Super Hard Critical Thinking Worksheet for 16-Year-Olds: English, Math, Science, Logic and Aptitude

Instructions

1. Treat this worksheet as an integrated investigation into a company designing a low-cost water-monitoring system.
2. Work independently unless your teacher or trainer allows collaboration.
3. Show every important step in mathematical, scientific, and logical problems. Unsupported answers may receive little credit.
4. For English questions, distinguish clearly between evidence, inference, assumption, and opinion.
5. Begin with the core questions, then complete the challenge questions. The challenge questions are intentionally demanding.
6. When a question has more than one defensible answer, support your answer with precise reasoning.
7. Suggested scoring: 5 points per core question, 8 points per challenge question. A complete solution should be more important than speed.

Useful reminders

• A valid argument can have a false conclusion if one of its premises is false.
• Correlation does not by itself prove causation.
• In probability, define the sample space before calculating.
• In optimization, test boundary points as well as intersection points.
• In scientific reasoning, identify variables, controls, sample size, and possible confounding factors.

Section 1: English—Critical Reading and Argument Analysis

Read the passage carefully.

The city should install algorithmic sensors in every public drinking fountain. The sensors would measure temperature, flow rate, and chemical indicators, then alert officials when contamination is suspected. Supporters claim that immediate warnings would prevent illness and reduce inspection costs. However, the proposal should not be approved in its current form. A sensor that detects an unusual chemical pattern does not necessarily identify a dangerous substance; false alarms could cause public panic, while missed alarms could create false confidence. In addition, the city has not explained who would audit the algorithm, how often the sensors would be calibrated, or what would happen to records showing where and when residents collected water. These questions are not arguments against technology itself. They are arguments for a limited pilot program with independent testing, published performance data, privacy protections, and a clear emergency protocol. If the pilot demonstrates that the sensors detect meaningful risks more accurately and more cheaply than existing inspections, expansion would be reasonable. If it does not, the city should reject the larger system rather than allowing its novelty to substitute for evidence.

1. State the author’s main claim in one precise sentence.

2. Identify two reasons the author gives for rejecting the proposal in its current form.

3. What important distinction does the author make between opposing the technology and opposing the proposal as currently designed?

4. Identify one unstated assumption made by supporters of the sensor system. Explain why it is an assumption rather than a proven fact.

5. Evaluate the reasoning. Does the author commit the fallacy of rejecting a technology merely because it is new? Use evidence from the passage.

6. Rewrite the supporters’ likely argument as a logical syllogism with two premises and a conclusion. Then identify one premise that requires evidence.

7. The sentence beginning with “If the pilot demonstrates...” is conditional. Explain exactly what evidence would make expansion reasonable and what evidence would make expansion unreasonable.

8. Challenge: Write a 150–200 word response arguing either for or against the pilot program. Your response must include a claim, two different kinds of evidence, one limitation or counterargument, and a rebuttal.

Section 2: Mathematics—Optimization, Probability, and Modeling

Problem 1: Production optimization

A company produces two water-monitoring kits, X and Y.

• Each X kit requires 3 hours of assembly and 2 units of testing material.
• Each Y kit requires 2 hours of assembly and 4 units of testing material.
• The company has at most 180 assembly hours and 240 units of testing material.
• Profit is $9 per X kit and $11 per Y kit.

1. Define variables and write all relevant inequalities.
2. Find every corner point of the feasible region.
3. Determine the number of each kit that maximizes profit.
4. Explain why checking only one constraint would produce an unreliable answer.

Problem 2: Conditional probability

A screening test is used to detect a contaminant.

• One percent of water samples actually contain the contaminant.
• The test correctly identifies contaminated samples 95% of the time.
• The test correctly gives a negative result for uncontaminated samples 90% of the time.

1. Find the probability that a randomly selected sample tests positive.
2. Find the probability that a sample actually contains the contaminant given that it tested positive.
3. Explain why the answer to part 2 is much lower than 95%.

Problem 3: Hypergeometric probability

A sealed box contains 7 functioning sensors and 3 defective sensors. Four sensors are selected without replacement.

1. Find the probability that exactly two selected sensors are defective.
2. Find the probability that at least one selected sensor is defective.
3. Explain why the binomial formula for independent trials is not the best model here.

Problem 4: Compound growth and decision-making

A training fund of $1,000 can be invested in either plan A or plan B.

• Plan A earns 6% annually, compounded once per year.
• Plan B earns 5.8% continuously.

