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.
Practice a medium GAT statistics question about finding the median after adding a new number.
Data Analysis and Statistics / Finding the median after inserting a new data valueThe numbers below are arranged from least to greatest:
6, 9, 12, 15, 21
If the number 18 is added to the list, what is the new median?
B.
13.5After adding 18, there will be 6 numbers, so the median is the average of the 3rd and 4th numbers.
Insert 18 into the ordered list:
6, 9, 12, 15, 18, 21
There are 6 numbers, so the median is the average of the 3rd and 4th values:
12 + 152 = 272 = 13.5
The median changes because the number of data values changes from odd to even. With an even number of values, the median is the average of the two middle values.
A common mistake is choosing 15 because it was near the middle after insertion. But with 6 values, there is no single middle value.
The trap is using the old median rule after the list size changes.
If this was missed, the issue may be median procedure under changing data conditions.
GAT median questions often test whether the list has odd or even count after a new value is inserted.
Always count the number of data points after the change, not before.
Insert first, count second, then find the middle.
Solve an SAT advanced algebra question about the percent decrease represented by an exponential decay factor.
Advanced Algebra / Interpreting the decay factor of an exponential function as a percent decreaseThe function f is defined by f(x) = 55(0.19)x. For any positive integer n, the value of f(n) is p% less than the value of f(n − 1). What is the value of p?
D.
81Compare f(n) to f(n−1).
For the exponential function:
f(n) = 55(0.19)n
and:
f(n−1) = 55(0.19)n−1
Therefore:
f(n)f(n−1) = 0.19
So f(n) is 19% of f(n−1), which means it is 81% less.
A decay factor of 0.19 means the new value keeps 19% of the previous value. The percent decrease is the missing part from 100%, which is 81%.
A common mistake is answering 19. That is the percent remaining, not the percent decrease.
The trap is confusing percent remaining with percent less.
If this was difficult, it may show a weak spot in interpreting exponential decay factors.
This SAT advanced algebra question tests exponential decay interpretation.
For exponential decay, percent decrease equals 100% minus the decay factor as a percent.
0.19 means 19% remains.
Practice an AP Calculus AB limit at infinity question involving a rational function.
Limits and Continuity / Evaluating limits at infinity using degree comparisonlimx→∞ 3x2 + 27x3 − 27 is
A.
0Compare the degree of the numerator with the degree of the denominator.
The numerator has degree 2, while the denominator has degree 3. As x→∞, the denominator grows faster than the numerator, so the quotient approaches 0.
For rational functions at infinity, if the degree of the denominator is greater than the degree of the numerator, the limit is 0.
A common mistake is comparing only the leading coefficients and choosing 3. Leading coefficients only determine the limit when the degrees are equal.
The trap is ignoring the different powers of x.
If this was difficult, it may reveal weakness in rational-function end behavior.
This AP Calculus AB limits question tests end behavior of rational functions.
Use degree comparison first, then leading coefficients only if the degrees match.
The denominator has the higher power of x.
Practice a hard GAT geometry question finding a central angle from a sector area fraction.
Geometry / Finding a sector angle from a fraction of circle areaA sector has area equal to 512 of the area of its circle. What is the central angle of the sector?
C.
150°The sector area fraction equals the same fraction of 360°.
The central angle is:
512 × 360°
= 5 × 30° = 150°
A sector is a fraction of the full circle. If the area is 512 of the circle, the angle is also 512 of the full 360°.
A common mistake is trying to use the circle area formula even though the radius is not needed.
The trap is looking for radius information that the question does not require.
If this felt hard, the weak point may be recognizing proportional sector relationships.
GAT circle sector questions often become fast if you recognize fraction-of-circle structure.
When sector area is given as a fraction of the circle, multiply that fraction by 360°.
Since 360 ÷ 12 = 30, this is quick mental math.
Solve an SAT algebra question about which region of the coordinate plane contains no solutions to a linear inequality.
Algebra / Analyzing regions of the coordinate plane that satisfy a linear inequalityIn the xy-plane, which of the following does NOT contain any points (x, y) that are solutions to the inequality −5x < 70y − 75?
B.
The region where x < 0 and y < 0Rewrite the inequality in terms of y.
