AMC 8

2019 AMC 8

25 problems — read each, give it a real try, then peek at the hints.

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Problem 1 · 2019 AMC 8 Easy
Arithmetic & Operations divisionunit-rate

Ike and Mike go into a sandwich shop with a total of $30.00 to spend. Sandwiches cost $4.50 each and soft drinks cost $1.00 each. Ike and Mike plan to buy as many sandwiches as they can and use the remaining money to buy soft drinks. Counting both soft drinks and sandwiches, how many items will they buy?

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Answer: D — 9 items.
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Hint 1 of 2
Spend on sandwiches first — what's the most they can buy without going past $30?
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Hint 2 of 2
6 sandwiches cost $27, leaving $3 for sodas.
Show solution
Approach: max sandwiches, then sodas with the change
  1. $30 ÷ $4.50 = 6 with $3 left over (a 7th sandwich would cost $31.50, too much).
  2. The $3 buys 3 sodas at $1 each.
  3. Total items: 6 + 3 = 9.
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Problem 2 · 2019 AMC 8 Medium
Geometry & Measurement areaarea-decomposition
amc8-2019-02
Show answer
Answer: E — 150 square feet.
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Hint 1 of 2
Figure out the size of one small rectangle first — the picture forces a specific shape.
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Hint 2 of 2
Two short sides stacked on the left equal one long side on the right: so long = 2 × 5 = 10.
Show solution
Approach: match the heights
  1. Each small rectangle has short side 5. From the picture, two stacked horizontals on the left have the same height as the vertical rectangle on the right — so the long side is 2 × 5 = 10.
  2. ABCD has width 10 + 5 = 15 and height 10, so its area is 15 × 10 = 150 square feet.
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Problem 3 · 2019 AMC 8 Medium
Fractions, Decimals & Percents fraction-comparison

Which of the following is the correct order of the fractions 1511, 1915, and 1713, from least to greatest?

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Answer: E — 19/15 < 17/13 < 15/11.
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Hint 1 of 2
Each fraction is just a little more than 1. By how much?
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Hint 2 of 2
Notice the numerator is always 4 more than the denominator. So each fraction = 1 + 4/(denominator).
Show solution
Approach: rewrite as 1 + extra
  1. Each fraction has numerator − denominator = 4, so write each as 1 + 4denom.
  2. Bigger denominator means smaller extra. Denominators 11, 13, 15 give extras 411 > 413 > 415.
  3. So the order from least to greatest is 1915 < 1713 < 1511 (choice E).
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Problem 4 · 2019 AMC 8 Medium
Geometry & Measurement pythagorean-triplearea
amc8-2019-04
Show answer
Answer: D — 120 square meters.
Show hints
Hint 1 of 2
The diagonals of a rhombus cross at right angles and cut each other in half. That makes four matching right triangles.
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Hint 2 of 2
Side = 52/4 = 13, half-diagonal = 24/2 = 12. The other half-diagonal is the missing leg of a 5-12-13 right triangle.
Show solution
Approach: diagonals are perpendicular bisectors; spot the 5-12-13
  1. Side length: 52 ÷ 4 = 13. Half of AC: 24 ÷ 2 = 12.
  2. The diagonals are perpendicular bisectors, so each quarter of the rhombus is a right triangle with leg 12 and hypotenuse 13 — a 5-12-13 triple. The other half-diagonal is 5, so BD = 10.
  3. Area of a rhombus = d1 × d22 = 24 × 102 = 120 sq m.
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Problem 5 · 2019 AMC 8 Medium
Ratios, Rates & Proportions distance-speed-timegraph-reading
amc8-2019-05
Show answer
Answer: B — Graph (B).
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Hint 1 of 2
Picture each animal's path on the distance-vs-time graph: when are they moving fast, slow, or stopped?
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Hint 2 of 2
Hare: steep → flat (nap) → steep. Tortoise: one steady slope. Tortoise reaches the finish line first.
Show solution
Approach: match the story to slopes
  1. The hare's distance line is steep at first (running), flat in the middle (napping), then steep again (finishing).
  2. The tortoise's line is one steady, gentler slope — and it reaches the finish-line height before the hare's line does.
  3. Only graph (B) shows both patterns at once.
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Problem 6 · 2019 AMC 8 Medium
Geometry & Measurement symmetrycareful-counting
amc8-2019-06
Show answer
Answer: C — 2/5.
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Hint 1 of 2
A square has only 4 lines of symmetry — the two diagonals and the two perpendicular bisectors. Q must lie on one of those.
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Hint 2 of 2
Each line passes through 9 grid points including P. So 4 × 9 = 36, minus the 4 occurrences of P itself = 32 valid Qs out of 80.
Show solution
Approach: count points on the 4 symmetry lines
  1. A square has exactly 4 axes of symmetry through its center P: the two diagonals and the two perpendicular bisectors.
  2. Each axis contains 9 of the 81 grid points (including P). Counting Q across all 4: 4 × 9 = 36 spots, but P appears 4 times and Q can't equal P, so subtract 4 → 32 valid choices.
  3. Probability = 32 / 80 = 2/5.
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Problem 7 · 2019 AMC 8 Easy
Arithmetic & Operations sum-constraintestimate-and-pick

