Percent · Ratio · Proportion
Q 01
Trap: "percent of original" vs "percent of new." Students confuse which value is the base (denominator).
🧠MEMO: Percent Change =
(New − Old) ÷ Old × 100 — always OLD as baseA jacket originally costs $80. It is first discounted by 25%, then the sale price is increased by 25%. What is the final price of the jacket?
Explanation
Step 1: Discount 25% → \$80 × 0.75 = \$60
Step 2: Increase 25% → \$60 × 1.25 = \$75
The trap: −25% then +25% does not return to the original. The second +25% is applied to the smaller base (\$60), not \$80. You lose 6.25% overall: \(0.75 × 1.25 = 0.9375\).
Step 2: Increase 25% → \$60 × 1.25 = \$75
The trap: −25% then +25% does not return to the original. The second +25% is applied to the smaller base (\$60), not \$80. You lose 6.25% overall: \(0.75 × 1.25 = 0.9375\).
Q 02
Trap: "ratio increases" doesn't mean you add the same amount to both parts. Find the exact unknown added.
🧠MEMO: RATIO TRAP → set up equation with variable, don't just scale ratio
A solution is 3 parts water and 2 parts juice. If 10 liters of water are added and the ratio of water to juice becomes 7 : 3, how many liters of juice are in the final solution?
Explanation
Let original total = 5k liters. Water = 3k, Juice = 2k.
After adding 10 L water: \(\dfrac{3k+10}{2k} = \dfrac{7}{3}\)
Cross multiply: \(3(3k+10) = 7(2k)\) → \(9k + 30 = 14k\) → \(k = 6\)
Juice = 2k = 12 liters... wait, check: water = 3(6)+10 = 28, juice = 12. Ratio = 28:12 = 7:3 ✓
But the question asks for liters in the final solution — juice stays at 12 liters. Answer is B?
Re-read: "how many liters of juice" → juice = 2k = 2×6 = 12. Answer B (12). (Note: if you picked C=15, you multiplied incorrectly — that's the trap.)
After adding 10 L water: \(\dfrac{3k+10}{2k} = \dfrac{7}{3}\)
Cross multiply: \(3(3k+10) = 7(2k)\) → \(9k + 30 = 14k\) → \(k = 6\)
Juice = 2k = 12 liters... wait, check: water = 3(6)+10 = 28, juice = 12. Ratio = 28:12 = 7:3 ✓
But the question asks for liters in the final solution — juice stays at 12 liters. Answer is B?
Re-read: "how many liters of juice" → juice = 2k = 2×6 = 12. Answer B (12). (Note: if you picked C=15, you multiplied incorrectly — that's the trap.)
Linear Equations · Systems
Q 03
Trap: The question asks for a value of a constant that makes the system have NO solution — not that you solve for x, y.
🧠MEMO: NO SOLUTION = same slope, different y-intercept → \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
The system below has no solution. What is the value of \(k\)?
\[3x - ky = 6\]\[6x - 8y = 10\]
Explanation
For no solution: slopes equal, intercepts different.
Divide row 2 by 2: \(3x - 4y = 5\).
Compare with row 1: \(3x - ky = 6\).
Coefficients of x match (3=3). For no solution: \(-k = -4\) → k = 4.
Check constant: \(6 \neq 5\) ✓ (different → no solution confirmed).
Trap: k=8 makes the original ratio 3/6 = k/8 → k=4, but students mistakenly write k=8 by not dividing.
Divide row 2 by 2: \(3x - 4y = 5\).
Compare with row 1: \(3x - ky = 6\).
Coefficients of x match (3=3). For no solution: \(-k = -4\) → k = 4.
Check constant: \(6 \neq 5\) ✓ (different → no solution confirmed).
Trap: k=8 makes the original ratio 3/6 = k/8 → k=4, but students mistakenly write k=8 by not dividing.
Q 04
Trap: Students solve for a specific value instead of interpreting what the slope MEANS in context.
🧠MEMO: SLOPE = rate of change = "for every 1 increase in x, y changes by slope"
A plumber charges a flat fee of $50 plus $45 per hour. The total cost \(C\) (in dollars) for \(h\) hours of work is \(C = 45h + 50\). What does the value 45 represent in this context?
Explanation
In \(C = 45h + 50\), the slope is 45 and the y-intercept is 50.
• 50 = flat fee (cost at h=0) → Option B describes the y-intercept, not the slope.
• 45 = slope = rate of change = cost increases by $45 for each additional hour.
This is a pure slope-interpretation question. The trap is choosing B (the flat fee) because the number 50 is mentioned in the problem.
