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Problem 1 · 2018 AMC 8
Easy
Ratios, Rates & Proportionsratioproportion
An amusement park has a collection of scale models, with a ratio of 1 : 20, of buildings and other sights from around the country. The height of the United States Capitol is 289 feet. What is the height in feet of its replica at this park, rounded to the nearest whole number?
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Answer: A — 14 feet.
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Hint 1 of 2
A scale of 1 : 20 means "the model is the smaller one." The real building is the big side (20), the replica is the small side (1) — so which way does 289 go, up or down?
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Hint 2 of 2
The technique: a ratio 1 : k shrinks the real size by dividing by k. Map the bigger number to the bigger part of the ratio so you never flip it by accident.
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Approach: divide by the scale factor
The replica is the "1" side and the real Capitol is the "20" side, so the replica is 1/20 of 289 — smaller, which is the sanity check that we're dividing (a model should be tiny).
289 ÷ 20 = 14.45, which rounds to 14 feet.
You'll see it again: any scale-model or map problem is just multiply or divide by the scale factor — the only decision is which way, and matching big-to-big settles it.
Unlike our digits, here position doesn't matter — a symbol is worth the same wherever it appears. So you only need to count how many of each kind.
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Hint 2 of 2
Tally each symbol type and add their table values, just like the example (three ∩ arches and two | strokes made 32). Notice there's no thousands symbol — what does that force the thousands digit to be?
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Approach: count each symbol type, then add the values
The key idea: in this Egyptian system a symbol's value doesn't depend on where it sits — you just count how many of each kind there are. So tally: one 10,000, four 100s, two 10s, three 1s.
Two hundred thousand times two hundred thousand equals
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Answer: E — forty billion.
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Hint 1 of 3
Split each big number into a small front number and its trailing zeros. Two hundred thousand = 2 followed by how many zeros?
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Hint 2 of 3
Handle the fronts and the zeros separately: multiply the 2 × 2, then just pile ALL the zeros together.
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Hint 3 of 3
Five zeros from each factor means ten zeros stack up. What does 4 with ten zeros after it spell out?
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Approach: separate the front digits from the trailing zeros, then recombine
Write each number as a front digit times its zeros: 200,000 = 2 with five 0's = 2 × 10⁵. Multiplying, the fronts give 2 × 2 = 4 and the zeros simply add up: 5 + 5 = 10 zeros.
So the product is 4 followed by ten zeros = 40,000,000,000 = forty billion.
Why this transfers: when multiplying round numbers, never line them up to multiply digit-by-digit — peel off the trailing zeros, multiply the small leftovers, and re-attach the combined zero count. (300 × 4000 = 12 with 5 zeros = 1,200,000.)
Sanity check on the name: a billion has 9 zeros; 40 billion is 4 followed by 10 zeros, which is exactly what we got. ✓
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?
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Answer: C — $140.
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Hint 1 of 2
The seven $2.50 extras didn't vanish — together they exactly paid off Judi's one share. So 7 × $2.50 is one person's share of the bill.
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Hint 2 of 2
Find the size of one equal part first, then scale up to the whole. One share × (number of shares) = total.
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Approach: find one share, then multiply by 8
The 7 extra payments covered exactly Judi's portion, so one person's share = 7 × $2.50 = $17.50. (That's the key reframe: the extras add up to one full share.)
Everyone owed the same share, and there are 8 people, so the total bill = 8 × $17.50 = $140.
Sanity check: $17.50 per person feels right for a restaurant, and 8 of them lands at $140 — matching a choice.
Which of the “checkerboards” illustrated below CANNOT be covered exactly and completely by a whole number of non-overlapping dominoes?
(A)
(B)
(C)
(D)
(E)
Show answer
Answer: B — 3 × 5.
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Hint 1 of 3
You don't need to attempt any actual covering. One domino always hides exactly 2 squares — so the squares always vanish two at a time. What must be true about the TOTAL number of squares for them all to disappear in pairs?
