Datasets:

Asilarkness
/

DiffuRefill-SFT

+{"prompt": "Question: A medical coding system assigns a unique 4-digit code to each of 200 identified jaw abnormality cases. The codes range from 0001 to 0200. A code is considered \"valid for reimbursement\" if the sum of its digits is divisible by 3. How many of the 200 codes satisfy this validity condition?\nAnswer:", "response": " A number's digit sum is divisible by 3 if and only if the number is divisible by 3.\nThe codes 0001 to 0200 correspond to the integers 1 to 200.\nWe count the multiples of 3 in the range 1 to 200.\nThe multiples are 3, 6, 9, ..., 198.\nThe count is given by floor(200 / 3).\n200 / 3 = 66.66...\nfloor(66.66...) = 66.\n#### 66", "source": "t38b_probability", "level": "general", "gv": 2}
+{"prompt": "Question: A software team is scheduling bug fixes for four modules (Alpha, Beta, Gamma, Delta). Each fix takes exactly 1 hour. The following dependencies exist: Beta cannot be fixed until Alpha is finished. Gamma cannot be fixed until both Beta and Delta are finished. There are no dependencies for Delta or Alpha. The team has 2 developers working in parallel, and they can only work on one module at a time. What is the minimum number of hours required to fix all four modules?\nAnswer:", "response": " The dependency chain Alpha → Beta → Gamma requires 3 sequential hours.\nTwo developers can fix Alpha and Delta in hour 1.\nBeta can be fixed in hour 2, after Alpha is complete.\nGamma can be fixed in hour 3, after Beta and Delta are complete.\nThe total time is 3 hours.\n\n#### 3", "source": "t38b_planning", "level": "general", "gv": 2}
+{"prompt": "Question: Five distinct sacred rivers—A, B, C, D, and E—are arranged in a specific order along the banks of the Ganges from upstream to downstream. The clues are as follows: River C flows immediately after River A. River D is located at the very end (downstream) of the sequence. River B is not immediately adjacent to River C. River E is the first river in the sequence. What is the complete correct order of the five rivers from upstream to downstream?\nAnswer:", "response": " Positions 1 through 5 are assigned to rivers E, A, B, C, D.\nPosition 1 is E.\nPosition 5 is D.\nThe remaining positions 2, 3, and 4 must be filled by A, B, and C.\nRiver C flows immediately after River A, so the pair (A, C) occupies two consecutive positions.\nThe possible consecutive positions for (A, C) are (2, 3) or (3, 4).\nIf A is at position 2 and C is at position 3, then B is at position 4.\nIn this case, B at position 4 is immediately adjacent to C at position 3, which violates the clue that B is not immediately adjacent to C.\nTherefore, A cannot be at position 2.\nIf A is at position 3 and C is at position 4, then B is at position 2.\nIn this case, B at position 2 is not immediately adjacent to C at position 4, satisfying the clue.\nThe sequence is E at position 1, B at position 2, A at position 3, C at position 4, and D at position 5.\nThe complete order from upstream to downstream is E, B, A, C, D.\n#### E, B, A, C, D", "source": "t38b_logic", "level": "general", "gv": 2}
+{"prompt": "Question: Summarize the following technical description into a single sentence that retains all causal relationships but omits all specific technical jargon: \"The resolution system queries the DOI handle service, which returns a 302 redirect to the publisher’s landing page; however, if the handle service times out after five seconds, the system falls back to a direct lookup of the DOI suffix in the central metadata repository, which then constructs the URL manually to ensure link integrity.\"\nAnswer:", "response": " The system first asks a central directory for the location of a document, and if that directory responds, it sends you to the publisher's page; if the directory is too slow to answer within five seconds, the system instead searches its own internal records to build the correct link directly, ensuring you still reach the right place.\n\n#### The system uses a primary directory for links but switches to internal records if the directory is too slow.", "source": "t38b_language", "level": "general", "gv": 2}
