For what value of $k$ does the system of equations $3x + ky = 12$ and $6x + 8y = 20$ have no solution?
$2$
$4$
$8$
$16$
Two linear equations have no solution when the lines are parallel but not identical — the coefficient ratios must match while the constant ratio does not. Here $\dfrac{3}{6} = \dfrac{k}{8}$, so $k = 4$. Checking the constants: $\dfrac{12}{20} = \dfrac{3}{5} \ne \dfrac{1}{2}$, confirming the lines never meet. A very common Korean-student error is setting $\dfrac{3}{6}=\dfrac{12}{k}$ (mixing up which pair of coefficients to compare), which produces the wrong answer $k=8$.
If $r$ and $s$ are the two solutions of $2x^2 - 8x + 3 = 0$, what is the value of $r + s + rs$?
$\dfrac{3}{2}$
$4$
$\dfrac{11}{2}$
$-\dfrac{11}{2}$
Use Vieta's formulas directly — no need to find the actual roots. For $ax^2+bx+c=0$: sum of roots $r+s = -\dfrac{b}{a} = -\dfrac{-8}{2}=4$, and product $rs=\dfrac{c}{a}=\dfrac{3}{2}$. So $r+s+rs = 4+\dfrac{3}{2}=\dfrac{11}{2}$. Students who forget the product term and only report the sum pick $4$; watch the sign of $b$ carefully.
A radioactive substance has an initial mass of $500$ mg and decays at a rate of $8\%$ per hour. Which value is closest to the mass, in mg, remaining after $5$ hours?
$300.0$
$358.2$
$329.5$
$460.0$
Exponential decay: $f(t) = 500(0.92)^t$. At $t=5$: $f(5)=500(0.92)^5 \approx 329.5$ mg. The trap answer $300.0$ comes from treating this as linear decay ($500 - 500(0.08)(5)$) instead of exponential — decay problems almost never behave linearly. $358.2$ is the value at $t=4$ (an off-by-one error), and $460.0$ is just one single $8\%$ decrease applied once.
If $f(x) = x^2 - 3x + 5$, what is $f(2x-1)$?
$4x^2 - 10x + 9$
$4x^2 - 6x + 9$
$2x^2 - 6x + 4$
$4x^2 - 10x + 7$
Substitute $(2x-1)$ everywhere $x$ appears: $(2x-1)^2 - 3(2x-1) + 5 = (4x^2-4x+1) - (6x-3) + 5 = 4x^2 -4x+1-6x+3+5 = 4x^2-10x+9$. The most common slip is expanding $(2x-1)^2$ incorrectly as $4x^2-1$ (forgetting the middle term), which leads to $4x^2-6x+9$.
What is the solution to $|2x - 5| = 3x + 1$?
$x=-6$
$x=\dfrac{4}{5}$
$x=\dfrac{6}{5}$
$x=-\dfrac{4}{5}$
Split into two cases. Case 1: $2x-5 = 3x+1 \Rightarrow x=-6$. Case 2: $2x-5=-(3x+1) \Rightarrow x=\dfrac{4}{5}$. You must check both in the original equation, because the right side $3x+1$ must be $\ge 0$ for absolute value equality to hold. At $x=-6$: right side $=3(-6)+1=-17<0$, so this is extraneous — reject it. At $x=\dfrac{4}{5}$: both sides equal $\dfrac{17}{5}$. Only $x=\dfrac{4}{5}$ is valid. Skipping the extraneous-root check is the single most common error on absolute-value equations.
What is the solution set of $\dfrac{3}{x-2} + 4 = \dfrac{x+1}{x-2}$?
$x=2$
No solution
$x=0$
All real numbers except $2$
Multiply every term by $(x-2)$: $3 + 4(x-2) = x+1 \Rightarrow 3+4x-8=x+1 \Rightarrow 4x-5=x+1 \Rightarrow x=2$. But $x=2$ makes the original denominators zero — it is excluded from the domain. Since the only algebraic solution is exactly the excluded value, the equation has no solution. This is a classic rational-equation trap: always cross off any solution that matches an excluded value before selecting your final answer.
