AJHSME · Test Mode

1986 AJHSME

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Problem 1 · 1986 AJHSME Easy
Ratios, Rates & Proportions per-unit

In July 1861, 366 inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?

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Answer: A — 366 ⁄ (31 × 24).
Show hints
Hint 1 of 2
Hours in July = 31 days × 24 hours/day.
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Hint 2 of 2
Average per hour = total ÷ hours.
Show solution
Approach: divide total by hours
  1. July has 31 × 24 hours, so the per-hour average is total inches divided by total hours.
  2. = 366 ⁄ (31 × 24).
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Problem 2 · 1986 AJHSME Easy
Fractions, Decimals & Percents reciprocal-monotonicity

Which of the following numbers has the largest reciprocal?

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Answer: A — 1⁄3.
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Hint 1
Smaller positive number ↔ larger reciprocal.
Show solution
Approach: smallest positive wins
  1. Among positives, the reciprocal is largest for the smallest number.
  2. 1⁄3 is the smallest of the choices, so its reciprocal (3) is largest.
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Problem 3 · 1986 AJHSME Easy
Arithmetic & Operations pick-smallest

The smallest sum one could get by adding three different numbers from the set {7, 25, −1, 12, −3} is

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Answer: C — 3.
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Hint 1
Pick the three smallest numbers in the set.
Show solution
Approach: sum the three smallest
  1. The three smallest are −3, −1, and 7.
  2. Their sum: −3 + (−1) + 7 = 3.
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Problem 4 · 1986 AJHSME Easy
Arithmetic & Operations round-and-multiply

The product (1.8)(40.3 + .07) is closest to

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Answer: C — 74.
Show hint
Hint 1
The small .07 barely changes the sum; round 40.37 ≈ 40.
Show solution
Approach: round before multiplying
  1. 1.8 × 40.37 ≈ 1.8 × 40 = 72.
  2. Closest of the choices is 74.
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Problem 5 · 1986 AJHSME Easy
Number Theory time-arithmetic

A contest began at noon one day and ended 1000 minutes later. At what time did the contest end?

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Answer: D — 4:40 a.m.
Show hints
Hint 1 of 2
Convert 1000 minutes to hours and minutes.
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Hint 2 of 2
1000 = 16·60 + 40.
Show solution
Approach: convert to hh:mm and add
  1. 1000 minutes = 16 hours 40 minutes.
  2. Noon + 16 h 40 m = 4:40 a.m. (the next day).
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Problem 6 · 1986 AJHSME Medium
Fractions, Decimals & Percents simplify-complex-fraction

2 ⁄ (1 − 2⁄3) =

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Answer: E — 6.
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Hint 1 of 2
Simplify the denominator first.
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Hint 2 of 2
1 − 2⁄3 = 1⁄3.
Show solution
Approach: simplify denominator, then divide
  1. Denominator: 1 − 2⁄3 = 1⁄3.
  2. 2 ÷ 1⁄3 = 2 × 3 = 6.
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Problem 7 · 1986 AJHSME Medium
Number Theory bound-square-roots

How many whole numbers are between √8 and √80?

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Answer: B — 6.
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Hint 1 of 2
√8 sits between 2 and 3; √80 sits between 8 and 9.
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Hint 2 of 2
Count the integers strictly between.
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Approach: bound each square root by the nearest perfect squares
  1. 4 < 8 < 9, so 2 < √8 < 3. 64 < 80 < 81, so 8 < √80 < 9.
  2. Whole numbers strictly between √8 and √80 are 3, 4, 5, 6, 7, 8 — that's 6.
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Problem 8 · 1986 AJHSME Medium
Number Theory last-digitguess-and-check
ajhsme-1986-08
Show answer
Answer: E — 8.
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Hint 1 of 2
The ones digit of the product is 2 · B, which must end in 6.
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Hint 2 of 2
Then check that B2 × 7B actually equals 6396.
Show solution
Approach: ones-digit clue then verify
  1. Product ends in 6, and 2 × B ends in 6 only for B = 3 or B = 8.
  2. Test: 82 × 78 = 6396. ✓ So B = 8.
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Problem 9 · 1986 AJHSME Medium
Counting & Probability count-pathsdirected-graph
ajhsme-1986-09
Show answer
Answer: E — 6.
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Hint 1 of 2
Trace each path from M to N, following arrow directions only.
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Hint 2 of 2
From each split, enumerate where you can go next.
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Approach: enumerate all directed paths
  1. Following the arrows carefully and listing each distinct path from M to N gives 6 routes.
  2. Answer: 6.
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Problem 10 · 1986 AJHSME Medium
Geometry & Measurement centering

A picture 3 feet across is hung in the center of a wall that is 19 feet wide. How many feet from the end of the wall is the nearest edge of the picture?