• 50 = flat fee (cost at h=0) → Option B describes the y-intercept, not the slope.
• 45 = slope = rate of change = cost increases by $45 for each additional hour.
This is a pure slope-interpretation question. The trap is choosing B (the flat fee) because the number 50 is mentioned in the problem.
Quadratics · Polynomials
Q 05
Trap: The question asks for the MAXIMUM VALUE of the function, not the x-value where it occurs.
🧠MEMO: Vertex \((h, k)\): x = h gives MAX/MIN, y = k IS the max/min VALUE
A ball is thrown upward. Its height \(H\) in meters after \(t\) seconds is:
\[H(t) = -4(t-3)^2 + 36\]
What is the maximum height the ball reaches, and at what time does it occur?
Explanation
The function is in vertex form: \(H(t) = a(t-h)^2 + k\).
Here \(a = -4\) (opens down → maximum), \(h = 3\), \(k = 36\).
• The vertex is at \((3, 36)\)
• Maximum height = k = 36 meters, occurring at t = 3 seconds.
Common trap: flipping h and k, or reporting t=3 as the maximum VALUE.
Here \(a = -4\) (opens down → maximum), \(h = 3\), \(k = 36\).
• The vertex is at \((3, 36)\)
• Maximum height = k = 36 meters, occurring at t = 3 seconds.
Common trap: flipping h and k, or reporting t=3 as the maximum VALUE.
Q 06
Trap: Students factor or use the quadratic formula unnecessarily — Vieta's formulas give the answer instantly.
🧠MEMO: VIETA: sum of roots = \(-b/a\), product of roots = \(c/a\)
If \(r\) and \(s\) are the two solutions of \(2x^2 - 10x + 9 = 0\), what is the value of \(r + s\)?
Explanation
By Vieta's formulas: \(r + s = \dfrac{-b}{a} = \dfrac{-(-10)}{2} = \dfrac{10}{2} = \mathbf{5}\).
Trap A: choosing 9/2 = product of roots (\(c/a\)), not the sum.
Trap C: forgetting the negative sign in \(-b/a\) and writing −5.
Trap D: writing just −b = 10 without dividing by a.
Trap A: choosing 9/2 = product of roots (\(c/a\)), not the sum.
Trap C: forgetting the negative sign in \(-b/a\) and writing −5.
Trap D: writing just −b = 10 without dividing by a.
Rates · Distance · Work
Q 07
Trap: Average speed ≠ average of two speeds. Use Total Distance ÷ Total Time.
🧠MEMO: AVG SPEED = Total D ÷ Total T (never average the speeds directly!)
Maya drives from City A to City B at 60 mph and returns at 40 mph. What is her average speed for the entire trip?
Explanation
Let distance A→B = \(d\).
Time going: \(d/60\). Time returning: \(d/40\).
Total distance = \(2d\). Total time = \(d/60 + d/40 = 2d/120 + 3d/120 = 5d/120\).
Average speed = \(\dfrac{2d}{5d/120} = \dfrac{2d \times 120}{5d} = \dfrac{240}{5} = \mathbf{48 \text{ mph}}\).
This is the harmonic mean: \(\dfrac{2 \times 60 \times 40}{60+40} = 48\).
Time going: \(d/60\). Time returning: \(d/40\).
Total distance = \(2d\). Total time = \(d/60 + d/40 = 2d/120 + 3d/120 = 5d/120\).
Average speed = \(\dfrac{2d}{5d/120} = \dfrac{2d \times 120}{5d} = \dfrac{240}{5} = \mathbf{48 \text{ mph}}\).
This is the harmonic mean: \(\dfrac{2 \times 60 \times 40}{60+40} = 48\).
Q 08
Trap: Add RATES (1/time), not the times themselves.
🧠MEMO: WORK RATE: \(\frac{1}{A} + \frac{1}{B} = \frac{1}{T}\) — add FRACTIONS of job per hour
Printer A takes 4 hours to print a report. Printer B takes 6 hours. How many hours does it take if both printers work simultaneously?
Explanation
Rate of A = \(1/4\) report/hr. Rate of B = \(1/6\) report/hr.
Combined rate = \(1/4 + 1/6 = 3/12 + 2/12 = 5/12\) report/hr.
Time = \(\dfrac{1}{5/12} = \dfrac{12}{5} = 2.4 = 2\tfrac{2}{5}\) hours.
Trap A: (4+6)/2 = 5 — simply averaging the times, which is wrong.
Combined rate = \(1/4 + 1/6 = 3/12 + 2/12 = 5/12\) report/hr.
