🧭 Strategy Mini Mock 25 — AMC 10/12 Techniques (Clean v2)
Recommended: 60–75 minutes. No calculator.
Each solution highlights a technique you can apply quickly on contest day.
Problems
1.
Tags: Mixture · Backsolve · Easy · source: Original (AMC-style)
A tank holds 25 L of a 32% acid solution. How many liters of water must be added so the mixture becomes 20% acid?
A) 15
B) 12
C) 10
D) 18
E) 14
Answer & Solution
Answer: A
Backsolve / One-equation. Acid amount is fixed: $0.32\cdot25=8$ L. For $20\%$, total must be $8/0.20=40$ L, so add $40-25=15$ L.
2.
Tags: Estimation & Bounds · Easy · source: Original (AMC-style)
Which is larger?
A) $14.1$
B) $\sqrt{200}$
C) equal
D) cannot be determined
E) $14$
Answer & Solution
Answer: B
Bounding squares. $14^2=196$ and $15^2=225$, so $\sqrt{200}\approx14.142>14.1$.
3.
Tags: Divisibility Rules · Backsolve · Easy · source: Original (AMC-style)
Which number is divisible by 45?
A) 83235
B) 54065
C) 72630
D) 91225
E) 70215
Answer & Solution
Answer: C
Rule-based scan. $45=9\cdot5$. Need last digit $0$ or $5$ and digit sum multiple of $9$. Only $72630$ ends with $0$ and has digit sum $7+2+6+3+0=18$.
4.
Tags: Complementary Counting · Medium · source: Original (AMC-style)
How many 4-digit integers have at least one zero?
A) 2340
B) 2169
C) 2430
D) 2439
E) 2000
Answer & Solution
Answer: D
Complement. Total $=9000$ ($1000$–$9999$). No zero: $9\cdot9\cdot9\cdot9=6561$. Hence $9000-6561=2439$.
5.
Tags: Parity/Mod · Easy · source: Original (AMC-style)
Which cannot be the sum of three consecutive integers?
A) 201
B) 204
C) 210
D) 222
E) 202
Answer & Solution
Answer: E
Mod $3$ check. Sum of three consecutive integers is a multiple of $3$; $202$ is not.
6.
Tags: Quadratic Vertex · Complete the Square · Easy · source: Original (AMC-style)
The maximum value of $-2x^2+8x+3$ is
A) 11
B) 12
C) 10
D) 9
E) 13
Answer & Solution
Answer: A
Vertex. $x=-\frac{b}{2a}=\frac{-8}{-4}=2$. Value $=-2(4)+16+3=11$.
7.
Tags: Last-Digit Cycle · Easy · source: Original (AMC-style)
The units digit of $3^{2025}$ is
A) 1
B) 3
C) 5
D) 7
E) 9
Answer & Solution
Answer: B
Cycle of length 4. $3,9,7,1$. Since $2025\equiv1\pmod4$, digit $=3$.
8.
Tags: Telescoping Sum · Partial Fractions · Easy · source: Original (AMC-style)
Evaluate $\displaystyle \sum_{k=1}^{20}\frac{1}{k(k+1)}$.
A) $\frac{19}{21}$
B) $1-\frac{1}{20}$
C) $\frac{20}{21}$
D) $\frac{21}{22}$
E) $1-\frac{1}{22}$
Answer & Solution
Answer: C
Telescoping. $\frac{1}{k(k+1)}=\frac1k-\frac1{k+1}$ ⇒ sum $=1-\frac1{21}=\frac{20}{21}$.
9.
Tags: Power of a Point · Medium · source: Original (AMC-style)
From external point $P$, a tangent has length $10$. A secant from $P$ meets the circle at $A$ then $B$ with $PA=6$. Find $PB$.
A) $16$
B) $16.5$
C) $17$
D) $\dfrac{50}{3}$
E) $18$
Answer & Solution
Answer: D
PoP. $PT^2=PA\cdot PB\Rightarrow 100=6\cdot PB\Rightarrow PB=50/3$.
10.
