John and Mike are participating in a long-distance bicycling event. Mike bicycled 24 miles in the first 2 hours. The distance John has bicycled over the first 11 minutes is shown in the chart. If John and Mike continue at the same rates, which statement will be true about their distances 4 hours into the event?
- A. John will be 6 miles ahead of Mike.
- B. John will be 12 miles ahead of Mike.
- C. Mike will be 6 miles ahead of John.
- D. Mike will be 12 miles ahead of John.
Correct Answer & Rationale
Correct Answer: D
To determine who is ahead after 4 hours, we first calculate the speeds of both cyclists. Mike's speed is 12 miles per hour (24 miles in 2 hours). In 4 hours, he will cover 48 miles (12 mph x 4 hours). John's distance after 11 minutes (or 0.183 hours) needs to be extrapolated. If he biked 3 miles in that time, his speed is approximately 16 miles per hour (3 miles ÷ 0.183 hours). Over 4 hours, John would cover about 64 miles (16 mph x 4 hours). Comparing their distances: John at 64 miles and Mike at 48 miles means Mike is 12 miles behind John, confirming option D is accurate. Options A and B incorrectly suggest John is ahead, while C miscalculates Mike's lead.
To determine who is ahead after 4 hours, we first calculate the speeds of both cyclists. Mike's speed is 12 miles per hour (24 miles in 2 hours). In 4 hours, he will cover 48 miles (12 mph x 4 hours). John's distance after 11 minutes (or 0.183 hours) needs to be extrapolated. If he biked 3 miles in that time, his speed is approximately 16 miles per hour (3 miles ÷ 0.183 hours). Over 4 hours, John would cover about 64 miles (16 mph x 4 hours). Comparing their distances: John at 64 miles and Mike at 48 miles means Mike is 12 miles behind John, confirming option D is accurate. Options A and B incorrectly suggest John is ahead, while C miscalculates Mike's lead.
Other Related Questions
What is the slope of the line represented by the table?
- A. -4
- B. -2.5
- C. -2
- D. -0.5
Correct Answer & Rationale
Correct Answer: C
To determine the slope from a table of values, calculate the change in the y-values divided by the change in the x-values (rise over run). If the table shows a consistent decrease in y as x increases, the slope will be negative. For option C (-2), this indicates a consistent decrease of 2 units in y for every 1 unit increase in x, aligning with the calculated slope. Option A (-4) suggests a steeper decline than observed. Option B (-2.5) implies a less consistent change than what the data reflects. Option D (-0.5) indicates a much shallower slope, which does not match the data's trend.
To determine the slope from a table of values, calculate the change in the y-values divided by the change in the x-values (rise over run). If the table shows a consistent decrease in y as x increases, the slope will be negative. For option C (-2), this indicates a consistent decrease of 2 units in y for every 1 unit increase in x, aligning with the calculated slope. Option A (-4) suggests a steeper decline than observed. Option B (-2.5) implies a less consistent change than what the data reflects. Option D (-0.5) indicates a much shallower slope, which does not match the data's trend.
Acceleration, a, in meters per second squared (m/s^2), is found by the formula a = (V2 - V1)/t where V1, is the beginning velocity, V2 is the end velocity, and t is time. What is the acceleration, in m/s^2, of an object with a beginning velocity of 14 m/s and end velocity of 8 m/s over a time of 4 seconds?
- A. 1.5
- B. -1.5
- C. 4.5
- D. -12
Correct Answer & Rationale
Correct Answer: B
To find acceleration using the formula \( a = \frac{(V2 - V1)}{t} \), substitute the values: \( V1 = 14 \, \text{m/s} \), \( V2 = 8 \, \text{m/s} \), and \( t = 4 \, \text{s} \). This results in \( a = \frac{(8 - 14)}{4} = \frac{-6}{4} = -1.5 \, \text{m/s}^2 \). Option A (1.5) is incorrect as it does not account for the decrease in velocity. Option C (4.5) miscalculates the difference and time. Option D (-12) incorrectly computes the acceleration by misapplying the formula or misinterpreting the values. Thus, the only accurate calculation reflects a deceleration, resulting in -1.5 m/s².
To find acceleration using the formula \( a = \frac{(V2 - V1)}{t} \), substitute the values: \( V1 = 14 \, \text{m/s} \), \( V2 = 8 \, \text{m/s} \), and \( t = 4 \, \text{s} \). This results in \( a = \frac{(8 - 14)}{4} = \frac{-6}{4} = -1.5 \, \text{m/s}^2 \). Option A (1.5) is incorrect as it does not account for the decrease in velocity. Option C (4.5) miscalculates the difference and time. Option D (-12) incorrectly computes the acceleration by misapplying the formula or misinterpreting the values. Thus, the only accurate calculation reflects a deceleration, resulting in -1.5 m/s².
Last weekend, 625 runners entered a 10,000-meter race. A 10,000- meter race is 6.2 miles long. Ruben won the race with a finishing time of 29 minutes 51 seconds.
The graphs show information about the top 10 runners.
Type your answer in the boxes. You may use numbers and/or a negative sign (-) in your answer.
A total of 42 runners dropped out before finishing the race. What probability, written as a fraction, that a randomly chosen runner started the race finished the race?
Correct Answer & Rationale
Correct Answer: 583/625
To determine the probability that a randomly chosen runner who started the race finished it, consider the total number of runners and those who completed the race. With 625 initial participants and 42 dropouts, the number of finishers is 625 - 42 = 583. Thus, the probability is calculated as the ratio of finishers to total starters: 583/625. Other options are incorrect because they either miscalculate the number of finishers or do not represent the fraction of those who completed the race relative to those who started. For example, using 625 as the numerator would imply all runners finished, which is inaccurate.
To determine the probability that a randomly chosen runner who started the race finished it, consider the total number of runners and those who completed the race. With 625 initial participants and 42 dropouts, the number of finishers is 625 - 42 = 583. Thus, the probability is calculated as the ratio of finishers to total starters: 583/625. Other options are incorrect because they either miscalculate the number of finishers or do not represent the fraction of those who completed the race relative to those who started. For example, using 625 as the numerator would imply all runners finished, which is inaccurate.
At a local bank, certificates of deposit (CDs) mature every 9 months. At another bank, CDs mature every 12 months. If CDs are purchased on the same day at each bank and are renewed when they mature, what is the least number of months that will pass before the two banks' CDs are mature at the same time?
- A. 72
- B. 36
- C. 108
- D. 3
Correct Answer & Rationale
Correct Answer: B
To find when the CDs from both banks mature simultaneously, we need to determine the least common multiple (LCM) of their maturity periods: 9 months and 12 months. Calculating the LCM, we see that the multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, and 81. The multiples of 12 are 12, 24, 36, 48, 60, 72, and 84. The smallest common multiple is 36 months. Option A (72) is incorrect as it’s not the smallest shared maturity. Option C (108) is also incorrect; it exceeds the LCM. Option D (3) is far too short, as it does not accommodate either maturity period. Thus, 36 months is the earliest point both CDs will mature together.
To find when the CDs from both banks mature simultaneously, we need to determine the least common multiple (LCM) of their maturity periods: 9 months and 12 months. Calculating the LCM, we see that the multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, and 81. The multiples of 12 are 12, 24, 36, 48, 60, 72, and 84. The smallest common multiple is 36 months. Option A (72) is incorrect as it’s not the smallest shared maturity. Option C (108) is also incorrect; it exceeds the LCM. Option D (3) is far too short, as it does not accommodate either maturity period. Thus, 36 months is the earliest point both CDs will mature together.