+{"prompt": "Question: A secret code consists of a sequence of 6 distinct symbols chosen from a set of 10 available symbols. How many such valid codes are there if the first symbol must be from the subset {A, B, C, D} and the last symbol must be an odd number from the set {1, 3, 5, 7, 9}?\nAnswer:", "response": " First symbol choices: 4\nLast symbol choices: 5\nFirst and last symbols are distinct\nRemaining symbols available: 10 - 2 = 8\nMiddle positions to fill: 4\nChoices for first middle position: 8\nChoices for second middle position: 7\nChoices for third middle position: 6\nChoices for fourth middle position: 5\nOrdered middle sequences: 8 × 7 × 6 × 5 = 1680\nTotal valid codes: 4 × 5 × 1680 = 33600\n#### 33600", "source": "t38b_probability", "level": "general", "gv": 2}
+{"prompt": "Question: A committee of 5 members is to be selected from a group of 8 men and 6 women. If the committee must contain at least 3 men, what is the probability that the committee contains exactly 2 women? Express your answer as a simplified fraction.\nAnswer:", "response": " Total number of ways to select 5 members from 8 men and 6 women is not the denominator because of the constraint.\nThe constraint is that the committee must contain at least 3 men.\nPossible valid compositions for a committee of 5 are:\n1. 3 men and 2 women\n2. 4 men and 1 woman\n3. 5 men and 0 women\n\nCalculate the number of ways for each valid composition:\nWays to choose 3 men from 8: $\\binom{8}{3} = \\frac{8 \\times 7 \\times 6}{3 \\times 2 \\times 1} = 56$\nWays to choose 2 women from 6: $\\binom{6}{2} = \\frac{6 \\times 5}{2 \\times 1} = 15$\nNumber of committees with 3 men and 2 women: $56 \\times 15 = 840$\n\nWays to choose 4 men from 8: $\\binom{8}{4} = \\frac{8 \\times 7 \\times 6 \\times 5}{4 \\times 3 \\times 2 \\times 1} = 70$\nWays to choose 1 woman from 6: $\\binom{6}{1} = 6$\nNumber of committees with 4 men and 1 woman: $70 \\times 6 = 420$\n\nWays to choose 5 men from 8: $\\binom{8}{5} = \\binom{8}{3} = 56$\nWays to choose 0 women from 6: $\\binom{6}{0} = 1$\nNumber of committees with 5 men and 0 women: $56 \\times 1 = 56$\n\nTotal number of valid committees (denominator):\n$840 + 420 + 56 = 1316$\n\nThe event \"committee contains exactly 2 women\" corresponds to the composition of 3 men and 2 women.\nNumber of favorable committees (numerator): $840$\n\nProbability:\n$\\frac{840}{1316}$\n\nSimplify the fraction:\nDivide numerator and denominator by 4:\n$840 \\div 4 = 210$\n$1316 \\div 4 = 329$\nFraction becomes $\\frac{210}{329}$\n\nCheck for common factors:\n$210 = 2 \\times 3 \\times 5 \\times 7$\n$329 = 7 \\times 47$\nCommon factor is 7.\n$210 \\div 7 = 30$\n$329 \\div 7 = 47$\nFraction becomes $\\frac{30}{47}$\n\n30 and 47 are coprime (47 is prime and does not divide 30).\n\n#### 30/47", "source": "t38b_probability", "level": "general", "gv": 2}
+{"prompt": "Question: A packing crate has dimensions of 48 cm by 36 cm by 24 cm. It must be packed with rectangular boxes of dimensions 12 cm by 6 cm by 4 cm. The boxes can be oriented in any way, but they must not overlap and must fit entirely within the crate. What is the maximum number of these boxes that can fit into the crate?\nAnswer:", "response": " Crate volume = 48 × 36 × 24 = 41472\nBox volume = 12 × 6 × 4 = 288\nVolume upper bound = 41472 / 288 = 144\nBoxes per 48 cm side = 48 / 12 = 4\nBoxes per 36 cm side = 36 / 6 = 6\nBoxes per 24 cm side = 24 / 4 = 6\nTotal boxes in grid = 4 × 6 × 6 = 144\nGrid packing achieves the volume upper bound\nMaximum number of boxes = 144\n#### 144", "source": "t38b_planning", "level": "general", "gv": 2}
+{"prompt": "Question: Consider two identical violin strings, each with a linear mass density of 0.005 kg/m, stretched under a tension of 100 N. If the left string is plucked at its center (1/2 of its length) and the right string is plucked at 1/4 of its length from one end, explain why the fundamental frequency produced by both strings is identical, but the relative amplitude of the second harmonic (the octave above the fundamental) differs significantly between the two plucking locations. Your explanation must account for the mode shapes of the standing waves and the node/antinode positions at the plucking points.\nAnswer:", "response": " The fundamental frequency is identical because it depends only on the string’s length, tension, and mass density, not on where the string is excited. For both strings, the fundamental mode (n=1) has antinodes at the center and nodes at the ends. Since the left string is plucked at the center, it maximally excites the fundamental mode. The right string is plucked at L/4, which is also an antinode for the fundamental mode, so it too strongly excites the fundamental. Thus, both strings produce the same fundamental frequency, determined by f1 = (1/2L) * sqrt(T/μ).