🧮 Algebra Practice — Mixed Set 01
Recommended: 60–75 minutes. No calculator.
Problems
1.
Tags: Factoring · Easy · source: AMC10 2019 #1
What is the value of $2^{0} - (2^{0})^{0} + (2^{0})^{0^{0}}$?
A) $-2$
B) $-1$
C) $0$
D) $1$
E) $2$
Answer & Solution
Answer: D
We have $2^0 = 1$, so the expression becomes $1 - 1^0 + 1^{0^0} = 1 - 1 + 1^0 = 1 - 1 + 1 = 1$.
2.
Tags: Quadratics · Easy · source: Original (AMC-style)
If $x^2 - 5x + 6 = 0$, what is the sum of the roots?
A) $3$
B) $5$
C) $6$
D) $7$
E) $8$
Answer & Solution
Answer: B
By Vieta's formulas, the sum of the roots of $x^2 - 5x + 6 = 0$ is $5$.
3.
Tags: Rational Equations · Easy · source: Original (AMC-style)
What is the value of $\frac{x+1}{x-1}$ when $x = 3$?
A) $1$
B) $2$
C) $3$
D) $4$
E) $5$
Answer & Solution
Answer: B
Substituting $x = 3$: $\frac{3+1}{3-1} = \frac{4}{2} = 2$.
4.
Tags: Absolute Value · Easy · source: Original (AMC-style)
What is $|3 - 7|$?
A) $-4$
B) $0$
C) $4$
D) $10$
E) $21$
Answer & Solution
Answer: C
$|3 - 7| = |-4| = 4$.
5.
Tags: Systems · Easy · source: Original (AMC-style)
If $x + y = 7$ and $x - y = 3$, what is $x$?
A) $2$
B) $3$
C) $4$
D) $5$
E) $6$
Answer & Solution
Answer: D
Adding the equations: $(x+y) + (x-y) = 7 + 3$, so $2x = 10$ and $x = 5$.
6.
Tags: Sequences · Easy · source: Original (AMC-style)
What is the 5th term of the arithmetic sequence $2, 5, 8, 11, \ldots$?
A) $14$
B) $15$
C) $16$
D) $17$
E) $18$
Answer & Solution
Answer: A
The common difference is $3$, so the 5th term is $11 + 3 = 14$.
7.
Tags: Exponents · Easy · source: Original (AMC-style)
What is $2^3 \cdot 2^4$?
A) $2^7$
B) $2^{12}$
C) $4^7$
D) $8^7$
E) $16^7$
Answer & Solution
Answer: A
Using the exponent rule: $2^3 \cdot 2^4 = 2^{3+4} = 2^7$.
8.
Tags: Radicals · Easy · source: Original (AMC-style)
What is $\sqrt{16} + \sqrt{9}$?
A) $5$
B) $7$
C) $10$
D) $13$
E) $25$
Answer & Solution
Answer: B
$\sqrt{16} + \sqrt{9} = 4 + 3 = 7$.
9.
Tags: Factoring · Easy · source: Original (AMC-style)
Factor $x^2 - 9$.
A) $(x-3)^2$
B) $(x+3)^2$
C) $(x-3)(x+3)$
D) $(x-9)(x+1)$
E) Cannot be factored
Answer & Solution
Answer: C
This is a difference of squares: $x^2 - 9 = x^2 - 3^2 = (x-3)(x+3)$.
10.
Tags: Quadratics · Easy · source: Original (AMC-style)
What is the vertex of the parabola $y = x^2 - 4x + 3$?
A) $(2, -1)$
B) $(2, 1)$
C) $(-2, -1)$
D) $(-2, 1)$
E) $(4, 3)$
Answer & Solution
Answer: A
Completing the square: $y = x^2 - 4x + 3 = (x-2)^2 - 4 + 3 = (x-2)^2 - 1$. The vertex is $(2, -1)$.
11.
Tags: Rational Equations · Medium · source: Original (AMC-style)
Solve for $x$: $\frac{2x+1}{x-2} = 3$.
A) $5$
B) $6$
C) $7$
D) $8$
E) $9$
Answer & Solution
Answer: C
Cross-multiplying: $2x+1 = 3(x-2) = 3x-6$. So $2x+1 = 3x-6$, giving $x = 7$. We must check that $x \neq 2$, which is satisfied.
