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 involving frequency tables, weighted means, and a negative constant.
Statistics and Data Analysis / Calculating means from frequency tables with algebraic valuesThe frequency tables represent data sets A and B, where c is a negative integer constant. The mean of data set A is r and the mean of data set B is q. What is the value of rq?
Data Set A
| Value | Frequency |
|---|---|
| c | 12 |
| 2c | 21 |
| 3c | 30 |
Data Set B
| Value | Frequency |
|---|---|
| c | 30 |
| 2c | 21 |
| 3c | 12 |
C.
43Compute each weighted mean. The value of c will cancel in the ratio.
For data set A, the total frequency is:
12 + 21 + 30 = 63
The weighted sum is:
12c + 21(2c) + 30(3c) = 12c + 42c + 90c = 144c
So:
r = 144c63 = 16c7
For data set B, the weighted sum is:
30c + 21(2c) + 12(3c) = 30c + 42c + 36c = 108c
So:
q = 108c63 = 12c7
Therefore:
rq = 16c712c7 = 43
The fact that c is negative does not change the ratio because c appears in both means and cancels. The two means are different because the frequencies for c and 3c are reversed.
A common mistake is thinking the ratio cannot be determined because c is unknown. Since both means are multiples of c, it cancels.
The trap is treating the unknown negative constant as preventing calculation.
If this was difficult, it may show a weak spot in weighted averages with symbolic values.
This SAT statistics and data analysis question tests weighted means and algebraic cancellation.
For frequency-table mean questions, multiply each value by its frequency before dividing by total frequency.
The total frequency is the same for both data sets, which makes the ratio easier to compare.
Practice an AP Calculus AB differentiation question using the inverse function derivative formula.
Differentiation / Using the inverse function derivative formulaLet f be differentiable and one-to-one. If
f(3)=8
and
f′(3)=4
then
(f−1)′(8)
is
A.
14Use the input that maps to 8.
The inverse derivative formula is (f−1)′(a)=1f′(f−1(a)). Since f(3)=8, f−1(8)=3. Therefore, (f−1)′(8)=1f′(3)=14.
The derivative of an inverse function at an output value uses the reciprocal of the original derivative at the corresponding input value.
A common mistake is using f′(8), which is not given and is not the correct input.
The trap is confusing the inverse input with the original input.
If this was difficult, it may reveal weakness in inverse derivative notation.
This AP Calculus AB differentiation question tests inverse function derivatives.
For inverse derivatives, first find the original input associated with the given inverse input.
Since f(3)=8, the relevant original input is 3.
Solve a medium GAT quantitative age word problem involving age difference and future total.
Word Problems and Logic / Translating age relationships into equationsAhmed is 4 years older than Sami. In 6 years, the sum of their ages will be 50. How old is Ahmed now?
B.
21Let Sami’s current age be x. Then Ahmed’s current age is x + 4.
Let Sami’s current age be x. Then Ahmed’s current age is x + 4.
In 6 years:
Sami = x + 6
Ahmed = x + 10
Their sum will be 50:
(x + 6) + (x + 10) = 50
2x + 16 = 50
2x = 34
x = 17
Ahmed is now:
17 + 4 = 21
The future sum includes both people aging by 6 years, so the total increases by 12 years. The current total is therefore 50 - 12 = 38. Since Ahmed is 4 years older, Ahmed is 21 and Sami is 17.
A common mistake is adding 6 years only once. Both people age by 6 years.
The trap is applying the future change to only one person.
If this was missed, the issue may be translating time-shifted word problems into equations.
GAT age questions often test whether you apply time changes to every person involved.
For age problems, track each person separately when moving into the future.
Since two people age by 6 years each, the total age increases by 12.
Solve an SAT advanced algebra question about the coefficient of x to the fifteenth in a polynomial product.
Advanced Algebra / Finding a specific coefficient in a product of polynomialsf(x) = 11x10 + 3x8
g(x) = −17x7 + 9x5
The polynomial p(x) is defined as the product of the given polynomials, f(x) and g(x). What is the coefficient of x15 in p(x)?
B.
48Only multiply term pairs whose exponents add to 15.
The x15 terms come from:
(11x10)(9x5) = 99x15
and:
(3x8)(−17x7) = −51x15
Add the coefficients:
99 + (−51) = 48
You do not need to expand the entire product. Focus only on the term pairs that create x15.
A common mistake is using only one contributing pair and choosing 99.
The trap is forgetting the negative contribution from 3x8 · −17x7.
If this was difficult, it may show a weak spot in polynomial products and coefficient tracking.
This SAT advanced algebra question tests polynomial multiplication and targeted coefficient extraction.
For coefficient questions, identify all exponent pairs that produce the target power.
Look for exponent sums of 15: 10+5 and 8+7.
Practice an AP Calculus AB functions question about zeros of a composite function.
Functions / Finding zeros of a composite functionLet
g(x)=f(x2−1)
If f(a)=0 only when a=3, then the zeros of g are
C.
x=−2 and x=2Set the input to f equal to the value that makes f zero.
For g(x)=0, we need f(x2−1)=0. Since f(a)=0 only when a=3, set x2−1=3. Then x2=4, so x=−2 or x=2.
The zeros of the composite function happen when the inside expression produces the zero-input of the outside function.
A common mistake is setting x2−1=0 instead of setting it equal to 3.
The trap is solving for zeros of the inside expression instead of zeros of the composite.
If this was difficult, it may reveal weakness in composite-function reasoning.
This AP Calculus AB functions question checks composite-function structure.
For composite-function zeros, identify the required input to the outer function.
The input x2−1 must equal 3.
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.