\n\nThe difference arises in the second harmonic (n=2), which has a wavelength of L and nodes at both ends and at the center (L/2). The left string is plucked at L/2, which is a node for the second harmonic. A node has zero displacement in that mode, so plucking there cannot excite the second harmonic; its amplitude is effectively zero. The right string is plucked at L/4, which is an antinode for the second harmonic. Plucking at an antinode maximally excites this mode, resulting in a strong second harmonic component. Therefore, while the fundamental frequency is the same, the right string produces a prominent octave above the fundamental, whereas the left string does not.\n\n#### Plucking at a harmonic node suppresses that harmonic, while plucking at its antinode maximizes it.", "source": "t38b_explain", "level": "general", "gv": 2}
+{"prompt": "Question: Consider a network of three routers (R1, R2, R3) connected in a triangle where R1 connects to R2 with a latency of 10ms, R2 connects to R3 with a latency of 20ms, and R3 connects to R1 with a latency of 5ms. A data packet originates at R1 and must reach R3. The routing protocol uses a greedy algorithm that always chooses the next hop with the lowest immediate link latency, but only allows one hop per decision cycle. If the packet cannot arrive at R3 in the first hop, it takes the best available second-hop path. Trace the packet's path and calculate the total end-to-end latency, assuming propagation time is the only factor and processing delays are negligible.\nAnswer:", "response": " The packet originates at R1 and must reach R3. The routing protocol uses a greedy algorithm, meaning R1 evaluates the immediate link latencies to its neighbors, R2 and R3. The direct link from R1 to R3 has a latency of 5ms, while the link from R1 to R2 has a latency of 10ms.\n\nBecause the direct path to the destination R3 is available and offers the lowest immediate latency (5ms) compared to the alternative (10ms), the greedy algorithm selects the direct link. The condition \"if the packet cannot arrive at R3 in the first hop\" is not triggered because R1 is directly connected to R3. Therefore, the packet does not need to take a second hop via R2.\n\nThe path is simply R1 to R3. The total end-to-end latency is the sum of the propagation times along this single link. Since processing delays are negligible, the total latency is exactly the link latency between R1 and R3.\n\nTotal Latency = 5ms\n\n#### The packet travels directly from R1 to R3 with a total latency of 5ms.", "source": "t38b_explain", "level": "general", "gv": 2}
+{"prompt": "Question: In a simplified genetic simulation, a population of 1,000 organisms has a gene with two alleles: A (dominant) and a (recessive). Initially, 60% of the population has genotype AA, 30% has Aa, and 10% has aa. A selection event occurs where only organisms with the dominant phenotype (AA or Aa) survive to reproduce. Assuming random mating among the survivors and no mutation or migration, calculate the new frequency of the a allele in the next generation (F1).\nAnswer:", "response": " Start by determining the number of survivors in the initial population of 1,000 organisms. The genotype AA constitutes 60% (600 individuals), and Aa constitutes 30% (300 individuals). The aa genotype (10% or 100 individuals) is eliminated by selection because only the dominant phenotype survives. Thus, 900 organisms survive to reproduce.\n\nNext, calculate the total allele count in this surviving group. Each organism contributes two alleles, so there are 1,800 alleles in total. To find the frequency of the recessive allele a, count the specific a alleles present among the survivors. The 600 AA individuals contribute zero a alleles. The 300 Aa individuals each carry one a allele, contributing a total of 300 a alleles.\n\nThe frequency of the a allele (q) in the post-selection, pre-reproduction population is the number of a alleles divided by the total alleles: 300 / 1,800 = 1/6.\n\nUnder the assumption of random mating and no mutation or migration, the Hardy-Weinberg principle dictates that allele frequencies remain constant in the next generation unless acted upon by evolutionary forces. Since the selection event has already occurred and removed the aa individuals, the surviving pool mixes randomly. The allele frequency in the F1 generation will be identical to the allele frequency in the breeding population of the P generation.\n\nTherefore, the new frequency of the a allele is 1/6, which is approximately 0.1667 or 16.67%.\n\n#### The frequency of the a allele in the F1 generation is 1/6.", "source": "t38b_explain", "level": "general", "gv": 2}
