🧮 Algebra Mini Mock 25 — AMC12 Level (25 AMC-Style Questions)
Recommended: 75 minutes. No calculator.
Problems
1.
Tags: Exponents & Radicals · Easy · source: Original (AMC-style)
Simplify $$ (27x^{6}y^{-3})^{\tfrac23}\cdot(9x^{-3}y^{6})^{\tfrac12}. $$
A) $9x^{\tfrac52}y$
B) $27x^{\tfrac32}y^{2}$
C) $27x^{\tfrac52}y$
D) $9x^{\tfrac72}y^{-1}$
E) $27x^{\tfrac52}y^{-1}$
Answer & Solution
Answer: C
$(27)^{2/3}=9$, $x^{6\cdot 2/3}=x^4$, $y^{-3\cdot2/3}=y^{-2}$; $(9)^{1/2}=3$, $x^{-3/2}$, $y^3$. Multiply: $9\cdot 3=27$, $x^{4-3/2}=x^{5/2}$, $y^{-2+3}=y$.
2.
Tags: Quadratics & Vieta · Easy · source: Original (AMC-style)
The roots of $x^2-5x+p=0$ satisfy $r^2+s^2=17$. Find $p$.
A) $2$
B) $3$
C) $5$
D) $6$
E) $4$
Answer & Solution
Answer: E
$r+s=5$, $rs=p$. Then $r^2+s^2=(r+s)^2-2rs=25-2p=17\Rightarrow p=4$.
3.
Tags: Logarithms · Easy · source: Original (AMC-style)
Solve for $x$: $$ \log_3(x-1)+\log_3(2x-5)=2. $$
A) $3$
B) $4$
C) $5$
D) $6$
E) $7$
Answer & Solution
Answer: B
$(x-1)(2x-5)=3^2=9\Rightarrow 2x^2-7x-4=0\Rightarrow x=\frac{7\pm 9}{4}$. Only $x=4$ fits the domain.
4.
Tags: Complex Numbers · Easy · source: Original (AMC-style)
Let $z=\dfrac{1+i\sqrt3}{1-i}$. Compute $|z|$.
A) $\sqrt2$
B) $\sqrt3$
C) $\sqrt6$
D) $2$
E) $2\sqrt2$
Answer & Solution
Answer: D
$|z|=\dfrac{|1+i\sqrt3|}{|1-i|}=\dfrac{\sqrt{1+3}}{\sqrt2}=\dfrac{2}{\sqrt2}=2/\sqrt2\cdot \sqrt2=\mathbf{2}$ (or multiply numerator out to confirm).
5.
Tags: Symmetric Sums · Easy · source: Original (AMC-style)
If $a+b=5$ and $a^2+b^2=13$, find $ab$.
A) $6$
B) $4$
C) $7$
D) $8$
E) $9$
Answer & Solution
Answer: A
$(a+b)^2=a^2+b^2+2ab\Rightarrow 25=13+2ab\Rightarrow ab=6$.
6.
Tags: Remainder Theorem · Easy · source: Original (AMC-style)
Find the remainder when $x^5-3x^4+2x^2-7$ is divided by $(x-2)$.
A) $-7$
B) $-3$
C) $3$
D) $15$
E) $-15$
Answer & Solution
Answer: E
Remainder is $p(2)=32-48+8-7=-15$.
7.
Tags: Absolute Value & Inequalities · Easy · source: Original (AMC-style)
How many integers satisfy $|2x-5|<x+1$?
A) $3$
B) $5$
C) $4$
D) $6$
E) $7$
Answer & Solution
Answer: C
Case $x\ge 2.5$: $2x-5<x+1\Rightarrow x<6$. Case $x<2.5$: $-2x+5<x+1\Rightarrow x>4/3$. Overall $(4/3,6)$; integers $\{2,3,4,5\}$.
8.
Tags: Recurrence & Limits · Easy · source: Original (AMC-style)
Let $a_1=2$ and $a_{n+1}=\dfrac{a_n+6}{2}$. Find $\lim_{n\to\infty}a_n$.
A) $4$
B) $6$
C) $8$
D) $12$
E) $3$
Answer & Solution
Answer: B
Fixed point $L=\frac{L+6}{2}\Rightarrow L=6$.
9.
Tags: Rational Equations · Easy · source: Original (AMC-style)
Solve $$ \frac{x+3}{x-2}=\frac{x-1}{x+1}. $$
A) $-2$
B) $-1$
C) $0$
D) $-\dfrac17$
E) $1$
Answer & Solution
Answer: D
Cross-multiply: $(x+3)(x+1)=(x-1)(x-2)\Rightarrow 4x+3=-3x+2\Rightarrow x=-\tfrac17$ (valid in domain).
10.
