AMC 8 · Test Mode

2013 AMC 8

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Problem 1 · 2013 AMC 8 Easy
Number Theory next-multiple

Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?

Show hint
Find the next multiple of 6 above 23.
Show solution
  1. Next multiple of 6 after 23 is 24.
  2. She needs 24 − 23 = 1 more car.
A 1 more car.
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Problem 2 · 2013 AMC 8 Easy
Fractions, Decimals & Percents percent-of-price

A sign at the fish market says, "50% off, today only: half-pound packages for just $3 per package." What is the regular price for a full pound of fish, in dollars? (Assume that there are no deals for bulk.)

Show hint
First scale half-pound to full-pound; then undo the 50% discount.
Show solution
  1. Sale price per pound: 2 × $3 = $6.
  2. 50% off means regular = 2 × sale: 2 × $6 = $12.
D $12.
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Problem 3 · 2013 AMC 8 Easy
Arithmetic & Operations pair-grouping

What is the value of 4 · (−1 + 2 − 3 + 4 − 5 + 6 − 7 + … + 1000)?

Show hint
Group consecutive pairs (−1 + 2), (−3 + 4), …. Each pair = 1.
Show solution
  1. Each pair (−k + (k+1)) equals 1; there are 500 such pairs (covering 1…1000).
  2. Sum inside: 500. Multiply by 4: 2000.
E 2000.
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Problem 4 · 2013 AMC 8 Easy
Algebra & Patterns find-share-then-total

Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?

Show hint
The 7 friends together covered Judi's share: 7 × $2.50. That's one-eighth of the total.
Show solution
  1. Judi's share = 7 × $2.50 = $17.50.
  2. Everyone paid the same, so total = 8 × $17.50 = $140.
C $140.
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Problem 5 · 2013 AMC 8 Easy
Arithmetic & Operations mean-vs-median

Hammie is in the 6th grade and weighs 106 pounds. His quadruplet sisters are tiny babies and weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?

Show hint
Order the 5 weights to find the median. Mean = total / 5. The single outlier (106) pulls the mean way up.
Show solution
  1. Sorted: 5, 5, 6, 8, 106 ⇒ median = 6.
  2. Mean = (5 + 5 + 6 + 8 + 106)/5 = 130/5 = 26.
  3. Mean exceeds median by 26 − 6 = 20.
E Average, by 20.
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Problem 6 · 2013 AMC 8 Easy
Arithmetic & Operations multiplication-pyramid

The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, 30 = 6 × 5. What is the missing number in the top row?

Show hint (soft nudge)
Bottom = left-middle × right-middle. So right-middle = 600 / 30.
Show hint (sharpest)
Right-middle = top-middle × top-right = 5 × (top-right).
Show solution
  1. Right-middle box = 600 / 30 = 20.
  2. Right-middle = 5 × (top-right), so top-right = 20 / 5 = 4.
C 4.
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Problem 7 · 2013 AMC 8 Easy
Ratios, Rates & Proportions proportion

Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?

Show hint
Set up a proportion: 6 cars per 10 sec. Convert 2 min 45 sec to seconds.
Show solution
  1. Total time: 2 · 60 + 45 = 165 seconds.
  2. Cars: (6/10) × 165 = 99 ≈ 100.
C About 100 cars.
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Problem 8 · 2013 AMC 8 Easy
Counting & Probability enumerate-outcomes

A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?

Show hint
Only 8 outcomes total. List the ones with HH appearing in consecutive positions.
Show solution
  1. Outcomes with two consecutive heads: HHH, HHT, THH. That's 3.
  2. Total outcomes: 23 = 8.
  3. Probability: 3/8.
C 3/8.
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Problem 9 · 2013 AMC 8 Easy
Algebra & Patterns powers-of-two

The Incredible Hulk can double the distance it jumps with each succeeding jump. If its first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will it first be able to jump more than 1 kilometer (1,000 meters)?