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Hint 2 of 3
Dominoes cover squares in twos, so a fillable board must have an EVEN count of squares. The answer is the lone board whose square-count is odd — and square-count is just rows × columns.
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Hint 3 of 3
Multiply rows × columns for each board and hunt for the odd one — odd × odd is the only way to get an odd total.
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Approach: a board fillable by dominoes must have an even number of squares
Each domino covers exactly 2 squares, so however you lay them, squares get used up two at a time. That means a board can be filled only if its total number of squares is even — no clever arrangement can save an odd-square board.
Count the squares (rows × columns): 3×4 = 12, 3×5 = 15, 4×4 = 16, 4×5 = 20, 6×3 = 18. Only 15 is odd.
An odd number of squares can never split into pairs, so 3 × 5 is the board that cannot be covered.
Why this transfers: any tile that covers a fixed number of cells forces the total to be a multiple of that number — a divisibility check often settles "can it be tiled?" instantly, before you ever try to fit a single piece.
Going deeper (the harder version): an even count is necessary but not always enough. Color the board like a checkerboard — each domino must cover one dark and one light square, so a board with unequal dark/light counts is impossible even when its total is even. Here, though, the simple even/odd count already names the answer.
On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?
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Answer: C — 80 minutes.
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Hint 1 of 2
The one fact you're handed (10 miles in 30 minutes) is secretly a speed. Pin that down and the highway speed comes free — it's just three times bigger.
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Hint 2 of 2
The technique: when given distance and time for one stretch, get a speed, then use that speed (or a scaled version) to find the missing time on the other stretch. Time = distance ÷ speed.
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Approach: find each leg's time separately
Coastal leg: 10 miles in 30 min, so his coastal speed is 1/3 mile per minute. Highway is 3× faster: 1 mile per minute.
Highway time: 50 miles ÷ 1 mile/min = 50 min.
Total trip: 30 + 50 = 80 minutes.
Another way — compare the legs directly (no speeds needed):
Going 3× faster on the highway means each highway mile takes 1/3 the time of a coastal mile. The highway is 5× longer (50 vs 10 miles) but 3× quicker per mile, so the highway takes 5/3 of the coastal time.
Coastal took 30 min, so highway takes (5/3)(30) = 50 min, and the total is 30 + 50 = 80 minutes — reached entirely by scaling the one known time.
How many integers between 2020 and 2400 have four distinct digits arranged in increasing order? (For example, 2347 is one integer.)
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Answer: C — 15 integers.
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Hint 1 of 2
“Increasing order” is a gift: once you pick which digits to use, there's only one way to arrange them. So you're choosing a set of digits, never ordering them.
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Hint 2 of 2
The first two digits are forced (2, then 3, to stay between 2020 and 2400 while increasing). So just count how many ways to pick the last two digits from {4, 5, 6, 7, 8, 9}.
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Approach: increasing ⇒ choosing a set, not arranging
Increasing means strictly ascending, so once the four digits are chosen, the order is automatic. The thousands digit is 2 (number is 2020–2400), and the next digit must beat 2 yet keep the number ≤ 2400, so it's 3.
That leaves the last two digits: any 2 distinct values from {4, 5, 6, 7, 8, 9}. Each pair makes exactly one valid number (smaller digit first), so just count the pairs: C(6, 2) = 15.
Why this transfers: whenever an arrangement is forced to be increasing (or decreasing), counting collapses from permutations to combinations — you pick the values and the order takes care of itself. That's the difference between C(6,2) = 15 and the larger 6×5 = 30 of ordered pairs.
Another way — list the pairs to see C(6,2):
Last two digits from {4,5,6,7,8,9}: pick the smaller, then a larger partner. 4 with {5,6,7,8,9}=5; 5 with {6,7,8,9}=4; 6→3; 7→2; 8→1.
5 + 4 + 3 + 2 + 1 = 15 — the triangular-number shape that always shows up when you count unordered pairs.