+{"prompt": "Question: Consider a cryptographic hash function used to verify data integrity, where even a single bit change results in a completely different hash output. Why does this property make \"zero defects\" in data transmission (i.e., no bit errors) a necessary but not sufficient condition for security, and how does the concept of \"collision resistance\" introduce a different type of defect that cannot be detected by simple checksums?\nAnswer:", "response": " A cryptographic hash function’s sensitivity to single-bit changes, known as the avalanche effect, ensures that any transmission error, such as a flipped bit, will produce a radically different digest. This makes \"zero defects\" in transmission necessary because if even one bit is corrupted, the receiver’s calculated hash will not match the sender’s original hash, immediately flagging the data as invalid. However, this condition is not sufficient for security because it only verifies that the data arrived exactly as sent, not that the data is authentic or unaltered by an attacker. An adversary could transmit the correct, uncorrupted data but still compromise security if they replaced the original file with a different one that produces the same hash.\n\nThis leads to the concept of collision resistance. A hash collision occurs when two different inputs produce the same output. If an attacker can find a malicious document that hashes to the same value as a legitimate document, they can substitute the malicious file without triggering the integrity check. Simple checksums, which are designed for error detection rather than security, are weak against this because they do not guarantee that a unique hash corresponds to a unique input. In cryptographic terms, a simple checksum might allow an attacker to modify data in a way that preserves the checksum, or it might not provide enough output bits to make finding a collision computationally infeasible. Thus, collision resistance addresses the \"defect\" of forgery, which simple checksums cannot detect, by ensuring that finding two distinct inputs with the same hash is practically impossible.\n\n#### <one-sentence takeaway>\nWhile bit-error detection ensures data integrity, collision resistance is required to prevent attackers from substituting malicious data that yields the same hash value.", "source": "t38b_explain", "level": "general", "gv": 2}
+{"prompt": "Question: In the context of machine learning model evaluation, a classifier that perfectly predicts the majority class (e.g., 99% non-fraudulent transactions) achieves 99% accuracy but fails the \"zero defects\" ideal for the minority class (fraud). Explain how the \"imbalance ratio\" in the dataset makes traditional accuracy metrics misleading, and why a metric like the F1-score is better suited to penalize the specific type of defect (false negatives) that is most costly in this scenario.\nAnswer:", "response": " In a dataset with a 99/1 imbalance, the base rate for the majority class is 99%. A trivial classifier that predicts every transaction as non-fraud achieves 99% accuracy, which appears excellent on a standard accuracy metric. However, this model has a 100% false negative rate for fraud cases, meaning it detects zero actual fraud. This demonstrates that accuracy is heavily biased toward the majority class and ignores the model's performance on the critical minority class.\n\nThe imbalance ratio, defined as the number of negative samples divided by the number of positive samples, quantifies this skew. When this ratio is high, the absolute number of misclassified minority instances becomes small relative to the total, masking the model's failure to identify them. For example, in one million transactions, 10,000 fraud cases exist. If the model misses all of them, it still reports 99% accuracy, hiding the fact that it failed 100% of the fraud detection task.\n\nThe F1-score addresses this by calculating the harmonic mean of precision and recall. Recall measures the proportion of actual fraud cases correctly identified, directly penalizing false negatives. Precision measures the proportion of predicted fraud cases that are actually fraudulent, penalizing false positives. By averaging these two, F1 forces the model to balance sensitivity and specificity. In the fraud scenario, a high F1-score requires the model to catch a significant portion of the fraud (high recall) without flagging too many legitimate transactions (high precision). Unlike accuracy, F1 does not reward the model for correctly classifying the vast majority of benign transactions, making it a far more robust metric for imbalanced, high-stakes classification problems.\n\n#### F1-score penalizes false negatives by balancing recall and precision, unlike accuracy which is skewed by the majority class.", "source": "t38b_explain", "level": "general", "gv": 2}
