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.
41 past weekly challenges found. Open one card at a time for the full solution review.
Solve an SAT statistics question comparing medians and standard deviations from two dot plots.
Statistics and Data Analysis / Comparing median and standard deviation from dot plotsThe dot plots represent the distributions of values in data sets A and B.
Which of the following statements must be true?
I. The median of data set A is equal to the median of data set B.
II. The standard deviation of data set A is equal to the standard deviation of data set B.
B.
I onlyCompare center and spread separately.
Data set A has counts:
8(1), 9(3), 10(4), 11(5), 12(4), 13(3), 14(1)
Data set B has counts:
8(2), 9(3), 10(4), 11(3), 12(4), 13(3), 14(2)
Both data sets have 21 values and are centered at 11, so their medians are both 11. Statement I is true.
However, data set B has more values at the extremes 8 and 14 and fewer values at the center 11. Therefore, data set B is more spread out, so the standard deviations are not equal. Statement II is false.
The correct answer is I only.
Both distributions are balanced around the same center, which makes the medians equal. But the second distribution places more weight farther from the center, increasing its standard deviation.
A common mistake is assuming that equal medians imply equal standard deviations.
The trap is confusing symmetry around the same value with equal variability.
If this was difficult, it may show a weak spot in interpreting distributions visually.
This SAT statistics and data analysis question tests median and standard deviation from dot plots.
For dot plot comparisons, analyze center and spread as separate features.
Same center does not automatically mean same spread.
Practice an AP Calculus AB definite integral question using the Fundamental Theorem of Calculus.
Definite Integrals / Using the Fundamental Theorem of Calculus with chain ruleLet
F(x)= x2 ∫ 1 cos(t) dt
What is F′(x)?
C.
2xcos(x2)Use the Fundamental Theorem of Calculus and multiply by the derivative of the upper limit.
By the Fundamental Theorem of Calculus, differentiating ∫1x2 cos(t)dt gives cos(x2) times the derivative of x2. Therefore, F′(x)=2xcos(x2).
The upper limit is not simply x, so the chain rule is required after applying the Fundamental Theorem of Calculus.
A common mistake is forgetting the factor 2x.
The trap is treating x2 as if it were x.
If this was difficult, it may reveal weakness in FTC with chain rule.
This AP Calculus AB definite integrals question tests FTC Part 1.
For accumulation functions with non-linear bounds, apply FTC plus chain rule.
The lower limit is constant, so it does not contribute a derivative term.
Practice a GAT quantitative timing question using least common multiple and endpoint counting.
Arithmetic and Multiples / Using least common multiple in repeated timing eventsTwo lights flash together at exactly 8:00:00. One light flashes every 18 seconds, and the other flashes every 24 seconds. How many times will they flash together from 8:00:00 through 8:06:00, including both endpoints?
C.
6They flash together every least common multiple of 18 and 24 seconds.
Find the least common multiple:
18 = 2 × 32
24 = 23 × 3
LCM = 23 × 32 = 72
They flash together every 72 seconds.
Six minutes equals 360 seconds. The together-flash times are:
0, 72, 144, 216, 288, 360
That is 6 times.
The phrase “including both endpoints” matters. Since the lights flash together at the start and again exactly 360 seconds later, both moments must be counted.
A common mistake is counting only after the starting flash, which gives 5 instead of 6.
The trap is forgetting the event at time 0.
If this felt easy but you missed it, the issue may be endpoint counting under time pressure.
GAT timing questions often test both LCM and careful reading.
For repeated-event timing questions, use LCM first, then check whether endpoints are included.
After finding 72, divide 360 ÷ 72 = 5, then add the starting moment.
Solve an SAT geometry question using similar cylinders, volume ratio, and surface area ratio.
Geometry and Trigonometry / Using volume scale factor and surface area scale factor for similar solidsThe table shows the volume of two similar solids, right circular cylinder A and right circular cylinder B. The radius of right circular cylinder A is 2 units. The surface area of right circular cylinder A is kπ square units, and the surface area of right circular cylinder B is nπ square units, where k and n are constants. What is the value of n − k? (The surface area of a right circular cylinder with radius r and height h is 2πr2 + 2πrh.)
| Volume (cubic units) | |
|---|---|
| Right circular cylinder A | 32π |
| Right circular cylinder B | 864π |
For similar solids, volume scale factor is the cube of the linear scale factor.
