PSAT Question of the Day

Rationale:
First, determine the rate at which Maria is saving money per week. From the table, the savings increase consistently:

• Week 2 - Week 1: $290 - $275 = $15
• Week 3 - Week 2: $305 - $290 = $15
• Week 4 - Week 3: $320 - $305 = $15

Maria is saving $15 per week. We can model this with a linear equation. Let S be the total savings and W be the week number. The general form is S = mW + b, where m is the rate ($15) and b is the initial amount (savings at week 0).

Using the data from Week 1 (W=1, S=275):
275 = 15(1) + b
275 = 15 + b
b = 260

So, the equation for Maria's savings is S = 15W + 260.

Next, we need to find the week W when Maria's savings S reach $400:
400 = 15W + 260
140 = 15W
W = 140 / 15 = 28 / 3 ≈ 9.33

This means Maria will have enough money sometime during the 9th week after week 0, which corresponds to week 9.33. Since the question asks for the week number *during* which she will reach her goal, we need to consider when her savings *first meet or exceed* $400.

• At the end of Week 9: S = 15(9) + 260 = 135 + 260 = $395 (not yet $400).
• At the end of Week 10: S = 15(10) + 260 = 150 + 260 = $410 (exceeds $400).

Therefore, Maria will reach her goal of $400 during Week 10.

Distractor Analysis:

• 9: This answer results from rounding down 9.33. While she is very close to $400 by the end of week 9 ($395), she has not yet reached it. The goal is met *during* the subsequent week.
• 8: This distractor could arise from misinterpreting the initial value or incorrectly calculating the number of additional weeks needed. For example, if a student calculates (400 - 275) / 15 = 125 / 15 = 8.33 and then incorrectly assumes this is the final week number without accounting for Week 1, or if they define Week 1 as Week 0 and round down 8.33.
• 11: This could result from an error in rounding up the additional weeks needed. If a student calculates that 80 more dollars are needed from Week 4 ($400 - $320 = $80), and $80 / $15 per week ≈ 5.33 additional weeks. Rounding 5.33 up to 6 additional weeks would mean 4 + 6 = 10. An error in rounding 5.33 up to 7 additional weeks would lead to 4 + 7 = 11.

Rationale:

A system of two linear equations has no solutions if the lines they represent are parallel but distinct, meaning they have the same slope but different y-intercepts. For the given first equation, 2x - 3y = 7, the equation 2x - 3y = 10 maintains the exact same coefficients for x and y (thus the same slope) but has a different constant term, ensuring the lines are parallel and distinct.

Conversely, an equation like 3x - 2y = 7 or 2x + 3y = 7 would create a system with one unique solution because their differing coefficients for x and y result in different slopes, causing the lines to intersect at a single point. If the second equation were 4x - 6y = 14, which is simply a multiple of the first equation (2(2x - 3y) = 2(7)), the system would have infinitely many solutions because both equations represent the exact same line.

Rationale:
The question asks for a conditional probability: the probability that a student is involved in clubs GIVEN that the student has a GPA above 3.0.

To calculate this, we first identify the total number of students who meet the condition (GPA above 3.0). From the table, there are 110 students with a GPA above 3.0. This will be the denominator of our probability fraction.

Next, we identify the number of students who meet both the condition AND the event (GPA above 3.0 AND involved in clubs). From the table, there are 70 students who have a GPA above 3.0 and are involved in clubs. This will be the numerator of our probability fraction.

So, the probability is 70/110. When simplified by dividing both the numerator and the denominator by 10, the fraction becomes 7/11.

Distractor 1 (7/10): This option represents the probability that a student has a GPA above 3.0 GIVEN that they are involved in clubs (70 students with GPA above 3.0 out of 100 students involved in clubs). This is a reversal of the conditional probability requested.

Distractor 2 (1/2): This option represents the overall probability that a randomly selected student is involved in clubs, without any specific GPA condition (100 students involved in clubs out of a total of 200 students).

Distractor 3 (4/11): This option represents the probability that a student is NOT involved in clubs GIVEN that the student has a GPA above 3.0 (40 students not involved in clubs out of 110 students with a GPA above 3.0).

Rationale:
Let $P$ represent the number of hours Alex spent on PSAT Math prep and $D$ represent the number of hours he spent on the debate club. The problem provides two pieces of information that can be translated into a system of two linear equations.

First, Alex dedicated a total of 15 hours to both activities: $P + D = 15$ (Equation 1)

Second, Alex tracked 4 'focus points' for each hour of PSAT Math prep and 2 'focus points' for each hour of debate, accumulating a total of 46 focus points: $4P + 2D = 46$ (Equation 2)

To solve this system, we can use the substitution method. From Equation 1, we can express $D$ in terms of $P$: $D = 15 - P$

Now substitute this expression for $D$ into Equation 2: $4P + 2(15 - P) = 46$ $4P + 30 - 2P = 46$ $2P + 30 = 46$ $2P = 46 - 30$ $2P = 16$ $P = 8$

Thus, Alex spent 8 hours on PSAT Math prep last week.

Distractor Analysis:

• 7: This is the number of hours Alex spent on the debate club. If $P=8$, then $D = 15 - 8 = 7$. A student might correctly solve the system but answer with the value for the wrong variable.
• 6: A student might arrive at this answer due to a common arithmetic error during solving the equation (e.g., if they miscalculated $46-30$ as $12$ instead of $16$, leading to $2P=12$, then $P=6$). Another possibility is if they misread the total points as 42, then $2P = 42-30=12$, which gives $P=6$.
• 10: If Alex spent 10 hours on PSAT prep, he would have spent 5 hours on debate ($10+5=15$). Calculating the total focus points for this scenario would be $4(10) + 2(5) = 40 + 10 = 50$. This value (50) is relatively close to the actual total of 46, making 10 a plausible but incorrect choice for a student who might be trying to estimate or made a small error in calculation or setup.

Rationale:
The problem describes two parallel lines, \(l\) and \(m\), cut by a transversal \(t\). One angle is specified as being above line \(l\) and to the right of transversal \(t\), while the other is above line \(m\) and to the right of transversal \(t\). Angles in these relative positions are defined as corresponding angles. A fundamental property of parallel lines intersected by a transversal is that corresponding angles are congruent (equal in measure). Therefore, setting their algebraic expressions equal to each other allows for the determination of \(x\).

Distractor 1 is incorrect because alternate interior angles are located between the parallel lines and on opposite sides of the transversal. The angles described are on the same side of the transversal and outside the region between the parallel lines relative to each other.

Distractor 2 is incorrect because consecutive interior angles (also known as same-side interior angles) are located between the parallel lines and on the same side of the transversal. These angles are supplementary, not congruent, when lines are parallel. The described angles are not consecutive interior angles, nor are they supplementary.

Distractor 3 is incorrect because vertical angles are formed by the intersection of two lines and are opposite each other at the vertex. The two angles described are at different vertices (one on line \(l\), one on line \(m\)) and are not opposite each other in this manner.