AAAnup Aryal

SAT • QUESTION & ASSESSMENT DEVELOPMENT

A good wrong answer should tell me why the student was wrong.

I develop SAT-style practice with difficulty, variation and distractor behavior in mind. The goal is not to make questions look strange. It is to prevent students from succeeding only because they recognize a template.

← SAT OverviewPractice Library
3Complete 22-question modules
66Questions across those modules
6Selected portfolio samples
2Target uses: tutoring + question work

BUILDING A MODULE

Twenty-two acceptable questions can still make a bad module.

I check the set as a whole, not only the individual questions.

I look for repeated structures, overused solution methods, weak distractors and difficulty that comes from unnecessary calculation rather than reasoning. I solve every question myself and revise questions when they feel too similar to material already used.

DIFFICULTY
→
PROBLEM STRUCTURE
→
FULL SOLUTION
→
LIKELY MISTAKES
→
DISTRACTORS
→
REVIEW / REVISE

THE DISTRACTOR LAB

Random wrong numbers waste information.

Where possible, each wrong option is built from a mistake I can realistically imagine a student making.

SIGN ERRORCorrect setup, wrong sign
WRONG FORMULAFamiliar rule applied in the wrong situation
STOPS EARLYIntermediate value selected as final answer
WRONG QUANTITYSolves for x when the question asks for 2x
ARITHMETIC SLIPConcept is right, calculation is not
MISREADS CONDITIONReasonable mathematics applied to the wrong interpretation

SAT-STYLE QUESTIONS I DEVELOPED

Six samples selected for range, not just difficulty.

These are self-developed practice questions written to reflect Digital SAT Math content and reasoning. They are not College Board questions.

01

Advanced Math

Rational Function Reconstruction

A function is defined by f(x) = (ax + b)/(x − 4), where a and b are constants and x ≠ 4. The graph passes through (0, −3) and (2, −5). It can also be written as f(x) = c + d/(x − 4). For the value of x at which f(x) = 7, what is x + c + d?
See reasoning & distractor design
WHY I DESIGNED IT

The student has to reconstruct the function, change representation, solve a new condition and keep track of the final requested expression.

LIKELY FAILURE POINTS

Common traps: stopping after finding a and b; sign errors in the rewrite; finding x but forgetting c + d.

SOLUTION

From (0, −3), b = 12. From (2, −5), a = −1. So f(x) = (−x + 12)/(x − 4) = −1 + 8/(x − 4). Thus c = −1 and d = 8. Solving f(x) = 7 gives x = 5, so x + c + d = 12.

02

Advanced Math

Function Composition Without the Function

The function f is linear. For all x, f(f(x)) = 9x + 8. If f(0) > 0, what is f(5)?
See reasoning & distractor design
WHY I DESIGNED IT

A short problem with hidden structure. The student has information about the composition, not the function itself.

LIKELY FAILURE POINTS

Common traps: treating 9x + 8 as f(x); choosing the negative slope without checking f(0); finding the slope but not the intercept.

SOLUTION

Let f(x) = mx + b. Then f(f(x)) = m²x + b(m + 1). Therefore m² = 9 and b(m + 1) = 8. The condition f(0) = b > 0 selects m = 3 and b = 2. Hence f(5) = 17.

03

Geometry + Advanced Math

A Tangent Line to a Circle

In the xy-plane, the line y = mx intersects the circle (x − 6)² + y² = 9 at exactly one point. There are two possible values of m. What is the sum of those values?
See reasoning & distractor design
WHY I DESIGNED IT

The phrase “exactly one point” has to be translated from geometry into an algebraic tangency condition.

LIKELY FAILURE POINTS

Common traps: solving for one slope only; treating the line as a radius; missing the symmetry that the two valid slopes have opposite signs.

SOLUTION

Substitute y = mx into the circle. The resulting quadratic in x must have exactly one real solution. The tangency condition gives m² = 1/3, so the two slopes are opposites. Their sum is 0.

04

Problem Solving & Data Analysis

Sampling, Margin of Error and Population Inference

A district has 18,000 students. Study A randomly samples 400 students from all schools: 46% report traveling by bus, with a margin of error of 5 percentage points. Study B samples 1,600 students only from the district’s high schools and finds 44%. Which conclusion is best supported?
See reasoning & distractor design
WHY I DESIGNED IT

The larger sample is deliberately tempting, but it does not represent the full district population as well as the smaller random district-wide sample.

LIKELY FAILURE POINTS

Common traps: assuming larger sample automatically means better estimate; treating 46% as exact; averaging the two percentages.

SOLUTION

Study A supports a district-wide interval from 41% to 51%. Applying that interval to 18,000 gives approximately 7,380 to 9,180 students.

05

Algebra + Applied Modeling

The Last Step Is Part of the Problem

A company sells 370 subscriptions. Premium subscriptions are 40 more than twice the number of basic subscriptions. Basic plans earn $25 and premium plans earn $40. The company refunds $5 for every premium subscription sold after the number of premium subscriptions exceeds 250. What is the final revenue?
See reasoning & distractor design
WHY I DESIGNED IT

The system of equations is only the first half. The question is designed so a correct intermediate revenue appears tempting before the refund condition is applied.

LIKELY FAILURE POINTS

Common traps: stopping at gross revenue; refunding every premium plan instead of only those above 250; reversing the basic/premium relationship.

SOLUTION

Basic + premium = 370 and premium = 2(basic) + 40, giving 110 basic and 260 premium. Gross revenue is $13,150. Ten premium subscriptions exceed 250, so the $50 refund reduces final revenue to $13,100.

06

Advanced Math

Tangency to a Parabola

The line y = mx + 4 is tangent to the parabola y = x² − 6x + 13. There are two possible values of m. What is the sum of these values?
See reasoning & distractor design
WHY I DESIGNED IT

The same one-solution idea appears in a different representation. Students must create the quadratic themselves and recognize tangency through its discriminant.

LIKELY FAILURE POINTS

Common traps: finding one tangent slope; setting the functions equal but not using the tangency condition; solving for x instead of the requested sum of slopes.

SOLUTION

Equating the line and parabola gives x² − (m + 6)x + 9 = 0. Tangency requires one real solution, so the discriminant is zero: (m + 6)² − 36 = 0. Thus m = 0 or −12, and the sum is −12.

DIFFICULTY PHILOSOPHY

The mathematics should remain SAT mathematics. The challenge should come from what the student has to notice.

NOT HARD

Ugly numbers

Long arithmetic does not automatically create meaningful difficulty.

NOT HARD

Unnecessary steps

A problem can be tedious without requiring stronger reasoning.

USEFUL HARD

Connected ideas

Difficulty becomes useful when familiar concepts must be recognized and combined in an unfamiliar arrangement.

HOW I EXPLAIN

One question. More than one way in.

These are examples of my worked explanations for students. The underlying practice questions in this section are not presented as questions I authored.

PARABOLA

Structure, algebra and Desmos

I use roots, vertex symmetry, coefficient relationships and graphing/regression when they genuinely offer different routes to the same result.

STATISTICS

Extreme possible means

I break histogram interval problems into lower and upper possible values so students can see why the maximum or minimum difference occurs.

RECTANGULAR PRISMS

Visualize before calculating

The systematic surface-area method is followed by the shorter observation: two original surface areas minus the two glued faces that disappear inside.

TEACHING MATERIALS

Questions are only one part of the preparation system.

I have also developed study notes, formula sheets, worksheets, mock and diagnostic material, solution manuals and recorded explanation videos.

The material is designed to support the same sequence I use in teaching: understand the concept, diagnose the mistake, build reliability and then reduce the time.