1. Calculate the value of each plan after 10 years, rounded to the nearest cent.
2. Which plan produces more money after 10 years?
3. Determine approximately how many years must pass before Plan A overtakes Plan B, if it does overtake it.

Problem 5: Challenge—Constrained redesign

A rectangular testing enclosure has one side against an existing wall, so only the other three sides require fencing. Exactly 80 meters of fencing are available. Safety rules require the width to be at least 12 meters and no more than 30 meters.

1. Express the area as a function of the width.
2. Determine the dimensions that maximize area under the safety rule.
3. Explain why the unconstrained maximum must still be checked against the allowed interval.

Section 3: Science—Experimental Design and Evidence Evaluation

Problem 6: Identifying variables and confounding

A student investigates whether a new fertilizer increases plant growth. Group 1 receives fertilizer and is placed beside a sunny window. Group 2 receives no fertilizer and is placed in a darker room. After four weeks, Group 1 is taller.

1. Identify the independent variable the student intended to test.
2. Identify the dependent variable.
3. Identify the major confounding variable.
4. Design a fairer experiment. Include a control group, at least one controlled variable, and a method for improving reliability.

Problem 7: Interpreting experimental results

Four groups of plants are grown for the same length of time. Their average height increases are:

• Fertilizer and light: 18 cm
• Fertilizer and darkness: 5 cm
• Water only and light: 14 cm
• Water only and darkness: 4 cm

1. What conclusion about light is most strongly supported?
2. What conclusion about fertilizer is most strongly supported?
3. Does the evidence prove that fertilizer works equally well under all lighting conditions? Explain.
4. Name one additional piece of information needed before making a strong scientific claim.

Problem 8: Enzyme reasoning

An enzyme works best at pH 7. At pH 2, its reaction rate falls sharply; at pH 11, its reaction rate also falls sharply.

1. Explain why extreme pH values can reduce enzyme activity.
2. Predict what may happen if the enzyme is exposed to pH 2 for a very long time and then returned to pH 7.
3. Design a test that distinguishes temporary inhibition from permanent denaturation.

Problem 9: Base rates and medical testing

A disease affects 1% of a population. A test has 95% sensitivity and 90% specificity.

1. Using a hypothetical population of 10,000 people, calculate the expected number of true positives, false positives, true negatives, and false negatives.
2. Calculate the positive predictive value of the test.
3. Explain why a positive result should not automatically be treated as proof that the person has the disease.

Problem 10: Challenge—Correlation, causation, and policy

A city notices that neighborhoods with more water-monitoring sensors report more contamination incidents. A council member concludes that sensors are causing contamination.

1. Explain why the conclusion is not justified.
2. Give two plausible alternative explanations for the correlation.
3. Propose a study that could better test whether sensor installation changes the number of detected incidents. Address comparison groups, time period, and possible bias.

Section 4: Reasoning and Logic

Problem 11: Truth-value system

Five analysts—A, B, C, D, and E—make the following statements. Each statement is either true or false.

• A says: “B’s statement is false.”
• B says: “C’s statement is true.”
• C says: “D’s statement is false.”
• D says: “A and E have the same truth value.”
• E says: “Exactly two of A, B, C, and D are true.”

Determine the truth value of every statement. Show a chain of reasoning rather than guessing.

Problem 12: Scheduling under constraints

Five tasks—A, B, C, D, and E—must be scheduled once each from Monday to Friday.

• A must occur before C.
• B must occur immediately before D.
• E must occur on Wednesday.
• A cannot occur on Monday.
• C must occur on Friday.

Determine the only valid schedule and explain why each alternative placement fails.

Problem 13: Argument validity

Consider this argument:

1. If a monitoring system is reliable, then it is calibrated regularly.

2. This system is calibrated regularly.

3. Therefore, this system is reliable.

4. Is the argument deductively valid?

5. Name the logical error, if one exists.

6. Create a concrete example showing why the conclusion does not necessarily follow.

7. Rewrite the argument so that the conclusion follows validly, while keeping the first premise.

Problem 14: Ranking deduction

Five projects—P, Q, R, S, and T—are ranked from first to fifth.

• P ranks higher than Q.
• R ranks immediately below S.
• T ranks higher than P but lower than S.
• Q does not rank fifth.

Find the complete ranking. Prove that your answer is unique.

Problem 15: Challenge—Set-based reasoning

In a group of 120 trainees:

• seventy-five study mathematics,
• sixty study science,
• fifty study English,
• thirty-five study both mathematics and science,
• twenty-eight study both mathematics and English,
• twenty-two study both science and English,
• fifteen study all three subjects.

