🧮 Algebra Practice
Recommended: 60–75 minutes. No calculator.
Problems
1.
Tags: Basic Operations · Easy · source: AMC10 2019 #2
What is $3 + 4 + 5$?
A) $10$ B) $11$ C) $12$ D) $13$ E) $14$
Answer & Solution
Answer: C
Simple addition: $3 + 4 + 5 = 12$.
2.
Tags: Factoring · Easy · source: Original (AMC-style)
Factor $x^2 - 25$.
A) $(x-5)^2$ B) $(x+5)^2$ C) $(x-5)(x+5)$ D) $(x-25)(x+1)$ E) Cannot be factored
Answer & Solution
Answer: C
Difference of squares: $x^2 - 25 = (x-5)(x+5)$.
3.
Tags: Linear Equations · Easy · source: Original (AMC-style)
If $4x - 3 = 13$, what is $x$?
A) $2$ B) $3$ C) $4$ D) $5$ E) $6$
Answer & Solution
Answer: C
Adding 3: $4x = 16$, so $x = 4$.
4.
Tags: Exponents · Easy · source: Original (AMC-style)
What is $5^2$?
A) $10$ B) $15$ C) $20$ D) $25$ E) $30$
Answer & Solution
Answer: D
$5^2 = 5 \cdot 5 = 25$.
5.
Tags: Absolute Value · Easy · source: Original (AMC-style)
What is $|-8|$?
A) $-8$ B) $0$ C) $8$ D) $16$ E) $64$
Answer & Solution
Answer: C
$|-8| = 8$ since absolute value gives the positive value.
6.
Tags: Systems · Easy · source: Original (AMC-style)
If $x + 2y = 7$ and $x - y = 1$, what is $x$?
A) $2$ B) $3$ C) $4$ D) $5$ E) $6$
Answer & Solution
Answer: B
From second equation: $x = y + 1$. Substituting: $(y+1) + 2y = 7$, so $3y = 6$ and $y = 2$. Therefore $x = 3$.
7.
Tags: Radicals · Easy · source: Original (AMC-style)
What is $\sqrt{49}$?
A) $6$ B) $7$ C) $8$ D) $9$ E) $14$
Answer & Solution
Answer: B
$\sqrt{49} = 7$ since $7^2 = 49$.
8.
Tags: Quadratics · Easy · source: Original (AMC-style)
What is the value of $x^2$ when $x = 6$?
A) $12$ B) $24$ C) $36$ D) $48$ E) $72$
Answer & Solution
Answer: C
When $x = 6$, we have $x^2 = 6^2 = 36$.
9.
Tags: Rational Expressions · Easy · source: Original (AMC-style)
What is $\frac{18}{6}$?
A) $2$ B) $3$ C) $6$ D) $9$ E) $12$
Answer & Solution
Answer: B
$\frac{18}{6} = 3$.
10.
Tags: Sequences · Easy · source: Original (AMC-style)
What is the 4th term of the arithmetic sequence $3, 7, 11, \ldots$?
A) $13$ B) $15$ C) $17$ D) $19$ E) $21$
Answer & Solution
Answer: B
Common difference is 4, so 4th term is $11 + 4 = 15$.
11.
Tags: Rational Equations · Medium · source: Original (AMC-style)
Solve for $x$: $\frac{4x-1}{x+2} = 3$.
A) $5$ B) $6$ C) $7$ D) $8$ E) $9$
Answer & Solution
Answer: C
Cross-multiplying: $4x-1 = 3(x+2) = 3x+6$. So $4x-1 = 3x+6$, giving $x = 7$. Check: $x \neq -2$ ✓.
12.
Tags: Systems · Medium · source: Original (AMC-style)
If $5x + 3y = 23$ and $3x + 2y = 14$, what is $x + y$?
A) $4$ B) $5$ C) $6$ D) $7$ E) $8$
Answer & Solution
Answer: B
Subtracting: $(5x+3y) - (3x+2y) = 23 - 14$, so $2x + y = 9$. Adding to second equation: $(2x+y) + (3x+2y) = 9 + 14$, so $5x + 3y = 23$. This gives $x + y = 5$.
13.
Tags: Absolute Value · Medium · source: Original (AMC-style)
How many solutions does $|x-3| = 6$ have?
A) $0$ B) $1$ C) $2$ D) $3$ E) Infinitely many
Answer & Solution
Answer: C
We have $x-3 = 6$ or $x-3 = -6$, giving $x = 9$ or $x = -3$. So there are 2 solutions.
14.
Tags: Sequences · Medium · source: Original (AMC-style)
The first term of a geometric sequence is 4, and the common ratio is 2. What is the 6th term?
A) $64$ B) $128$ C) $256$ D) $512$ E) $1024$
Answer & Solution
Answer: B
The nth term is $a_n = 4 \cdot 2^{n-1}$. So $a_6 = 4 \cdot 2^{6-1} = 4 \cdot 2^5 = 4 \cdot 32 = 128$.
15.
Tags: Exponents · Medium · source: Original (AMC-style)
What is $\log_2(16) + \log_2(8)$?
A) $4$ B) $5$ C) $6$ D) $7$ E) $8$
Answer & Solution
Answer: D
$\log_2(16) + \log_2(8) = \log_2(16 \cdot 8) = \log_2(128) = \log_2(2^7) = 7$.
16.
Tags: Quadratics · Medium · source: Original (AMC-style)
For what value of $k$ does the equation $x^2 + kx + 25 = 0$ have exactly one real solution?
A) $-10$ B) $-5$ C) $0$ D) $5$ E) $10$
Answer & Solution
Answer: A
For exactly one real solution, the discriminant must be 0: $k^2 - 4(1)(25) = 0$, so $k^2 = 100$ and $k = \pm 10$. The answer is $k = -10$.
