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 quantitative digit logic question using digit sum and digit difference.
Word Problems and Logic / Using multiple number conditions to identify a valueA two-digit number has digits whose sum is 11. If the tens digit is 3 more than the ones digit, what is the number?
A.
74Let the ones digit be x. Then the tens digit is x + 3.
Let the ones digit be x. Then the tens digit is x + 3.
The sum of the digits is 11:
x + (x + 3) = 11
2x + 3 = 11
2x = 8
x = 4
So the ones digit is 4, and the tens digit is 7. The number is 74.
The phrase “tens digit is 3 more” means the larger digit is in the tens place. Once the ones digit is found, the place value determines the actual number.
A common mistake is finding the digits 7 and 4 but writing 47. Place value matters.
The trap is reversing the digits after solving.
If this was missed, the issue may be place-value interpretation in digit problems.
GAT number-logic questions often test place value after the equation is solved.
For digit questions, solve the digit values first, then place them correctly.
After finding the digits, check which one belongs in the tens place.
Solve an SAT free-response quadratic modeling question using a maximum value and another point.
Advanced Algebra / Using vertex form of a quadratic modelA quadratic function gives the estimated length of daylight d(t), in hours, in a certain city t months after March 1, where 0 ≤ t ≤ 7. According to the function, the estimated length of daylight is 12.66 hours 6 months after March 1 and the maximum estimated length of daylight is 13.91 hours 3.5 months after March 1. Based on this function, what is the estimated length of daylight, in hours, on March 1?
Use vertex form because the maximum value and when it occurs are given.
The maximum occurs at t = 3.5, and the maximum value is 13.91, so write:
d(t) = a(t − 3.5)2 + 13.91
Use d(6) = 12.66:
12.66 = a(6 − 3.5)2 + 13.91
12.66 = 6.25a + 13.91
−1.25 = 6.25a
a = −0.2
March 1 corresponds to t = 0:
d(0) = −0.2(0 − 3.5)2 + 13.91
d(0) = −0.2(12.25) + 13.91
d(0) = 11.46
The maximum point gives the vertex of the quadratic: (3.5, 13.91). A second point, (6, 12.66), determines the coefficient a. Once the model is known, substitute t = 0 because March 1 is zero months after March 1.
A common mistake is using t = 1 for March 1. The problem defines t as months after March 1, so March 1 is t = 0.
The trap is not recognizing that the maximum gives the vertex.
If this was difficult, it may show a weak spot in quadratic modeling and vertex-form substitution.
This is a SAT modeling question involving vertex form and a real-world context.
When a quadratic maximum is given, vertex form is usually the most direct setup.
Desmos can help visualize the downward-opening parabola, but vertex form is the cleanest exact method.
Start with d(t) = a(t − 3.5)2 + 13.91.
Practice an AP Calculus AB differentials question about possible error in the volume of a cube.
Applications of Differential Calculus / Using differentials to approximate error in volumeThe edge of a cube has length 10 in., with a possible error of 1%. The possible error, in cubic inches, in the volume of the cube is
D.
30Use V=s3 and approximate error with dV.
The volume of a cube is V=s3. With side length s=10, a 1% possible error is ds=0.01(10)=0.1. Since dV=3s2ds,
dV=3(10)2(0.1)=30
The possible error in volume is approximated by the differential of the volume function. The derivative 3s2 measures how sensitive volume is to a small side-length error.
A common mistake is using 1% of the volume, which gives 10, instead of using the differential.
The trap is taking 1% of 103 directly.
If this was difficult, it may reveal weakness in using differentials for error estimates.
This AP Calculus AB applications question tests differentials and measurement error.
Use the derivative of volume with respect to side length for error propagation.
The side-length error is 0.1 inch.
Solve a hard GAT quantitative exponent equation using powers of 2 with detailed solution and common trap.
Arithmetic and Exponents / Rewriting powers with a common baseIf 2a × 8a - 1 = 256, what is the value of a?
C.
114Rewrite 8 and 256 as powers of 2.
Since 8 = 23 and 256 = 28, rewrite the equation:
2a × (23)a - 1 = 28
2a × 23a - 3 = 28
24a - 3 = 28
So:
4a - 3 = 8
4a = 11
a = 114
The exponent on 8 affects the entire power. Since 8 = 23, the exponent becomes 3(a - 1), not just 3a.
A common mistake is writing 8a - 1 as 23a - 1. The correct exponent is 3a - 3.
The trap is losing the parentheses around a - 1.
If this felt hard, it may reveal weakness in exponent laws, especially powers raised to expressions.