Show hint
Jump n is 2n−1 meters. Find the smallest n with 2n−1 > 1000.
Show solution
  1. Jump n = 2n−1 m.
  2. 29 = 512 (10th jump), 210 = 1024 (11th jump).
  3. First > 1000 is the 11th.
C 11th jump.
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Problem 10 · 2013 AMC 8 Medium
Number Theory lcm-gcd prime-factorization

What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?

Show hint
lcm(a, b) / gcd(a, b) = ab / gcd(a, b)2. Or factor and compare exponents.
Show solution
  1. 180 = 22 · 32 · 5. 594 = 2 · 33 · 11.
  2. gcd: 2 · 32 = 18. (Take the minimum exponent of each shared prime.)
  3. lcm: 22 · 33 · 5 · 11 = 4 · 27 · 55 = 5940.
  4. Ratio: 5940 / 18 = 330.
C 330.
Another way: shortcut
  1. lcm(a, b) · gcd(a, b) = a · b, so lcm/gcd = ab/gcd2.
  2. 180 · 594 = 106,920. gcd = 18 ⇒ lcm/gcd = 106,920 / 324 = 330.
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Problem 11 · 2013 AMC 8 Medium
Ratios, Rates & Proportions time-equals-distance-over-rate

Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?

Show hint
Time for 2 miles at rate r is 2/r hours. Convert each to minutes and compare with the all-4-mph plan.
Show solution
  1. Monday: (2/5) hr = 24 min. Wednesday: (2/3) hr = 40 min. Friday: (2/4) hr = 30 min. Total: 94 min.
  2. All-4-mph plan: 3 × 30 = 90 min.
  3. Saved: 94 − 90 = 4 minutes.
D 4 minutes.
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Problem 12 · 2013 AMC 8 Easy
Fractions, Decimals & Percents percent-savings

At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the $150 regular price did he save?

Show hint
Compute the dollar saved on the 2nd and 3rd pairs (the 1st is full price). Then divide by $150.
Show solution
  1. 2nd pair saves 40% of $50 = $20. 3rd pair saves half of $50 = $25.
  2. Total saved: $45. Regular price: $150.
  3. Percentage saved: 45/150 = 30%.
B 30%.
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Problem 13 · 2013 AMC 8 Medium
Number Theory place-value-difference divisibility-by-9

When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?

Show hint
If a score's last two digits are 10a + b, reversing gives 10b + a. The difference is 9(ab): always a multiple of 9.
Show solution
  1. (10a + b) − (10b + a) = 9(ab).
  2. The sum's error must therefore be a multiple of 9.
  3. Among the choices, only 45 = 9 · 5 qualifies.
A 45.
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Problem 14 · 2013 AMC 8 Easy
Counting & Probability independent-events

Abe holds 1 green and 1 red jelly bean in his hand. Bob holds 1 green, 1 yellow, and 2 red jelly beans in his hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?

Show hint
Match in two ways: both green, or both red. Multiply each person's color probability per case, then add.
Show solution
  1. Both green: (1/2)(1/4) = 1/8.
  2. Both red: (1/2)(2/4) = 1/4.
  3. Total: 1/8 + 1/4 = 1/8 + 2/8 = 3/8.
C 3/8.
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Problem 15 · 2013 AMC 8 Medium
Algebra & Patterns powers-by-inspection

If 3p + 34 = 90, 2r + 44 = 76, and 53 + 6s = 1421, what is the product of p, r, and s?

Show hint (soft nudge)
Subtract the known term from each equation, then identify the power.
Show hint (sharpest)
9 = 32, 32 = 25, 1296 = 64.
Show solution
  1. 3p = 90 − 81 = 9 = 32p = 2.
  2. 2r = 76 − 44 = 32 = 25r = 5.
  3. 6s = 1421 − 125 = 1296 = 64s = 4.
  4. Product: 2 · 5 · 4 = 40.
B 40.
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Problem 16 · 2013 AMC 8 Medium
Ratios, Rates & Proportions lcm-for-ratios

A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of 8th-graders to 6th-graders is 5 : 3, and the ratio of 8th-graders to 7th-graders is 8 : 5. What is the smallest number of students that could be participating in the project?