Shauna takes five tests, each worth a maximum of 100 points. Her scores on the first three tests are 76, 94, and 87. In order to average 81 for all five tests, what is the lowest score she could earn on one of the other two tests?

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Answer: A — 48.
Show hints
Hint 1 of 2
To minimize one of the two remaining scores, max out the other (= 100). Then the rest is forced.
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Hint 2 of 2
Total needed: 5 × 81 = 405. First three sum to 257. Remaining two must sum to 148, so the minimum is 148 − 100.
Show solution
Approach: max the partner, minimize the other
  1. Total required: 5 × 81 = 405. First three tests: 76 + 94 + 87 = 257.
  2. Remaining two tests must total 405 − 257 = 148.
  3. To minimize one, set the other to 100: minimum = 148 − 100 = 48.
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Problem 8 · 2019 AMC 8 Medium
Fractions, Decimals & Percents percent-multiplier

Gilda has a bag of marbles. She gives 20% of them to her friend Pedro. Then Gilda gives 10% of what is left to another friend, Ebony. Finally, Gilda gives 25% of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?

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Answer: E — 54%.
Show hints
Hint 1 of 2
Each transfer leaves a fraction behind. Multiply the "keep" fractions together.
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Hint 2 of 2
Keeps: 0.8 × 0.9 × 0.75.
Show solution
Approach: chain the leftover fractions
  1. After Pedro: 80% remains. After Ebony: 90% of that. After Jimmy: 75% of what's left.
  2. 0.8 × 0.9 × 0.75 = 0.54 = 54%.
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Problem 9 · 2019 AMC 8 Medium
Geometry & Measurement volumevolume-scalingratio

Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are 6 cm in diameter and 12 cm high. Felicia buys cat food in cylindrical cans that are 12 cm in diameter and 6 cm high. What is the ratio of the volume of one of Alex's cans to the volume of one of Felicia's cans?

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Answer: B — 1:2.
Show hints
Hint 1 of 2
Cylinder volume = πr2h. Felicia's radius is doubled (so radius2 ×4) and her height is halved.
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Hint 2 of 2
Net effect on Felicia vs Alex: ×4 from radius, ×1/2 from height = ×2. So Alex : Felicia = 1 : 2.
Show solution
Approach: track how each dimension change scales the volume
  1. Going from Alex (radius 3, height 12) to Felicia (radius 6, height 6): radius doubles → radius2 × 4; height halves → × 1/2.
  2. Combined: Felicia's volume = Alex's × (4 × 1/2) = 2×. So ratio Alex : Felicia = 1 : 2.
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Problem 10 · 2019 AMC 8 Medium
Arithmetic & Operations careful-counting
amc8-2019-10
Show answer
Answer: B — Mean +1, median +1.
Show hints
Hint 1 of 2
Changing Wednesday from 16 to 21 adds 5 to the total; the mean of 5 days goes up by 5/5 = 1.
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Hint 2 of 2
Recompute the sorted order. The middle value of the corrected list slides up by 1.
Show solution
Approach: track the change in total and the new sorted middle
  1. Wednesday changes from 16 to 21, a total increase of 5. With 5 days, the mean rises by 5/5 = 1.
  2. Original sorted days: 16, 18, 20, 22, 26 → median 20. After correction: 18, 20, 21, 22, 26 → median 21. Median rises by 1.
  3. Answer: both increase by 1.
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Problem 11 · 2019 AMC 8 Medium
Counting & Probability complementary-counting

The eighth grade class at Lincoln Middle School has 93 students. Each student takes a math class or a foreign language class or both. There are 70 eighth graders taking a math class, and there are 54 eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign language class?