12.
Tags: Systems · Medium · source: Original (AMC-style)
If $2x + 3y = 13$ and $3x + 2y = 12$, what is $x + y$?
A) $3$
B) $4$
C) $5$
D) $6$
E) $7$
Answer & Solution
Answer: C
Adding the equations: $(2x+3y) + (3x+2y) = 13 + 12$, so $5x + 5y = 25$. Dividing by 5: $x + y = 5$.
13.
Tags: Absolute Value · Medium · source: Original (AMC-style)
How many solutions does $|x-3| = 5$ have?
A) $0$
B) $1$
C) $2$
D) $3$
E) Infinitely many
Answer & Solution
Answer: C
We have $x-3 = 5$ or $x-3 = -5$, giving $x = 8$ or $x = -2$. So there are 2 solutions.
14.
Tags: Sequences · Medium · source: Original (AMC-style)
The first term of a geometric sequence is 3, and the common ratio is 2. What is the 4th term?
A) $12$
B) $18$
C) $24$
D) $48$
E) $96$
Answer & Solution
Answer: C
The nth term is $a_n = a_1 \cdot r^{n-1} = 3 \cdot 2^{n-1}$. So $a_4 = 3 \cdot 2^{4-1} = 3 \cdot 2^3 = 3 \cdot 8 = 24$.
15.
Tags: Exponents · Medium · source: Original (AMC-style)
What is $\log_2(8) + \log_2(4)$?
A) $3$
B) $4$
C) $5$
D) $6$
E) $7$
Answer & Solution
Answer: C
$\log_2(8) + \log_2(4) = \log_2(8 \cdot 4) = \log_2(32) = \log_2(2^5) = 5$.
16.
Tags: Quadratics · Medium · source: Original (AMC-style)
For what value of $k$ does the equation $x^2 + kx + 9 = 0$ have exactly one real solution?
A) $-6$
B) $-3$
C) $0$
D) $3$
E) $6$
Answer & Solution
Answer: A
For exactly one real solution, the discriminant must be 0: $k^2 - 4(1)(9) = 0$, so $k^2 = 36$ and $k = \pm 6$. The answer is $k = -6$.
17.
Tags: Radicals · Medium · source: Original (AMC-style)
What is $\sqrt{2} \cdot \sqrt{8}$?
A) $2$
B) $4$
C) $8$
D) $16$
E) $32$
Answer & Solution
Answer: B
$\sqrt{2} \cdot \sqrt{8} = \sqrt{2 \cdot 8} = \sqrt{16} = 4$.
18.
Tags: Factoring · Medium · source: Original (AMC-style)
Factor $x^3 - 8$.
A) $(x-2)^3$
B) $(x+2)^3$
C) $(x-2)(x^2+2x+4)$
D) $(x+2)(x^2-2x+4)$
E) Cannot be factored
Answer & Solution
Answer: C
This is a difference of cubes: $x^3 - 8 = x^3 - 2^3 = (x-2)(x^2+2x+4)$.
19.
Tags: Systems · Medium · source: Original (AMC-style)
If $x^2 + y^2 = 25$ and $x + y = 7$, what is $xy$?
A) $6$
B) $8$
C) $10$
D) $12$
E) $14$
Answer & Solution
Answer: D
We have $(x+y)^2 = x^2 + 2xy + y^2 = 49$. Since $x^2 + y^2 = 25$, we get $25 + 2xy = 49$, so $2xy = 24$ and $xy = 12$.
20.
Tags: Rational Equations · Medium · source: Original (AMC-style)
What is the sum of all solutions to $\frac{1}{x} + \frac{1}{x+1} = \frac{1}{2}$?
A) $-1$
B) $0$
C) $1$
D) $2$
E) $3$
Answer & Solution
Answer: A
Multiplying by $2x(x+1)$: $2(x+1) + 2x = x(x+1)$, so $4x + 2 = x^2 + x$. Rearranging: $x^2 - 3x - 2 = 0$. By Vieta's formulas, the sum of roots is $3$, but we need to check for extraneous solutions. The solutions are $x = \frac{3 \pm \sqrt{17}}{2}$, and their sum is $3$. However, we must verify these don't make any denominator zero, which they don't. The answer is $3$, but since the question asks for the sum and we have $x^2 - 3x - 2 = 0$, the sum is $3$. Wait, let me recalculate: $\frac{1}{x} + \frac{1}{x+1} = \frac{x+1+x}{x(x+1)} = \frac{2x+1}{x(x+1)} = \frac{1}{2}$. So $4x+2 = x^2+x$, giving $x^2-3x-2 = 0$. The sum of roots is $3$.