Time = \(\dfrac{1}{5/12} = \dfrac{12}{5} = 2.4 = 2\tfrac{2}{5}\) hours.
Trap A: (4+6)/2 = 5 — simply averaging the times, which is wrong.
Functions · Graphs · Transformations
Q 09
Trap: \(f(g(x))\) means plug g(x) INTO f — not multiply f and g.
🧠MEMO: COMPOSITION: \(f(g(x))\) = substitute g(x) wherever you see x in f
Let \(f(x) = 3x + 1\) and \(g(x) = x^2 - 2\).
What is \(f(g(3))\)?
What is \(f(g(3))\)?
Explanation
Step 1: Evaluate the inner function first: \(g(3) = 3^2 - 2 = 9 - 2 = 7\)
Step 2: Plug into f: \(f(7) = 3(7) + 1 = 21 + 1 = \mathbf{22}\)
Trap A (28): computing \(g(f(3))\) instead — reversed order.
Trap B (10): computing \(f(3) = 10\) and stopping there.
Step 2: Plug into f: \(f(7) = 3(7) + 1 = 21 + 1 = \mathbf{22}\)
Trap A (28): computing \(g(f(3))\) instead — reversed order.
Trap B (10): computing \(f(3) = 10\) and stopping there.
Q 10
Trap: \(f(x - 3)\) shifts RIGHT (not left), and \(f(x+3)\) shifts LEFT — opposite of the sign!
🧠MEMO: SHIFT: \(f(x - h)\) → RIGHT by h; \(f(x + h)\) → LEFT by h (the sign is INSIDE, so it's opposite)
The graph of \(y = f(x)\) passes through the point \((2, 5)\). The graph of \(y = f(x - 3) + 1\) passes through which point?
Explanation
Original: \((2, 5)\). Transformation: \(f(x-3)+1\).
\(x - 3\) shifts RIGHT by 3: new x = 2 + 3 = 5.
\(+1\) shifts UP by 1: new y = 5 + 1 = 6.
Answer: \((5, 6)\).
Trap A: students subtract 3 (shift left) instead of right — x becomes 2−3=−1.
\(x - 3\) shifts RIGHT by 3: new x = 2 + 3 = 5.
\(+1\) shifts UP by 1: new y = 5 + 1 = 6.
Answer: \((5, 6)\).
Trap A: students subtract 3 (shift left) instead of right — x becomes 2−3=−1.
Statistics · Data Interpretation
Q 11
Trap: Adding an outlier heavily shifts the MEAN but barely changes the MEDIAN.
🧠MEMO: OUTLIER → distorts MEAN (average) more than MEDIAN (middle value)
A class of 5 students scored: 70, 72, 74, 76, 78. A 6th student joins with a score of 10. Which of the following is true?
Explanation
Original mean = 74. New mean = (70+72+74+76+78+10)/6 = 380/6 ≈ 63.3. Decrease ≈ 10.7.
Original median = 74 (middle of 5). New sorted list: {10, 70, 72, 74, 76, 78}. New median = (72+74)/2 = 73. Decrease = 1.
Mean drops by ~10.7; median drops by only 1. Mean decreases more.
Original median = 74 (middle of 5). New sorted list: {10, 70, 72, 74, 76, 78}. New median = (72+74)/2 = 73. Decrease = 1.
Mean drops by ~10.7; median drops by only 1. Mean decreases more.
Q 12
Trap: Use the ROW or COLUMN total as the denominator — not the grand total — for conditional probability.
🧠MEMO: "GIVEN THAT" = conditional → denominator is the GIVEN group's total
The table shows survey results of 200 students:
\[\begin{array}{|c|c|c|c|}\hline & \text{Cats} & \text{Dogs} & \text{Total}\\\hline \text{Grade 9} & 30 & 50 & 80\\\hline \text{Grade 10} & 60 & 60 & 120\\\hline \text{Total} & 90 & 110 & 200\\\hline\end{array}\]
What is the probability that a student prefers cats, given that the student is in Grade 10?
Explanation
"Given Grade 10" → restrict to the Grade 10 row (total = 120).
Cats in Grade 10 = 60.
P(cats | Grade 10) = \(\dfrac{60}{120} = \dfrac{1}{2}\).
Trap A: uses grand total 200 — ignores the conditional.
Trap D: uses column total for cats (90) — wrong denominator.
Cats in Grade 10 = 60.
P(cats | Grade 10) = \(\dfrac{60}{120} = \dfrac{1}{2}\).
Trap A: uses grand total 200 — ignores the conditional.
Trap D: uses column total for cats (90) — wrong denominator.
Q 13
Trap: Predicting outside the data range = extrapolation = less reliable (SAT often asks about this).
