- The table shows a runner's split times for a 5 km race. Using the table shown, during which kilometer did the runner's pace most decrease (slow down the most) compared to the previous kilometer?
Answer: Kilometer 4
Split times: K1=4:30, K2=4:45 (+0:15), K3=4:50 (+0:05), K4=5:20 (+0:30), K5=5:05 (−0:15). The largest increase in split time (biggest slowdown) is Kilometer 4 at +30 seconds. (A) shows a 15-second increase. (B) shows only 5 seconds. (D) is actually a speed-up.
- The frequency table shows the number of hours per week a sample of 80 employees spend in meetings. Based on the table shown, what is the best estimate of the mean number of meeting hours per week?
Answer: 7.5 hours
Use the midpoint of each interval times its frequency. Midpoints × frequency: (2)(10)=20; (5)(24)=120; (8)(28)=224; (11)(12)=132; (14)(6)=84. Sum = 580. Mean = 580/80 = 7.25 hours. The closest estimate is 7.5 hours. (A) and (B) underestimate by using values too far below the calculated mean; (D) overestimates. - A scatter plot analysis of patient recovery time versus dosage of a new drug produced a line of best fit with the equation y=−2.5x+30, where y is the recovery time in days and x is the dosage in milligrams (mg). The original data shows that a patient who received a 6 mg dosage had an actual recovery time of 13 days. What is the absolute difference between this patient's actual recovery time and the time predicted by the line of best fit?
Answer: 2.0 days
This question requires calculating a predicted value from an equation and then finding the residual. First, use the line of best fit equation to find the predicted recovery time (y) for a dosage (x) of 6 mg: y=−2.5(6)+30. This simplifies to y=−15+30=15 days. Second, identify the patient's actual recovery time, which is given as 13 days. Finally, calculate the absolute difference between the actual and predicted values: ∣13−15∣=∣−2∣=2.0 days. - The frequency distribution of ages for 40 attendees at a conference is described as follows: 8 attendees are in the range 20-29, 18 are in the range 30-39, 10 are in the range 40-49, and 4 are in the range 50-59. To estimate the mean age, one would use the midpoint of each range. Based on the data, the estimated mean age is 36.5 years. The median age falls within the 30-39 range. Which of the following is the most accurate interpretation of these data?
Answer: The mean is greater than the median, suggesting the age distribution is skewed to the right.
This question tests the conceptual understanding of how mean, median, and skewness relate. The median age falls in the 30-39 range, so the median value is less than 40. The estimated mean age is given as 36.5. Because the median falls in the 30-39 range (midpoint 34.5), the mean (36.5) is greater than the median. When the mean is pulled to the right (higher value) of the median, it indicates that the distribution is skewed to the right by higher-value outliers (in this case, the attendees in their 40s and 50s).
- The graph below shows the populations of two bacteria cultures, A and B, over 8 hours. Using the graph shown, approximately when do the two cultures first have equal populations?
Answer: Between 4 and 5 hours
Culture A grows linearly: starts at 200, adds ~100 per hour. Culture B grows exponentially from 50, doubling every 1.5 hours approximately. Reading the graph, the curves intersect between hour 4 (A≈600, B≈500) and hour 5 (A≈700, B≈800). The crossing occurs between 4 and 5 hours. (A) and (B) are too early — B is still below A. (D) is after the crossing.
- The population of a city was recorded over three decades. In 1990, the population was 20,000. In 2000, it had grown to 28,000. In 2010, it was 36,400. In 2020, it reached 43,680. During which 10-year period was the percentage growth rate the greatest?
Answer: 1990 to 2000
This question requires calculating and comparing the percentage growth rate for three distinct periods. The formula for percentage growth is OldNew−Old×100%. For 1990-2000: 20,00028,000−20,000=20,0008,000=0.40, or 40%. For 2000-2010: 28,00036,400−28,000=28,0008,400=0.30, or 30%. For 2010-2020: 36,40043,680−36,400=36,4007,280=0.20, or 20%. The greatest percentage growth rate was 40%, which occurred from 1990 to 2000. - A clinical trial with 600 patients tracks medication adherence. Initial data shows 50% of patients take the medication 'always', 30% take it 'sometimes', and 20% take it 'rarely'. After an intervention, the number of patients in the 'always' group increases by 20%. These new 'always' patients were previously in the 'sometimes' and 'rarely' groups, in proportion to the original sizes of those two groups. After this change, how many patients are in the 'sometimes' group?