+{"prompt": "Question: In nuclear reactor safety design, the goal is \"zero release\" of radioactive material, achieved through multiple layers of containment (defense in depth). If a single layer has a failure probability of $10^{-5}$ per year, and there are four independent layers, calculate the combined annual failure probability of the entire system, and explain why the assumption of \"independence\" is a critical theoretical flaw when common cause failures (such as a simultaneous loss of power and cooling) are considered.\nAnswer:", "response": " The combined annual failure probability is calculated by multiplying the individual probabilities of each independent layer failing. With four layers, each having a failure probability of $10^{-5}$, the system failure probability is $(10^{-5})^4$, which equals $10^{-20}$ per year. This astronomically low number illustrates the mathematical power of defense in depth: each additional layer drastically reduces the likelihood that radioactive material escapes containment.\n\nHowever, the assumption of independence is a critical theoretical flaw because it ignores common cause failures (CCFs). In reality, external events like earthquakes, severe storms, or grid-wide power outages can compromise multiple layers simultaneously. For instance, a total loss of off-site power might disable the primary cooling systems and the backup diesel generators at the same time. If these failures are correlated, the probability of the system failing is not the product of the individual probabilities but is instead dominated by the probability of the shared external stressor occurring. If the probability of a common cause event affecting all four layers is $10^{-3}$, the system failure probability becomes $10^{-3}$, which is nineteen orders of magnitude higher than the independent calculation. Therefore, safety designs must explicitly model CCFs and introduce diverse, physically separated backup systems (such as gravity-driven cooling or battery-backed controls) to break the correlation between layers.\n\n#### <one-sentence takeaway>", "source": "t38b_explain", "level": "general", "gv": 2}
+{"prompt": "Question: Four residents of a historic villa—Aldo, Bruno, Carla, and Davide—each occupy one of the four main rooms: the Library, the Conservatory, the Solarium, and the Archive. Each resident prefers a different beverage: tea, coffee, lemonade, or wine. The clues are as follows: Aldo does not live in the Library and does not drink wine. The resident who drinks coffee lives in the Solarium. Bruno lives in the Archive and drinks tea. Carla drinks lemonade. Davide does not live in the Conservatory. Based on these constraints, which beverage does the resident of the Library drink?\nAnswer:", "response": " Bruno lives in the Archive and drinks tea.\nCarla drinks lemonade.\nThe resident of the Solarium drinks coffee.\nAldo does not live in the Library.\nAldo does not drink wine.\nDavide does not live in the Conservatory.\nThe remaining rooms for Aldo, Carla, and Davide are Library, Conservatory, and Solarium.\nCarla cannot live in the Solarium because she drinks lemonade, not coffee.\nAldo cannot live in the Conservatory because the only beverage available there would be wine (tea is taken by Bruno, lemonade by Carla, coffee by the Solarium resident), and Aldo does not drink wine.\nAldo must live in the Solarium and drink coffee.\nThe remaining rooms for Carla and Davide are Library and Conservatory.\nDavide cannot live in the Conservatory, so Davide lives in the Library.\nCarla lives in the Conservatory.\nThe remaining beverage for Davide is wine.\nThe resident of the Library drinks wine.\n#### wine", "source": "t38b_logic", "level": "general", "gv": 2}
