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.
60 past weekly challenges found. Open one card at a time for the full solution review.
Solve a medium GAT geometry question using similar triangles and proportional side lengths.
Geometry / Using proportional sides in similar trianglesTwo triangles are similar. In the smaller triangle, a side of length 6 corresponds to a side of length 15 in the larger triangle. If another side of the smaller triangle is 10, what is the corresponding side length in the larger triangle?
C.
25Find the scale factor from the smaller triangle to the larger triangle.
The scale factor is:
156 = 52
So the corresponding side is:
10 × 52 = 25
The triangles are similar, so corresponding sides have the same ratio. Since 6 became 15, every smaller-triangle side is multiplied by 52.
A common mistake is adding the difference 15 - 6 = 9 to the other side and getting 19. Similarity uses multiplication, not addition.
The trap is using the difference between sides rather than the scale factor.
If this was missed, the issue may be confusing additive and multiplicative relationships.
GAT similarity questions usually reward ratio thinking instead of additive thinking.
For similar figures, use scale factor, not difference.
Convert 6 → 15 into a multiplier before touching the side 10.
Solve an SAT algebra word problem about a bee colony growth model and remaining population needed.
Algebra / Writing a linear function from a piecewise growth contextA beekeeper's initial observation of the population of a certain bee colony was 1,800 bees. The beekeeper set a goal to increase the population to 3,300 bees. The beekeeper uses a model that predicts the population of this bee colony begins at 1,800 and increases by 120 bees per week in the first two weeks after the initial observation, and then increases by 180 bees per week until the beekeeper's goal is reached. According to this model, at the end of week w after the initial observation, where w > 2, which of the following functions gives the predicted number of bees still needed to reach the beekeeper's goal?
C.
p(w) = 1,620 − 180wFind the population after the first two weeks, then model the remaining amount.
After the first two weeks, the population increases by:
2(120) = 240
So at the end of week 2, the population is:
1,800 + 240 = 2,040
For w > 2, the colony then increases by 180 bees per week. The number of weeks after week 2 is w − 2, so the population is:
2,040 + 180(w − 2)
= 2,040 + 180w − 360 = 1,680 + 180w
The number still needed to reach 3,300 is:
3,300 − (1,680 + 180w) = 1,620 − 180w
The model changes after the first two weeks, so the function for w > 2 must include the progress already made during those first two weeks.
A common mistake is using 3,300 − 180w, which ignores the first two weeks of growth at 120 bees per week.
The trap is applying the 180-bee rate from week 0 instead of after week 2.
If this was difficult, it may show a weak spot in translating piecewise verbal models into formulas.
This SAT algebra question tests linear modeling in a multi-stage context.
For piecewise contexts, account for the earlier interval before writing the later formula.
At week 2, the colony has already gained 240 bees.
Practice an AP Calculus AB limit question about one-sided limits and removable discontinuity.
Limits and Continuity / Using one-sided limits to identify a removable discontinuitySuppose limx→−3− f(x)=−1, limx→−3+ f(x)=−1, and f(−3) is not defined. Which of the following statements is (are) true?
I. limx→−3 f(x)=−1
II. f is continuous everywhere except at x=−3
III. f has a removable discontinuity at x=−3
C.
I and III onlyEqual one-sided limits give a two-sided limit, but they do not describe continuity everywhere else.
Since the left-hand and right-hand limits at x=−3 are both −1, the two-sided limit exists and equals −1. So statement I is true. Since f(−3) is not defined while the limit exists, f has a removable discontinuity at x=−3. So statement III is true. Statement II is not guaranteed because no information is given about continuity at other values of x.
The information is local to x=−3. It determines the limit and the type of discontinuity there, but it cannot prove continuity everywhere else.
A common mistake is assuming statement II is true because the problem only mentions x=−3.
The trap is overgeneralizing from one point.
If this was difficult, it may reveal weakness in local versus global continuity claims.
This AP Calculus AB limits question tests one-sided limits and removable discontinuities.
Do not infer global continuity from local limit information.
Statement II uses the word “everywhere,” which requires more information than given.
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).