The volume ratio is:
864π32π = 27
So the linear scale factor from cylinder A to cylinder B is:
∛27 = 3
For cylinder A, radius r = 2 and volume is 32π:
πr2h = 32π
π(22)h = 32π
4h = 32
h = 8
Surface area of cylinder A:
2π(22) + 2π(2)(8) = 8π + 32π = 40π
So k = 40.
Surface area scales by the square of the linear scale factor. Since the scale factor is 3, surface area scale factor is 9.
nπ = 9(40π) = 360π
So n = 360, and:
n − k = 360 − 40 = 320
The volume ratio gives a linear scale factor of 3. Surface area scales by the square of that, so the surface area of cylinder B is 9 times the surface area of cylinder A.
A common mistake is multiplying the surface area by 27 instead of by 9. Volume scales cubically, but surface area scales quadratically.
The trap is using the volume scale factor directly on surface area.
If this was difficult, it may show a weak spot in dimensional scaling.
This SAT geometry question tests scale factors for similar solids.
For similar solids, length scales by s, area by s2, and volume by s3.
Once the volume ratio is 27, the linear scale factor is 3.
Practice an AP Calculus AB implicit differentiation question at a point.
Differentiation / Finding dy/dx by implicit differentiationIf
x2+xy+y2=12
then at the point (2,2), dydx is
B.
−1Use product rule on xy.
Differentiating implicitly gives 2x+xdydx+y+2ydydx=0. Grouping derivative terms gives (x+2y)dydx=−(2x+y). Thus dydx=−2x+yx+2y. At (2,2), this is −66=−1.
The term xy requires product rule because both variables depend on x implicitly.
A common mistake is differentiating xy as only xdydx, forgetting the +y term.
The trap is missing product rule on xy.
If this was difficult, it may reveal weakness in implicit differentiation.
This AP Calculus AB differentiation question tests implicit differentiation.
For implicit differentiation, treat y as a function of x.
After differentiating, plug in (2,2).
Practice a GAT algebra system of equations question using fast elimination.
Algebra / Solving a two-variable system using eliminationIf 2x + 3y = 19 and 4x - 3y = 17, what is the value of x?
C.
6Add the two equations. The y-terms cancel.
Add the equations:
(2x + 3y) + (4x - 3y) = 19 + 17
6x = 36
x = 6
The equations are designed for elimination because +3y and -3y cancel immediately. Once they cancel, only 6x = 36 remains.
A common mistake is trying substitution first and making the question longer than necessary.
The trap is over-solving when the equations are already aligned.
If this felt slow, it may reveal that you are not checking for elimination opportunities first.
GAT systems often have a short route. Look for cancellation before doing full substitution.
Before solving a system, check whether adding or subtracting cancels one variable quickly.
The fastest move is direct addition.
Solve an SAT geometry question about finding the height of a right square pyramid from its surface area.
Geometry and Trigonometry / Using surface area and a right-triangle relationship to find pyramid heightA right square pyramid has a surface area of 100 + 20√146 square inches, which includes a base area of 100 square inches. What is the height, in inches, of this pyramid?
Subtract the base area first to find the lateral area.
The total surface area is 100 + 20√146, and the base area is 100, so the lateral area is 20√146.
For a right square pyramid, lateral area equals 12Pl, where P is the base perimeter and l is the slant height.
The base area is 100, so the square base has side length 10. Thus the perimeter is 40.
12(40)l = 20√146
20l = 20√146
l = √146
The slant height, the pyramid height, and half the base side form a right triangle:
h2 + 52 = (√146)2
h2 + 25 = 146
h2 = 121
h = 11
After removing the base area, the remaining area is the sum of the four triangular faces. That gives the slant height. Then use the right triangle formed by the pyramid height, half of the base side, and the slant height to find the height.
A common mistake is using the full base side length 10 instead of half the side length 5 in the right triangle.
The trap is using the total surface area directly as if it were only lateral area.
If this was difficult, it may show a weak spot in geometric modeling with pyramids.
This SAT geometry question combines surface area, lateral area, and the Pythagorean theorem.
For square pyramids, separate base area from lateral area before applying the pyramid surface-area formula.
Base area 100 means the square side length is 10.
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.