Liz spent 1/2, 1/3, 1/4, $15 left. Birthday money?
- A. $360
- B. $180
- C. $120
- D. $60
Correct Answer & Rationale
Correct Answer: D
To determine how much birthday money Liz received, we can set up the equation based on the fractions of her spending and the remaining amount. Let \( x \) represent the total birthday money. She spent \( \frac{1}{2}x + \frac{1}{3}x + \frac{1}{4}x + 15 = x \). Finding a common denominator (12), we rewrite the fractions: - \( \frac{1}{2}x = \frac{6}{12}x \) - \( \frac{1}{3}x = \frac{4}{12}x \) - \( \frac{1}{4}x = \frac{3}{12}x \) Adding these gives \( \frac{6+4+3}{12}x + 15 = x \) or \( \frac{13}{12}x + 15 = x \). Rearranging yields \( 15 = x - \frac{13}{12}x \), simplifying to \( 15 = \frac{1}{12}x \). Therefore, \( x = 180 \). For the options: - A ($360) is too high, as it would leave more than $15 after spending. - B ($180) results in no remaining amount after spending. - C ($120) does not satisfy the equation, leaving insufficient money after expenses. - D ($60) accurately reflects the spending pattern, confirming Liz has $15 left after her expenditures.
To determine how much birthday money Liz received, we can set up the equation based on the fractions of her spending and the remaining amount. Let \( x \) represent the total birthday money. She spent \( \frac{1}{2}x + \frac{1}{3}x + \frac{1}{4}x + 15 = x \). Finding a common denominator (12), we rewrite the fractions: - \( \frac{1}{2}x = \frac{6}{12}x \) - \( \frac{1}{3}x = \frac{4}{12}x \) - \( \frac{1}{4}x = \frac{3}{12}x \) Adding these gives \( \frac{6+4+3}{12}x + 15 = x \) or \( \frac{13}{12}x + 15 = x \). Rearranging yields \( 15 = x - \frac{13}{12}x \), simplifying to \( 15 = \frac{1}{12}x \). Therefore, \( x = 180 \). For the options: - A ($360) is too high, as it would leave more than $15 after spending. - B ($180) results in no remaining amount after spending. - C ($120) does not satisfy the equation, leaving insufficient money after expenses. - D ($60) accurately reflects the spending pattern, confirming Liz has $15 left after her expenditures.
Other Related Questions
Quickly multiply 24x16?
- A. 20x20-4x4
- B. 20x20
- C. 20x10+4x6
- D. 25x10+4x15
Correct Answer & Rationale
Correct Answer: A
Option A, 20x20 - 4x4, effectively utilizes the difference of squares method. It simplifies the multiplication by recognizing that 24 can be expressed as 20 + 4 and 16 as 20 - 4, leading to a calculation of (20+4)(20-4). Option B, 20x20, underestimates the value of 24 and 16, yielding only 400 instead of the correct 384. Option C, 20x10 + 4x6, inaccurately breaks down the multiplication, leading to 200 + 24, which totals 224. Option D, 25x10 + 4x15, misrepresents the factors, resulting in 250 + 60, totaling 310. Thus, option A is the most accurate approach for this multiplication.
Option A, 20x20 - 4x4, effectively utilizes the difference of squares method. It simplifies the multiplication by recognizing that 24 can be expressed as 20 + 4 and 16 as 20 - 4, leading to a calculation of (20+4)(20-4). Option B, 20x20, underestimates the value of 24 and 16, yielding only 400 instead of the correct 384. Option C, 20x10 + 4x6, inaccurately breaks down the multiplication, leading to 200 + 24, which totals 224. Option D, 25x10 + 4x15, misrepresents the factors, resulting in 250 + 60, totaling 310. Thus, option A is the most accurate approach for this multiplication.
Measure changed?
- A. Mean
- B. Median
- C. Mode
- D. Range
Correct Answer & Rationale
Correct Answer: A
The mean is sensitive to changes in data values, as it considers all values in a dataset. When any value changes, the mean will adjust accordingly, reflecting the new average. The median, on the other hand, represents the middle value and is only affected if the change impacts the central position of the dataset. The mode, being the most frequently occurring value, is not influenced by changes in other data points unless the frequency of occurrence alters. Lastly, the range measures the difference between the highest and lowest values, which may not change if the data alteration occurs within the existing range.
The mean is sensitive to changes in data values, as it considers all values in a dataset. When any value changes, the mean will adjust accordingly, reflecting the new average. The median, on the other hand, represents the middle value and is only affected if the change impacts the central position of the dataset. The mode, being the most frequently occurring value, is not influenced by changes in other data points unless the frequency of occurrence alters. Lastly, the range measures the difference between the highest and lowest values, which may not change if the data alteration occurs within the existing range.
Prime numbers? Select ALL.
- A. 21
- B. 23
- C. 25
- D. 27
- E. 29
Correct Answer & Rationale
Correct Answer: B,E
Prime numbers are defined as natural numbers greater than 1 that have no positive divisors other than 1 and themselves. - **Option A: 21** is not prime because it can be divided by 1, 3, 7, and 21. - **Option B: 23** is prime; it has no divisors other than 1 and 23. - **Option C: 25** is not prime as it can be divided by 1, 5, and 25. - **Option D: 27** is not prime since it can be divided by 1, 3, 9, and 27. - **Option E: 29** is prime; it has no divisors other than 1 and 29. Thus, 23 and 29 are the only prime numbers in the list.
Prime numbers are defined as natural numbers greater than 1 that have no positive divisors other than 1 and themselves. - **Option A: 21** is not prime because it can be divided by 1, 3, 7, and 21. - **Option B: 23** is prime; it has no divisors other than 1 and 23. - **Option C: 25** is not prime as it can be divided by 1, 5, and 25. - **Option D: 27** is not prime since it can be divided by 1, 3, 9, and 27. - **Option E: 29** is prime; it has no divisors other than 1 and 29. Thus, 23 and 29 are the only prime numbers in the list.
Answerable?
- A. 4.5 pounds?
- B. At least 15?
- C. Less than 8?
- D. 6-12 pounds?
Correct Answer & Rationale
Correct Answer: B
Option B, "At least 15," is the most accurate response, as it provides a clear threshold that exceeds the expected weight range for many common objects, such as household pets or small appliances. Option A, "4.5 pounds," is too low for many items, making it an unreliable estimate. Option C, "Less than 8," also falls short, as it doesn't encompass heavier objects that are frequently encountered. Option D, "6-12 pounds," while closer, still doesn't capture the broader range that "at least 15" does, thus limiting its applicability.
Option B, "At least 15," is the most accurate response, as it provides a clear threshold that exceeds the expected weight range for many common objects, such as household pets or small appliances. Option A, "4.5 pounds," is too low for many items, making it an unreliable estimate. Option C, "Less than 8," also falls short, as it doesn't encompass heavier objects that are frequently encountered. Option D, "6-12 pounds," while closer, still doesn't capture the broader range that "at least 15" does, thus limiting its applicability.