+{"prompt": "Question: In the basement of Villa Gabrielli, three archivists—Elena, Frank, and Gina—store exactly three distinct types of documents: maps, ledgers, and deeds. Each archivist stores exactly one type, and each type is stored by exactly one archivist. Elena states, \"I do not store the maps.\" Frank states, \"I store the ledgers.\" Gina states, \"Frank does not store the ledgers.\" It is known that exactly one of the three archivists tells the truth, while the other two lie. Who stores the maps?\nAnswer:", "response": " Elena, Frank, and Gina each store a distinct document type from maps, ledgers, deeds.\nElena's statement: \"I do not store the maps\" is true if and only if Elena does not store maps.\nFrank's statement: \"I store the ledgers\" is true if and only if Frank stores ledgers.\nGina's statement: \"Frank does not store the ledgers\" is true if and only if Frank does not store ledgers.\nFrank's statement and Gina's statement are logical negations of each other.\nTherefore, exactly one of Frank and Gina tells the truth in any valid assignment.\nThe problem states that exactly one of the three archivists tells the truth.\nSince exactly one of Frank and Gina is truthful, Elena must be the one who lies.\nElena's statement \"I do not store the maps\" is false.\nTherefore, Elena stores the maps.\n#### Elena", "source": "t38b_logic", "level": "general", "gv": 2}
+{"prompt": "Question: Four project leads at the Irving Group—Hale, Iris, Jaden, and Kira—are ranked from 1st (highest) to 4th (lowest) in performance. The following facts are true: Jaden is ranked higher than Kira. Iris is ranked immediately above Jaden. Hale is not ranked 1st. The person ranked 2nd is not Iris. Determine the exact ranking order of the four leads from 1st to 4th.\nAnswer:", "response": " Iris is immediately above Jaden, so the pair (Iris, Jaden) occupies positions (1,2), (2,3), or (3,4).\nThe person ranked 2nd is not Iris, so the pair cannot be (2,3).\nJaden is ranked higher than Kira, so Jaden cannot be 4th, eliminating the pair (3,4).\nTherefore, Iris is 1st and Jaden is 2nd.\nJaden is ranked higher than Kira, so Kira must be 3rd or 4th.\nHale takes the remaining position, which is 4th or 3rd respectively.\nHale is not ranked 1st, which is satisfied in both cases.\nThe valid rankings are Iris, Jaden, Kira, Hale and Iris, Jaden, Hale, Kira.\nThe exact ranking order is not uniquely determined.\n#### Not uniquely determined", "source": "t38b_logic", "level": "general", "gv": 2}
+{"prompt": "Question: A senior legislative clerk earns an annual taxable income of $95,000. In a hypothetical jurisdiction with a progressive tax system defined as follows: 15% on the first $40,000, 25% on the next $40,000, and 35% on any amount exceeding $80,000. Calculate the clerk’s effective tax rate for the year, expressed as a percentage rounded to two decimal places.\nAnswer:", "response": " Tax on first $40,000: 0.15 × 40,000 = 6,000\nTax on next $40,000: 0.25 × 40,000 = 10,000\nTax on remaining $15,000: 0.35 × (95,000 − 80,000) = 5,250\nTotal tax: 6,000 + 10,000 + 5,250 = 21,250\nEffective tax rate: 21,250 / 95,000 = 0.2236842105\nPercentage: 22.36842105%\nRounded to two decimal places: 22.37%\n#### 22.37%", "source": "t38b_finance", "level": "general", "gv": 2}
+{"prompt": "Question: A Roman estate manager is calculating the long-term cost of labor. He considers two options for harvesting a vineyard: hiring free temporary workers at a market wage that fluctuates with seasonal demand, or maintaining a permanent slave labor force. Explain, using economic concepts of fixed versus variable costs and the impact of labor scarcity, why the estate manager might prefer slaves during a period of high inflation and labor shortage, but why the high upfront capital cost of purchasing slaves could make them a worse investment compared to free labor if the market wage for free workers drops significantly.\nAnswer:", "response": " The estate manager faces a classic trade-off between fixed and variable costs. Slaves represent a high fixed cost because their purchase price is a large, upfront capital expenditure that does not change with production volume. Free temporary workers represent a variable cost, as the manager only pays wages when labor is actually hired, and the total cost scales directly with the amount of work required.\n\nDuring a period of high inflation and labor shortage, the market wage for free workers spikes due to scarcity. If the daily wage for a free harvester rises above the effective daily cost of a slave (calculated by amortizing the slave’s purchase price and maintenance over their working life), the variable cost of free labor becomes prohibitively expensive. In this scenario, the fixed cost of slaves becomes the more predictable and potentially cheaper option. The manager avoids the volatility of wage inflation and secures a stable workforce when free labor is scarce or unreliable.