Show hint (soft nudge)
Both ratios link to the 8th-graders. The number of 8th-graders must be a multiple of 5 (first ratio) and 8 (second ratio).
Show hint (sharpest)
Smallest such number is lcm(5, 8) = 40.
Show solution
  1. 8th-graders divisible by both 5 and 8 ⇒ smallest is 40.
  2. 6th-graders: 40 · 3/5 = 24.
  3. 7th-graders: 40 · 5/8 = 25.
  4. Total: 40 + 24 + 25 = 89.
E 89 students.
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Problem 17 · 2013 AMC 8 Easy
Algebra & Patterns consecutive-integers

The sum of six consecutive positive integers is 2013. What is the largest of these six integers?

Show hint
Sum of 6 consecutive integers starting at x is 6x + 15.
Show solution
  1. Let smallest = x. Then 6x + 15 = 2013 ⇒ 6x = 1998 ⇒ x = 333.
  2. Largest = 333 + 5 = 338.
B 338.
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Problem 18 · 2013 AMC 8 Medium
Geometry & Measurement volume-difference shell-counting

Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?

Show hint
Outer block-volume = 12 · 10 · 5. Subtract the hollow inside. Inside shrinks by 1 in each wall direction and 1 in the floor (no ceiling).
Show solution
  1. Outer: 12 × 10 × 5 = 600 blocks (as if solid).
  2. Hollow interior: length 10 (2 walls shaved), width 8 (2 walls shaved), height 4 (floor shaved, no ceiling) ⇒ 10 × 8 × 4 = 320 empty.
  3. Blocks used: 600 − 320 = 280.
B 280 blocks.
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Problem 19 · 2013 AMC 8 Medium
Logic & Word Problems logic-puzzle

Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest score?

Show hint
Both Cassie and Bridget saw Hannah's score. Why can each be sure of her own ranking only after seeing Hannah's?
Show solution
  1. Cassie's claim is only certain if her score > Hannah's (otherwise Hannah's lower score below Cassie's could be the lowest, and Cassie wouldn't know). So Cassie > Hannah.
  2. Bridget's claim is only certain if her score < Hannah's. So Hannah > Bridget.
  3. Combine: Cassie > Hannah > Bridget.
D Cassie, Hannah, Bridget.
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Problem 20 · 2013 AMC 8 Medium
Geometry & Measurement inscribed-rectangle pythagorean

A 1 × 2 rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?

Show hint (soft nudge)
The two upper corners of the rectangle lie on the semicircle. Use the center of the diameter as the origin: each upper corner is at (±1, 1).
Show hint (sharpest)
Its distance from the center is the radius.
Show solution
  1. Place the semicircle's center at the midpoint of the diameter. Upper corners of the rectangle are at (1, 1) and (−1, 1).
  2. Radius2 = 12 + 12 = 2 ⇒ r = √2.
  3. Semicircle area = (1/2)πr2 = (1/2)π(2) = π.
C &pi;.
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Problem 21 · 2013 AMC 8 Hard
Counting & Probability lattice-paths multiplication-principle

Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?

Show hint (soft nudge)
Three independent legs: home → SW corner, the unique diagonal through the park, NE corner → school. Count lattice paths for each leg and multiply.
Show hint (sharpest)
Home to SW corner: 2 E + 1 N. NE corner to school: 2 E + 2 N.
Show solution
  1. Home → SW corner: choose 1 of 3 step-orderings = C(3, 1) = 3.
  2. Diagonal through the park: 1 way.
  3. NE corner → school: C(4, 2) = 6.
  4. Total: 3 × 1 × 6 = 18.
E 18 routes.
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Problem 22 · 2013 AMC 8 Medium
Counting & Probability grid-of-segments

Toothpicks are used to make a grid that is 60 toothpicks long and 32 toothpicks wide. How many toothpicks are used altogether?

