A scale drawing of a truck has a length of 3 inches (in.), as shown below. The actual truck has a length of 18 feet (ft). What scale was used for the drawing?
- A. 6 in. = 1 ft
- B. 1 in. = 15 ft
- C. 1 in. = 6 ft
- D. 15 in. = 1 ft
Correct Answer & Rationale
Correct Answer: C
To determine the scale used for the drawing, we first convert the actual truck length from feet to inches. Since 1 foot equals 12 inches, an 18-foot truck is 216 inches long (18 ft x 12 in/ft). The scale drawing shows a length of 3 inches. To find the scale, we set up the ratio of the drawing length to the actual length: 3 in. (drawing) to 216 in. (actual). Simplifying this gives us a scale of 1 in. = 72 in., which translates to 1 in. = 6 ft (since 72 in. ÷ 12 in/ft = 6 ft). Option A (6 in. = 1 ft) is incorrect; it implies a much larger drawing. Option B (1 in. = 15 ft) underestimates the actual size. Option D (15 in. = 1 ft) greatly exaggerates the scale, making the drawing too small.
To determine the scale used for the drawing, we first convert the actual truck length from feet to inches. Since 1 foot equals 12 inches, an 18-foot truck is 216 inches long (18 ft x 12 in/ft). The scale drawing shows a length of 3 inches. To find the scale, we set up the ratio of the drawing length to the actual length: 3 in. (drawing) to 216 in. (actual). Simplifying this gives us a scale of 1 in. = 72 in., which translates to 1 in. = 6 ft (since 72 in. ÷ 12 in/ft = 6 ft). Option A (6 in. = 1 ft) is incorrect; it implies a much larger drawing. Option B (1 in. = 15 ft) underestimates the actual size. Option D (15 in. = 1 ft) greatly exaggerates the scale, making the drawing too small.
Other Related Questions
Solve the inequality for x: -4/3 x + 4 ? 16
- A. x??9
- B. x??9
- C. x??9
- D. x?9
Correct Answer & Rationale
Correct Answer: A
To solve the inequality \(-\frac{4}{3}x + 4 < 16\), first isolate \(x\) by subtracting 4 from both sides, resulting in \(-\frac{4}{3}x < 12\). Next, multiply both sides by \(-\frac{3}{4}\), remembering to reverse the inequality sign, yielding \(x > 9\). Options B and C incorrectly suggest \(x < 9\), which contradicts our solution. Option D, stating \(x \leq 9\), also misrepresents the inequality since it does not include values greater than 9. Thus, only option A accurately reflects the solution \(x > 9\).
To solve the inequality \(-\frac{4}{3}x + 4 < 16\), first isolate \(x\) by subtracting 4 from both sides, resulting in \(-\frac{4}{3}x < 12\). Next, multiply both sides by \(-\frac{3}{4}\), remembering to reverse the inequality sign, yielding \(x > 9\). Options B and C incorrectly suggest \(x < 9\), which contradicts our solution. Option D, stating \(x \leq 9\), also misrepresents the inequality since it does not include values greater than 9. Thus, only option A accurately reflects the solution \(x > 9\).
Ricardo has two bank accounts. Each month, he will withdraw a certain amount of money from the first account and deposit a different amount of money into the second account. The inequality 8,000 – 200x ? 5,000 + 300x can be solved to find the number of months, x, for which the account has more money than the second account. What is the solution to this inequality?
- A. x ? 6
- B. x ? 30
- C. x ? 30
- D. x ? 6
Correct Answer & Rationale
Correct Answer: D
To solve the inequality \( 8,000 - 200x > 5,000 + 300x \), we first isolate \( x \). Rearranging gives \( 8,000 - 5,000 > 300x + 200x \), simplifying to \( 3,000 > 500x \). Dividing by 500 results in \( x < 6 \). Thus, the solution indicates that for \( x \) to ensure the first account has more money, it must be less than 6 months. Option A incorrectly states \( x \geq 6 \), which contradicts the solution. Options B and C mistakenly suggest \( x \geq 30 \), which is not relevant to the problem.
