AMC 8 · Test Mode

2011 AMC 8

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Problem 1 · 2011 AMC 8 Easy
Arithmetic & Operations subtraction

Margie bought 3 apples at a cost of 50 cents per apple. She paid with a 5-dollar bill. How much change did Margie receive?

Show hint
Total cost = 3 × $0.50; change = $5 − that.
Show solution
  1. Cost: 3 · $0.50 = $1.50.
  2. Change: $5.00 − $1.50 = $3.50.
E $3.50.
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Problem 2 · 2011 AMC 8 Easy
Geometry & Measurement area-comparison

Karl's rectangular vegetable garden is 20 feet by 45 feet, and Makenna's is 25 feet by 40 feet. Which of the following statements are true?

Show hint
Compute each area; subtract.
Show solution
  1. Karl: 20 × 45 = 900 sq ft.
  2. Makenna: 25 × 40 = 1000 sq ft.
  3. Difference: 1000 − 900 = 100 ⇒ Makenna's is larger by 100 sq ft.
E Makenna's garden is larger by 100 square feet.
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Problem 3 · 2011 AMC 8 Easy
Geometry & Measurement border-count

Extend the square pattern of 8 black and 17 white square tiles by attaching a border of black tiles around the square. What is the ratio of black tiles to white tiles in the extended pattern?

Show hint
Border around an n × n square has 4n + 4 tiles. White count is unchanged.
Show solution
  1. Original is 5 × 5 = 25 tiles (8 black + 17 white). New 7 × 7 has 49 tiles.
  2. Added border tiles: 49 − 25 = 24, all black. New black total: 8 + 24 = 32. White still 17.
  3. Ratio: 32 : 17.
D 32 : 17.
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Problem 4 · 2011 AMC 8 Easy
Arithmetic & Operations mean-median-mode

Here is a list of the numbers of fish that Tyler caught in nine outings last summer: 2, 0, 1, 3, 0, 3, 3, 1, 2. Which statement about the mean, median, and mode is true?

Show hint
Sort the list. Mode = most common value, median = middle (5th of 9), mean = sum/9.
Show solution
  1. Sorted: 0, 0, 1, 1, 2, 2, 3, 3, 3.
  2. Mode = 3, median = 2, mean = 15/9 = 5/3 ≈ 1.67.
  3. 5/3 < 2 < 3 ⇒ mean < median < mode.
C mean &lt; median &lt; mode.
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Problem 5 · 2011 AMC 8 Easy
Arithmetic & Operations time-conversion

What time was it 2011 minutes after midnight on January 1, 2011?

Show hint
Divide 2011 by 60 to convert to hours and minutes. Then subtract 24 hours to advance to the next day.
Show solution
  1. 2011 / 60 = 33 hr 31 min (since 33 · 60 = 1980 and 2011 − 1980 = 31).
  2. 33 hr = 24 hr + 9 hr ⇒ one day later, 9 hr 31 min after midnight = 9:31 AM on January 2.
D January 2 at 9:31 AM.
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Problem 6 · 2011 AMC 8 Easy
Counting & Probability complement

In a town of 351 adults, every adult owns a car, motorcycle, or both. If 331 adults own cars and 45 adults own motorcycles, how many of the car owners do not own a motorcycle?

Show hint
Everyone owns at least one. So car-only count = total − motorcycle-owners.
Show solution
  1. Each non-motorcycle-owner must own a car (since every adult has at least one).
  2. Car-only = 351 − 45 = 306.
D 306.
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Problem 7 · 2011 AMC 8 Easy
Geometry & Measurement fraction-of-area

Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is partially bolded. What percent of the total area is partially bolded?

Show hint
Compute each square's shaded fraction; add; divide by 4 (since there are 4 squares of equal area).
Show solution
  1. Shaded fractions of the four squares: 1/4, 1/8, 3/8, 1/4.
  2. Sum: 1/4 + 1/8 + 3/8 + 1/4 = 2/8 + 1/8 + 3/8 + 2/8 = 8/8 = 1.
  3. Of 4 total squares: 1 / 4 = 25%.
C 25%.
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Problem 8 · 2011 AMC 8 Easy
Counting & Probability enumerate-sums

Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips?

