5. Now we must determine the second person for Tuesday. Let's check the total days worked so far: H=2, F=2, G=1. This satisfies all frequency rules. The last open slot is Tuesday's second person. The only person available is G.
6. So, the Tuesday pair must be {F,G}.
7. The complete, unique schedule is: M:{G,H}, T:{F,G}, W:{F,H}. Let's double check it. Freq: F=2, G=2, H=2 (OK). Consec: {G,H}\(\neq\){F,G}, {F,G}\(\neq\){F,H} (OK). Conditional: H on Mon, G is there (OK). The schedule is valid.
8. Based on this fixed schedule, let's evaluate the options:
- (A) Feng and Gómez staff the booth on Tuesday. Our schedule shows T:{F,G}. This must be true.
- (B) Feng and Hull staff the booth on Monday. False, it's {G,H}.
- (C) Feng and Hull staff the booth on Tuesday. False, it's {F,G}.
- (D) Gómez and Hull staff the booth on Tuesday. False, it's {F,G}.
- (E) Gómez and Hull staff the booth on Wednesday. False, it's {F,H}.
Step 4: Final Answer:
The initial condition forces a unique schedule where Feng and Gómez must staff the booth together on Tuesday.
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Question: 4
If Hull staffs the booth on only one of the days, which of the following can be true?
If Hull staffs the booth on only one of the days, which of the following can be true?
Show Hint
For "can be true" questions with complex initial conditions, try to construct just one valid example. Don't worry about finding all possibilities. If your example makes one of the answer choices true, you have found your answer.
Updated On: Oct 4, 2025
- Feng and Hull staff the booth on Monday.
- Feng and Hull staff the booth on Wednesday.
- Gómez and Hull staff the booth on Monday.
- Gómez and Hull staff the booth on Tuesday.
- Gómez and Hull staff the booth on Wednesday.
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
We are given a new condition about the total number of days H works. We need to find a possible arrangement that is consistent with this condition.
Step 2: Key Formula or Approach:
First, determine the distribution of days worked among the employees given the new constraint. Then, build a valid schedule based on that distribution and test the options.
Step 3: Detailed Explanation:
1. Hull (H) works exactly one day.
2. The total number of employee slots is 6. We know F\(\geq\)2 and G\(\geq\)1. With H=1, the remaining 5 slots must be filled by F and G.
3. The only possible distributions are (F=4, G=1), (F=3, G=2), or (F=2, G=3). Since no employee can work on more than 3 days, F=4 is impossible. So the distributions must be (F=3, G=2, H=1) or (F=2, G=3, H=1).
4. Let's try to build a schedule. Let's test the distribution F=2, G=3, H=1. This means G works every day.
- Schedule framework: M:{G,?}, T:{G,?}, W:{G,?}.
- The remaining three slots must be filled by F, F, and H.
- Let's place them: M:{G,F}, T:{G,H}, W:{G,F}.
- Let's check this schedule: - \textit{Frequency:} F=2, G=3, H=1. (OK)
- \textit{Conditional:} H is not on Monday. (OK)
- \textit{Consecutive Pairs:} {G,F} \(\neq\) {G,H}. {G,H} \(\neq\) {G,F}. (OK)
- This is a valid schedule. Now let's see which option it makes true.
- (A) F and H staff Mon? No, {G,F}.
- (B) F and H staff Wed? No, {G,F}.
- (C) G and H staff Mon? No, {G,F}.
- (D) G and H staff Tue? Yes, our valid schedule has {G,H} on Tuesday. This can be true.
- (E) G and H staff Wed? No, {G,F}.
Step 4: Final Answer:
We found a valid scenario (M:{G,F}, T:{G,H}, W:{G,F}) consistent with Hull working only one day. In this scenario, Gómez and Hull staff the booth on Tuesday. Therefore, this statement "can be true."
We are given a new condition about the total number of days H works. We need to find a possible arrangement that is consistent with this condition.
Step 2: Key Formula or Approach:
First, determine the distribution of days worked among the employees given the new constraint. Then, build a valid schedule based on that distribution and test the options.
Step 3: Detailed Explanation:
1. Hull (H) works exactly one day.
2. The total number of employee slots is 6. We know F\(\geq\)2 and G\(\geq\)1. With H=1, the remaining 5 slots must be filled by F and G.
3. The only possible distributions are (F=4, G=1), (F=3, G=2), or (F=2, G=3). Since no employee can work on more than 3 days, F=4 is impossible. So the distributions must be (F=3, G=2, H=1) or (F=2, G=3, H=1).
4. Let's try to build a schedule. Let's test the distribution F=2, G=3, H=1. This means G works every day.
- Schedule framework: M:{G,?}, T:{G,?}, W:{G,?}.
- The remaining three slots must be filled by F, F, and H.
- Let's place them: M:{G,F}, T:{G,H}, W:{G,F}.
- Let's check this schedule: - \textit{Frequency:} F=2, G=3, H=1. (OK)
- \textit{Conditional:} H is not on Monday. (OK)
- \textit{Consecutive Pairs:} {G,F} \(\neq\) {G,H}. {G,H} \(\neq\) {G,F}. (OK)
- This is a valid schedule. Now let's see which option it makes true.
- (A) F and H staff Mon? No, {G,F}.
- (B) F and H staff Wed? No, {G,F}.
- (C) G and H staff Mon? No, {G,F}.
- (D) G and H staff Tue? Yes, our valid schedule has {G,H} on Tuesday. This can be true.
- (E) G and H staff Wed? No, {G,F}.
Step 4: Final Answer:
We found a valid scenario (M:{G,F}, T:{G,H}, W:{G,F}) consistent with Hull working only one day. In this scenario, Gómez and Hull staff the booth on Tuesday. Therefore, this statement "can be true."
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