Question

Which of the following MUST be true?

• A. The contamination level at P10 was recorded as high.
• B. The contamination level at P13 was recorded as low.
• C. The contamination level at P20 was recorded as medium.
• D. The contamination level at P12 was recorded as high.
• A. Exactly 8
• B. More than 4
• C. At least 8
• D. At most 9
• A. The contamination level at P12 was recorded as high.
• B. The contamination level at P14 was recorded as medium.
• C. The contamination level at P15 was recorded as medium.
• D. The contamination level at P18 was recorded as low.
• A. Contamination level at P14 was recorded to be higher than that at P15.
• B. Contamination levels at P10 and P14 were recorded as the same.
• C. Contamination levels at P13 and P17 were recorded as the same.
• D. Contamination levels at P11 and P16 were recorded as the same.

Answer 1

The Correct Option is A

To determine which statement must be true, let's analyze the given conditions: 

Pump	Contamination Level
P1 - P5	3 High
P6	Low
P7-P8	Same Level
P16-P20	No High

1. The number of pumps with high contamination levels is twice the number with low levels.

Let's consider:

• From condition 1, three pumps among P1-P5 are recorded as high. Thus, P1-P5 has at least three high contamination levels.
• P6 is the only pump among P1-P10 with low contamination, implying P7-P10 cannot be low.
• P7 and P8 must have the same level, and since P6 is low, P7 and P8 are either both high or both medium.
• Given P16-P20 have no high levels, these must be medium or low.
• The total number of high contamination is twice the total number of low ones.

We need to accommodate the constraints, especially that P6 is the only low among P1-P10, and P16-P20 can't be high:

1. If we have 4 low (P6 + 3 others), then there are 8 highs. P16-P20 must all be medium because 3 lows occupy remaining pumps in P1-P15.
2. In this arrangement, P10 being high satisfies 3 highs in P1-P5 and the overall high level count.

Conclusion: Given this breakdown, The contamination level at P10 was recorded as high must be true. It fits all conditions without contradiction.

Answer 2

To solve the problem, we need to determine the number of pumps at which contamination levels were recorded as medium. Let's use the given information to deduce the number of pumps with each contamination level:

There are 20 pumps in total: P1 to P20. 

From statement 1, three of the pumps (P1-P5) have high contamination.

From statement 2, P6 is the only pump from P1 to P10 with low contamination, leaving the rest with either high or medium contamination.

Statement 3 indicates P7 and P8 have the same contamination level; since P6 is low and not adjacent to P7 and P8, these two must be either high or medium. Based on statement 1, neither can be high, so both must be medium.

Statement 4 says no high contamination at P16-P20.

Statement 5 states that high contamination pumps are twice the number of low contamination pumps. Therefore, let x be the number of low contamination pumps. Then, there are 2x high contamination pumps. Only one low contamination pump is given (P6), so x=1, making high contamination pumps 2x=2.

Based on the information:

High contamination pumps: 2 from P1-P5 and undistributed elsewhere due to limitation at P16-P20.

Low contamination pump: P6.

From the calculation above, we have identified the following contamination distribution: 2 high, 1 low (P6), and two medium (P7 and P8). Therefore, the remaining ambiguity should be calculated: Medium contamination pumps = Total pumps - High pumps - Low pumps = 20 - 6 - 2 = 8 medium contamination pumps.

Thus, the number of pumps at which contamination levels were recorded as medium is exactly 8.

Answer 3

To determine which statement must be true if the contamination level at P11 was recorded as low, we will analyze the given conditions one by one: 

1. Contamination levels at three pumps among P1 – P5 were recorded as high.
   Since exact pumps are unspecified, there could be any distribution of high contamination across these pumps.
2. P6 was the only pump among P1 – P10 where the contamination level was recorded as low.
   This means P11 can have a low contamination level without contradiction, as it is not within P1–P10 range.
3. P7 and P8 were the only two consecutively numbered pumps where the same levels of contamination were recorded.
4. High contamination levels were not recorded at any of the pumps P16 – P20.
   This confines high contamination occurrences to P1–P15 range.
5. The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.
   This implies the equation: 2n_high = n_low.

Assuming P11 has a low contamination level, let's examine how this affects the rest, especially P14:

• If P11 is low, and with P6 already low, there's a need to check the balance of the high contamination levels being twice that of low.
• We know P6 is low, and P11 was just assumed low, thus we have 2 low recordings now. To maintain condition (5), there should be 2*2 = 4 high recordings.
• Considering no high contamination in P16-P20 and limited high possibilities in P1–P5, this somewhat affects middle pumps like P14.
• The condition at P14 becomes more apparent in satisfying this condition since P14 can't be high (too few unknowns); medium becomes a logical placement when adhering to constraints and balancing conditions.

Hence, the statement that must be true is:

The contamination level at P14 was recorded as medium.

Answer 4

To solve this problem, we must determine which statement MUST be FALSE given that the contamination level at P15 is recorded as medium.

The constraints given are:

1. High levels at 3 pumps among P1 – P5. 
2. P6 is the only low level among P1 – P10.
3. P7 and P8 have the same levels.
4. No high levels at P16 – P20.
5. High levels are twice the low levels.

Let's examine the options:

1. Contamination level at P14 was recorded to be higher than that at P15.
   If P15 is medium, P14 cannot be high (as it would contradict no high levels in P16-P20) and must be low or medium, not necessarily false.
2. Contamination levels at P10 and P14 were recorded as the same.
   If both are the same, they can be either medium or low. The specific levels are undetermined, meaning it is not a definitive contradiction.
3. Contamination levels at P13 and P17 were recorded as the same.
   Possible if both are medium or low, it does not create a definitive contradiction.
4. Contamination levels at P11 and P16 were recorded as the same.
   If P11 is medium or high (to satisfy the twice high over low condition), and since no high levels can be at P16 and P20, P16 must be low. Hence, they cannot be the same. Therefore, this must be FALSE based on the given information.

Thus, the only statement that cannot be true is the contamination levels at P11 and P16 were recorded as the same.