\n\nHowever, if the market wage for free workers drops significantly, the variable cost of hiring them falls below the effective daily cost of maintaining slaves. In this case, the high upfront capital cost of slaves becomes a financial burden. The manager would be locked into a high fixed cost while a cheaper variable alternative exists. Furthermore, slaves require constant care, feeding, and security regardless of whether the vineyard is in peak harvest or off-season, whereas free labor costs can be reduced to zero during idle periods. If the initial investment in slaves was made during a high-wage period but wages subsequently plummet, the return on that capital investment deteriorates, making free labor the superior economic choice.\n\n#### <one-sentence takeaway>", "source": "t38b_explain", "level": "general", "gv": 2}
+{"prompt": "Question: A small business has fixed annual costs of $120,000, variable costs of $42 per unit, and sells its product for $75 per unit. The business operates in a jurisdiction with a 25% corporate tax rate on profits. If the business aims to achieve an after-tax net income of $50,000, calculate the break-even quantity of units that must be sold to meet this target profit level, rounded up to the nearest whole unit.\nAnswer:", "response": " Contribution margin per unit = 75 - 42 = 33\nRequired pre-tax profit = 50,000 / (1 - 0.25) = 66,666.6667\nTotal required contribution = 120,000 + 66,666.6667 = 186,666.6667\nQuantity = 186,666.6667 / 33 = 5,656.5657\nRounded up to nearest whole unit = 5,657\n#### 5657", "source": "t38b_finance", "level": "general", "gv": 2}
+{"prompt": "Question: A corporation issues a 10-year bond with a face value of $1,000 and a coupon rate of 7.5%, paid semi-annually. The current market interest rate for equivalent-risk bonds is 8.0%. Calculate the current market price of the bond, rounded to the nearest cent.\nAnswer:", "response": " Semiannual coupon = 1000 × 0.075 / 2 = 37.50\nNumber of periods = 10 × 2 = 20\nSemiannual market rate = 0.08 / 2 = 0.04\n(1.04)^20 = 2.1911231430334193\n(1.04)^(-20) = 1 / 2.1911231430334193 = 0.45638694620129\nAnnuity factor = (1 - 0.45638694620129) / 0.04 = 13.59032634496775\nPresent value of coupons = 37.50 × 13.59032634496775 = 509.63723793629065\nPresent value of face value = 1000 × 0.45638694620129 = 456.38694620129\nTotal price = 509.63723793629065 + 456.38694620129 = 966.02418413758065\nRounded to the nearest cent = 966.02\n#### $966.02", "source": "t38b_finance", "level": "general", "gv": 2}
+{"prompt": "Question: A military supply depot tracks the inventory of three types of standard-issue gear packs: Type A, Type B, and Type C. The depot starts the week with 120 Type A packs, 80 Type B packs, and 60 Type C packs. During the week, 15% of the Type A packs are issued out, while the number of Type B packs issued is equal to one-fourth of the remaining Type A packs. The number of Type C packs issued is twice the number of Type B packs issued. If the depot then receives a new shipment of 50 Type A packs, 30 Type B packs, and 20 Type C packs, what is the ratio of the total number of Type B packs to the total number of Type C packs remaining in the depot?\nAnswer:", "response": " Type A issued: 15% of 120 = 18\nRemaining Type A: 120 - 18 = 102\nType B issued: 1/4 of 102 = 25.5\nType C issued: 2 × 25.5 = 51\nType B remaining after issuance: 80 - 25.5 = 54.5\nType C remaining after issuance: 60 - 51 = 9\nType B after shipment: 54.5 + 30 = 84.5\nType C after shipment: 9 + 20 = 29\nRatio of Type B to Type C: 84.5 : 29\nMultiply both terms by 2 to clear the decimal: 169 : 58\n\n#### 169:58", "source": "t38b_wordproblem", "level": "general", "gv": 2}
+{"prompt": "Question: A logistics team is planning a supply drop. They have 4500 liters of water to distribute among three platoons: Alpha, Bravo, and Charlie. Alpha receives 20% more water than Bravo. Charlie receives 15% less water than the combined amount received by Alpha and Bravo. If each soldier in Alpha, Bravo, and Charlie receives exactly 2 liters of water per day, and the supply is intended to last for 3 days, what is the total number of soldiers in the three platoons combined?\nAnswer:", "response": " Total water available: 4500 L\nWater per soldier per day: 2 L\nDays of supply: 3\nTotal water per soldier: 2 L/day × 3 days = 6 L\nTotal soldiers: 4500 L ÷ 6 L/soldier = 750 soldiers\n#### 750", "source": "t38b_wordproblem", "level": "general", "gv": 2}