The big trap: those bars are not seven numbers to average. Each bar's height is how many students reported that many days — so a tall bar should count much more heavily than a short one.
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Hint 2 of 2
The technique is a weighted average: total up (day-value × how many students), then divide by the total number of students — never just average the labels 1 through 7.
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Approach: weighted mean from the bar heights
Read the bar heights as student counts for 1 through 7 days: 1, 3, 2, 6, 8, 3, 2. Add them: 1+3+2+6+8+3+2 = 25 students total.
Mean = 109 ÷ 25 = 4.36. Sanity check: the data piles up around 4–5 days, so an average near 4.4 fits — and the naive (wrong) average of 1–7 would be 4.00, which isn't even an option.
You'll see it again: any "mean from a frequency table or bar graph" is a weighted average — sum of (value × frequency) over total frequency.
To complete the grid below, each of the digits 1 through 4 must occur once in each row and once in each column. What number will occupy the lower right-hand square?
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Answer: B — 2.
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Hint 1 of 2
The question asks about one cell, so don't solve the whole puzzle — only follow the row and column that pass through that corner.
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Hint 2 of 2
Sudoku logic: attack the most-filled line first. A cell forbids every digit already in its row or column, so the tightest spots get forced one at a time, and each forced cell tightens the next.
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Approach: force the target cell using only its own column
Aim at column 4, the column through the corner we want. It already holds a 4 (row 3), so no other cell in it can be 4 — that's the lever.
Row 1's two blanks are 3 and 4, but column 4 has banned 4, so row 1 col 4 = 3. Same move on row 2 (blanks 1 and 4): col 4 = 1.
Column 4 now shows 3, 1, 4, leaving only one digit for the bottom-right square: 2.
Transfers everywhere: in any Latin-square / Sudoku cell, when three of four values are accounted for in a line, the fourth is forced — chase eliminations, not the whole grid.
Jorge's teacher asks him to plot all the ordered pairs (w, l) of positive integers for which w is the width and l is the length of a rectangle with area 12. What should his graph look like?
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Answer: A — Graph A.
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Hint 1 of 2
Don't reason about the graph abstractly — just LIST the actual rectangles. Which whole-number widths and lengths multiply to 12?
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Hint 2 of 2
Two clues separate the choices: (1) only whole numbers work, so you get a few separate dots, not a solid line; (2) wl = 12 means as w grows, l shrinks — an inverse relationship, which curves rather than going straight.
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Approach: list the integer points and read their shape
Find every pair of positive integers multiplying to 12: (1,12), (2,6), (3,4), (4,3), (6,2), (12,1). That's six separate dots — not a continuous line.
As w goes 1, 2, 3, 4, 6, 12 the length l drops 12, 6, 4, 3, 2, 1: a falling pattern, but the drops get smaller and smaller (a curve that bends, not a straight slant).
Six dots, falling and curving — that's graph A.
How to rule out look-alikes: a straight falling line (C) would need equal-sized drops, but inverse proportion (l = 12/w) bends. Constant l (D) or rising dots (B) ignore the rule entirely. Listing the points makes the right shape obvious.
How many two-digit numbers have digits whose sum is a perfect square?
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Answer: C — 17.
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Hint 1 of 2
Flip the question: instead of testing all 90 two-digit numbers, first ask which digit SUMS are even allowed. The smallest sum is 1 (from 10) and the largest is 18 (from 99), so only the perfect squares 1, 4, 9, 16 can occur.
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Hint 2 of 2
Now handle each target sum separately. The tens digit must be 1–9 (no leading zero), but the units digit may be 0–9 — that asymmetry is what makes the counts uneven, so watch the edges.
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Approach: first pin down the possible digit sums, then count each case
A two-digit number's digit sum runs from 1 (for 10) up to 18 (for 99). The perfect squares in that window are 1, 4, 9, and 16 — only four cases to check.
Sum = 16: 79, 88, 97 ⇒ 3 numbers (only these fit, since each digit caps at 9 — big sums leave very few splits).