Start with:
−5x < 70y − 75
Add 75 to both sides:
−5x + 75 < 70y
Divide by 70:
y > −114x + 1514
If x < 0, then −114x is positive, so the right-hand side is greater than 1514, which is positive. Therefore y must be positive. That means no point with x < 0 and y < 0 can satisfy the inequality.
The line y = −114x + 1514 has a positive y-intercept and a slight negative slope. For negative x-values, the boundary lies above 1, so any solution must have positive y. That rules out the region where both x and y are negative.
A common mistake is testing only one or two quadrants without rewriting the inequality into slope-intercept form.
The trap is assuming every quadrant must contain some solution without analyzing the boundary.
If this was difficult, it may show a weak spot in graphing and interpreting linear inequalities.
This SAT algebra question tests inequality interpretation in the coordinate plane.
For region questions, rewrite the inequality in terms of y and think about sign constraints.
Once the inequality is written as y > mx + b, check whether an entire region is impossible.
Practice an AP Calculus AB differential equation question about restricted exponential growth behavior.
Differential Equations / Interpreting restricted growth from a differential equationA population P(t) grows according to the differential equation
dPdt=k(500−P)
where k>0 and P(0)=100. Which of the following must be true?
C.
P(t) increases toward 500.Find the equilibrium value and check the sign of dPdt when P=100.
The equilibrium occurs when 500−P=0, so P=500. Since P(0)=100<500, 500−P>0, and because k>0, dPdt>0. Therefore the population increases. As P approaches 500, the growth rate approaches 0, so the population approaches 500.
This is restricted growth because the rate depends on the remaining gap between the population and 500.
A common mistake is treating the model as unrestricted exponential growth.
The trap is thinking a positive derivative means the population increases forever.
If this was difficult, it may reveal weakness in restricted growth and equilibrium behavior.
This AP Calculus AB differential equations question tests qualitative solution behavior.
For differential equation behavior, identify equilibrium and sign of the derivative.
The equilibrium is P=500.
Solve a hard GAT geometry question involving similar solids and volume scale factor.
Geometry / Using scale factor to compare volumes of similar solidsTwo similar solids have corresponding side lengths in the ratio 2:3. If the volume of the smaller solid is 40, what is the volume of the larger solid?
D.
135Volume scales by the cube of the side-length scale factor.
The side-length ratio is:
2:3
The volume ratio is:
23:33 = 8:27
The smaller volume corresponds to 8 parts:
8 parts = 40
1 part = 5
The larger volume is:
27 × 5 = 135
For similar solids, the side ratio is not the volume ratio. Since volume is three-dimensional, the scale factor is cubed. That changes 2:3 into 8:27.
A common mistake is multiplying 40 by 32 and getting 60. That uses the length scale factor, not the volume scale factor.
The trap is using the side ratio directly on volume.
If this was missed, the weak point is dimensional scaling.
GAT similarity questions often test whether the quantity is length, area, or volume.
Length scale factor: first power. Area: square. Volume: cube.
Convert the side ratio 2:3 into volume ratio 8:27.
Solve an SAT advanced algebra question by rewriting a quadratic in completed-square form.
Advanced Algebra / Rewriting a quadratic expression by completing the squarey = 4x2 − bx − 5
Which of the following equations is equivalent to the given equation, where b is a positive constant?
A.
y = 4(x − b8)2 − 5 − b216Factor out the 4 from the quadratic and linear terms first.
Start with:
y = 4x2 − bx − 5
Factor out 4 from the first two terms:
y = 4(x2 − b4x) − 5
Complete the square inside the parentheses:
x2 − b4x = (x − b8)2 − b264
Substitute back:
y = 4[(x − b8)2 − b264] − 5
y = 4(x − b8)2 − b216 − 5
The completed-square form must keep the expression equivalent, so the square-completion correction term must also be accounted for outside the parentheses. Because the linear term is negative, the square uses (x − b8), not a plus sign.
A common mistake is omitting the subtraction of b216.
The trap is matching only the linear term and ignoring the constant adjustment.
If this was difficult, it may show a weak spot in completing the square.
This SAT advanced algebra question tests structure and quadratic rewriting.
When completing the square, factor out the leading coefficient first if it is not 1.
Inside the parentheses, half of −b4 is −b8.
Practice an AP Calculus AB definite integral question involving a rational expression and logarithm.
