Past SAT Math, GAT / Qudurat, and AP Calculus AB weekly challenges collected for focused review, solution analysis, and smarter practice.
Each archived challenge was previously published through the StudyGlitch Weekly Math Challenge system. Use the archive to revisit solution paths, compare trap answers, and connect the result to diagnostic-style SAT, GAT, and AP math practice.
10 past weekly challenges found. Open one card at a time for the full solution review.
Practice an AP Calculus AB application question about concavity and second derivative sign changes.
Applications of Differential Calculus / Determining inflection behavior from second derivative sign changesIf
f″(x)=x(x−4)2
then the graph of f changes concavity at
A.
x=0 onlyA change in concavity requires f″ to change sign.
The expression f″(x)=x(x−4)2 has zeros at x=0 and x=4. The factor x changes sign at 0. The factor (x−4)2 does not change sign at 4. Therefore, the graph changes concavity only at x=0.
A zero of the second derivative is only an inflection point if the sign of the second derivative changes.
A common mistake is assuming every solution of f″(x)=0 is an inflection point.
The trap is listing both zeros of f″ automatically.
If this was difficult, it may reveal weakness in second derivative sign analysis.
This AP Calculus AB applications question tests concavity and inflection reasoning.
Use sign changes, not just zeros, to identify concavity changes.
The squared factor at x=4 does not change sign.
Solve a medium GAT quantitative percentage question involving consecutive increase and decrease with full explanation and common mistake.
Arithmetic and Percentages / Handling consecutive percentage increase and decreaseA price is increased by 30%, then the new price is decreased by 20%. If the final price is 624 riyals, what was the original price?
B.
600Do not subtract 30% - 20%. Apply each percentage to the current value.
Let the original price be x. After a 30% increase, the price becomes 1.3x. After a 20% decrease, it becomes 0.8(1.3x).
0.8(1.3x) = 624
1.04x = 624
x = 600
The key is that the decrease happens after the increase, so the decrease is taken from the larger price, not from the original price. Multiplying the percentage factors gives 1.3 × 0.8 = 1.04. The final price is therefore 104% of the original price. Since 624 is 104% of the original value, the original value is 600.
A common mistake is treating the changes as a net 10% increase. That would be wrong because the 20% decrease is applied after the price has already changed.
The trap is assuming opposite percentages cancel. They do not unless they are applied to the same base.
If this felt slow, the issue may be percentage structure recognition. A diagnostic can show whether percentage traps repeat across your GAT quantitative work.
In GAT percentage questions, the fastest path is often multiplier thinking instead of long percentage calculations.
For consecutive percentage changes, convert each change into a multiplier first.
Use 1.3 × 0.8 immediately. This avoids slower percentage-by-percentage working.
Solve an SAT question about the signs of x- and y-intercepts of a transformed cubic function.
Advanced Algebra / Determining signs of intercepts from a transformed functionThe function f is defined by f(x) = 11x3. The graph of y = f(−x) + c in the xy-plane, where c is a positive integer constant, has an x-intercept at (r, 0) and a y-intercept at (0, t), where r and t are constants. Which of the following must be true about r and t?
C.
r > 0 and t > 0First rewrite f(−x).
Since f(x) = 11x3, then:
f(−x) = 11(−x)3 = −11x3
So the graph is:
y = −11x3 + c
The y-intercept occurs when x = 0:
t = c
Since c is positive, t > 0.
The x-intercept occurs when y = 0:
−11x3 + c = 0
11x3 = c
Since c > 0, x > 0. Therefore, r > 0 and t > 0.
The transformation f(−x) turns 11x3 into −11x3. Adding a positive constant shifts the graph up, giving a positive y-intercept and a positive x-intercept.
A common mistake is forgetting that (−x)3 = −x3.
The trap is treating f(−x) as if it were still 11x3.
If this was difficult, it may show a weak spot in odd functions and graph transformations.
This is a SAT function-transformation question involving intercept signs.
For transformed functions, rewrite the equation before analyzing intercepts.
Desmos can help visualize the transformed cubic, but the sign conclusion follows directly from the equation.
Compute f(−x) before thinking about intercepts.
Solve an AP Calculus AB continuity question involving a parameter in a piecewise function.
Limits and Continuity / Using one-sided limits to make a piecewise function continuousFor what value of k is the function f continuous at x=2?
f(x)=
{ kx2−3x, x<2
{ x+k, x≥2
C.
83Set the left-hand limit equal to the function value from the right-hand piece at x=2.
Continuity at x=2 requires limx→2− f(x)=f(2). The left-hand value is k(2)2−3(2)=4k−6. The function value is from the second piece: f(2)=2+k. Set them equal: 4k−6=2+k. Then 3k=8, so k=83.
The point x=2 belongs to the second piece because of the condition x≥2. The left expression only determines the left-hand limit.
A common mistake is using the first piece to compute f(2), even though the first piece is only for x<2.
The trap is using the wrong piece at the endpoint.
If this was difficult, it may reveal weakness in piecewise continuity conditions.
This AP Calculus AB continuity question tests one-sided limits and piecewise definitions.
For piecewise continuity, compare the one-sided limit with the actual defined value.
The equality to solve is 4k−6=2+k.
Solve a hard GAT quantitative work-rate problem involving two pipes filling a tank together.
Word Problems and Logic / Combining work rates to find time working togetherA pipe can fill a tank in 6 hours, and another pipe can fill the same tank in 9 hours. If both pipes work together, how many hours are needed to fill the tank?
A.
185Use rates: one pipe fills 16 of the tank per hour, and the other fills 19 per hour.
The combined hourly rate is:
16 + 19
Use common denominator 18:
318 + 218 = 518
Together, they fill 518 of the tank per hour.