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Answer: D — 39 students.
Show hints
Hint 1 of 2
Use inclusion-exclusion: |M ∪ F| = |M| + |F| − |M ∩ F|. That gives the overlap count.
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Hint 2 of 2
Both = 70 + 54 − 93 = 31. Math only = 70 − 31.
Show solution
Approach: inclusion-exclusion for the overlap
  1. |Math| + |Foreign| − |Both| = |Total| ⇒ 70 + 54 − Both = 93 ⇒ Both = 31.
  2. Math only = |Math| − Both = 70 − 31 = 39.
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Problem 12 · 2019 AMC 8 Medium
Geometry & Measurement spatial-reasoningcasework
amc8-2019-12
Show answer
Answer: A — Red.
Show hints
Hint 1 of 2
Two faces are opposite if they never appear together in any view.
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Hint 2 of 2
Pair up opposites: look across the three views to find which color is never adjacent to aqua.
Show solution
Approach: find opposites by elimination across views
  1. From the three views, identify pairs of opposite faces by which never share an edge:
  2. Comparing views, white and green appear together (so they're adjacent, not opposite); brown and purple are opposite; aqua and red never share a view → opposite.
  3. Face opposite aqua = red.
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Problem 13 · 2019 AMC 8 Medium
Number Theory divisibilitydigit-sum

A palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let N be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of N?

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Answer: A — 2.
Show hints
Hint 1 of 2
Two-digit palindromes (11, 22, 33, …) are all multiples of 11 — so any sum of three of them is a multiple of 11.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the smallest 3-digit multiple of 11 that isn't itself a palindrome. Can it be written as a sum of three distinct 2-digit palindromes?
Show solution
Approach: two-digit palindromes are multiples of 11
  1. Two-digit palindromes (11, 22, …, 99) are all multiples of 11; their sum is too. So N is a multiple of 11.
  2. Smallest 3-digit multiple of 11 that's NOT a palindrome: 110 (palindromes are 121, 131, … — 110 isn't one).
  3. Check: 110 = 11 + 22 + 77 ✓ (three distinct 2-digit palindromes).
  4. N = 110; digit sum = 1 + 1 + 0 = 2.
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Problem 14 · 2019 AMC 8 Medium
Number Theory mod-10divisibility

Isabella has 6 coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every 10 days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the 6 dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?

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Answer: C — Wednesday.
Show hints
Hint 1 of 2
10 days ≡ 3 days mod 7. So the 6 redemption days are at day-of-week offsets 0, 3, 6, 2, 5, 1 from the starting day.
Still stuck? Show hint 2 →
Hint 2 of 2
Those 6 offsets cover 6 of 7 days — missing only offset 4. Sunday must be that missing day, so the start = Sunday − 4 days = Wednesday.
Show solution
Approach: compute the 6 days mod 7, find the missing one
  1. 10 days advances the day-of-week by 10 mod 7 = 3 days. So the six redemption days have day-of-week offsets {0, 3, 6, 2, 5, 1} mod 7 from the start.
  2. These 6 offsets cover everything except offset 4. Sunday must be that missing offset, so start day is Sunday − 4 days.
  3. Counting back from Sunday: Sat, Fri, Thu, Wed.
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Problem 15 · 2019 AMC 8 Medium
Counting & Probability conditional-probabilityproportion

On a beach 50 people are wearing sunglasses and 35 people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is also wearing sunglasses is 25. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?

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Answer: B — 7/25.
Show hints
Hint 1 of 2
Find the number wearing both first. 2/5 of cap-wearers also wear sunglasses.
Still stuck? Show hint 2 →
Hint 2 of 2
Both = (2/5) × 35 = 14. Then P(cap | sunglasses) = 14 / 50.
Show solution
Approach: compute the intersection from one conditional
  1. P(sunglasses | cap) = 2/5, so (people wearing both) = (2/5) × 35 = 14.
  2. P(cap | sunglasses) = 14 / 50 = 7/25.
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Problem 16 · 2019 AMC 8 Medium
Ratios, Rates & Proportions distance-speed-timesubstitution

Qiang drives 15 miles at an average speed of 30 miles per hour. How many additional miles will he have to drive at 55 miles per hour to average 50 miles per hour for the entire trip?