Tags: AM–GM & Optimization · Easy · source: Original (AMC-style)
For $x>0$, the minimum value of $x+\dfrac{9}{x}$ equals
A) $6$
B) $5$
C) $3\sqrt2$
D) $2\sqrt6$
E) $9$
Answer & Solution
Answer: A
AM–GM: $x+\frac{9}{x}\ge 2\sqrt{9}=6$, equality at $x=3$.
11.
Tags: Quadratic Parameter & Discriminant · Medium · source: Original (AMC-style)
For which real $k$ does $x^2+kx+(1-k)=0$ have exactly one real solution?
A) $k=2$
B) $k=-2$
C) $k=-2\pm\sqrt{2}$
D) $k=2\pm\sqrt{2}$
E) $k=-2\pm 2\sqrt{2}$
Answer & Solution
Answer: E
Discriminant $k^2-4(1-k)=k^2+4k-4=0\Rightarrow k=-2\pm 2\sqrt2$.
12.
Tags: Functional Equations (Polynomial) · Medium · source: Original (AMC-style)
Let $f$ be a quadratic polynomial satisfying $$ f(x)+f(1-x)=x^2+(1-x)^2\quad\text{for all }x. $$ Find $f!\left(\tfrac13\right)$.
A) $\tfrac19$
B) $\tfrac19$
C) $\tfrac29$
D) $\tfrac13$
E) $\tfrac23$
Answer & Solution
Answer: B
Let $f(x)=ax^2+bx+c$. Then $f(1-x)=a(1-2x+x^2)+b(1-x)+c$. Sum gives $2ax^2+(-2a-b)x+(a+b+2c)=2x^2-2x+1$. Thus $a=1$, $b=0$, $c=0$, so $f(x)=x^2$ and $f(1/3)=1/9$.
13.
Tags: Vieta & Reciprocal Roots · Medium · source: Original (AMC-style)
If the roots of $x^2-(m+3)x+(2m-3)=0$ satisfy $r=\dfrac1s$, find $m$.
A) $0$
B) $1$
C) $2$
D) $3$
E) $4$
Answer & Solution
Answer: C
$rs=1=2m-3\Rightarrow m=2$.
14.
Tags: Logarithmic Inequality · Medium · source: Original (AMC-style)
Solve: $$ \log_2(x-1)>\log_4(2x). $$
A) $x>\sqrt{3}$
B) $x>1+\sqrt{2}$
C) $x>2$
D) $x>2+\sqrt{3}$
E) $x>3$
Answer & Solution
Answer: D
Domain $x>1$. $\log_4(2x)=\frac{\log_2(2x)}{2}=\frac{1+\log_2 x}{2}$. Inequality $\Rightarrow 2\log_2(x-1)>1+\log_2 x\Rightarrow \log_2((x-1)^2)>\log_2(2x)\Rightarrow (x-1)^2>2x\Rightarrow x^2-4x+1>0\Rightarrow x>2+\sqrt3$ (since $x>1$).
15.
Tags: Radicals & Existence · Medium · source: Original (AMC-style)
Solve for real $x$: $$ \sqrt{x+3}+\sqrt{7-x}=6. $$
A) No real solution
B) $x=2$
C) $x=3$
D) $x=5$
E) $x=6$
Answer & Solution
Answer: A
Let $a=\sqrt{x+3}$, $b=\sqrt{7-x}$. Then $a,b\ge0$, $a^2+b^2=10$, but $a+b\le 2\sqrt{\frac{a^2+b^2}{2}}=2\sqrt5<6$. Impossible.
16.
Tags: Optimization (AM–GM) · Medium · source: Original (AMC-style)
For $x\ne 0$, the minimum value of $$ x^2+\frac{4}{x^2} $$ is
A) $2$
B) $3$
C) $4$
D) $2\sqrt{2}$
E) $8$
Answer & Solution
Answer: C
AM–GM: $x^2+\frac{4}{x^2}\ge 2\sqrt{4}=4$, equality when $x^2=2$.
17.
Tags: Telescoping Series · Medium · source: Original (AMC-style)
Evaluate $$ \sum_{k=1}^{\infty}\frac{3}{(k+1)(k+2)}. $$
A) $1$
B) $\tfrac32$
C) $\tfrac43$
D) $2$
E) $\tfrac54$
Answer & Solution
Answer: B
$\frac{3}{(k+1)(k+2)}=3\left(\frac{1}{k+1}-\frac{1}{k+2}\right)$. Telescopes to $3\cdot \frac12=\tfrac32$.
18.
Tags: Complex Roots of Unity · Medium · source: Original (AMC-style)
Let the roots of $z^3=1$ be $1,\omega,\omega^2$ with $\omega\ne 1$. Compute $$ (1+1)(1+\omega)(1+\omega^2). $$
A) $0$
B) $1$
C) $\sqrt3$
D) $3$
E) $2$
Answer & Solution
Answer: E
$(1+\omega)(1+\omega^2)=1+\omega+\omega^2+\omega^3=2+(\omega+\omega^2)=2-1=1$. Product is $2\cdot 1=2$.
19.