Tags: Slope/Perpendicular · Intercept · Easy · source: Original (AMC-style)
A line perpendicular to $y=\tfrac{3}{4}x-2$ passes through $(2,5)$. Its $y$-intercept is
A) $5$
B) $4$
C) $3$
D) $\tfrac{7}{3}$
E) $\tfrac{17}{3}$
Answer & Solution
Answer: E
Perpendicular slope. $m=-\tfrac{4}{3}$. $y-5=-\tfrac{4}{3}(x-2)\Rightarrow y=-\tfrac43 x+\tfrac{8}{3}+5=\tfrac{17}{3}$.
11.
Tags: Weighted Average · One-step Update · Easy · source: Original (AMC-style)
A class average is $80$ for $20$ students. One new score is added and the average becomes $81$. The new score is
A) 101
B) 99
C) 102
D) 100
E) 105
Answer & Solution
Answer: A
Sum-shift. Old sum $=1600$. New sum $=21\cdot 81=1701$. Added score $=101$.
12.
Tags: Casework · Divisible by 5 · Distinct Digits · Medium · source: Original (AMC-style)
How many 3-digit numbers with distinct digits are divisible by 5?
A) 120
B) 136
C) 128
D) 112
E) 108
Answer & Solution
Answer: B
Cases on last digit. Last digit $0$: $9\cdot 8=72$. Last digit $5$: hundreds $\in\{1,2,3,4,6,7,8,9\}$ ($8$ choices), tens any of remaining $8$ (including $0$) $=64$. Total $72+64=136$.
13.
Tags: Complementary Probability · Dice · Easy/Medium · source: Original (AMC-style)
Two fair dice are rolled. Probability that at least one shows a 6 is
A) $\dfrac{1}{6}$
B) $\dfrac{5}{18}$
C) $\dfrac{11}{36}$
D) $\dfrac{7}{36}$
E) $\dfrac{13}{36}$
Answer & Solution
Answer: C
Complement. $1-(5/6)^2=1-25/36=11/36$.
14.
Tags: Expected Value · Linearity/Variance Trick · Medium · source: Original (AMC-style)
Let $X$ be the result of one fair die roll. Compute $\mathbb{E}!\left[(X-3)^2\right]$.
A) $3$
B) $\dfrac{7}{3}$
C) $2$
D) $\dfrac{19}{6}$
E) $\dfrac{10}{3}$
Answer & Solution
Answer: D
Variance shortcut. $\mathbb E[X]=3.5$, $\mathrm{Var}(X)=\tfrac{35}{12}$. Then $\mathbb E[(X-3)^2]=\mathrm{Var}(X)+(\mathbb E X-3)^2=\tfrac{35}{12}+\tfrac{1}{4}=\tfrac{19}{6}$.
15.
Tags: Trig Angle Addition · Medium · source: Original (AMC-style)
Compute $\cos 75^\circ$.
A) $\dfrac{\sqrt3}{4}$
B) $\dfrac{\sqrt2}{4}$
C) $\dfrac{\sqrt6+\sqrt2}{4}$
D) $\dfrac{\sqrt3-1}{2}$
E) $\dfrac{\sqrt6-\sqrt2}{4}$
Answer & Solution
Answer: E
Angle addition. $\cos(45^\circ+30^\circ)=\cos45\cos30-\sin45\sin30=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6-\sqrt2}{4}$.
16.
Tags: Base-Change Pairing · Medium · source: Original (AMC-style)
Evaluate $\log_2 9\cdot \log_3 4$.
A) 4
B) 3
C) 2
D) 6
E) 8
Answer & Solution
Answer: A
Base change chain. $(\ln9/\ln2)\cdot(\ln4/\ln3)=(2\ln3/\ln2)\cdot(2\ln2/\ln3)=4$.
17.