Assume the pairwise totals include those who study all three.

1. How many study at least one subject?
2. How many study exactly two subjects?
3. How many study exactly one subject?
4. How many study none of the three subjects?
5. Explain why subtracting the three pairwise totals directly from the subject totals would cause an error.

Section 5: Aptitude—Advanced Quantitative and Verbal Problem Solving

Problem 16: Work-rate analysis

Worker A can complete a task alone in 12 days. Worker B can complete it alone in 18 days.

1. What fraction of the task do they complete together in one day?
2. If both work for 3 days and A then leaves, how many additional days does B need?
3. Find the total time required.

Problem 17: Arrangements with repeated letters

The word ALGEBRA contains seven letters, with A repeated twice.

1. How many distinct arrangements are possible?
2. How many arrangements have the two A letters together?
3. How many arrangements have the two A letters separated?
4. Explain why treating the two A letters as distinct would overcount the answer.

Problem 18: Data sufficiency

For a positive integer n, determine whether n is divisible by 6.

Statement 1: n is divisible by 2.

Statement 2: n is divisible by 3.

Choose one conclusion and justify it:

• Statement 1 alone is sufficient.
• Statement 2 alone is sufficient.
• Both statements together are sufficient, but neither alone is sufficient.
• Each statement alone is sufficient.
• Even together, the statements are insufficient.

Problem 19: Cryptarithm

Each letter represents a different digit, and no number begins with zero.

Solve:

S E N D

• M O R E

  M O N E Y

Find the digits represented by all five letters and verify the addition numerically.

Problem 20: Challenge—Decision threshold

A company can either buy a sophisticated sensor or continue using a basic inspection method.

• The sophisticated sensor costs $18,000 and reduces the expected annual loss from undetected contamination from $30,000 to $9,000.
• The basic method has an annual operating cost of $4,000.
• The sophisticated sensor has an annual operating cost of $7,000.
• Ignore discounting and assume the risk estimates remain constant.

1. Calculate the annual cost of the basic method, including expected loss.
2. Calculate the annual cost of the sophisticated method, including expected loss.
3. Ignoring the purchase cost, calculate the annual savings from using the sophisticated sensor.
4. Calculate the payback period for the purchase cost.
5. Identify one nonfinancial factor that could rationally change the decision.

Answer Key

Section 1 Answers

1. The city should not install the sensor system throughout the city yet; it should first conduct a limited, independently evaluated pilot with privacy and emergency safeguards.

2. Possible answers include: unusual chemical patterns may create false alarms or missed alarms; the algorithm’s auditing process is unspecified; calibration procedures are unclear; privacy risks have not been addressed; and the city has not explained its emergency protocol.

3. The author is not claiming that all sensor technology is harmful or useless. The author is claiming that the current proposal lacks sufficient testing, oversight, privacy protection, and operational planning.

4. One assumption is that detecting an unusual chemical pattern will reliably identify a dangerous contaminant. This is an assumption because the passage provides no performance data proving that the pattern corresponds accurately to actual danger.

5. No. The author does not reject the technology because it is new. The author proposes a pilot program and says expansion would be reasonable if evidence shows that the system is accurate and cost-effective. The argument is evidence-based rather than anti-technology.

6. A possible syllogism is: Premise 1—If sensors provide rapid and accurate warnings, they can reduce illness and inspection costs. Premise 2—The proposed sensors provide rapid and accurate warnings. Conclusion—The city should install them. The second premise requires evidence because the passage says accuracy has not yet been demonstrated.

7. Expansion would be reasonable if the pilot shows that the sensors detect meaningful risks accurately and at lower cost than current inspections. Expansion would be unreasonable if the pilot shows poor accuracy, excessive false alarms, unacceptable cost, inadequate privacy protection, or no meaningful improvement.

8. Answers will vary. A strong answer must contain a clear position, two relevant and different forms of support, a limitation or counterargument, and a logical rebuttal. A model position is that the pilot should proceed because it limits risk while producing evidence, but only if independent auditing and privacy controls are mandatory.

Section 2 Answers

1. Let x be the number of X kits and y be the number of Y kits. The constraints are 3x + 2y <= 180, 2x + 4y <= 240, x >= 0, and y >= 0. Profit is P = 9x + 11y.

2. Rewrite the second constraint as x + 2y <= 120. The corner points are (0, 0), (60, 0), (0, 60), and the intersection of 3x + 2y = 180 and x + 2y = 120. Subtracting gives 2x = 60, so x = 30; then y = 45. The intersection is (30, 45).