Total: 1 + 4 + 9 + 3 = 17.
Why the counts shrink at the top: for sum 16 the digits are nearly maxed out (each at most 9), so few splits fit; for middling sums like 9 there's lots of freedom. Narrowing to the four legal sums first is what keeps this from being a 90-number slog.
You'll never compute 132012 — and you don't have to. When you multiply, only the units digit of each factor affects the units digit of the answer, so 132012 ends in the same digit as 32012.
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Hint 2 of 2
Now list a few powers of 3 and watch the last digit: 3, 9, 7, 1, then 3 again. It cycles with period 4 — so the answer depends only on where 2012 lands in that cycle of 4.
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Approach: units digits repeat in a short cycle — find the position
The units digit of a product depends only on the units digits being multiplied, so 132012 ends in the same digit as 32012. The big base is irrelevant.
List the last digits of powers of 3: 31→3, 32→9, 33→7, 34→1, 35→3… They loop every 4 powers: (3, 9, 7, 1).
So divide the exponent by 4 and look at the leftover: 2012 = 4 × 503 with leftover 0, meaning we land exactly on the end of a cycle — the 4th spot, which is 1.
This transfers to every units-digit problem: last digits always cycle (period 1, 2, or 4 for single digits); find the cycle length, then reduce the exponent by that length. A leftover of 0 lands on the last entry, not the first.
The lengths of the sides of a triangle in inches are three consecutive integers. The length of the shortest side is 30% of the perimeter. What is the length of the longest side?
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Answer: E — 11 inches.
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Hint 1 of 2
Consecutive integers are evenly spaced, so the middle side is exactly the average — meaning the perimeter is 3 times the middle side. That single fact replaces a lot of algebra.
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Hint 2 of 2
When three numbers are evenly spaced, their sum is 3 × the middle one. Now ‘shortest = 30% of perimeter’ becomes a clean comparison.
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Approach: use the evenly-spaced middle term
Sides are m−1, m, m+1 around a middle value m. Their sum (the perimeter) is exactly 3m.
The shortest side is 30% of the perimeter: m−1 = 0.30 · 3m = 0.9m. So 0.1m = 1, giving m = 10.
Longest = m+1 = 11.
Why this transfers: for any evenly-spaced list, swapping in ‘sum = (count) × middle’ collapses messy sums to one variable. It's the same idea behind averaging a run of consecutive numbers.
Another way — test the answer choices:
Longest options are 7…11; try the longest = 11, so sides are 9, 10, 11 with perimeter 30.
Is the shortest 30% of 30? 0.30 · 30 = 9 = shortest. It fits, so the longest side is 11 — a fast check when you'd rather verify than solve.
◇ and △ are whole numbers and ◇ × △ = 36. The largest possible value of ◇ + △ is
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Answer: E — 37.
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Hint 1 of 2
With the product locked at 36, you only get to choose which factor pair to use. To make the sum as *big* as possible, do you want the two numbers close together or far apart?
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Hint 2 of 2
For a fixed product, the sum grows as the factors spread apart and shrinks as they bunch up. So go straight for the most lopsided pair.
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Approach: for a fixed product, spread the factors apart
The factor pairs of 36 are (1,36), (2,18), (3,12), (4,9), (6,6), with sums 37, 20, 15, 13, 12. The most spread-out pair, 1 and 36, gives the biggest sum.
1 + 36 = 37.
Why this transfers: when a product is fixed, balanced factors (like 6 × 6) give the *smallest* sum and the most lopsided factors (1 × 36) give the *largest*. Knowing which end you want lets you jump to the answer without testing every pair.
Number TheoryCounting & ProbabilityArithmetic & Operationssymmetryorganizing-datalogical-reasoning
Find the digit-sum of every number from 1 to 999, then add all those digit-sums together. (The digit-sum of 254 is \(2 + 5 + 4 = 11\).) What is the grand total?