Tags: Fixed Point / Heron Iteration · Medium · source: Original (AMC-style)
Define $x_{1}>0$ and $x_{n+1}=\dfrac{x_n+4/x_n}{2}$. Then $\lim_{n\to\infty}x_n$ equals
A) diverges
B) 2
C) 1
D) 3
E) 0
Answer & Solution
Answer: B
Fixed point. $L=\frac{L+4/L}{2}\Rightarrow L^2=4\Rightarrow L=2$ (positive sequence). This is Heron’s method for $\sqrt{4}$.
18.
Tags: Tangent Circles to a Line · Geometry Trick · Medium · source: Original (AMC-style)
Two circles with radii $4$ and $25$ are externally tangent to each other and to the same line. The distance between their tangency points on the line is
A) 16
B) 18
C) 20
D) 22
E) 24
Answer & Solution
Answer: C
Right triangle / difference of squares. Horizontal gap $=\sqrt{(R+r)^2-(R-r)^2}=2\sqrt{Rr}=2\sqrt{100}=20$.
19.
Tags: Area · Shoelace/Base–Height · Easy/Medium · source: Original (AMC-style)
Find the area of $\triangle$ with vertices $(0,0)$, $(5,0)$, $(2,6)$.
A) 12
B) 13
C) 14
D) 15
E) 16
Answer & Solution
Answer: D
Base–height. Base $=5$, height $=6$ (vertical). $A=\frac12\cdot5\cdot6=15$.
20.
Tags: Symmetric Sums · Identity · Easy/Medium · source: Original (AMC-style)
If $x+y=10$ and $xy=13$, then $x^2+y^2=$
A) 64
B) 68
C) 72
D) 78
E) 74
Answer & Solution
Answer: E
Identity. $(x+y)^2=x^2+2xy+y^2\Rightarrow x^2+y^2=100-26=74$.
21.
Tags: Stars & Bars with Parity · Medium · source: Original (AMC-style)
Number of nonnegative integer solutions to $a+b+c=10$ with $a$ even is
A) 36
B) 34
C) 32
D) 38
E) 30
Answer & Solution
Answer: A
Substitute. Let $a=2k$, $k=0,\dots,5$. For each $k$, $b+c=10-2k$ → $(10-2k)+1$ solutions. Sum: $\sum_{k=0}^{5}(11-2k)=66-30=36$.
22.
Tags: Product Rule · Distinct Digits · Medium · source: Original (AMC-style)
A random 3-digit integer (from 100–999) has all digits distinct with probability
A) $\dfrac{17}{25}$
B) $\dfrac{18}{25}$
C) $\dfrac{19}{25}$
D) $\dfrac{4}{5}$
E) $\dfrac{3}{5}$
Answer & Solution
Answer: B
Multiplication principle. First digit $9$ ways; second $9$ (exclude first, allow $0$); third $8$. Total $=9\cdot9\cdot8=648$. Out of $900$ numbers ⇒ $648/900=18/25$.
23.
Tags: AM–GM / Max with Fixed Sum · Easy/Medium · source: Original (AMC-style)
For positive $x,y$ with $x+y=10$, the maximum of $xy$ is
A) 20
B) 21
C) 25
D) 24
E) 26
Answer & Solution
Answer: C
AM–GM / symmetry. Max at $x=y=5$ ⇒ $xy=25$.
24.
Tags: Telescoping (Finite Differences) · Easy/Medium · source: Original (AMC-style)
Compute $\displaystyle \sum_{k=1}^{50} \big[(k+1)^2-k^2\big]$.
A) 2500
B) 2550
C) 2600
D) 2650
E) 2700
Answer & Solution
Answer: C
Telescoping. Sum collapses to $(51^2-1^2)=2601-1=2600$.
25.
Tags: Complex Modulus · Ratio Trick · Medium · source: Original (AMC-style)
Compute $\left|\dfrac{3+4i}{1-2i}\right|$.
A) 1
B) $\sqrt2$
C) $\sqrt5$
D) 2
E) 5
Answer & Solution
Answer: C
Modulus of a quotient. $|3+4i|/|1-2i|=5/\sqrt5=\sqrt5$.
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 | A | B | C | D | E | A | B | C | D | E | A | B | C | D | E | A | B | C | D | E | A | B | C | D | E |