Show hint
All sums are odd + even = odd. List the 9 sums and count distinct values.
Show solution
  1. Possible sums: 1+2, 1+4, 1+6, 3+2, 3+4, 3+6, 5+2, 5+4, 5+6 = 3, 5, 7, 5, 7, 9, 7, 9, 11.
  2. Distinct: {3, 5, 7, 9, 11} ⇒ 5 values.
B 5 different values.
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Problem 9 · 2011 AMC 8 Easy
Ratios, Rates & Proportions average-speed

Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour?

Show hint
Average speed = total distance / total time. The graph's endpoints give you both.
Show solution
  1. Total: 35 miles in 7 hours.
  2. Average speed: 35 / 7 = 5 mph.
E 5 mph.
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Problem 10 · 2011 AMC 8 Medium
Fractions, Decimals & Percents piecewise-rate

The taxi fare in Gotham City is $2.40 for the first 12 mile and additional mileage charged at the rate $0.20 for each additional 0.1 mile. You plan to give the driver a $2 tip. How many miles can you ride for $10?

Show hint
Subtract the $2 tip from $10. Then subtract the $2.40 flag-drop for the first half-mile. Convert the rest at $0.20 per 0.1 mile = $2 per mile.
Show solution
  1. Available for fare: $10 − $2 = $8.
  2. After the first 1/2 mile costing $2.40: $8 − $2.40 = $5.60 left.
  3. Additional rate: $0.20 / 0.1 mile = $2 per mile. So $5.60 buys 2.80 miles.
  4. Total: 0.5 + 2.80 = 3.3 miles.
C 3.3 miles.
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Problem 11 · 2011 AMC 8 Easy
Arithmetic & Operations average-of-differences

The graph shows the number of minutes studied by both Asha (black bar) and Sasha (grey bar) in one week. On the average, how many more minutes per day did Sasha study than Asha?

Show hint
Add the daily differences (Sasha − Asha) and divide by 5.
Show solution
  1. Daily Sasha − Asha: +10, −10, +20, +30, −20. Sum: 30.
  2. Average over 5 days: 30 / 5 = 6.
A 6 minutes.
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Problem 12 · 2011 AMC 8 Easy
Counting & Probability fix-one-position

Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other?

Show hint
Fix Angie's seat. Carlos lands in any of the 3 remaining seats with equal probability; only 1 is opposite.
Show solution
  1. Fix Angie in any seat. Carlos has 3 equally likely seats among the remaining 3.
  2. Exactly 1 is opposite Angie ⇒ probability 1/3.
B 1/3.
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Problem 13 · 2011 AMC 8 Medium
Geometry & Measurement overlap-region area-ratio

Two congruent squares, ABCD and PQRS, have side length 15. They overlap to form the 15 by 25 rectangle AQRD shown. What percent of the area of rectangle AQRD is shaded?

Show hint (soft nudge)
Together the two squares cover area 2 · 152, but the rectangle is only 25 · 15 — the overlap is counted twice.
Show hint (sharpest)
Overlap area = (sum of both squares) − (rectangle).
Show solution
  1. Each square has area 225, total 450. Rectangle area = 25 × 15 = 375.
  2. Overlap = 450 − 375 = 75.
  3. Fraction of rectangle: 75 / 375 = 1/5 = 20%.
C 20%.
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Problem 14 · 2011 AMC 8 Medium
Fractions, Decimals & Percents ratio-totals

There are 270 students at Colfax Middle School, where the ratio of boys to girls is 5 : 4. There are 180 students at Winthrop Middle School, where the ratio of boys to girls is 4 : 5. The two schools hold a dance and all students from both schools attend. What fraction of the students at the dance are girls?