3. Profits are: (0, 0) = $0; (60, 0) = $540; (0, 60) = $660; (30, 45) = $765. The maximum occurs at 30 X kits and 45 Y kits, producing a profit of $765.

4. Both assembly time and testing material limit production. A solution that satisfies only one constraint may violate the other, making it impossible to produce.

5. The probability of a positive result is (0.01)(0.95) + (0.99)(0.10) = 0.0095 + 0.099 = 0.1085, or 10.85%.

6. The probability of actual contamination given a positive result is 0.0095 / 0.1085 = approximately 0.0876, or 8.76%.

7. The disease is rare, so the large number of uncontaminated samples creates many false positives. Although the test is sensitive, a positive result includes both true positives and false positives.

8. Exactly two defective sensors: C(3,2)C(7,2) / C(10,4) = (3 x 21) / 210 = 63/210 = 3/10.

9. The probability of at least one defective sensor is 1 minus the probability of selecting four functioning sensors: 1 - C(7,4)/C(10,4) = 1 - 35/210 = 175/210 = 5/6, approximately 83.33%.

10. The selections are without replacement, so the probability changes after each selection. The trials are not independent, which is an assumption of the ordinary binomial model.

11. Plan A: 1000(1.06)^10 = approximately $1,790.85. Plan B: 1000e^(0.058 x 10) = approximately $1,786.65. Plan A is higher after 10 years.

12. Set 1000(1.06)^t = 1000e^(0.058t). Taking natural logarithms gives t ln(1.06) = 0.058t. Since ln(1.06) is approximately 0.058268, Plan A is slightly better for every positive t under this model. There is no positive crossover time; Plan A begins slightly ahead and remains ahead.

13. If the width is w, the two widths use 2w meters, leaving 80 - 2w meters for the length. Thus A(w) = w(80 - 2w) = 80w - 2w^2.

14. The unconstrained vertex occurs at w = -b/(2a) = -80/(2(-2)) = 20. The length is 80 - 40 = 40, so the maximum area is 800 square meters. Since 20 is between 12 and 30, the constrained maximum is 20 meters by 40 meters.

15. The unconstrained maximum must be checked because safety restrictions can exclude the vertex. If the vertex were outside the allowed interval, the maximum would occur at one of the interval endpoints instead.

Section 3 Answers

16. The intended independent variable is fertilizer use. The dependent variable is plant growth or final height. The major confounding variable is light exposure. A fair experiment would randomly assign similar plants to fertilizer and no-fertilizer groups while keeping light, water, soil, temperature, pot size, and time constant. Several plants should be used in each group, and the experiment should be repeated.

17. Light appears to increase growth because both light groups are taller than their corresponding dark groups: 18 versus 5 with fertilizer, and 14 versus 4 with water only.

18. Fertilizer appears to increase growth because the fertilizer group is taller under both lighting conditions: 18 versus 14 in light, and 5 versus 4 in darkness. The effect is larger in light, but the difference in darkness is small.

19. No. The data suggest that fertilizer has different-sized effects under different lighting conditions, but they do not prove that it works equally or unequally in all conditions. Replication, variation, sample sizes, and statistical analysis are needed.

20. Useful additional information includes the number of plants in each group, individual measurements, variation or standard deviation, the number of repeated trials, and whether the groups were randomly assigned.

21. Extreme pH can change the charges and shapes of amino acids in the enzyme. This can alter the active site so the substrate no longer binds effectively. Very extreme conditions can disrupt the enzyme’s three-dimensional structure.

22. After brief exposure, activity may return if the pH change caused temporary inhibition. After prolonged exposure, the enzyme may remain inactive because it has been permanently denatured.

23. Expose equal enzyme samples to pH 2 for different lengths of time, return all samples to pH 7, and measure their reaction rates against an untreated control. Recovery of activity indicates temporary inhibition; failure to recover indicates likely permanent denaturation.

24. In 10,000 people, 100 have the disease. Of those, 95 are true positives and 5 are false negatives. Of the 9,900 without the disease, 90% or 8,910 are true negatives and 10% or 990 are false positives.

25. The positive predictive value is 95 / (95 + 990) = 95/1085 = approximately 8.76%.

26. A positive result is not proof because false positives outnumber true positives in this low-prevalence population. The result should usually be confirmed with additional testing and interpreted in context.