Definite Integrals / Evaluating a definite integral by simplifying the integrand∫12 3x − 13x dx =
B.
1 − 13ln 2Simplify the integrand before integrating.
3x−13x = 1 − 13x. Therefore,
∫12(1 − 13x)dx = [x − 13ln x]12
This equals 2 − 13ln 2 − 1 = 1 − 13ln 2.
The integrand is easier after splitting the fraction into 1 minus a reciprocal term. The natural logarithm appears because the integral contains 1x.
A common mistake is integrating the fraction directly without simplifying, which often causes sign or coefficient errors.
The trap is missing the coefficient 13 on the logarithm.
If this was difficult, it may reveal weakness in simplifying integrands before integrating.
This AP Calculus AB definite integral question tests algebraic simplification and logarithmic integration.
Split rational expressions when the denominator is a simple monomial.
Rewrite the integrand as 1 − 13x.
Solve a hard GAT quantitative mixture ratio question where water is added and the ratio changes.
Ratios and Proportions / Updating a ratio after adding to one partA mixture contains water and juice in the ratio 3:5. If 12 liters of water are added, the ratio becomes 5:5. How many liters of juice are in the mixture?
C.
30Let the original water and juice amounts be 3k and 5k.
Original amounts:
Water = 3k
Juice = 5k
After adding 12 liters of water:
Water = 3k + 12
The new ratio is 5:5, so water equals juice:
3k + 12 = 5k
12 = 2k
k = 6
Juice amount:
5k = 5 × 6 = 30
The ratio 5:5 means equal amounts, not necessarily 5 liters and 5 liters. The juice amount stayed constant while water increased.
A common mistake is thinking the final amounts are exactly 5 and 5. Ratios describe relative amounts, not fixed liters.
The trap is treating ratio numbers as actual quantities.
If this was difficult, the issue may be ratio modeling after a change.
GAT mixture-ratio questions often test whether you track what changed and what stayed the same.
When one part changes and the other stays fixed, use variables for the original ratio parts.
Write original amounts as 3k and 5k. Then update only water.
Solve an SAT algebra question involving f(x), g(x), h(x), and the coefficient of x.
Algebra / Multiplying linear functions and identifying a coefficientf(x) = 2x + 3
g(x) = 7x − 2
h(x) = 5x + 6
The functions f, g, and h are defined as shown. If f(x) · g(x) − h(x) = ax2 + bx + c where a, b, and c are constants, what is the value of b?
B.
12First multiply f(x) and g(x), then subtract h(x).
Multiply:
(2x + 3)(7x − 2) = 14x2 − 4x + 21x − 6 = 14x2 + 17x − 6
Now subtract h(x):
14x2 + 17x − 6 − (5x + 6)
= 14x2 + 12x − 12
So b = 12.
The expression must be fully simplified into the form ax2 + bx + c. The coefficient of x after simplification is 12.
A common mistake is forgetting to distribute the subtraction across both terms of h(x).
The trap is subtracting only 5x and not subtracting the constant 6.
If this was difficult, it may show a weak spot in multiplying expressions and combining like terms.
This SAT algebra question tests function operations and coefficient identification.
When subtracting a function, use parentheses before combining like terms.
Only the coefficient of x is needed, but full expansion prevents sign errors.
Solve an AP Calculus AB limit question using the squeeze theorem after dividing by x squared.
Limits and Continuity / Using the squeeze theorem after dividing by x squaredIf
3x2−2x3 ≤ f(x) ≤ 3x2+5x4
for all x near 0, then
limx→0 f(x)x2
is
C.
3Divide all three parts of the inequality by x2.
For x≠0, x2>0, so the inequality direction stays the same. Dividing gives 3−2x ≤ f(x)x2 ≤ 3+5x2. As x→0, both bounding expressions approach 3. Therefore, by the squeeze theorem, the limit is 3.
The squeeze theorem applies after transforming the bounds into bounds on the exact expression whose limit is requested.
A common mistake is applying the squeeze theorem to f(x) directly rather than to f(x)/x2.
The trap is not dividing the bounds by x2.
If this was difficult, it may reveal weakness in squeeze theorem transformations.
This AP Calculus AB limits question tests squeeze theorem setup.
When the target expression includes a denominator, divide the whole inequality carefully.
Because x2 is positive, the inequality signs do not reverse.