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.
21 past weekly challenges found. Open one card at a time for the full solution review.
Solve a hard GAT algebra parameter question using a given solution value.
Algebra / Finding a parameter from a given solutionThe equation kx - 5 = 3x + 7 has solution x = 4. What is the value of k?
B.
6Substitute x = 4 into the equation, then solve for k.
Substitute x = 4:
4k - 5 = 3(4) + 7
4k - 5 = 12 + 7
4k - 5 = 19
4k = 24
k = 6
The question gives the solution, so use it directly. Substitute 4 for x, then solve the remaining equation for k.
A common mistake is trying to solve for x even though the value of x is already given.
The trap is solving the equation in the wrong direction.
If this felt confusing, the issue may be interpreting parameter questions quickly.
GAT parameter questions test whether you understand what a solution means.
When a problem says “has solution,” substitute that value into the equation.
Use x = 4 immediately.
Solve an SAT percent-change question involving a 179 percent increase followed by a 27 percent decrease.
Statistics and Data Analysis / Determining net percent change after an increase and a decreaseThe value of a painting increased by 179% from the end of 2017 to the end of 2018 and then decreased by 27% from the end of 2019. What was the net percentage increase in the value of the painting from the end of 2017 to the end of 2019?
D.
103.67%Use multipliers for successive percentage changes.
An increase of 179% means the value is multiplied by 2.79. A decrease of 27% means the value is multiplied by 0.73.
2.79 × 0.73 = 2.0367
This means the final value is 203.67% of the original value. Therefore, the net percentage increase is:
203.67% − 100% = 103.67%
Forensic note: The visible wording says “and then decreased by 27% from the end of 2019.” That wording is preserved exactly in the question text. Based on the answer choices and the intended successive-change structure, the calculation is 2.79 × 0.73 = 2.0367, giving a net increase of 103.67%.
A common mistake is subtracting 27% from 179%. Successive percentage changes must be multiplied, not combined by simple subtraction.
The trap is treating the two percentage changes as if they have the same base.
If this was difficult, it may show a weak spot in compound percent-change reasoning.
This is a SAT percent-change question involving two successive changes.
For sequential percent changes, convert each change into a multiplier first.
Use 1 + 1.79 = 2.79 and 1 − 0.27 = 0.73.
Practice an AP Calculus AB continuity question involving removable and nonremovable discontinuities.
Limits and Continuity / Classifying continuity after removable and nonremovable discontinuitiesSuppose
f(x)=3x(x−1)x2−3x+2 for x ≠ 1, 2
f(1)=−3
f(2)=4
Then f(x) is continuous
B.
except at x=2Factor the denominator and simplify where possible.
The denominator factors as x2−3x+2=(x−1)(x−2). For x≠1,2, f(x)=3x(x−1)(x−1)(x−2)=3xx−2. At x=1, the limit is 3(1)1−2=−3, which equals f(1). So f is continuous at x=1. At x=2, the simplified expression has a vertical asymptote, so f is not continuous there. Therefore, f is continuous except at x=2.
The factor x−1 cancels, creating a removable discontinuity that has been filled correctly by f(1)=−3. The factor x−2 remains in the denominator, so x=2 is nonremovable.
A common mistake is saying the function is discontinuous at both 1 and 2 just because the original formula excludes both values.
The trap is not checking the separately defined function values.
If this was difficult, it may reveal weakness in removable versus nonremovable discontinuities.
This AP Calculus AB continuity question tests rational-function discontinuities.
Check whether each excluded value is removable and whether its assigned value fills the hole.
Simplify to 3xx−2 after canceling x−1.
Practice a medium GAT algebra question reading roots from a factored quadratic equation.
Algebra / Reading roots from a factored quadratic expressionIf (x - 4)(x + 7) = 0, what is the sum of the possible values of x?
B.
-3Set each factor equal to zero.
From x - 4 = 0, we get x = 4.
From x + 7 = 0, we get x = -7.
The sum is:
4 + (-7) = -3
The possible values are the roots of the equation. The signs reverse when solving each factor: x - 4 = 0 gives 4, and x + 7 = 0 gives -7.
A common mistake is taking the values as -4 and 7 directly from the parentheses.
The trap is copying signs directly instead of solving each factor.
If this was missed, the issue is likely sign interpretation in factored expressions.
GAT quadratic questions often test whether you read factored form correctly.
In factored form, remember that each factor equals zero, so the signs switch.
Do not expand. The factored form already gives the roots.
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.