Show hint
Each ratio uses 9 parts. Compute girls at each school, then total girls / total students.
Show solution
  1. Colfax: girls = (4/9)(270) = 120.
  2. Winthrop: girls = (5/9)(180) = 100.
  3. Total girls: 220 of 450 students ⇒ 22/45.
C 22/45.
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Problem 15 · 2011 AMC 8 Easy
Number Theory exponent-rewrite

How many digits are in the product 45 · 510?

Show hint
Rewrite 45 as 210. Then 210 · 510 = 1010.
Show solution
  1. 45 · 510 = (22)5 · 510 = 210 · 510 = 1010.
  2. 1010 = 1 followed by ten zeros ⇒ 11 digits.
D 11 digits.
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Problem 16 · 2011 AMC 8 Medium
Geometry & Measurement isosceles-altitude pythagorean-triple

Let A be the area of the triangle with sides of length 25, 25, and 30. Let B be the area of the triangle with sides of length 25, 25, and 40. What is the relationship between A and B?

Show hint (soft nudge)
Each triangle is isosceles. Drop the altitude to the unequal side and use the Pythagorean theorem.
Show hint (sharpest)
Look for 15-20-25 (3-4-5 scaled) in both pictures.
Show solution
  1. Triangle with base 30: half-base 15, hypotenuse 25 ⇒ height = √(252 − 152) = 20. Area A = (1/2)(30)(20) = 300.
  2. Triangle with base 40: half-base 20, hypotenuse 25 ⇒ height = √(252 − 202) = 15. Area B = (1/2)(40)(15) = 300.
  3. A = B.
C A = B.
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Problem 17 · 2011 AMC 8 Medium
Number Theory prime-factorization

Let w, x, y, and z be whole numbers. If 2w · 3x · 5y · 7z = 588, then what does 2w + 3x + 5y + 7z equal?

Show hint
Factor 588 into primes: 588 = 4 · 147 = 4 · 3 · 49.
Show solution
  1. 588 = 22 · 31 · 50 · 72w = 2, x = 1, y = 0, z = 2.
  2. 2w + 3x + 5y + 7z = 4 + 3 + 0 + 14 = 21.
A 21.
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Problem 18 · 2011 AMC 8 Medium
Counting & Probability symmetry probability

A fair 6-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?

Show hint
P(first > second) = P(second > first) by symmetry. P(equal) = 6/36 = 1/6.
Show solution
  1. P(first = second) = 6/36 = 1/6.
  2. By symmetry, P(first > second) = P(second > first) = (1 − 1/6)/2 = 5/12.
  3. P(first ≥ second) = 1/6 + 5/12 = 2/12 + 5/12 = 7/12.
D 7/12.
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Problem 19 · 2011 AMC 8 Medium
Counting & Probability systematic-counting

How many rectangles are in this figure?

Show hint
Label the small atomic regions, then list every combination of adjacent atoms that together form a rectangle.
Show solution
  1. Label the atomic rectangles produced by the three overlapping rectangles. Systematically list all rectangular unions of adjacent atoms.
  2. Counting all rectangles (single atoms and rectangular unions of two, three, or four adjacent atoms) gives 11.
D 11 rectangles.
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Problem 20 · 2011 AMC 8 Medium
Geometry & Measurement trapezoid-area drop-altitudes

Quadrilateral ABCD is a trapezoid, AD = 15, AB = 50, BC = 20, and the altitude is 12. What is the area of the trapezoid?

Show hint (soft nudge)
Drop altitudes from A and B to DC. Each leg becomes the hypotenuse of a right triangle of height 12.
Show hint (sharpest)
Use 9-12-15 and 12-16-20 (3-4-5 scaled) to find the base extensions.
Show solution
  1. Drop altitudes from A and B. On the left: 15-12-? right triangle ⇒ horizontal piece 9. On the right: 20-12-? ⇒ horizontal piece 16.
  2. Longer base DC = 9 + 50 + 16 = 75.
  3. Area = (1/2)(50 + 75)(12) = (1/2)(125)(12) = 750.
D 750.
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Problem 21 · 2011 AMC 8 Medium
Logic & Word Problems constraint-satisfaction

Students guess that Norb's age is 24, 28, 30, 32, 36, 38, 41, 44, 47, and 49. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?