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Answer: 13,500
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Hint 1 of 4
Don't add number by number. Count how many times each digit 1, 2, ..., 9 appears in total across 1 to 999. (Zeros add nothing.)
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Hint 2 of 4
By symmetry, every nonzero digit appears the exact same number of times. So just count how often, say, the digit 3 shows up.
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Hint 3 of 4
Count the digit 3 in the ones place, the tens place, and the hundreds place separately. In each place it appears 100 times.
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Approach: Count digit appearances by symmetry
Adding digit-sums is the same as counting how often each digit appears, weighted by its value. Zeros add nothing, so only digits 1 through 9 matter, and by symmetry each appears equally often.
Count the digit 3 across 1 to 999 (think 000 to 999, three places): ones place 100 times, tens place 100 times, hundreds place 100 times — so 300 times total.
Every nonzero digit likewise appears 300 times, so total \(= 300 (1 + 2 + \cdots + 9) = 300 \times 45 = 13{,}500\).
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?
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Answer: E — 89 students.
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Hint 1 of 2
The 8th-graders appear in both ratios — they're the shared hinge. In ratio 5:3 their count is a multiple of 5; in ratio 8:5 it's a multiple of 8. Make those two pictures agree first.
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Hint 2 of 2
To merge two ratios that share a quantity, force the shared term to a common value — the smallest is the LCM. Then every group count comes out whole.
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Approach: make the shared 8th-grader count an LCM
8th-graders link both ratios, so their count must be a multiple of 5 (from 5:3) and of 8 (from 8:5). Smallest such count = lcm(5, 8) = 40 — pick that to keep everyone whole.
6th-graders: 40 × 3/5 = 24 (from 5:3).
7th-graders: 40 × 5/8 = 25 (from 8:5).
Total = 40 + 24 + 25 = 89.
Why this transfers: any "chain" of ratios sharing a common term is stitched together by setting that term to the LCM — the same trick scales recipe ratios, gear ratios, and unit conversions.
Each corner cut takes 5 off both ends of a side, so the base shrinks by 10 in each direction: 20−10 and 30−10. The 5-wide flaps that fold up become the box's height of 5.
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Hint 2 of 2
An open box has a bottom and 4 walls but NO lid. Add those five faces; don't include a top.
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Approach: unfold the box into bottom + four walls (no top)
Cutting a 5×5 square from each corner removes 5 from both ends of every side, so the base is (20−5−5) × (30−5−5) = 10 × 20 = 200. The flaps that fold upward are 5 tall, so the box height is 5.
Four walls: two are 10×5 and two are 20×5, totaling 2(50) + 2(100) = 300. The interior is the bottom plus those four walls (no lid): 200 + 300 = 500.
Trap to dodge: 'open box' means skip the top — count 5 faces, not 6. And remember the corner cuts subtract from both ends, so the base loses 2×5 in each direction, not just 5. Choice E (1000) is what you'd get from the full original 20×30 sheet's worth of double-counting.
The four L-shapes plus the center square fill the whole big square. So instead of measuring the center directly, find what FRACTION of the square the four L's eat up — the center is whatever fraction is left.
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Hint 2 of 2
Work in area first; only turn area into side length at the very end with a square root.
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Approach: leftover area (whole − the four L's), then square-root to a side
Don't try to read the center square's side off the picture. Use the whole: the four L-regions and the center together make the full square (fraction 1). Each L is 316, so four of them cover 4 × 316 = 1216 = 34.
That leaves the center square as 1 − 34 = 14 of the area.
The big square is 100 × 100 = 10000 sq in, so the center is 14 × 10000 = 2500 sq in. Its side is √2500 = 50 inches.
Why this transfers: for 'what's left in the middle' figures, fractions of area beat measuring lengths — and a center square that's exactly 14 the area has side exactly 12 the big side (since √(1/4) = 1/2). Sure enough, 50 is half of 100.
The average (arithmetic mean) of 10 different positive whole numbers is 10. The largest possible value of any of these numbers is
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Answer: C — 55.