Tags: Symmetric System · Medium · source: Original (AMC-style)
Positive $a,b$ satisfy $a+\dfrac1b=3$ and $b+\dfrac1a=4$. Let $P=ab$. Find the sum of all possible values of $P$.
A) $10$
B) $9$
C) $8$
D) $6$
E) $5$
Answer & Solution
Answer: A
From $b=1/(3-a)$ and $1/(3-a)+1/a=4\Rightarrow a(3-a)=\tfrac34$. Then $P=ab=a/(3-a)=\frac{4a^2}{3}$. Using $4a^2-12a+3=0$, the two $a$-values give $P=4a-1$, so $P=5\pm 2\sqrt6$. Sum $=10$.
20.
Tags: Rational Functions & Involution · Medium · source: Original (AMC-style)
Let $f(x)=\dfrac{ax+b}{x+1}$ satisfy $f(f(x))=x$ for all $x\ne -1$, and $f(0)=2$. Find $a+b$.
A) $-1$
B) $0$
C) $2$
D) $1$
E) $3$
Answer & Solution
Answer: D
Composing gives $f(f(x))=\dfrac{(a^2+b)x+(ab+b)}{(a+1)x+(b+1)}$. For identity to hold, coefficient of $x^2$ on $x[(a+1)x+(b+1)]$ must vanish, so $a+1=0\Rightarrow a=-1$. Then $f(0)=b=2$, so $a+b=1$.
21.
Tags: Exponential vs Quadratic (Roots Count) · Hard · source: Original (AMC-style)
How many real solutions does $$ 2^x=x^2-3x+5 $$ have?
A) $0$
B) $1$
C) $2$
D) $3$
E) $4$
Answer & Solution
Answer: B
RHS is $(x-\tfrac32)^2+ \tfrac{11}{4}\ge 2.75$. At $x=1$, $2^x=2<3$; at $x=2$, $2^x=4>3$. Since $2^x$ is increasing and after crossing it stays above the upward-opening quadratic, there is exactly one intersection.
22.
Tags: Log Domain & Inequality · Hard · source: Original (AMC-style)
Solve for real $x$: $$ \log_{,x+1}(x^2-1)\ge 1. $$
A) $x\ge 1$
B) $x>1$
C) $x\ge 2$
D) $x\ge 0$
E) $x\ge 3$
Answer & Solution
Answer: C
Base: $x+1>0$ and $\ne 1\Rightarrow x>-1,\ x\ne 0$. Argument: $x^2-1>0\Rightarrow x>1$ or $x<-1$. Combine $\Rightarrow x>1$ so base $>1$. Then monotonicity gives $x^2-1\ge x+1\Rightarrow x^2-x-2\ge 0\Rightarrow (x-2)(x+1)\ge 0$. With $x>1$, this is $x\ge 2$.
23.
Tags: Symmetric Power Sums · Hard · source: Original (AMC-style)
If $x+\dfrac{1}{x}=3$ with $x\ne 0$, compute $x^3+\dfrac{1}{x^3}$.
A) $9$
B) $15$
C) $12$
D) $16$
E) $18$
Answer & Solution
Answer: E
$(x+\frac1x)^3=x^3+\frac{1}{x^3}+3\left(x+\frac1x\right)$. Thus $27=x^3+\frac{1}{x^3}+9\Rightarrow x^3+\frac{1}{x^3}=18$.
24.
Tags: Absolute Value Equation (Parameter) · Hard · source: Original (AMC-style)
For which real $t$ does the equation $|x-t|+|x+t|=10$ have at least one real solution $x$?
A) $-5\le t\le 5$
B) $t\ge 0$
C) $|t|\ge 5$
D) $-10\le t\le 10$
E) all real $t$
Answer & Solution
Answer: A
For fixed $t$, the minimum of $|x-t|+|x+t|$ occurs at $x=0$ and equals $2|t|$. Values $\ge 2|t|$ are achievable by varying $x$. Thus a solution exists iff $10\ge 2|t| \Rightarrow |t|\le 5$.
25.
Tags: Polynomials & Multiplicity · Hard · source: Original (AMC-style)
Let $y=x^2+bx+c$. Suppose $x=2$ is a double root of $$ (y-2)(y-3)=0. $$ Find $b$.
A) $-6$
B) $-5$
C) $-3$
D) $-4$
E) $-2$
Answer & Solution
Answer: D
$F(x)=(y-2)(y-3)$. A double root at $x=2$ requires $F(2)=0$ and $F'(2)=0$. Since $F'(x)=y'(x)\,(2y-5)$ with $y'=2x+b$, and $y(2)\in\{2,3\}$ makes $(2y-5)\ne 0$, we need $y'(2)=0\Rightarrow 4+b=0\Rightarrow b=-4$.
Answer Key
| # | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Ans | C | E | B | D | A | E | C | B | D | A | E | B | C | D | A | C | B | E | A | D | B | C | E | A | D |