Show hint
Count horizontals and verticals separately. A grid 60 long × 32 wide has 33 horizontal lines (each 60 toothpicks) and 61 vertical lines (each 32 toothpicks).
Show solution
  1. Horizontal lines: 32 + 1 = 33, each made of 60 toothpicks ⇒ 33 × 60 = 1980.
  2. Vertical lines: 60 + 1 = 61, each made of 32 toothpicks ⇒ 61 × 32 = 1952.
  3. Total: 1980 + 1952 = 3932.
E 3932 toothpicks.
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Problem 23 · 2013 AMC 8 Hard
Geometry & Measurement semicircle-area-arc pythagorean-triple

Angle ABC of ▵ABC is a right angle. The sides of ▵ABC are the diameters of semicircles as shown. The area of the semicircle on AB equals 8π, and the arc of the semicircle on AC has length 8.5π. What is the radius of the semicircle on BC?

Show hint (soft nudge)
Recover AB from the semicircle area, and AC from the semicircle arc length. Then use the Pythagorean theorem to find BC, then halve it.
Show hint (sharpest)
Semicircle area = (1/2)πr2. Semicircle arc length = πr. So radius is straightforward from each.
Show solution
  1. Semicircle on AB: (1/2)πr2 = 8π ⇒ r = 4 ⇒ AB = 8.
  2. Semicircle on AC: πr = 8.5π ⇒ r = 8.5 ⇒ AC = 17.
  3. Right angle at B: BC = √(172 − 82) = √225 = 15. (8-15-17 triple.)
  4. Radius of semicircle on BC = 15 / 2 = 7.5.
B Radius 7.5.
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Problem 24 · 2013 AMC 8 Hard
Geometry & Measurement coordinate-bash shoelace

Squares ABCD, EFGH, and GHIJ are equal in area. Points C and D are the midpoints of sides IH and HE, respectively. What is the ratio of the area of the shaded pentagon AJICB to the sum of the areas of the three squares?

Show hint (soft nudge)
Set each square's side to 1. Place coordinates: F = (0, 0), E = (0, 1), G = (1, 0), H = (1, 1), J = (2, 0), I = (2, 1).
Show hint (sharpest)
Then D = (0.5, 1), C = (1.5, 1), A = (0.5, 2), B = (1.5, 2). Use the shoelace formula on pentagon AJICB.
Show solution
  1. Side length 1. Vertices: A = (0.5, 2), J = (2, 0), I = (2, 1), C = (1.5, 1), B = (1.5, 2).
  2. Shoelace: ½ |(0.5·0 + 2·1 + 2·1 + 1.5·2 + 1.5·2) − (2·2 + 0·2 + 1·1.5 + 1·1.5 + 2·0.5)| = ½ |10 − 8| = 1.
  3. Total square area = 3 · 1 = 3. Ratio = 1 / 3 = 1/3.
C 1/3.
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Problem 25 · 2013 AMC 8 Hard
Geometry & Measurement rolling-ball-path arc-length

A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are R1 = 100 inches, R2 = 60 inches, and R3 = 80 inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?

Show hint (soft nudge)
The ball's center traces a path parallel to the track at distance = ball's radius (2 inches). On the outside of an arc the center's arc has radius R − 2; on the inside it has radius R + 2.
Show hint (sharpest)
Arc 1 (outside): radius 98. Arc 2 (inside): radius 62. Arc 3 (outside): radius 78.
Show solution
  1. Ball radius = 2 inches.
  2. Center radii along the three arcs: 100 − 2 = 98, 60 + 2 = 62, 80 − 2 = 78.
  3. Half-circumferences sum: π(98 + 62 + 78) = 238π.
A 238&pi;.
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