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Hint 1 of 3
Average 10 over 10 numbers means a FIXED total — the ten numbers always add to 100, a fixed pie. To give one number the biggest possible slice, what should the other nine slices be?
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Hint 2 of 3
The total is locked at 100. One number is huge only if the other nine are as TINY as possible — and they must be different positive whole numbers. What are the nine smallest such numbers?
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Hint 3 of 3
The nine smallest different positive whole numbers are 1, 2, 3, …, 9. Take their sum away from 100.
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Approach: fix the total, then starve the other nine numbers
Average 10 across 10 numbers means the total is fixed: 10 × 10 = 100. With a fixed total, one number is biggest exactly when the rest are smallest.
The nine others must be different positive whole numbers, so the smallest they can be is 1, 2, 3, …, 9. That sum is (9 × 10) ÷ 2 = 45.
The largest number takes whatever's left: 100 − 45 = 55.
Why this transfers: a fixed average is a fixed total — a budget. To maximize one part of a fixed budget, push every other part to its allowed minimum (and use the "different" rule to make them 1, 2, 3, …). The same "minimize the rest" move solves countless extremal problems.
Trap to dodge: 91 (= 100 − 9, leaving 1's) ignores "different"; the values must be distinct, forcing 1 through 9, not nine copies of 1.
A 1 × 2 rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?
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Answer: C — π.
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Hint 1 of 2
By symmetry, the center of the semicircle sits at the middle of the diameter — right below the rectangle's center. A radius drawn to an upper corner is the hypotenuse of a little right triangle. What are its two legs?
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Hint 2 of 2
Don't chase the radius directly — build a right triangle from the center to a point on the circle and use the Pythagorean theorem. Also note area only needs r2, so you never have to simplify √2.
Show solution
Approach: right triangle from the center to a top corner
Put the center at the midpoint of the diameter. The rectangle (long side 2 on the diameter, height 1) reaches a top corner that is 1 across and 1 up from the center.
That corner lies on the circle, so the radius is its distance: r2 = 12 + 12 = 2. (No need to take the square root.)
Semicircle area = ½πr2 = ½π(2) = π.
Worth keeping: area formulas use r2, so stop the moment you know r2 — squaring back a messy radius is wasted work.
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?
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Answer: E — 18 routes.
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Hint 1 of 2
The forced diagonal through the park splits the trip into three separate legs that don't interfere with each other. Count each leg's shortest routes on its own, then combine — how do independent choices multiply?
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Hint 2 of 2
A shortest grid route is just an arrangement of fixed moves (so many easts, so many norths). Number of routes = C(total steps, steps in one direction). And independent stages multiply.
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Approach: count each leg, then multiply (independent stages)
Leg 1, home → SW corner: 2 easts and 1 north in some order. Choosing where the single north goes among 3 steps gives C(3, 1) = 3 routes.
Leg 2, the diagonal through the park: a single forced path = 1 way (this is what cleanly separates the two grids).
Leg 3, NE corner → school: 2 easts and 2 norths, C(4, 2) = 6 routes.
The legs are independent, so multiply: 3 × 1 × 6 = 18.
You'll see this again: shortest-path counts on a grid are always "choose which steps are north," and independent stages of a journey multiply — the multiplication principle.
Tom's Hat Shoppe increased all original prices by 25%. Now the shoppe is having a sale where all prices are 20% off these increased prices. Which statement best describes the sale price of an item?
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Answer: E — The sale price is the same as the original price.
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Hint 1 of 2
The 20% off isn't taken off the original price — it's taken off the *already-raised* price, so the two percents can't simply cancel. Turn each change into a 'multiply by' factor instead.
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Hint 2 of 2
Up 25% means ×1.25; then 20% off means ×0.80. Multiply those two factors together — what do you get?
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Approach: turn each percent change into a multiplier and multiply
Raising by 25% multiplies the price by 1.25; taking 20% off multiplies by 0.80. Doing both means 1.25 × 0.80 = 1.00 — so the final price equals the original. The sale price is the same as the original price.
