Rounds to 87.5 in tenths?
- A. 88
- B. 87.56
- C. 87.459
- D. 87.05
Correct Answer & Rationale
Correct Answer: C
When rounding to the nearest tenth, the digit in the hundredths place determines whether to round up or down. For 87.5, the first digit after the decimal is 5, indicating that we round up. Option A (88) rounds to the nearest whole number, not the nearest tenth. Option B (87.56) rounds to 87.6, which is higher than 87.5. Option D (87.05) rounds to 87.1, which is lower. Only option C (87.459) rounds to 87.5 when considering the tenths place, making it the only valid choice for rounding to 87.5 in tenths.
When rounding to the nearest tenth, the digit in the hundredths place determines whether to round up or down. For 87.5, the first digit after the decimal is 5, indicating that we round up. Option A (88) rounds to the nearest whole number, not the nearest tenth. Option B (87.56) rounds to 87.6, which is higher than 87.5. Option D (87.05) rounds to 87.1, which is lower. Only option C (87.459) rounds to 87.5 when considering the tenths place, making it the only valid choice for rounding to 87.5 in tenths.
Other Related Questions
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.
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.
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.
P=2(L+W), P=48, W=L-4. Width?
- A. 10
- B. 12
- C. 20
- D. 24
Correct Answer & Rationale
Correct Answer: A
To find the width (W), start with the given perimeter formula \( P = 2(L + W) \). Substituting \( P = 48 \) gives \( 48 = 2(L + W) \), which simplifies to \( L + W = 24 \). Given \( W = L - 4 \), substitute this into the equation: \( L + (L - 4) = 24 \). This simplifies to \( 2L - 4 = 24 \), leading to \( 2L = 28 \) and \( L = 14 \). Thus, \( W = 14 - 4 = 10 \). Option B (12) does not satisfy the perimeter equation. Option C (20) and Option D (24) also do not fit the derived equations, confirming that W must be 10.
To find the width (W), start with the given perimeter formula \( P = 2(L + W) \). Substituting \( P = 48 \) gives \( 48 = 2(L + W) \), which simplifies to \( L + W = 24 \). Given \( W = L - 4 \), substitute this into the equation: \( L + (L - 4) = 24 \). This simplifies to \( 2L - 4 = 24 \), leading to \( 2L = 28 \) and \( L = 14 \). Thus, \( W = 14 - 4 = 10 \). Option B (12) does not satisfy the perimeter equation. Option C (20) and Option D (24) also do not fit the derived equations, confirming that W must be 10.