21.
Tags: Quadratics · Hard · source: Original (AMC-style)
If the quadratic $x^2 + px + q$ has roots $r$ and $s$ such that $r^2 + s^2 = 10$ and $rs = 3$, what is $p^2 - 4q$?
A) $4$
B) $8$
C) $12$
D) $16$
E) $20$
Answer & Solution
Answer: A
We have $r + s = -p$ and $rs = q$. Since $r^2 + s^2 = 10$ and $rs = 3$, we have $(r+s)^2 = r^2 + 2rs + s^2 = 10 + 2(3) = 16$. So $r+s = \pm 4$, giving $p = \mp 4$. The discriminant is $p^2 - 4q = 16 - 4(3) = 4$.
22.
Tags: Exponents · Hard · source: Original (AMC-style)
If $2^x = 3^y = 6^z$, what is $\frac{1}{x} + \frac{1}{y}$ in terms of $z$?
A) $\frac{1}{z}$
B) $\frac{2}{z}$
C) $\frac{3}{z}$
D) $\frac{1}{2z}$
E) $\frac{1}{3z}$
Answer & Solution
Answer: A
Let $2^x = 3^y = 6^z = k$. Then $x = \log_2 k$, $y = \log_3 k$, and $z = \log_6 k$. We have $\frac{1}{x} + \frac{1}{y} = \frac{1}{\log_2 k} + \frac{1}{\log_3 k} = \log_k 2 + \log_k 3 = \log_k(2 \cdot 3) = \log_k 6 = \frac{1}{\log_6 k} = \frac{1}{z}$.
23.
Tags: Sequences · Hard · source: Original (AMC-style)
The sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = 1$ and $a_{n+1} = 2a_n + 1$ for $n \geq 1$. What is $a_{10}$?
A) $511$
B) $1023$
C) $2047$
D) $4095$
E) $8191$
Answer & Solution
Answer: B
We have $a_{n+1} + 1 = 2(a_n + 1)$. Let $b_n = a_n + 1$. Then $b_{n+1} = 2b_n$ with $b_1 = 2$. So $b_n = 2^n$, giving $a_n = 2^n - 1$. Therefore $a_{10} = 2^{10} - 1 = 1024 - 1 = 1023$.
24.
Tags: Absolute Value · Hard · source: Original (AMC-style)
How many integer solutions does $|x-1| + |x-3| = 2$ have?
A) $0$
B) $1$
C) $2$
D) $3$
E) Infinitely many
Answer & Solution
Answer: E
For $x \leq 1$: $|x-1| + |x-3| = (1-x) + (3-x) = 4-2x = 2$, so $x = 1$. For $1 < x < 3$: $|x-1| + |x-3| = (x-1) + (3-x) = 2$, which is always true. For $x \geq 3$: $|x-1| + |x-3| = (x-1) + (x-3) = 2x-4 = 2$, so $x = 3$. Therefore, all $x$ with $1 \leq x \leq 3$ are solutions, giving infinitely many integer solutions.
25.
Tags: Systems · Hard · source: Original (AMC-style)
If $x + y + z = 6$, $x^2 + y^2 + z^2 = 14$, and $xyz = 6$, what is $x^3 + y^3 + z^3$?
A) $18$
B) $24$
C) $36$
D) $48$
E) $72$
Answer & Solution
Answer: C
We have $(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx) = 36$. Since $x^2 + y^2 + z^2 = 14$, we get $xy + yz + zx = 11$. Now $x^3 + y^3 + z^3 = (x+y+z)^3 - 3(x+y+z)(xy+yz+zx) + 3xyz = 6^3 - 3(6)(11) + 3(6) = 216 - 198 + 18 = 36$.
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 | D | B | B | C | D | A | A | B | C | A | C | C | C | C | C | A | B | C | D | A | A | A | B | E | C |