17.
Tags: Radicals · Medium · source: Original (AMC-style)
What is $\sqrt{5} \cdot \sqrt{20}$?
A) $5$ B) $10$ C) $15$ D) $20$ E) $100$
Answer & Solution
Answer: B
$\sqrt{5} \cdot \sqrt{20} = \sqrt{5 \cdot 20} = \sqrt{100} = 10$.
18.
Tags: Factoring · Medium · source: Original (AMC-style)
Factor $x^3 - 64$.
A) $(x-4)^3$ B) $(x+4)^3$ C) $(x-4)(x^2+4x+16)$ D) $(x+4)(x^2-4x+16)$ E) Cannot be factored
Answer & Solution
Answer: C
This is a difference of cubes: $x^3 - 64 = x^3 - 4^3 = (x-4)(x^2+4x+16)$.
19.
Tags: Systems · Medium · source: Original (AMC-style)
If $x^2 + y^2 = 17$ and $x + y = 7$, what is $xy$?
A) $8$ B) $12$ C) $16$ D) $20$ E) $24$
Answer & Solution
Answer: C
We have $(x+y)^2 = x^2 + 2xy + y^2 = 49$. Since $x^2 + y^2 = 17$, we get $17 + 2xy = 49$, so $2xy = 32$ and $xy = 16$.
20.
Tags: Rational Equations · Medium · source: Original (AMC-style)
What is the sum of all solutions to $\frac{3}{x} + \frac{4}{x+1} = 2$?
A) $-2$ B) $-1$ C) $0$ D) $1$ E) $2$
Answer & Solution
Answer: A
Multiplying by $x(x+1)$: $3(x+1) + 4x = 2x(x+1)$, so $7x + 3 = 2x^2 + 2x$. Rearranging: $2x^2 - 5x - 3 = 0$. By Vieta's formulas, the sum of roots is $\frac{5}{2}$.
21.
Tags: Quadratics · Hard · source: Original (AMC-style)
If the quadratic $x^2 + px + q$ has roots $r$ and $s$ such that $r^2 + s^2 = 26$ and $rs = 5$, what is $p^2 - 4q$?
A) $6$ B) $12$ C) $18$ D) $24$ E) $30$
Answer & Solution
Answer: A
We have $r + s = -p$ and $rs = q$. Since $r^2 + s^2 = 26$ and $rs = 5$, we have $(r+s)^2 = r^2 + 2rs + s^2 = 26 + 2(5) = 36$. So $r+s = \pm 6$, giving $p = \mp 6$. The discriminant is $p^2 - 4q = 36 - 4(5) = 16$.
22.
Tags: Exponents · Hard · source: Original (AMC-style)
If $2^x = 5^y = 10^z$, 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 $2^x = 5^y = 10^z = k$. Then $x = \log_2 k$, $y = \log_5 k$, and $z = \log_{10} k$. We have $\frac{1}{x} + \frac{1}{y} = \frac{1}{\log_2 k} + \frac{1}{\log_5 k} = \log_k 2 + \log_k 5 = \log_k(2 \cdot 5) = \log_k 10 = \frac{1}{\log_{10} k} = \frac{1}{z}$.
23.
Tags: Sequences · Hard · source: Original (AMC-style)
The sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = 3$ and $a_{n+1} = 4a_n - 2$ for $n \geq 1$. What is $a_6$?
A) $1022$ B) $2046$ C) $4094$ D) $8190$ E) $16382$
Answer & Solution
Answer: C
We have $a_{n+1} - \frac{2}{3} = 4(a_n - \frac{2}{3})$. Let $b_n = a_n - \frac{2}{3}$. Then $b_{n+1} = 4b_n$ with $b_1 = \frac{7}{3}$. So $b_n = \frac{7}{3} \cdot 4^{n-1}$, giving $a_n = \frac{7 \cdot 4^{n-1} + 2}{3}$. Therefore $a_6 = \frac{7 \cdot 4^5 + 2}{3} = \frac{7 \cdot 1024 + 2}{3} = \frac{7170}{3} = 2390$.
24.
Tags: Absolute Value · Hard · source: Original (AMC-style)
How many integer solutions does $|x-1| + |x-5| = 4$ have?
A) $0$ B) $1$ C) $2$ D) $3$ E) Infinitely many
Answer & Solution
Answer: E
For $x \leq 1$: $|x-1| + |x-5| = (1-x) + (5-x) = 6-2x = 4$, so $x = 1$. For $1 < x < 5$: $|x-1| + |x-5| = (x-1) + (5-x) = 4$, which is always true. For $x \geq 5$: $|x-1| + |x-5| = (x-1) + (x-5) = 2x-6 = 4$, so $x = 5$. Therefore, all $x$ with $1 \leq x \leq 5$ are solutions, giving infinitely many integer solutions.
25.
Tags: Systems · Hard · source: Original (AMC-style)
If $x + y + z = 12$, $x^2 + y^2 + z^2 = 50$, and $xyz = 24$, what is $x^3 + y^3 + z^3$?
A) $72$ B) $108$ C) $144$ D) $180$ E) $216$
Answer & Solution
Answer: B
We have $(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx) = 144$. Since $x^2 + y^2 + z^2 = 50$, we get $xy + yz + zx = 47$. Now $x^3 + y^3 + z^3 = (x+y+z)^3 - 3(x+y+z)(xy+yz+zx) + 3xyz = 12^3 - 3(12)(47) + 3(24) = 1728 - 1692 + 72 = 108$.
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 | C | C | D | C | B | B | C | B | B | C | B | C | B | D | A | B | C | C | A | A | A | C | E | B |