A car travels at a constant speed of $45$ miles per hour. What is the car's speed in feet per second? ($1$ mile $=5280$ feet)
$30$
$33$
$66$
$75.6$
Convert using both factors at once: $45 \dfrac{\text{mi}}{\text{hr}} \times \dfrac{5280\text{ ft}}{1\text{ mi}} \times \dfrac{1\text{ hr}}{3600\text{ s}} = \dfrac{45 \times 5280}{3600} = 66$ ft/s. A frequent shortcut error is dividing by $3600$ but forgetting to multiply by $5280$ first (or vice versa) — always write out both conversion factors explicitly rather than trying to do it from memory.
What is the radius of the circle given by $x^2+y^2-6x+4y-3=0$?
$3$
$\sqrt{13}$
$4$
$16$
Complete the square for both variables: $(x^2-6x+9) + (y^2+4y+4) = 3+9+4 \Rightarrow (x-3)^2+(y+2)^2=16$. So $r^2=16$, meaning $r=4$ — not $16$. Selecting $16$ (forgetting to take the square root of $r^2$) is the most common mistake on circle-equation questions.
In triangle $ABC$, segment $DE$ is parallel to $BC$, with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$. If $AD=4$, $DB=6$, and $AE=5$, what is the length of $EC$?
$4.8$
$6$
$7.5$
$9$
Because $DE \parallel BC$, triangle $ADE \sim$ triangle $ABC$, so corresponding side ratios are equal along each transversal: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. Substituting: $\dfrac{4}{6}=\dfrac{5}{EC} \Rightarrow EC = \dfrac{5\times 6}{4}=7.5$. A very common error is inverting the ratio (using $\dfrac{AD}{AB}=\dfrac{AE}{EC}$ instead of $\dfrac{AD}{DB}=\dfrac{AE}{EC}$), which gives the wrong answer $6$.
If $\sin\theta = \dfrac{3}{5}$ and $\theta$ is in Quadrant II, what is $\cos\theta$?
$\dfrac{4}{5}$
$-\dfrac{4}{5}$
$-\dfrac{3}{5}$
$\dfrac{3}{4}$
From the Pythagorean identity, $\cos^2\theta = 1-\sin^2\theta = 1-\dfrac{9}{25}=\dfrac{16}{25}$, so $\cos\theta = \pm\dfrac{4}{5}$. In Quadrant II, sine is positive but cosine is negative, so $\cos\theta=-\dfrac{4}{5}$. Forgetting the sign rule for Quadrant II (cosine negative, sine positive) is the most common error here.
A survey of $200$ students recorded whether each student passed a placement exam. Results: $70$ males passed, $30$ males failed, $90$ females passed, $10$ females failed. If one student who passed is selected at random, what is the probability that student is female?
$\dfrac{9}{20}$
$\dfrac{9}{16}$
$\dfrac{9}{10}$
$\dfrac{1}{2}$
This is a conditional probability — restrict your denominator to only the "passed" group, not all $200$ students. Total who passed $=70+90=160$. Females who passed $=90$. So $P(\text{Female}\mid\text{Pass}) = \dfrac{90}{160}=\dfrac{9}{16}$. The trap answer $\dfrac{9}{20}$ comes from dividing by the full $200$ students instead of restricting to the $160$ who passed — always re-read what group the "given" condition restricts you to.
A data set of $5$ numbers has a mean of $20$. A sixth number, $32$, is added to the set. What is the mean of the new set of $6$ numbers?
$20$
$21$
$22$
$26.4$
The original sum is $5\times 20=100$. Adding $32$ gives a new sum of $132$, spread over $6$ numbers: $\dfrac{132}{6}=22$. A common error is dividing by the old count of $5$ instead of the new count of $6$, giving $26.4$ — always update the denominator whenever the group size changes.
When the polynomial $f(x) = x^3 - 4x^2 + kx - 6$ is divided by $(x-2)$, the remainder is $4$. What is the value of $k$?
$5$
$9$
$14$
$-9$
By the Remainder Theorem, dividing by $(x-2)$ gives a remainder equal to $f(2)$. Compute: $f(2)=8-16+2k-6=2k-14$. Set this equal to the given remainder: $2k-14=4 \Rightarrow 2k=18 \Rightarrow k=9$. Sign errors while simplifying $8-16-6$ are the usual source of mistakes here — work one term at a time.
What is the solution set of the inequality $x^2 - 5x + 6 < 0$?