Answer: 144
This is a complex, multi-step problem. First, calculate the initial number of patients in each group: Always: 0.50×600=300. Sometimes: 0.30×600=180. Rarely: 0.20×600=120. Second, calculate the number of patients who move to the 'always' group: 0.20×300=60 patients. Third, determine the proportion of these 60 patients who came from 'sometimes' vs 'rarely'. The original ratio was 180:120, which simplifies to 3:2. So, 53 of the movers came from 'sometimes'. Fourth, calculate the number who left the 'sometimes' group: 53×60=36. Finally, calculate the new number in the 'sometimes' group: 180−36=144. - A company sells two models of a medical device, Model A and Model B. Sales data is reported from two regions. In the Eastern region, sales were $80,000 for Model A and $120,000 for Model B. In the Western region, total sales for both models were $250,000, with Model B accounting for 60% of that amount. What proportion of the company's total sales is represented by sales of Model A?
Answer: 40.0%
This is a multi-step problem. First, determine the sales of each model in the Western region. Model B sales: 0.60×$250,000=$150,000. Model A sales: $250,000−$150,000=$100,000. Second, calculate the total sales for the company: Eastern region total ($80,000 + $120,000\) + Western region total ($250,000\) = $200,000 + $250,000 = $450,000\). Third, calculate the total sales of Model A from both regions: $80,000 (East) + $100,000 (West) = $180,000\). Finally, find the proportion: $450,000$180,000=4518=52=0.40, or 40.0%. - Refer to the graph below. Which month experienced the greatest increase in average high temperature compared with the previous month?
Answer: May
The graph shows the average high temperatures each month. March to April increases by 6 °F, April to May increases by 10 °F, May to June increases by 4 °F. Therefore, the largest increase is between April and May, making May the correct answer. Distractors: March (incorrect, smaller increase), April (is previous month not latest), June (increase is smaller).
- The graph below shows the velocity (in m/s) of an object as a function of time (in seconds). Based on the graph shown, what is the total distance traveled by the object from t=0 to t=10 seconds?
Answer: 90 m
Distance equals the area under the velocity-time curve. Break into regions: from t=0 to 2 (triangle, 0→10 m/s): ½(2)(10)=10; t=2 to 6 (rectangle at 10 m/s): 4·10=40; t=6 to 8 (trapezoid 10→15): ½(2)(10+15)=25; t=8 to 10 (rectangle at... wait, triangle 15→0): ½(2)(15)=15. Total = 10+40+25+15 = 90 m. (A) uses only one region. (B) forgets the trapezoid. (D) double-counts a segment. - A dataset of 25 patient recovery times, in days, has been analyzed. The five-number summary is described as follows: the minimum is 5, the first quartile (Q1) is 12, the median (Q2) is 18, the third quartile (Q3) is 22, and the maximum is 45. Using the 1.5 × IQR rule, what is the largest recovery time that would not be considered an outlier?
Answer: 37.0 days
This question requires using the five-number summary to calculate the upper boundary for outliers. First, calculate the Interquartile Range (IQR): IQR=Q3−Q1=22−12=10. Second, use the 1.5 × IQR rule to find the outlier threshold. The upper fence is calculated as Q3+(1.5×IQR). Third, plug in the values: 22+(1.5×10)=22+15=37. Any data point above 37 is considered an outlier. Therefore, the largest recovery time that would not be considered an outlier is 37.0 days. - The table shows the number of students scoring in each grade category across three schools. Refer to the table shown. Which school has the highest percentage of students scoring either an A or a B?
Answer: School Y
Percent A or B: School X = (40+60)/250 = 100/250 = 40.0%. School Y = (54+72)/300 = 126/300 = 42.0%. School Z = (30+50)/200 = 80/200 = 40.0%. School Y has the highest percentage. (A) and (C) both have 40%. (D) ignores School Y's higher rate.
- The scatter plot below shows the relationship between hours studied and exam score for 10 students, along with the line of best fit. Based on the scatter plot shown, which statement is best supported by the data?
Answer: Each additional hour of study predicts approximately an 8-point increase in score.
The line of best fit passes approximately through (1, 60) and (6, 100), giving a slope of (100-60)/(6-1) = 40/5 = 8 points per hour. (A) is incorrect: the y-intercept is approximately 52, not 55. (C) is incorrect: the correlation is clearly positive since the line has positive slope. (D) is incorrect: one student who studied 5.5 hours scored 85, which is below 90.
- A nutritional label states that a 100g serving of a food contains 20g of fat, 40g of carbohydrates, and 15g of protein. The caloric values are 9 calories per gram of fat, 4 calories per gram of carbohydrates, and 4 calories per gram of protein. If a person consumes a 150g serving of this food, what percentage of their total caloric intake from this serving comes from protein, rounded to the nearest whole number?