GAT exponent questions often test exponent distribution more than calculation.
When powers have related bases, convert everything to the same base before solving.
The fastest route is common-base conversion: 8 → 23 and 256 → 28.
Solve an SAT advanced algebra question using signs of a factored quadratic to locate roots.
Advanced Algebra / Using signs of a factored quadratic to locate integer rootsf(x) = (x − a)(x − b)
The function f is defined by the given equation, where a and b are integer constants. If f(32) > 0, f(35) < 0, and f(38) > 0, which of the following could be the value of a + b?
D.
69A factored quadratic is positive outside its roots and negative between its roots.
The function f(x) = (x − a)(x − b) has roots a and b. Since f(35) < 0, the value 35 must be between the two roots.
Since f(32) > 0 and f(38) > 0, the roots must lie between 32 and 38, with one root less than 35 and one root greater than 35.
Because a and b are integers, possible roots include one from 33 or 34, and one from 36 or 37. One possible pair is 33 and 36, whose sum is:
33 + 36 = 69
So 69 could be the value of a + b.
The sign pattern tells where the roots are. A positive-leading-coefficient quadratic is negative between its roots and positive outside them. The values at 32, 35, and 38 locate the roots around 35.
A common mistake is trying to solve for exact roots from the answer choices without using the sign intervals.
The trap is forgetting that the roots are integer constants, which restricts the possible locations.
If this was difficult, it may show a weak spot in interpreting factored quadratic sign behavior.
This SAT advanced algebra question tests roots, intervals, and sign behavior.
For factored quadratics, use the sign pattern around the roots.
The value 35 must be between the two roots.
Practice an AP Calculus AB derivative question involving product rule and chain rule.
Differentiation / Using product rule and chain rule togetherIf
f(x)=x2sin(3x)
then f′(x) is
B.
2xsin(3x)+3x2cos(3x)Use product rule, then apply chain rule to sin(3x).
Using product rule, f′(x)=2xsin(3x)+x2·3cos(3x). Therefore, f′(x)=2xsin(3x)+3x2cos(3x).
This derivative requires both the product rule and the chain rule. The derivative of sin(3x) is 3cos(3x).
A common mistake is forgetting the factor 3 from the chain rule.
The trap is applying product rule but missing chain rule.
If this was difficult, it may reveal weakness in combined differentiation rules.
This AP Calculus AB differentiation question tests product rule and chain rule fluency.
When differentiating a product, differentiate both factors and preserve the inner derivative.
The derivative of sin(3x) is not just cos(3x).
Solve a medium GAT quantitative averages question using total sums and removed values.
Arithmetic and Averages / Using total sums to find removed valuesThe average of 8 numbers is 45. If two numbers are removed, the average of the remaining 6 numbers is 42. What is the average of the two removed numbers?
C.
54Convert each average into a total sum first.
The total of the original 8 numbers is:
8 × 45 = 360
The total of the remaining 6 numbers is:
6 × 42 = 252
So the sum of the two removed numbers is:
360 - 252 = 108
Their average is:
108 ÷ 2 = 54
Averages become easier when converted to totals. The removed numbers are not found by subtracting the averages. They are found by subtracting the total sums.
A common mistake is doing 45 - 42 = 3 and trying to use that difference directly. Average differences do not directly give removed values.
The trap is comparing averages instead of comparing totals.
If this felt slow, the pattern to improve is converting averages into sums before doing anything else.
Qudurat average questions often reward switching from average to total quickly.
For average problems, total sum is usually the cleanest route.
Use total = average × count. This keeps the question short.
Solve an SAT advanced algebra question involving common factors and difference of squares.
Advanced Algebra / Factoring by common factor and difference of squaresWhich expression is a factor of y2(x − 3) − 25(x − 3)3?
D.
y + 5x − 15Factor out the common expression (x−3) first.
Start with:
y2(x − 3) − 25(x − 3)3
Factor out (x−3):
(x − 3)[y2 − 25(x − 3)2]
The bracketed expression is a difference of squares:
y2 − [5(x − 3)]2
= [y − 5(x − 3)][y + 5(x − 3)]
One factor is:
y + 5(x − 3) = y + 5x − 15
The expression has a common factor and then a difference-of-squares structure. The listed factor comes from expanding y + 5(x−3).
A common mistake is stopping after factoring out (x−3) and missing the difference of squares.
The trap is not expanding y + 5(x−3) to match the answer choice.
If this was difficult, it may show a weak spot in factoring with grouped expressions.
This SAT advanced algebra question tests factoring structure.
For multi-step factoring, remove the common factor before applying special products.