27. The council’s conclusion is not justified because neighborhoods with more sensors may receive more scrutiny, have larger populations, or have had more suspected contamination before installation. Sensors may improve detection rather than cause contamination.

28. Alternative explanations include: sensor-rich neighborhoods may already have higher contamination risk; inspectors may investigate more often where sensors exist; or residents in those neighborhoods may report incidents more frequently.

29. A stronger study could randomly select comparable neighborhoods for early installation and delayed installation, measure incidents for an equal period before and after installation, use the same reporting rules, and adjust for population, industrial activity, weather, and inspection frequency. Researchers should distinguish actual contamination from detected or reported contamination.

Section 4 Answers

30. The truth values are A true, B false, C false, D true, and E true. Reasoning: If D were false, C would be true, B would be true, and A would be false. D’s statement would then require A and E to have the same value, forcing E to be false, but exactly two of A through D would still be true, creating a contradiction. Therefore D is true; C is false; B is false; A is true; and exactly two of A through D are true, so E is true.

31. The only schedule is Monday B, Tuesday D, Wednesday E, Thursday A, and Friday C. B and D must be consecutive. They cannot occupy Tuesday-Wednesday or Wednesday-Thursday because E must be on Wednesday. Therefore B-D must be Monday-Tuesday. C is fixed on Friday, and A must occur before C but cannot be Monday, leaving Thursday.

32. The argument is invalid. It commits the fallacy of affirming the consequent. From reliable -> calibrated and calibrated, we cannot conclude reliable; another unreliable system could also be calibrated regularly.

33. Example: All certified laboratories calibrate their thermometers regularly, but a poorly designed laboratory may also calibrate its thermometer regularly and still produce unreliable results. Calibration alone does not guarantee total reliability.

34. A valid rewrite is: If a monitoring system is reliable, then it is calibrated regularly. This system is reliable. Therefore, this system is calibrated regularly. This uses the valid form of affirming the antecedent.

35. The ranking is S, T, P, Q, R. Since R is immediately below S, the pair S-R must be consecutive. T must be below S and above P, while P is above Q. Testing the possible positions shows that S must be first, T second, P third, Q fourth, and R fifth. The conditions eliminate every other arrangement.

36. At least one subject = 75 + 60 + 50 - 35 - 28 - 22 + 15 = 115.

37. Exactly two subjects = (35 - 15) + (28 - 15) + (22 - 15) = 20 + 13 + 7 = 40.

38. Exactly one subject = (75 - 35 - 28 + 15) + (60 - 35 - 22 + 15) + (50 - 28 - 22 + 15) = 27 + 18 + 15 = 60.

39. None = 120 - 115 = 5.

40. Direct subtraction double-counts or miscounts people in all three groups. Someone in all three is included in every subject total and every pairwise overlap, so inclusion-exclusion is needed to correct the repeated counting.

Section 5 Answers

41. A completes 1/12 of the task per day and B completes 1/18. Together they complete 1/12 + 1/18 = 5/36 of the task per day.

42. In 3 days they complete 3(5/36) = 15/36 = 5/12. The remaining work is 7/12. B needs (7/12) / (1/18) = 10.5 additional days.

43. Total time = 3 + 10.5 = 13.5 days.

44. The total number of distinct arrangements is 7!/2! = 2,520.

45. Treating AA as one block gives 6! = 720 arrangements.

46. Separated arrangements = 2,520 - 720 = 1,800.

47. If the two As were treated as different, every arrangement would be counted twice: once with the first A in a position and once with the second A in that position. Dividing by 2! corrects this overcount.

48. Both statements together are sufficient, but neither alone is sufficient. Divisibility by 6 requires divisibility by both 2 and 3. A number divisible by 2 need not be divisible by 3, and a number divisible by 3 need not be divisible by 2.

49. The solution is S = 9, E = 5, N = 6, D = 7, M = 1, O = 0, R = 8, and Y = 2. Verification: 9567 + 1085 = 10652. Therefore SEND = 9567, MORE = 1085, and MONEY = 10652.

50. Basic method annual cost = $4,000 + $30,000 = $34,000.

51. Sophisticated method annual cost = $7,000 + $9,000 = $16,000, excluding the one-time purchase cost.

52. Annual savings = $34,000 - $16,000 = $18,000.

53. Payback period = $18,000 / $18,000 per year = 1 year.

54. Possible nonfinancial factors include public safety, the consequences of a missed contamination event, privacy, reliability during emergencies, public trust, maintenance complexity, and the availability of trained staff. Any such factor could justify choosing the sensor even if the financial advantage were smaller.