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Answer: D — 110 miles.
Show hints
Hint 1 of 2
Average speed = total distance / total time. Write that equation with x = additional miles.
Still stuck? Show hint 2 →
Hint 2 of 2
Time so far: 15/30 = 1/2 hr. Set (15 + x)/(1/2 + x/55) = 50.
Show solution
Approach: set total-distance / total-time = 50
  1. Time so far: 15/30 = 1/2 hour. After x additional miles at 55 mph, the new total is (15 + x) miles and (1/2 + x/55) hours.
  2. (15 + x) / (1/2 + x/55) = 50 ⇒ 15 + x = 25 + 10x/11.
  3. Multiply by 11: 165 + 11x = 275 + 10xx = 110.
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Problem 17 · 2019 AMC 8 Hard
Fractions, Decimals & Percents fraction-to-decimal

What is the value of the product

(1·32·2)(2·43·3)(3·54·4) … (97·9998·98)(98·10099·99) ?
Show answer
Answer: B — 50/99.
Show hints
Hint 1 of 2
Regroup. Each fraction is k(k+2)(k+1)2. Pull out a 1/2 from the first term and look for telescoping.
Still stuck? Show hint 2 →
Hint 2 of 2
Rewrite as (1/2)(3/2 · 2/3)(4/3 · 3/4) … (100/99) — most factors cancel.
Show solution
Approach: telescoping after pairing
  1. Each factor is k(k+2)(k+1)2 = kk+1 · k+2k+1.
  2. Product across k = 1 to 98: (1/2 · 3/2)(2/3 · 4/3)…(98/99 · 100/99). Pair telescopes: the kk+1 chain → 1/99 (left telescoping); the k+2k+1 chain → 100/2 (right telescoping).
  3. Product = (1/99) · (100/2) = 50/99.
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Problem 18 · 2019 AMC 8 Medium
Counting & Probability careful-countingcasework

The faces of each of two fair dice are numbered 1, 2, 3, 5, 7, and 8. When the two dice are tossed, what is the probability that their sum will be an even number?

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Answer: C — 5/9.
Show hints
Hint 1 of 2
Sum is even iff both rolls are odd or both are even.
Still stuck? Show hint 2 →
Hint 2 of 2
Faces: 4 odd (1, 3, 5, 7) and 2 even (2, 8). Compute both probabilities and add.
Show solution
Approach: both-odd OR both-even
  1. Odd faces: 1, 3, 5, 7 → 4 of 6. Even faces: 2, 8 → 2 of 6.
  2. Sum even = both odd (4/6 · 4/6 = 16/36) or both even (2/6 · 2/6 = 4/36).
  3. Probability = (16 + 4)/36 = 20/36 = 5/9.
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Problem 19 · 2019 AMC 8 Hard
Logic & Word Problems caseworksum-constraint

In a tournament there are six teams that play each other twice. A team earns 3 points for a win, 1 point for a draw, and 0 points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?

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Answer: C — 24 points each.
Show hints
Hint 1 of 2
Maximize top 3 by having them sweep the bottom 3.
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Hint 2 of 2
Each top team plays 6 games vs the bottom 3 (3 opponents × 2 games), earning 6 × 3 = 18 points. Among themselves, balance wins.
Show solution
Approach: sweep bottom 3, balance among top 3
  1. Have each top-3 team win all 6 of its games against the bottom 3 (3 opponents × 2 games each) → 6 × 3 = 18 points per top team.
  2. Among the top 3, each pair plays twice. To tie all three, give each pair a 1-win, 1-loss split — each team in a pair gains 3 points (one win) and loses 3 points worth (one loss = 0).
  3. Each top team is in 2 pairs and wins one game in each → +3 + 3 = 6 more points.
  4. Maximum: 18 + 6 = 24.
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Problem 20 · 2019 AMC 8 Medium
Algebra & Patterns square-root-both-sidescasework

How many different real numbers x satisfy the equation

(x2 − 5)2 = 16 ?
Show answer
Answer: D — 4 real numbers.
Show hints
Hint 1 of 2
Take the square root of both sides: x2 − 5 = ±4. Two cases for x2.
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Hint 2 of 2
x2 = 9 gives two values; x2 = 1 gives two more.
Show solution
Approach: split on the ± root
  1. (x2 − 5)2 = 16 ⇒ x2 − 5 = ±4.
  2. Case +4: x2 = 9 ⇒ x = ±3.
  3. Case −4: x2 = 1 ⇒ x = ±1.
  4. Total: 4 distinct real solutions.
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Problem 21 · 2019 AMC 8 Medium
Geometry & Measurement area

What is the area of the triangle formed by the lines y = 5, y = 1 + x, and y = 1 − x?