🧠MEMO: INTERPOLATION (inside data) = valid; EXTRAPOLATION (outside data) = unreliable
A line of best fit for a dataset shows the equation \(\hat{y} = 2.5x + 4\), where \(x\) = months of training and \(y\) = push-ups completed. The data was collected for months 1 through 12. A student uses this equation to predict push-ups for month 30. Which best describes this prediction?
Explanation
Month 30 is far outside the observed range of months 1–12. This is extrapolation — applying the model beyond its data range.
The linear trend may not continue past month 12 (e.g., physical limits, plateaus). Extrapolation is unreliable.
Predicting within months 1–12 would be interpolation — acceptable and relatively reliable.
The linear trend may not continue past month 12 (e.g., physical limits, plateaus). Extrapolation is unreliable.
Predicting within months 1–12 would be interpolation — acceptable and relatively reliable.
Geometry · Word Problems with Shapes
Q 14
Trap: Area scales by the SQUARE of the scale factor, not by the scale factor itself.
🧠MEMO: SCALE k → length ×k, area ×k², volume ×k³
Two similar triangles have corresponding side lengths in a ratio of 3 : 5. If the area of the smaller triangle is 27 cm², what is the area of the larger triangle?
Explanation
Area ratio = \(\left(\dfrac{3}{5}\right)^2 : 1 = \dfrac{9}{25}\).
So \(\dfrac{27}{\text{large area}} = \dfrac{9}{25}\) → Large area = \(\dfrac{27 \times 25}{9} = 3 \times 25 = \mathbf{75}\) cm².
Trap A: multiplies by 5/3 = linear scale factor (gets 45) — forgets to square.
Trap D: multiplies 27 × 5 = 135 — uses ratio 3:5 directly.
So \(\dfrac{27}{\text{large area}} = \dfrac{9}{25}\) → Large area = \(\dfrac{27 \times 25}{9} = 3 \times 25 = \mathbf{75}\) cm².
Trap A: multiplies by 5/3 = linear scale factor (gets 45) — forgets to square.
Trap D: multiplies 27 × 5 = 135 — uses ratio 3:5 directly.
Q 15
Trap: Confusing arc LENGTH (part of circumference) with sector AREA (part of circle area).
🧠MEMO: Arc length = \(\frac{\theta}{360} \times 2\pi r\); Sector area = \(\frac{\theta}{360} \times \pi r^2\)
A sector of a circle has a central angle of 120° and a radius of 9 cm. What is the arc length of the sector?
Explanation
Arc length = \(\dfrac{120}{360} \times 2\pi(9) = \dfrac{1}{3} \times 18\pi = \mathbf{6\pi}\) cm.
Trap A (\(27\pi\) cm²): this is the SECTOR AREA = \(\frac{1}{3}\pi(9)^2 = 27\pi\), but the units cm² give it away — the question asks for arc length (units: cm).
Trap D: forgetting to multiply by 1/3.
Trap A (\(27\pi\) cm²): this is the SECTOR AREA = \(\frac{1}{3}\pi(9)^2 = 27\pi\), but the units cm² give it away — the question asks for arc length (units: cm).
Trap D: forgetting to multiply by 1/3.
Q 16
Trap: Cone volume has a 1/3 factor. Students often use the cylinder formula instead.
🧠MEMO: Cone = \(\frac{1}{3}\pi r^2 h\); Cylinder = \(\pi r^2 h\) — cone is exactly 1/3 of cylinder with same r, h
A cone-shaped cup has a radius of 3 cm and a height of 8 cm. It is filled with water. Another cylindrical cup has the same radius and height. What fraction of the cylindrical cup is filled if all the water from the cone is poured into it?
Explanation
Cone volume = \(\dfrac{1}{3}\pi(3)^2(8) = \dfrac{1}{3}\pi(72) = 24\pi\)
Cylinder volume = \(\pi(3)^2(8) = 72\pi\)
Fraction = \(\dfrac{24\pi}{72\pi} = \dfrac{1}{3}\).
This is always 1/3 regardless of dimensions — a cone is always 1/3 of the cylinder with same r and h.
Cylinder volume = \(\pi(3)^2(8) = 72\pi\)
Fraction = \(\dfrac{24\pi}{72\pi} = \dfrac{1}{3}\).
This is always 1/3 regardless of dimensions — a cone is always 1/3 of the cylinder with same r and h.
Q 17
Trap: The problem gives extra information (e.g., total distance traveled) to confuse — draw a diagram and identify the right triangle.