Answer: 15%
The serving size of 150g is extraneous information, as the percentage of calories from each macronutrient is constant regardless of the serving size. First, calculate the calories from each macronutrient per 100g: Fat: 20g×9 cal/g=180 calories. Carbohydrates: 40g×4 cal/g=160 calories. Protein: 15g×4 cal/g=60 calories. Second, calculate the total calories per 100g: 180+160+60=400 calories. Finally, calculate the percentage of calories from protein: 400 total calories60 protein calories×100%=15%. - The histogram below shows the distribution of test scores for 50 students. Based on the histogram shown, which of the following could be the median score?
Answer: 78
Cumulative frequencies: 50–59: 4 (cum 4); 60–69: 8 (cum 12); 70–79: 14 (cum 26); 80–89: 16 (cum 42); 90–99: 8 (cum 50). The median (average of 25th and 26th values) falls within the 70–79 bin. Only 78 lies in that bin. (A) and (B) fall in earlier bins; (D) falls in the 80–89 bin. - The double-line graph below shows monthly revenue (in thousands) for two stores over six months. Refer to the graph shown. In how many months was Store B's revenue at least 20% greater than Store A's revenue?
Answer: 3
Calculate the ratio B/A for each month: Jan: 30/25 = 1.20 (exactly 20% greater, which qualifies as 'at least'). Feb: 36/30 = 1.20. Mar: 40/35 ≈ 1.14. Apr: 50/38 ≈ 1.32. May: 45/40 = 1.125. Jun: 48/42 ≈ 1.14. Months with ratio ≥ 1.20: January, February, and April = 3 months. - The bar graph below shows quarterly sales (in thousands of dollars) for a company across four years. Using the graph shown, which of the following statements is true?
Answer: The largest quarter-to-quarter increase within a single year occurred in Year 2.
In Year 2, sales rise from Q2 ($45k) to Q3 ($80k), a $35k jump — the largest single-quarter increase shown. (A) is false: Year 3 mean is $62.5k, Year 4 mean is $65k. (C) is false: Q4 dropped from Year 2 ($70k) to Year 3 ($60k). (D) is false: Year 4 totals $260k vs. Year 3's $250k.
- The box plot below summarizes daily high temperatures (°F) in two cities over one month. Refer to the box plots shown. Which statement must be true?
Answer: The interquartile range of City X is greater than that of City Y.
City X: Q1=62, Q3=78, IQR=16. City Y: Q1=70, Q3=82, IQR=12. So 16 > 12 — City X has the larger IQR. (A) Box plots do not show mean. (C) City X's max is 88, City Y's Q3 is 82, so less than 25% of Y exceeds X's max. (D) Cannot be determined from quartiles alone — both medians and Q3 suggest Y has more warm days, but exact counts are unknown.
- A lab tracks the number of samples processed per day: Monday (150), Tuesday (165), Wednesday (155), Thursday (170), Friday (180). On the following Monday, 198 samples are processed. What is the absolute change in the 5-day moving average from the period ending on Friday to the period ending on the following Monday?
Answer: 9.6
This problem requires calculating two moving averages and their difference. First, calculate the 5-day moving average for the period ending Friday: 5150+165+155+170+180=5820=164.0. Second, calculate the 5-day moving average for the new period ending the following Monday. This period includes Tuesday, Wednesday, Thursday, Friday, and the new Monday. The original Monday's data (150) is dropped: 5165+155+170+180+198=5868=173.6. Finally, calculate the absolute change between the two averages: ∣173.6−164.0∣=9.6. - A patient's temperature is recorded over an 8-hour period. At 8:00 AM, the temperature is 99.5°F. It rises to a peak of 101.5°F at 12:00 PM. By 4:00 PM, it has decreased to 100.0°F. The final reading at 8:00 PM is 99.0°F. What is the average rate of temperature change, in degrees Fahrenheit per hour, from the time of the peak temperature to the final reading?
Answer: -0.3125 °F/hr
This is a multi-step problem requiring data extraction and rate calculation. First, identify the peak temperature and time (101.5°F at 12:00 PM) and the final temperature and time (99.0°F at 8:00 PM). Second, calculate the total change in temperature: 99.0°F−101.5°F=−2.5°F. Third, calculate the elapsed time: 8:00 PM is 8 hours after 12:00 PM. Fourth, calculate the average rate of change by dividing the temperature change by the time change: 8 hours−2.5°F=−0.3125°F/hr.