To solve the inequality \( 8,000 - 200x > 5,000 + 300x \), we first isolate \( x \). Rearranging gives \( 8,000 - 5,000 > 300x + 200x \), simplifying to \( 3,000 > 500x \). Dividing by 500 results in \( x < 6 \). Thus, the solution indicates that for \( x \) to ensure the first account has more money, it must be less than 6 months. Option A incorrectly states \( x \geq 6 \), which contradicts the solution. Options B and C mistakenly suggest \( x \geq 30 \), which is not relevant to the problem.
A bag of dog food weighs 40 pounds. The amount of food in the bag is more than 3 times the amount needed to feed a dog for one week. Which inequality can be used to determine the possible values for p, the pounds of food needed to feed the dog for one week?
- A. p < 3(40)
- B. 3p < 40
- C. p > 3(40)
- D. 3p > 40
Correct Answer & Rationale
Correct Answer: D
To find the amount of food needed for one week, we know that the total weight of the dog food (40 pounds) is more than three times the weekly requirement (3p). Therefore, the relationship can be expressed as 3p < 40, indicating that the total food exceeds three times the weekly amount. Option A (p < 3(40)) incorrectly suggests that the weekly requirement is less than three times the total weight, which is not supported by the problem statement. Option B (3p < 40) misrepresents the relationship, as it implies the total food is less than three times the weekly need, contradicting the given information. Option C (p > 3(40)) inaccurately states that the weekly requirement exceeds three times the total weight, which is impossible given the context. Thus, the correct inequality is 3p > 40, indicating the total food is indeed more than three times the weekly requirement.
To find the amount of food needed for one week, we know that the total weight of the dog food (40 pounds) is more than three times the weekly requirement (3p). Therefore, the relationship can be expressed as 3p < 40, indicating that the total food exceeds three times the weekly amount. Option A (p < 3(40)) incorrectly suggests that the weekly requirement is less than three times the total weight, which is not supported by the problem statement. Option B (3p < 40) misrepresents the relationship, as it implies the total food is less than three times the weekly need, contradicting the given information. Option C (p > 3(40)) inaccurately states that the weekly requirement exceeds three times the total weight, which is impossible given the context. Thus, the correct inequality is 3p > 40, indicating the total food is indeed more than three times the weekly requirement.
Fix It Fast is an auto repair shop that employs 10 mechanics. Each day, the shop owner randomly picks 1 mechanic to receive a free lunch. What is the probability the shop owner will pick the same mechanic to receive a free lunch 2 days in a row?
- A. 1\20
- B. 1/100
- C. 1\5
- D. 1\10
Correct Answer & Rationale
Correct Answer: B
To determine the probability of picking the same mechanic two days in a row, we start by recognizing that there are 10 mechanics. On the first day, any mechanic can be chosen, which does not affect the overall probability. On the second day, to pick the same mechanic again, there is only 1 favorable outcome (the chosen mechanic) out of 10 possible mechanics. Thus, the probability of selecting that same mechanic on the second day is 1/10. Since the first day's choice does not influence this, we multiply the probabilities: (1/10) * (1/10) = 1/100. - Option A (1/20) is incorrect as it miscalculates the favorable outcomes. - Option C (1/5) incorrectly assumes a higher likelihood without considering the second day's requirement. - Option D (1/10) only reflects the probability of picking a mechanic on day two, not the two-day scenario.
To determine the probability of picking the same mechanic two days in a row, we start by recognizing that there are 10 mechanics. On the first day, any mechanic can be chosen, which does not affect the overall probability. On the second day, to pick the same mechanic again, there is only 1 favorable outcome (the chosen mechanic) out of 10 possible mechanics. Thus, the probability of selecting that same mechanic on the second day is 1/10. Since the first day's choice does not influence this, we multiply the probabilities: (1/10) * (1/10) = 1/100. - Option A (1/20) is incorrect as it miscalculates the favorable outcomes. - Option C (1/5) incorrectly assumes a higher likelihood without considering the second day's requirement. - Option D (1/10) only reflects the probability of picking a mechanic on day two, not the two-day scenario.