Concrete check: start at $100 → up 25% → $125 → 20% off $125 is $25 off → $100. Right back where we started.
Trap to avoid: +25% then −20% does NOT make +5%. The 20% is a slice of the larger $125, not of the original $100, so the bigger 'off' exactly undoes the increase. Percent changes combine by multiplying their factors, never by adding the percents.
Speed = distance ÷ time, and every choice covers the same one hour. So 'fastest hour' just means 'the hour where the distance climbed the most' — you're comparing rises, not doing division.
Still stuck? Show hint 2 →
Hint 2 of 2
On a distance–time graph, the steeper the line over an hour, the faster the plane went that hour. Don't read exact numbers — just eyeball where the curve shoots up most sharply.
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Approach: steepest one-hour segment = fastest hour (read the slope by eye)
Average speed over an hour = (miles gained) ÷ (1 hour) = the miles gained that hour. Since every option is a 1-hour stretch, the fastest hour is simply the one with the biggest jump in distance — the steepest part of the graph.
Scan the curve: between hours 1 and 2 it leaps from about 400 to about 900 miles — roughly a 500-mile climb, clearly steeper than any other single hour (the later hours flatten out).
So the largest average speed is during the second hour (1-2).
*Worth keeping:* on a distance–time graph, steepness *is* speed — comparing slopes by eye beats computing every segment, and a flattening curve means slowing down.
Every triangle on the top half lands on exactly one on the bottom — nothing vanishes. So just track one half and watch each color get used up.
Still stuck? Show hint 2 →
Hint 2 of 2
Work color by color: count how many reds and blues are "spent" by the given pairs, see what's forced to pair with white, and the leftover whites pair with each other.
Show solution
Approach: account for each color on one half (conservation)
One half has 3 red, 5 blue, 8 white. The 2 red-red pairs spend 2 reds, leaving 1 red; the 3 blue-blue pairs spend 3 blues, leaving 2 blue.
Now the leftovers must pair with whites. The 2 red-white pairs spend that last red and 1 white. The 2 leftover blues can't make a 4th blue-blue pair (only 3 are allowed), so each pairs with a white — using 2 more whites.
Whites spent: 1 (with red) + 2 (with blue) = 3, leaving 8 − 3 = 5 whites per half. Those face each other as 5 white-white pairs.
The engine here is conservation: a folded figure pairs everything one-to-one, so once you know how each "colored" piece is consumed, the remainder of any one color is forced.
Which of the following sets of whole numbers has the largest average?
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Answer: D — multiples of 5 between 1 and 101.
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Hint 1 of 3
You do NOT have to add up dozens of numbers. Each list is evenly spaced, so its values are perfectly balanced around their middle — the average is just the midpoint of the first and last terms.
Still stuck? Show hint 2 →
Hint 2 of 3
So for each set you only need two numbers: its smallest multiple and its largest multiple (up to 100). Average = (first + last) ⁄ 2.
Still stuck? Show hint 3 →
Hint 3 of 3
The winner will tend to be the one whose last term reaches closest to the top, 100.
Show solution
Approach: average of an evenly-spaced list = (first + last) ⁄ 2
Each set is an evenly-spaced (arithmetic) list, and such lists are symmetric about their center — every term below the middle is matched by one equally far above. So the average equals the midpoint of the first and last terms; no long addition needed.
Just grab first and last of each: 2's → (2 + 100)⁄2 = 51; 3's → (3 + 99)⁄2 = 51; 4's → (4 + 100)⁄2 = 52; 5's → (5 + 100)⁄2 = 52.5; 6's → (6 + 96)⁄2 = 51.
The largest is 52.5, from the multiples of 5.
Why 5 wins: its last multiple under 101 is 100 (hitting the ceiling) while its first term, 5, is small — so its first-and-last midpoint sits highest. Whenever you must average an evenly spaced list, reach for the (first + last)⁄2 shortcut.
