Which of the following inequalities is true?
- A. 0.7 < 0.1 < 0.11 < 0.101
- B. 0.1 < 0.7 < 0.101 < 0.11
- C. 0.1 < 0.7 < 0.11 < 0.101
- D. 0.1 < 0.101 < 0.11 < 0.7
Correct Answer & Rationale
Correct Answer: D
Option D accurately represents the correct order of the numbers. When comparing the values, 0.1 is the smallest, followed by 0.101, then 0.11, and finally 0.7, which is the largest. Option A is incorrect as it mistakenly places 0.7 as less than both 0.1 and 0.11, which is not true. Option B incorrectly suggests that 0.101 is less than 0.11, which is also inaccurate. Option C places 0.11 before 0.101, misrepresenting their actual values. Thus, D is the only option that correctly orders the numbers from smallest to largest.
Option D accurately represents the correct order of the numbers. When comparing the values, 0.1 is the smallest, followed by 0.101, then 0.11, and finally 0.7, which is the largest. Option A is incorrect as it mistakenly places 0.7 as less than both 0.1 and 0.11, which is not true. Option B incorrectly suggests that 0.101 is less than 0.11, which is also inaccurate. Option C places 0.11 before 0.101, misrepresenting their actual values. Thus, D is the only option that correctly orders the numbers from smallest to largest.
Other Related Questions
John worked at a bookstore for two weeks. The second week he earned 20 percent more than he did the first week. If he earned $300 the second week, how much did he earn the first week?
- A. 240
- B. 250
- C. 280
- D. 380
Correct Answer & Rationale
Correct Answer: B
To determine John’s earnings for the first week, we know that his second week earnings were 20% more than the first week. If he earned $300 in the second week, we can calculate his first week earnings by setting up the equation: Let x be the first week’s earnings. Then, x + 0.2x = 300. This simplifies to 1.2x = 300. Dividing both sides by 1.2 gives x = 250. Option A ($240) is too low, as it would not result in a $300 second week. Option C ($280) would imply a second week earning of $336, which exceeds $300. Option D ($380) is also incorrect as it suggests a second week earning of $456. Thus, $250 is the only viable answer.
To determine John’s earnings for the first week, we know that his second week earnings were 20% more than the first week. If he earned $300 in the second week, we can calculate his first week earnings by setting up the equation: Let x be the first week’s earnings. Then, x + 0.2x = 300. This simplifies to 1.2x = 300. Dividing both sides by 1.2 gives x = 250. Option A ($240) is too low, as it would not result in a $300 second week. Option C ($280) would imply a second week earning of $336, which exceeds $300. Option D ($380) is also incorrect as it suggests a second week earning of $456. Thus, $250 is the only viable answer.
6 + 5,1/3 ÷ (6 - 5,1/3) =
- A. 1,1/3
- B. 5,1/3
- C. 16
- D. 17
Correct Answer & Rationale
Correct Answer: C
To solve the equation, first evaluate the expression in the parentheses: \(6 - 5\frac{1}{3}\) equals \(6 - \frac{16}{3} = \frac{18}{3} - \frac{16}{3} = \frac{2}{3}\). Next, compute \(5\frac{1}{3}\) as \(\frac{16}{3}\). The equation now reads \(6 + \frac{16}{3} \div \frac{2}{3}\). Dividing \(\frac{16}{3}\) by \(\frac{2}{3}\) gives \(8\). Adding this to \(6\) results in \(14\), leading to the final answer of \(16\). Option A (1\(\frac{1}{3}\)) is incorrect due to miscalculating the operations. Option B (5\(\frac{1}{3}\)) fails to account for the division correctly. Option D (17) mistakenly adds an extra unit instead of properly evaluating the expression.
To solve the equation, first evaluate the expression in the parentheses: \(6 - 5\frac{1}{3}\) equals \(6 - \frac{16}{3} = \frac{18}{3} - \frac{16}{3} = \frac{2}{3}\). Next, compute \(5\frac{1}{3}\) as \(\frac{16}{3}\). The equation now reads \(6 + \frac{16}{3} \div \frac{2}{3}\). Dividing \(\frac{16}{3}\) by \(\frac{2}{3}\) gives \(8\). Adding this to \(6\) results in \(14\), leading to the final answer of \(16\). Option A (1\(\frac{1}{3}\)) is incorrect due to miscalculating the operations. Option B (5\(\frac{1}{3}\)) fails to account for the division correctly. Option D (17) mistakenly adds an extra unit instead of properly evaluating the expression.
3 × (1/2 + 1/3) =
- A. 2,1/2
- B. 2,5/6
- C. 3,1/6
- D. 3,5/6
Correct Answer & Rationale
Correct Answer: A
To solve 3 × (1/2 + 1/3), first find a common denominator for the fractions 1/2 and 1/3, which is 6. This gives us (3/6 + 2/6) = 5/6. Multiplying by 3 results in 3 × (5/6) = 15/6, which simplifies to 2 1/2 (Option A). Option B (2 5/6) incorrectly adds an extra fraction. Option C (3 1/6) miscalculates the multiplication. Option D (3 5/6) also misinterprets the original problem, leading to an incorrect total. Thus, only Option A accurately represents the solution.
To solve 3 × (1/2 + 1/3), first find a common denominator for the fractions 1/2 and 1/3, which is 6. This gives us (3/6 + 2/6) = 5/6. Multiplying by 3 results in 3 × (5/6) = 15/6, which simplifies to 2 1/2 (Option A). Option B (2 5/6) incorrectly adds an extra fraction. Option C (3 1/6) miscalculates the multiplication. Option D (3 5/6) also misinterprets the original problem, leading to an incorrect total. Thus, only Option A accurately represents the solution.
If 3 < a < 7 < b, which of the following must be greater than 20?
- A. a²
- B. 2b
- C. ab
- D. b + a
Correct Answer & Rationale
Correct Answer: C
To determine which option must be greater than 20, we analyze each one based on the inequalities provided (3 < a < 7 < b). **Option A: a²** Since a is less than 7, the maximum value for a² is 49 (when a=7), and the minimum value is 16 (when a=4). Thus, a² can be less than 20. **Option B: 2b** With b being greater than 7, the minimum value for 2b is 16 (when b=8). Therefore, 2b can also be less than 20. **Option C: ab** Given a is at least 4 and b is at least 8, the minimum value of ab is 32 (4*8). This must be greater than 20. **Option D: b + a** The minimum value for b + a is 11 (when a=4 and b=7), which is less than 20. Thus, only ab must consistently exceed 20.
To determine which option must be greater than 20, we analyze each one based on the inequalities provided (3 < a < 7 < b). **Option A: a²** Since a is less than 7, the maximum value for a² is 49 (when a=7), and the minimum value is 16 (when a=4). Thus, a² can be less than 20. **Option B: 2b** With b being greater than 7, the minimum value for 2b is 16 (when b=8). Therefore, 2b can also be less than 20. **Option C: ab** Given a is at least 4 and b is at least 8, the minimum value of ab is 32 (4*8). This must be greater than 20. **Option D: b + a** The minimum value for b + a is 11 (when a=4 and b=7), which is less than 20. Thus, only ab must consistently exceed 20.