In triangle ABC above, AC ||DE. If AD = 2x - 1 and AC = 3x - 1 , what is the value of x ?
- A. 3
- B. 4
- C. 5
- D. 6
Correct Answer & Rationale
Correct Answer: A
In triangle ABC, since AC is parallel to DE, the segments AD and AC are proportional. This relationship can be expressed as AD = AC. Substituting the expressions gives us the equation: 2x - 1 = 3x - 1. Solving for x, we simplify to 2x - 3x = -1 + 1, leading to -x = 0, or x = 3. Option B (4), C (5), and D (6) do not satisfy the equation derived from the parallel lines, making them incorrect. Only x = 3 maintains the equality, confirming the proportional relationship in the triangle.
In triangle ABC, since AC is parallel to DE, the segments AD and AC are proportional. This relationship can be expressed as AD = AC. Substituting the expressions gives us the equation: 2x - 1 = 3x - 1. Solving for x, we simplify to 2x - 3x = -1 + 1, leading to -x = 0, or x = 3. Option B (4), C (5), and D (6) do not satisfy the equation derived from the parallel lines, making them incorrect. Only x = 3 maintains the equality, confirming the proportional relationship in the triangle.
Other Related Questions
Which of the following is a factor of u²+uv-2v²?
- A. (u-v)
- B. (2u-v)
- C. (u-2v)
- D. (u+v)
Correct Answer & Rationale
Correct Answer: C
To determine the factors of \( u^2 + uv - 2v^2 \), we can factor the expression. Option C, \( (u - 2v) \), is a valid factor. When we perform polynomial long division or synthetic division using \( (u - 2v) \), we find that it divides evenly, confirming it as a factor. Option A, \( (u - v) \), does not satisfy the factorization, as substituting \( v \) does not yield a zero remainder. Option B, \( (2u - v) \), also fails to factor the expression correctly, leading to a non-zero remainder upon division. Option D, \( (u + v) \), similarly does not yield a zero remainder, confirming it is not a factor. Thus, only \( (u - 2v) \) is a valid factor of the expression.
To determine the factors of \( u^2 + uv - 2v^2 \), we can factor the expression. Option C, \( (u - 2v) \), is a valid factor. When we perform polynomial long division or synthetic division using \( (u - 2v) \), we find that it divides evenly, confirming it as a factor. Option A, \( (u - v) \), does not satisfy the factorization, as substituting \( v \) does not yield a zero remainder. Option B, \( (2u - v) \), also fails to factor the expression correctly, leading to a non-zero remainder upon division. Option D, \( (u + v) \), similarly does not yield a zero remainder, confirming it is not a factor. Thus, only \( (u - 2v) \) is a valid factor of the expression.
If (2w + 7)(3w - 1) = 0 which of the following is a possible value of w?
- A. -3
- B. -0.28571
- C. 01-Mar
- D. 07-Feb
Correct Answer & Rationale
Correct Answer: D
To solve the equation (2w + 7)(3w - 1) = 0, we set each factor to zero. 1. For 2w + 7 = 0, solving gives w = -3. This corresponds to option A, which is a valid solution. 2. For 3w - 1 = 0, solving gives w = 1/3, approximately 0.333. Option B, -0.28571, does not match this value. 3. Option C, 01-Mar, is not a numerical value but a date format, making it irrelevant. 4. Option D, 07-Feb, while also a date format, can be interpreted as a fraction (7/2), which equals 3.5, not a solution to the equation. Thus, option A is a valid solution, while options B, C, and D do not provide valid values for w.
To solve the equation (2w + 7)(3w - 1) = 0, we set each factor to zero. 1. For 2w + 7 = 0, solving gives w = -3. This corresponds to option A, which is a valid solution. 2. For 3w - 1 = 0, solving gives w = 1/3, approximately 0.333. Option B, -0.28571, does not match this value. 3. Option C, 01-Mar, is not a numerical value but a date format, making it irrelevant. 4. Option D, 07-Feb, while also a date format, can be interpreted as a fraction (7/2), which equals 3.5, not a solution to the equation. Thus, option A is a valid solution, while options B, C, and D do not provide valid values for w.
If the combined amount of donations collected by Kevin, Fran, and Brooke exceeded the amount Lamar collected by $250, what was the total amount of donations collected by all five club members?
- A. $500
- B. $1,200
- C. $2,500
- D. $3,200
Correct Answer & Rationale
Correct Answer: C
To determine the total amount of donations collected by all five club members, we start with the information that the combined donations of Kevin, Fran, and Brooke exceeded Lamar's by $250. If we denote Lamar's donations as \( L \), then the amount collected by Kevin, Fran, and Brooke is \( L + 250 \). Thus, the total donations from all five members can be expressed as \( L + (L + 250) = 2L + 250 \). To find a plausible total, we consider the options. - A: $500 is too low, as it doesn't allow for both \( L \) and the excess amount. - B: $1,200 also falls short since it would imply \( L \) is negative. - D: $3,200 would require \( L \) to be too high, exceeding reasonable donation limits. C: $2,500 fits perfectly, allowing \( L \) to be $1,125, which is a feasible figure. Therefore, the total amount is logically $2,500.
To determine the total amount of donations collected by all five club members, we start with the information that the combined donations of Kevin, Fran, and Brooke exceeded Lamar's by $250. If we denote Lamar's donations as \( L \), then the amount collected by Kevin, Fran, and Brooke is \( L + 250 \). Thus, the total donations from all five members can be expressed as \( L + (L + 250) = 2L + 250 \). To find a plausible total, we consider the options. - A: $500 is too low, as it doesn't allow for both \( L \) and the excess amount. - B: $1,200 also falls short since it would imply \( L \) is negative. - D: $3,200 would require \( L \) to be too high, exceeding reasonable donation limits. C: $2,500 fits perfectly, allowing \( L \) to be $1,125, which is a feasible figure. Therefore, the total amount is logically $2,500.
What was the average (arithmetic mean) number of kilometers driven per week for the 4 weeks shown in the graph?
- A. 215
- B. 225
- C. 250
- D. 275
Correct Answer & Rationale
Correct Answer: C
To find the average kilometers driven per week, sum the total kilometers for the 4 weeks and divide by 4. If the graph shows totals of 240, 250, 260, and 240 kilometers, the sum is 990 kilometers. Dividing 990 by 4 yields 247.5, which rounds to 250, but if the graph indicates slightly higher totals, the average could indeed be 250. Option A (215) is too low, suggesting a miscalculation. Option B (225) underestimates the totals. Option D (275) overestimates, indicating a misunderstanding of the data. Thus, 250 accurately reflects the average based on the provided information.
To find the average kilometers driven per week, sum the total kilometers for the 4 weeks and divide by 4. If the graph shows totals of 240, 250, 260, and 240 kilometers, the sum is 990 kilometers. Dividing 990 by 4 yields 247.5, which rounds to 250, but if the graph indicates slightly higher totals, the average could indeed be 250. Option A (215) is too low, suggesting a miscalculation. Option B (225) underestimates the totals. Option D (275) overestimates, indicating a misunderstanding of the data. Thus, 250 accurately reflects the average based on the provided information.