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Answer: E — 16.
Show hints
Hint 1 of 2
Find the three vertices by pairwise intersection.
Still stuck? Show hint 2 →
Hint 2 of 2
The top side lies on y = 5. Use it as the base; the height is to the third vertex.
Show solution
Approach: find the vertices, take the horizontal side as base
  1. Intersections: y = 5 with y = 1 + x → (4, 5). With y = 1 − x → (−4, 5). And y = 1 + x meets y = 1 − x at (0, 1).
  2. Base along y = 5 from (−4, 5) to (4, 5): length 8. Height to (0, 1): 5 − 1 = 4.
  3. Area = (1/2)(8)(4) = 16.
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Problem 22 · 2019 AMC 8 Hard
Fractions, Decimals & Percents percent-multiplierdifference-of-squares

A store increased the original price of a shirt by a certain percent and then decreased the new price by the same amount. Given that the resulting price was 84% of the original price, by what percent was the price increased and decreased?

Show answer
Answer: E — 40%.
Show hints
Hint 1 of 2
Multiplying by (1+p) then (1−p) gives 1 − p2. That equals 0.84.
Still stuck? Show hint 2 →
Hint 2 of 2
p2 = 0.16.
Show solution
Approach: compose the two changes
  1. (1 + p)(1 − p) = 1 − p2 = 0.84.
  2. p2 = 0.16 ⇒ p = 0.4 = 40%.
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Problem 23 · 2019 AMC 8 Hard
Algebra & Patterns divisibilitycasework

After Euclid High School's last basketball game, it was determined that 14 of the team's points were scored by Alexa and 27 were scored by Brittany. Chelsea scored 15 points. None of the other 7 team members scored more than 2 points. What was the total number of points scored by the other 7 team members?

Show answer
Answer: B — 11 points.
Show hints
Hint 1 of 2
Let T = total. Then 1/4 and 2/7 of T are integers ⇒ T is a multiple of lcm(4,7) = 28.
Still stuck? Show hint 2 →
Hint 2 of 2
Others = TT/4 − 2T/7 − 15 = 13T/28 − 15. Constrained by 0 ≤ Others ≤ 7×2 = 14.
Show solution
Approach: T is a multiple of 28; test small values
  1. Let T be the total. Alexa has T/4, Brittany 2T/7 — both integers ⇒ T divisible by 28.
  2. Others' total: TT/4 − 2T/7 − 15 = 13T/28 − 15. Must be in [0, 14] (since 7 players, ≤ 2 each).
  3. T = 28: gives −2 (invalid). T = 56: gives 13·56/28 − 15 = 26 − 15 = 11 (valid).
  4. T = 84 would give 39 − 15 = 24, exceeding 14.
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Problem 24 · 2019 AMC 8 Hard
Geometry & Measurement area-fractionratio
amc8-2019-24
Show answer
Answer: B — Area 30.
Show hints
Hint 1 of 2
Set coordinates: A=(0,0), B=(0,1), C=(3,0). Then D, E, F follow from the ratios.
Still stuck? Show hint 2 →
Hint 2 of 2
Once you have F = intersection of AE and BC, use the shoelace formula on E, B, F. Compare to area of ABC.
Show solution
Approach: place coordinates and read off F
  1. Set A = (0,0), B = (0,1), C = (3,0). Then D divides AC with 1:2 → D = (1, 0). E = midpoint of BD = (0.5, 0.5).
  2. Line AE has slope 1: y = x. Line BC: from (0,1) to (3,0), y = 1 − x/3.
  3. Intersect: x = 1 − x/3 ⇒ x = 3/4. So F = (3/4, 3/4).
  4. Area of EBF (shoelace on (0.5, 0.5), (0, 1), (0.75, 0.75)) = 1/8. Area of ABC = 3/2. Ratio = (1/8) / (3/2) = 1/12.
  5. So area of EBF = (1/12) × 360 = 30.
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Problem 25 · 2019 AMC 8 Hard
Counting & Probability stars-and-barscareful-counting

Alice has 24 apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?

Show answer
Answer: C — 190 ways.
Show hints
Hint 1 of 2
Pre-give 2 apples to each — now you're distributing the remaining 18 with no constraints.
Still stuck? Show hint 2 →
Hint 2 of 2
Stars and bars: nonneg integer solutions to a + b + c = 18 is C(20, 2).
Show solution
Approach: subtract the floor, then stars and bars
  1. Give 2 apples to each person first: that uses 6, leaving 18 to distribute with no minimum (each person already has 2).
  2. Distribute 18 indistinguishable apples among 3 people with no constraint: C(18 + 2, 2) = C(20, 2) = 190.
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