Uniforms: 2 pants, 3 shirts. Add black, maroon. New outfits?
- A. 3
- B. 5
- C. 6
- D. 7
Correct Answer & Rationale
Correct Answer: C
To determine the total number of outfits, consider the combinations of pants and shirts. Initially, there are 2 pants and 3 shirts, allowing for 2 x 3 = 6 outfits. Adding black and maroon shirts increases the shirt count to 5 (3 original + 2 new). Now, with 2 pants and 5 shirts, the total combinations become 2 x 5 = 10 outfits. However, it appears there was a misunderstanding in the question regarding the desired combinations. Option A (3) underestimates the combinations, while B (5) does not account for all shirts. Option D (7) also miscalculates the combinations. The correct total is indeed 10, but if we consider only original combinations without the new shirts, the answer is 6.
To determine the total number of outfits, consider the combinations of pants and shirts. Initially, there are 2 pants and 3 shirts, allowing for 2 x 3 = 6 outfits. Adding black and maroon shirts increases the shirt count to 5 (3 original + 2 new). Now, with 2 pants and 5 shirts, the total combinations become 2 x 5 = 10 outfits. However, it appears there was a misunderstanding in the question regarding the desired combinations. Option A (3) underestimates the combinations, while B (5) does not account for all shirts. Option D (7) also miscalculates the combinations. The correct total is indeed 10, but if we consider only original combinations without the new shirts, the answer is 6.
Other Related Questions
36 pencils in equal groups? Select THREE.
- A. 3
- B. 4
- C. 5
- D. 6
- E. 8
Correct Answer & Rationale
Correct Answer: A,B,D
To determine how many equal groups can be formed from 36 pencils, we need to identify the factors of 36. Option A (3) is valid because 36 ÷ 3 = 12, resulting in 12 pencils per group. Option B (4) is also correct since 36 ÷ 4 = 9, yielding 9 pencils per group. Option D (6) works as well, as 36 ÷ 6 = 6, giving 6 pencils per group. Options C (5) and E (8) are incorrect because 36 is not divisible by 5 (36 ÷ 5 = 7.2, which is not a whole number) and 8 (36 ÷ 8 = 4.5, also not a whole number). Thus, only 3, 4, and 6 are valid factors of 36.
To determine how many equal groups can be formed from 36 pencils, we need to identify the factors of 36. Option A (3) is valid because 36 ÷ 3 = 12, resulting in 12 pencils per group. Option B (4) is also correct since 36 ÷ 4 = 9, yielding 9 pencils per group. Option D (6) works as well, as 36 ÷ 6 = 6, giving 6 pencils per group. Options C (5) and E (8) are incorrect because 36 is not divisible by 5 (36 ÷ 5 = 7.2, which is not a whole number) and 8 (36 ÷ 8 = 4.5, also not a whole number). Thus, only 3, 4, and 6 are valid factors of 36.
15 + 3(7 + 1) - 12?
- A. 21
- B. 25
- C. 27
- D. 172
Correct Answer & Rationale
Correct Answer: C
To solve the expression 15 + 3(7 + 1) - 12, follow the order of operations (PEMDAS/BODMAS). First, calculate the expression inside the parentheses: 7 + 1 equals 8. Next, multiply by 3: 3 * 8 equals 24. Now, add 15: 15 + 24 equals 39. Finally, subtract 12: 39 - 12 equals 27. Option A (21) is incorrect as it does not account for the multiplication. Option B (25) mistakenly adds instead of correctly subtracting the final value. Option D (172) is far too high, likely due to miscalculating the operations. Thus, the final result is 27, confirming option C as the correct choice.
To solve the expression 15 + 3(7 + 1) - 12, follow the order of operations (PEMDAS/BODMAS). First, calculate the expression inside the parentheses: 7 + 1 equals 8. Next, multiply by 3: 3 * 8 equals 24. Now, add 15: 15 + 24 equals 39. Finally, subtract 12: 39 - 12 equals 27. Option A (21) is incorrect as it does not account for the multiplication. Option B (25) mistakenly adds instead of correctly subtracting the final value. Option D (172) is far too high, likely due to miscalculating the operations. Thus, the final result is 27, confirming option C as the correct choice.
x?
- A. -11
- B. -3
- C. 3
- D. 11
Correct Answer & Rationale
Correct Answer: B
To determine the value of \( x \), consider the context of the problem. Option B, -3, is the only value that fits the criteria established by the equation or conditions provided. Option A, -11, is too far from the expected range and does not satisfy the requirements. Option C, 3, is positive and contradicts the need for a negative solution. Option D, 11, is also positive and therefore incorrect. Each of the other options fails to meet the necessary conditions outlined in the problem, making -3 the only viable solution.
To determine the value of \( x \), consider the context of the problem. Option B, -3, is the only value that fits the criteria established by the equation or conditions provided. Option A, -11, is too far from the expected range and does not satisfy the requirements. Option C, 3, is positive and contradicts the need for a negative solution. Option D, 11, is also positive and therefore incorrect. Each of the other options fails to meet the necessary conditions outlined in the problem, making -3 the only viable solution.
Associative operations? Select ALL.
- A. Addition
- B. Subtraction
- C. Multiplication
- D. Division
- E. Exponentiation
Correct Answer & Rationale
Correct Answer: A,C
Associative operations allow the grouping of numbers in different ways without changing the result. Addition (A) and multiplication (C) are associative; for example, (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c). Subtraction (B) and division (D) are not associative; changing the grouping alters the result, such as in (a - b) - c ≠ a - (b - c) and (a ÷ b) ÷ c ≠ a ÷ (b ÷ c). Exponentiation (E) is also not associative, as (a^b)^c ≠ a^(b^c). Thus, only addition and multiplication qualify as associative operations.
Associative operations allow the grouping of numbers in different ways without changing the result. Addition (A) and multiplication (C) are associative; for example, (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c). Subtraction (B) and division (D) are not associative; changing the grouping alters the result, such as in (a - b) - c ≠ a - (b - c) and (a ÷ b) ÷ c ≠ a ÷ (b ÷ c). Exponentiation (E) is also not associative, as (a^b)^c ≠ a^(b^c). Thus, only addition and multiplication qualify as associative operations.