$2 < x < 3$
$x<2$ or $x>3$
$x\le 2$ or $x\ge 3$
All real numbers except $2$ and $3$
Factor first: $x^2-5x+6=(x-2)(x-3)$. This parabola opens upward with roots at $x=2$ and $x=3$, so it is below zero (negative) only between the roots: $20$ instead.
For what value of $k$ does the line $y = -2x + 1$ intersect the parabola $y = x^2 - 4x + k$ at exactly one point?
$-2$
$1$
$2$
$6$
Set the equations equal: $x^2-4x+k = -2x+1 \Rightarrow x^2-2x+(k-1)=0$. Exactly one intersection point means this quadratic has exactly one solution, so its discriminant must equal zero: $(-2)^2-4(1)(k-1)=0 \Rightarrow 4-4k+4=0 \Rightarrow 8=4k \Rightarrow k=2$. Sign errors while distributing the $-4(k-1)$ term are the most common source of mistakes on tangency questions.
A price is increased by $20\%$ and then the new price is decreased by $25\%$. What is the overall percent change from the original price?
$0\%$ (no change)
$-10\%$
$+50\%$
$-45\%$
Never subtract percent changes directly — multiply the growth factors instead: $(1.20)(0.75)=0.90$, which represents a net multiplier of $0.90$, i.e. a $10\%$ decrease from the original price. The trap answer $0\%$ comes from wrongly assuming a $20\%$ increase and a $25\%$ decrease "cancel out" — they don't, because the second percentage is taken from a larger base.
Line $\ell$ is given by $2x+3y=6$. A second line passes through the point $(4,-1)$ and is perpendicular to $\ell$. What is the $y$-intercept of the second line?
$-7$
$7$
$-1$
$1$
Rewrite $\ell$ in slope-intercept form: $y=-\dfrac{2}{3}x+2$, so its slope is $-\dfrac{2}{3}$. A perpendicular line has slope equal to the negative reciprocal: $\dfrac{3}{2}$. Using point-slope form through $(4,-1)$: $y-(-1)=\dfrac{3}{2}(x-4) \Rightarrow y = \dfrac{3}{2}x - 6 - 1 = \dfrac{3}{2}x - 7$. So the $y$-intercept is $-7$. Forgetting to flip and negate the slope (using $-\dfrac{2}{3}$ again instead of $\dfrac{3}{2}$) is the classic error on perpendicular-line questions.
A geometric sequence has first term $3$ and common ratio $2$. What is the sum of the first $6$ terms?
$96$
$189$
$192$
$378$
Use the finite geometric sum formula: $S_n = a_1\dfrac{r^n-1}{r-1} = 3\cdot\dfrac{2^6-1}{2-1}=3(63)=189$. You can double-check by listing all six terms directly: $3,6,12,24,48,96$, which sum to $189$. The trap answer $192$ comes from computing $3\cdot 2^6=192$ and forgetting to subtract $1$ inside the formula before dividing.
Pipe A can fill a tank in $6$ hours working alone. Pipe B can fill the same tank in $4$ hours working alone. If both pipes work together, how many hours will it take to fill the tank?
$2$
$2.4$
$5$
$10$
Add the individual work rates (tanks per hour), not the times: rate of A $=\dfrac{1}{6}$, rate of B $=\dfrac{1}{4}$. Combined rate $=\dfrac{1}{6}+\dfrac{1}{4}=\dfrac{2}{12}+\dfrac{3}{12}=\dfrac{5}{12}$ tank per hour. Time $=\dfrac{1}{5/12}=\dfrac{12}{5}=2.4$ hours. Averaging the two times directly ($\dfrac{6+4}{2}=5$) is a very common but incorrect shortcut — combined work problems always require adding rates, never times.
How many liters of a $30\%$ acid solution must be mixed with $10$ liters of a $70\%$ acid solution to produce a mixture that is $50\%$ acid?
$5$
$10$
$14$
$20$
Let $x$ be the liters of $30\%$ solution needed. Set total acid before mixing equal to total acid after mixing: $0.30x + 0.70(10) = 0.50(x+10)$. Expanding: $0.30x+7=0.50x+5 \Rightarrow 2=0.20x \Rightarrow x=10$. A quick sanity check: mixing equal $10$-liter amounts of $30\%$ and $70\%$ solutions should land exactly halfway at $50\%$ — which matches, confirming $x=10$.