Show hint (soft nudge)
"At least half too low" with 10 guesses means age > 5th-smallest guess = 36, so age ≥ 37.
Show hint (sharpest)
"Two off by one" means age is squeezed between two guesses that differ by 2. The only such pair above 36 is 36 and 38, or 47 and 49.
Show solution
  1. Sorted guesses: 24, 28, 30, 32, 36, 38, 41, 44, 47, 49. "At least half too low" ⇒ age > 36.
  2. "Two are off by one" ⇒ age sits between two guesses 2 apart. Candidates: 37 (between 36 and 38) or 48 (between 47 and 49).
  3. Age is prime ⇒ 37 (prime) wins; 48 isn't prime.
  4. Norb is 37.
C 37.
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Problem 22 · 2011 AMC 8 Hard
Number Theory tens-digit-cycle mod-100

What is the tens digit of 72011?

Show hint (soft nudge)
Tens digit depends only on the value mod 100. Compute 7k mod 100 for small k and look for a cycle.
Show hint (sharpest)
71 = 07, 72 = 49, 73 = 343 (43), 74 = 2401 (01). Cycle length 4.
Show solution
  1. Last two digits cycle: 07, 49, 43, 01, repeating with period 4.
  2. 2011 = 4 · 502 + 3 ⇒ 72011 ≡ 73 ≡ 43 (mod 100).
  3. Tens digit = 4.
D Tens digit 4.
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Problem 23 · 2011 AMC 8 Hard
Counting & Probability casework permutations-with-restrictions

How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?

Show hint (soft nudge)
All digits come from {0, 1, 2, 3, 4, 5}, with 5 present (it's the largest). Units digit is 0 or 5 (divisible by 5).
Show hint (sharpest)
Split into two cases by units digit.
Show solution
  1. Case A: units = 0. The remaining three slots contain 5 and two distinct digits chosen from {1, 2, 3, 4}: C(4, 2) = 6 ways to pick the other two; 3! = 6 ways to arrange them. Subtotal: 6 × 6 = 36.
  2. Case B: units = 5. The remaining three slots use three distinct digits from {0, 1, 2, 3, 4}. Choose and arrange: 5 · 4 · 3 = 60. Subtract leading-zero arrangements: 4 · 3 = 12 with 0 first. Subtotal: 60 − 12 = 48.
  3. Total: 36 + 48 = 84.
D 84.
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Problem 24 · 2011 AMC 8 Medium
Number Theory parity primes

In how many ways can 10001 be written as the sum of two primes?

Show hint
Odd sum ⇒ one prime must be even ⇒ that prime is 2. Then check 10001 − 2 = 9999.
Show solution
  1. 10001 is odd; primes summing to odd require one to be even, so it must be 2.
  2. Then the other is 10001 − 2 = 9999 = 3 · 3333 (digit sum 36 divisible by 3). Not prime.
  3. So 0 representations.
A 0 ways.
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Problem 25 · 2011 AMC 8 Hard
Geometry & Measurement inscribed-circumscribed approximation

A circle with radius 1 is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?

Show hint (soft nudge)
Inner square has diagonal = diameter = 2, so its side is √2 and area is 2.
Show hint (sharpest)
Shaded area inside the circle = π · 12 − 2 = π − 2. Area between the two squares = outer area − inner area = 4 − 2 = 2.
Show solution
  1. Outer square: side 2, area 4. Inner square: diagonal = 2 (diameter), so side √2 and area 2.
  2. Shaded (inside circle, outside inner square): π(1)2 − 2 = π − 2.
  3. Between the squares: 4 − 2 = 2.
  4. Ratio: (π − 2)/2 ≈ (3.14 − 2)/2 ≈ 0.57. Closest to 1/2.
A Closest to 1/2.
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