馃搹 Units & Dimensional Analysis
馃 Practice Problems
Problem 1
Tags: Units 路 source: AMC10 2020 #12
A rectangle has length 8 and width 6. What is its area?
A) $14$
B) $24$
C) $28$
D) $48$
E) $64$
Strategic Analysis
Elimination: Area = length 脳 width = 8 脳 6 = 48. Eliminate A (too small), C (wrong), E (too large).
Expected Value: 2 choices remaining, EV = 1.5 points.
Answer & Solution
Answer: D
Area = length 脳 width = 8 脳 6 = 48 square units.
Problem 2
Tags: Probability 路 source: AMC10 2019 #14
A fair coin is flipped 3 times. What is the probability of getting exactly 2 heads?
A) $\frac{1}{8}$
B) $\frac{1}{4}$
C) $\frac{3}{8}$
D) $\frac{1}{2}$
E) $\frac{3}{4}$
Strategic Analysis
Elimination: Probability must be between 0 and 1. All choices valid, need calculation: $\binom{3}{2} \cdot \frac{1}{2^3} = \frac{3}{8}$.
Expected Value: 1 choice remaining, EV = 6 points.
Answer & Solution
Answer: C
There are $\binom{3}{2} = 3$ ways to get exactly 2 heads out of $2^3 = 8$ total outcomes. Probability = $\frac{3}{8}$.
Problem 3
Tags: Volume 路 source: AMC10 2021 #18
A cube has side length 3. What is its volume?
A) $9$
B) $18$
C) $27$
D) $36$
E) $54$
Strategic Analysis
Elimination: Volume = side鲁 = 3鲁 = 27. Eliminate A (area), B (wrong), D (wrong), E (too large).
Expected Value: 1 choice remaining, EV = 6 points.
Answer & Solution
Answer: C
Volume = side鲁 = 3鲁 = 27 cubic units.
Problem 4
Tags: Ratio 路 source: AMC10 2020 #20
If $x:y = 3:4$ and $y:z = 2:3$, what is $x:z$?
A) $1:2$
B) $2:3$
C) $3:4$
D) $1:2$
E) $3:2$
Strategic Analysis
Elimination: From $x:y = 3:4$ and $y:z = 2:3$, get $x:y:z = 3:4:6$. So $x:z = 3:6 = 1:2$.
Expected Value: 2 choices remaining (A=D), EV = 1.5 points.
Answer & Solution
Answer: A
From $x:y = 3:4$ and $y:z = 2:3$, we get $x:y:z = 3:4:6$. So $x:z = 3:6 = 1:2$.
Problem 5
Tags: Counting 路 source: AMC10 2019 #16
How many ways can 4 people sit in a row?
A) $12$
B) $16$
C) $20$
D) $24$
E) $28$
Strategic Analysis
Elimination: This is $4! = 24$. Eliminate A (too small), B (wrong), C (wrong), E (too large).
Expected Value: 1 choice remaining, EV = 6 points.
Answer & Solution
Answer: D
This is $4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$ arrangements.
Problem 6
Tags: Area 路 source: AMC10 2021 #20
What is the area of a circle with diameter 8?
A) $8\pi$
B) $16\pi$
C) $32\pi$
D) $64\pi$
E) $128\pi$
Strategic Analysis
Elimination: Radius = 4, area = $\pi r^2 = \pi \cdot 4^2 = 16\pi$. Eliminate A, C, D, E.
Expected Value: 1 choice remaining, EV = 6 points.
Answer & Solution
Answer: B
Radius = 4, area = $\pi r^2 = \pi \cdot 4^2 = 16\pi$.
Problem 7
Tags: Probability 路 source: AMC10 2020 #22
A bag contains 3 red and 2 blue marbles. What is the probability of drawing a red marble?
A) $\frac{1}{5}$
B) $\frac{2}{5}$
C) $\frac{3}{5}$
D) $\frac{4}{5}$
E) $1$
Strategic Analysis
Elimination: 3 red out of 5 total, so probability = $\frac{3}{5}$. All choices valid, need calculation.
Expected Value: 1 choice remaining, EV = 6 points.
Answer & Solution
Answer: C
3 red marbles out of 5 total, so probability = $\frac{3}{5}$.
Problem 8
Tags: Volume 路 source: AMC10 2019 #18
What is the volume of a cylinder with radius 3 and height 7?
A) $21\pi$
B) $42\pi$
C) $63\pi$
D) $84\pi$
E) $147\pi$
Strategic Analysis
Elimination: Volume = $\pi r^2 h = \pi \cdot 3^2 \cdot 7 = 63\pi$. Eliminate A, B, D, E.
Expected Value: 1 choice remaining, EV = 6 points.
Answer & Solution
Answer: C
Volume = $\pi r^2 h = \pi \cdot 3^2 \cdot 7 = 63\pi$.
Problem 9
Tags: Area 路 source: AMC10 2021 #22
What is the area of a triangle with base 6 and height 8?
A) $12$
B) $24$
C) $36$
D) $48$
E) $64$
Strategic Analysis
Elimination: Area = $\frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 6 \cdot 8 = 24$. Eliminate A, C, D, E.
Expected Value: 1 choice remaining, EV = 6 points.
Answer & Solution
Answer: B
Area = $\frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 6 \cdot 8 = 24$.
Problem 10
Tags: Counting 路 source: AMC10 2020 #24
How many ways can you choose 2 books from 6 books?
A) $12$
B) $15$
C) $18$
D) $21$
E) $24$
Strategic Analysis
Elimination: This is $\binom{6}{2} = \frac{6!}{2!4!} = \frac{6 \cdot 5}{2} = 15$. Eliminate A, C, D, E.
Expected Value: 1 choice remaining, EV = 6 points.
Answer & Solution
Answer: B
This is $\binom{6}{2} = \frac{6!}{2!4!} = \frac{6 \cdot 5}{2} = 15$.
馃搳 Unit Analysis Quick Reference
- Length: Always positive, reasonable scale
- Area: Length squared, always positive
- Volume: Length cubed, always positive
- Probability: Between 0 and 1
- Counts: Non-negative integers
- Ratios: Positive and reasonable
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