Cost of 3 cans of peaches is $2.67. Cost of 8 cans?
- A. $5.34
- B. $7.12
- C. $8.01
- D. $21.36
Correct Answer & Rationale
Correct Answer: B
To determine the cost of 8 cans of peaches, first calculate the cost per can. The cost of 3 cans is $2.67, so the cost per can is $2.67 ÷ 3 = $0.89. To find the cost of 8 cans, multiply the cost per can by 8: $0.89 × 8 = $7.12. Option A ($5.34) incorrectly assumes a lower total based on miscalculated per can pricing. Option C ($8.01) slightly overestimates the total, likely from rounding errors. Option D ($21.36) suggests a misunderstanding of basic multiplication, as it implies a much higher price than calculated. Thus, $7.12 accurately reflects the cost for 8 cans.
To determine the cost of 8 cans of peaches, first calculate the cost per can. The cost of 3 cans is $2.67, so the cost per can is $2.67 ÷ 3 = $0.89. To find the cost of 8 cans, multiply the cost per can by 8: $0.89 × 8 = $7.12. Option A ($5.34) incorrectly assumes a lower total based on miscalculated per can pricing. Option C ($8.01) slightly overestimates the total, likely from rounding errors. Option D ($21.36) suggests a misunderstanding of basic multiplication, as it implies a much higher price than calculated. Thus, $7.12 accurately reflects the cost for 8 cans.
Other Related Questions
Point (-3,-6) quadrant?
- A. I
- B. II
- C. III
- D. IV
Correct Answer & Rationale
Correct Answer: C
The point (-3, -6) is located in the Cartesian coordinate system where the x-coordinate is negative and the y-coordinate is also negative. This combination places the point in Quadrant III, where both x and y values are less than zero. Option A (I) is incorrect as Quadrant I contains positive x and y values. Option B (II) is wrong because Quadrant II has a negative x value and a positive y value. Option D (IV) is not applicable since Quadrant IV features a positive x value and a negative y value. Thus, the only quadrant that matches the coordinates (-3, -6) is Quadrant III.
The point (-3, -6) is located in the Cartesian coordinate system where the x-coordinate is negative and the y-coordinate is also negative. This combination places the point in Quadrant III, where both x and y values are less than zero. Option A (I) is incorrect as Quadrant I contains positive x and y values. Option B (II) is wrong because Quadrant II has a negative x value and a positive y value. Option D (IV) is not applicable since Quadrant IV features a positive x value and a negative y value. Thus, the only quadrant that matches the coordinates (-3, -6) is Quadrant III.
Square side 5(1/2)cm. Area?
Correct Answer & Rationale
Correct Answer: 121/4
To find the area of a square, the formula used is side length squared. Here, the side length is 5(1/2) cm, which converts to 5.5 cm or 11/2 cm. Squaring this value gives (11/2)² = 121/4 cm², confirming the correct area. The other options are incorrect because: - If calculated as 5 cm, the area would be 25 cm², neglecting the fractional part. - If 5.5 cm is incorrectly squared as 30.25 cm², it miscalculates the area. - Any other value derived from misinterpretation of the side length will not yield the correct area.
To find the area of a square, the formula used is side length squared. Here, the side length is 5(1/2) cm, which converts to 5.5 cm or 11/2 cm. Squaring this value gives (11/2)² = 121/4 cm², confirming the correct area. The other options are incorrect because: - If calculated as 5 cm, the area would be 25 cm², neglecting the fractional part. - If 5.5 cm is incorrectly squared as 30.25 cm², it miscalculates the area. - Any other value derived from misinterpretation of the side length will not yield the correct area.
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.
Which inequality?
- A. 2(x+1)<x
- B. x+2(x+1)>-1
- C. x<2x-1
- D. 2(x/2+1)<1
Correct Answer & Rationale
Correct Answer: C
Option C, \( x < 2x - 1 \), simplifies to \( x - 2x < -1 \), leading to \( -x < -1 \) or \( x > 1 \). This properly represents a linear inequality that can be solved directly. Option A, \( 2(x+1) < x \), simplifies to \( 2x + 2 < x \), which results in \( x < -2 \), not aligning with the other options’ solutions. Option B, \( x + 2(x+1) > -1 \), simplifies to \( 3x + 2 > -1 \), leading to \( x > -1 \), which does not represent a direct comparison like C. Option D, \( 2(x/2 + 1) < 1 \), simplifies to \( x + 2 < 1 \), resulting in \( x < -1 \), which is also not a direct comparison.
Option C, \( x < 2x - 1 \), simplifies to \( x - 2x < -1 \), leading to \( -x < -1 \) or \( x > 1 \). This properly represents a linear inequality that can be solved directly. Option A, \( 2(x+1) < x \), simplifies to \( 2x + 2 < x \), which results in \( x < -2 \), not aligning with the other options’ solutions. Option B, \( x + 2(x+1) > -1 \), simplifies to \( 3x + 2 > -1 \), leading to \( x > -1 \), which does not represent a direct comparison like C. Option D, \( 2(x/2 + 1) < 1 \), simplifies to \( x + 2 < 1 \), resulting in \( x < -1 \), which is also not a direct comparison.