🧠MEMO: DRAW IT FIRST — identify the hypotenuse vs legs before computing
A ladder leans against a wall. The base of the ladder is 5 feet from the wall, and the ladder reaches 12 feet up the wall. A second ladder of the same length is placed with its base 6 feet from the wall. How high up the wall does the second ladder reach?
Explanation
Ladder length (hypotenuse): \(\sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13\) ft.
Second ladder: same length 13 ft, base = 6 ft.
Height = \(\sqrt{13^2 - 6^2} = \sqrt{169 - 36} = \sqrt{133}\) ft.
Trap C (11): subtracting 1 from 12 since base increased by 1 — ignores the nonlinear relationship.
Note: \(\sqrt{133} \approx 11.53\) ft.
Second ladder: same length 13 ft, base = 6 ft.
Height = \(\sqrt{13^2 - 6^2} = \sqrt{169 - 36} = \sqrt{133}\) ft.
Trap C (11): subtracting 1 from 12 since base increased by 1 — ignores the nonlinear relationship.
Note: \(\sqrt{133} \approx 11.53\) ft.
Q 18
Trap: "doubles every 3 hours" means the exponent is t/3, NOT t×3 or just t.
🧠MEMO: DOUBLING: \(A = A_0 \cdot 2^{t/d}\) where d = doubling period
A bacteria population starts at 500 and doubles every 4 hours. Which equation gives the population \(P\) after \(t\) hours?
Explanation
At t=0: P = 500 ✓. At t=4: P should = 1000 (one doubling).
Check C: \(500 \cdot 2^{4/4} = 500 \cdot 2^1 = 1000\) ✓
Check A: \(500 \cdot 2^{4(4)} = 500 \cdot 2^{16}\) ✗ — way too large.
Check B: \(500 \cdot 4^4 = 500 \cdot 256\) ✗
The exponent must be \(t/4\) so that every 4 hours adds exactly 1 to the exponent (one doubling).
Check C: \(500 \cdot 2^{4/4} = 500 \cdot 2^1 = 1000\) ✓
Check A: \(500 \cdot 2^{4(4)} = 500 \cdot 2^{16}\) ✗ — way too large.
Check B: \(500 \cdot 4^4 = 500 \cdot 256\) ✗
The exponent must be \(t/4\) so that every 4 hours adds exactly 1 to the exponent (one doubling).
Q 19
Trap: "At least" = ≥, "No more than" = ≤. Translating the inequality direction wrong kills the whole problem.
🧠MEMO: AT LEAST ≥, AT MOST ≤, EXCEEDS >, LESS THAN < — underline these words first!
A phone plan costs $25 per month plus $0.10 per text message. Jordan wants to spend no more than $40 per month. What is the maximum number of text messages Jordan can send?
Explanation
Set up: \(25 + 0.10t \leq 40\)
\(0.10t \leq 15\)
\(t \leq 150\)
Maximum = 150 messages.
Trap A (400): solves \(0.10t = 40\) ignoring the $25 base fee.
The key is subtracting the flat fee first before dividing.
\(0.10t \leq 15\)
\(t \leq 150\)
Maximum = 150 messages.
Trap A (400): solves \(0.10t = 40\) ignoring the $25 base fee.
The key is subtracting the flat fee first before dividing.
Q 20
Trap: This problem combines percent AND function AND word setup. Don't rush — re-read which quantity is asked.
🧠MEMO: RE-READ the final question! Many students solve for the wrong variable at the last step.
A store marks up items by 40% from wholesale price. During a sale, all items are discounted 20% from the marked-up price. If the final sale price of a jacket is $84, what was the original wholesale price?
Explanation
Let wholesale = \(w\).
After 40% markup: \(w \times 1.4\)
After 20% discount: \(w \times 1.4 \times 0.8 = 1.12w\)
Set equal to $84: \(1.12w = 84\)
\(w = \dfrac{84}{1.12} = \mathbf{\$75}\).
Trap A ($70): mistakenly reverses the percent: \(84 / 1.2 = 70\) (only undoes discount).
Trap D: multiplies instead of divides.
The chain \(1.4 \times 0.8 = 1.12\) is the key insight — a net 12% increase from wholesale.
After 40% markup: \(w \times 1.4\)
After 20% discount: \(w \times 1.4 \times 0.8 = 1.12w\)
Set equal to $84: \(1.12w = 84\)
\(w = \dfrac{84}{1.12} = \mathbf{\$75}\).
Trap A ($70): mistakenly reverses the percent: \(84 / 1.2 = 70\) (only undoes discount).
Trap D: multiplies instead of divides.
The chain \(1.4 \times 0.8 = 1.12\) is the key insight — a net 12% increase from wholesale.
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