Time to fill one full tank:
1 ÷ 518 = 185
Work-rate problems require adding rates, not adding times. The pipes do not take 6 + 9 hours together; together they work faster than either pipe alone.
A common mistake is averaging 6 and 9 or adding them. Work rates must be added as fractions of the job per hour.
The trap is adding or averaging the hours directly.
If this was difficult, the weak point may be rate thinking in work problems.
GAT work-rate questions often trap students who operate on times instead of rates.
For “working together” questions, convert time to rate first.
Rate first: 16 + 19.
Solve an SAT statistics question about how increasing sample size affects margin of error.
Statistics and Data Analysis / Understanding how sample size affects margin of errorA researcher is designing a study to investigate the average number of hours students at a high school spend reading per day. The researcher will report an estimated average number of hours students at the high school spend reading per day with an associated margin of error. The researcher is considering using a random sample of either 115 or 230 students from the high school. Which of the following would be the most likely effect of using the larger random sample compared to the smaller random sample?
A.
The reported margin of error would be lower.A larger random sample usually gives a more precise estimate.
Increasing the random sample size from 115 to 230 would most likely reduce sampling variability. A lower sampling variability means the associated margin of error would be lower.
The larger sample size affects the precision of the estimate, not the direction of the estimated average itself. The reported average could be higher or lower depending on the sample, but the margin of error is expected to decrease.
A common mistake is thinking a larger sample must change the reported average in a specific direction.
The trap is confusing margin of error with the sample mean.
If this was difficult, it may show a weak spot in sampling concepts.
This SAT statistics and data analysis question tests sampling and margin of error.
For margin-of-error questions, remember that larger sample size generally means smaller margin of error.
More data generally means a more precise estimate.
Practice an AP Calculus AB differentials question about the approximate change in area of a square.
Applications of Differential Calculus / Using differentials to approximate change in areaIf the side e of a square is increased by 1%, then the area is increased approximately
B.
0.02e2Use A=e2 and approximate the change with dA.
The area of the square is A=e2. A 1% increase in side length means de=0.01e. Since dA=2e,de,
dA=2e(0.01e)=0.02e2
The approximate change in area is found by multiplying the derivative of area with respect to side length by the small change in side length.
A common mistake is choosing 0.01e2, which treats the percent change in side length as the percent change in area.
The trap is forgetting that area changes approximately twice as fast percentage-wise as side length.
If this was difficult, it may reveal weakness in differential approximation.
This AP Calculus AB applications question tests linear approximation with differentials.
For small changes, use differentials: dA=A′(e)de.
A 1% change in e means de=0.01e.
Solve a medium GAT geometry question using the Pythagorean theorem and a 7-24-25 triangle.
Geometry / Using the Pythagorean theorem to find a missing sideA right triangle has hypotenuse 25 and one leg 7. What is the length of the other leg?
C.
24Use a2 + b2 = c2, where c is the hypotenuse.
Let the missing leg be x.
72 + x2 = 252
49 + x2 = 625
x2 = 576
x = 24
This is the classic 7-24-25 right triangle. Recognizing the triple makes the question very fast.
A common mistake is adding 7 and 25 or subtracting them directly. Side lengths in a right triangle are related through squares.
The trap is treating the hypotenuse like an ordinary side in addition/subtraction.
If this was slow, the improvement area is common right-triangle triples.
GAT right-triangle questions often become faster if you know common Pythagorean triples.
Recognize common triples like 3-4-5, 5-12-13, and 7-24-25.
If you recognize 7-24-25, solve instantly.
Solve an SAT data analysis question by choosing the best linear model for a scatterplot.
Statistics and Data Analysis / Choosing an appropriate linear model from a scatterplotWhich of the following equations is the most appropriate linear model for the data shown?
A.
d = −48.1 + 2.02tUse a visible point on the trend line and test the model.
The graph shows a positive slope of about 2, so all choices have the same reasonable slope 2.02. To choose the model, test the intercept using a visible point. Around t = 230, the line is near d = 416.
For choice A:
d = −48.1 + 2.02(230)
d = −48.1 + 464.6 = 416.5
This matches the graph closely, so the appropriate model is d = −48.1 + 2.02t.
Because each answer choice has the same slope, the decision depends on the intercept. Choice A gives a predicted value near the line at the left side of the graph, while the other choices produce values much too high.
A common mistake is choosing the option with the largest intercept because the graph values are large. The variable t is also large, so the intercept must be interpreted through substitution.
The trap is reading the intercept as if t = 0 were visible on the same scale.
If this was difficult, it may show a weak spot in connecting graphs to equations.
This SAT statistics and data analysis question tests interpreting scatterplots and linear models.
For model-choice questions, plug in a visible coordinate from the graph.
Since all choices have slope 2.02, test only one visible point.
Practice an AP Calculus AB particle motion question about finding maximum speed from a velocity graph.
Applications of Differential Calculus / Interpreting speed as the absolute value of velocityThe object attains its maximum speed when t=
D.
3Speed is the absolute value of velocity.
The graph shows velocity, not speed. The speed is |v(t)|. The largest magnitude of velocity shown occurs at t=3, where v=-10. The speed there is 10, which is greater than the speed at the other listed times.
Although the velocity is negative at t=3, the speed is positive and equals the magnitude of velocity.
A common mistake is choosing where velocity is greatest rather than where speed is greatest.
The trap is ignoring negative velocity values when comparing speed.
If this was difficult, it may reveal weakness in particle motion vocabulary.
This AP Calculus AB applications question tests velocity and speed interpretation.
For particle motion, remember that speed is |v|.
Look for the point farthest from the t-axis.