Recognize 25(x−3)2 as [5(x−3)]2.
Practice an AP Calculus AB definite integral question using parametric equations to change variables.
Definite Integrals / Changing variables in a definite integral using parametric equationsIf x=4cos θ and y=3sin θ, then ∫24 xy dx is equivalent to
D.
48∫0π/3 sin2θ cos θ dθReplace x, y, and dx in terms of θ.
Since x=4cosθ, dx=−4sinθ dθ. Also y=3sinθ, so xy=(4cosθ)(3sinθ)=12sinθcosθ. Thus xy dx = 12sinθcosθ(−4sinθ)dθ = −48sin2θcosθ dθ.
When x=2, 4cosθ=2, so θ=π/3. When x=4, θ=0. Therefore,
∫24xy dx = ∫π/30−48sin2θcosθ dθ = 48∫0π/3sin2θcosθ dθ
The negative from dx is handled by reversing the bounds.
A common mistake is replacing x and y but forgetting to replace dx.
The trap is keeping the original x-bounds after switching to θ.
If this was difficult, it may reveal weakness in changing variables in definite integrals.
This AP Calculus AB definite integral question tests parametric substitution.
For parametric substitutions, convert the bounds and the differential.
The bounds in θ go from π/3 to 0, then reverse them to remove the negative sign.
Practice a hard GAT algebra question simplifying an algebraic fraction using difference of squares.
Algebra / Simplifying algebraic fractions by factoringFor x ≠ 3, simplify:
x2 - 9x - 3
B.
x + 3Factor the numerator using difference of squares.
Factor the numerator:
x2 - 9 = (x - 3)(x + 3)
So:
x2 - 9x - 3 = (x - 3)(x + 3)x - 3
Since x ≠ 3, cancel x - 3:
x + 3
The numerator is a difference of squares, not a simple subtraction. Once it is factored, the common factor x - 3 cancels.
A common mistake is canceling the x terms or subtracting 9 - 3. Only factors can be canceled.
The trap is canceling pieces that are not full factors.
If this was missed, the weak point may be factoring before simplification.
GAT algebra simplification questions often test whether you cancel factors or terms.
In algebraic fractions, factor before canceling.
Recognize x2 - 9 as (x - 3)(x + 3).
Solve an SAT statistics question comparing the median and standard deviation of two dot plots.
Statistics and Data Analysis / Interpreting dot plots and comparing measures of center and spreadThe 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 onlyFirst determine the median of each dot plot, then compare the spread.
Count the values in Data Set A:
4(1), 5(4), 6(2), 7(3), 8(2), 9(4), 10(1)
This is 17 values total, so the median is the 9th value. Because the distribution is centered at 7, the median is 7.
Data Set B has counts:
4(2), 5(4), 6(2), 7(1), 8(2), 9(4), 10(2)
This is also 17 values total, so the median is again the 9th value, which is 7. So Statement I is true.
However, Data Set B has more values farther from the center and fewer values at the center than Data Set A, so Data Set B has a larger standard deviation. Therefore, Statement II is false.
The correct answer is I only.
Both dot plots are symmetric about 7, so they share the same median. But the second distribution places more values at the extremes 4 and 10 and fewer at the center 7, which increases the spread. That means the standard deviations are not equal.
A common mistake is thinking that because both distributions are symmetric and centered at the same value, their standard deviations must also be equal.
The trap is assuming the same center automatically means the same spread.
If this was difficult, it may show a weak spot in comparing measures of center and variability.
This SAT statistics and data analysis question tests median and standard deviation from dot plots.
When comparing dot plots, treat center and spread as separate questions.
For an odd number of data points, the median is the middle position after ordering.
Practice an AP Calculus AB question using the graph of f prime to compare function values.
Applications of Differential Calculus / Using the sign of a derivative graph to compare function valuesFrom the graph it follows that
D.
f(2)<f(3)A function increases where its derivative is positive.
The graph shown is f′. Since f′(x)>0 between x=2 and x=3, the function f is increasing on that interval. Therefore, f(2)<f(3).
The derivative is positive from 0 to 5, so f increases there. The graph does not imply discontinuity of f at 4. Also, f is not decreasing on the entire interval 4<x<7 because f′ is still positive from 4 to 5.
A common mistake is reading the graph as f instead of f′.
The trap is claiming f decreases before f′ becomes negative.
If this was difficult, it may reveal weakness in interpreting derivative graphs.
This AP Calculus AB applications question tests derivative graph interpretation.
Use the sign of f′ to determine where f is increasing or decreasing.
On (2,3), the derivative is positive.