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Answer: B — 8.
Show hint
Hint 1
The leftover wall is split evenly on both sides of the picture.
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Approach: split the leftover
  1. Leftover wall = 19 − 3 = 16 feet, split into two equal margins.
  2. Each margin = 16 ⁄ 2 = 8 feet.
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Problem 11 · 1986 AJHSME Hard
Algebra & Patterns custom-operation

If A ✶ B means (A + B) ⁄ 2, then (3 ✶ 5) ✶ 8 is

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Answer: A — 6.
Show hints
Hint 1 of 2
Apply the inner operation first, then the outer.
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Hint 2 of 2
3 ✶ 5 is the average of 3 and 5.
Show solution
Approach: apply the operation twice
  1. 3 ✶ 5 = (3 + 5)⁄2 = 4. Then 4 ✶ 8 = (4 + 8)⁄2 = 6.
  2. = 6.
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Problem 12 · 1986 AJHSME Hard
Fractions, Decimals & Percents diagonal-sum
ajhsme-1986-12
Show answer
Answer: D — 40%.
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Hint 1 of 2
Students with the same grade on both tests sit on the table's diagonal (A-A, B-B, …).
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Hint 2 of 2
Add those five diagonal cells, then divide by 30.
Show solution
Approach: sum the diagonal
  1. Diagonal: 2 (A-A) + 4 (B-B) + 5 (C-C) + 1 (D-D) + 0 (F-F) = 12.
  2. 12 ⁄ 30 = 40%.
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Problem 13 · 1986 AJHSME Hard
Geometry & Measurement perimeter-equals-bounding-box
ajhsme-1986-13
Show answer
Answer: C — 28.
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Hint 1 of 2
The inner step has matching horizontal and vertical pieces that line up with the outer corners — so the perimeter equals that of the bounding rectangle.
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Hint 2 of 2
Bounding rectangle: 8 wide, 6 tall.
Show solution
Approach: slide step pieces — perimeter unchanged
  1. Slide each piece of the inward step to the matching outer edge: the horizontal step segment fits onto the top edge, the vertical step segment onto the right edge.
  2. Perimeter = 2(8 + 6) = 28.
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Problem 14 · 1986 AJHSME Hard
Fractions, Decimals & Percents maximize-fraction

If 200 ≤ a ≤ 400 and 600 ≤ b ≤ 1200, then the largest value of the quotient b ⁄ a is

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Answer: C — 6.
Show hint
Hint 1
Maximize a fraction by maximizing the numerator and minimizing the denominator.
Show solution
Approach: biggest top, smallest bottom
  1. Take b = 1200 (the maximum) and a = 200 (the minimum).
  2. b ⁄ a = 1200 ⁄ 200 = 6.
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Problem 15 · 1986 AJHSME Hard
Fractions, Decimals & Percents multiply-discount-factors

Sale prices at the Ajax Outlet Store are 50% below original prices. On Saturdays an additional discount of 20% off the sale price is given. What is the Saturday price of a coat whose original price is $180?

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Answer: B — $72.
Show hint
Hint 1
Apply each discount as a multiplier: 0.5 then 0.8.
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Approach: multiply the two discount factors
  1. Sale price = $180 × 0.5 = $90. Saturday price = $90 × 0.8.
  2. = $72.
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Problem 16 · 1986 AJHSME Hard
Fractions, Decimals & Percents percent-of-total
ajhsme-1986-16
Show answer
Answer: A — 2.5.
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Hint 1 of 2
Fall = 25% of total means total = 4 × Fall.
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Hint 2 of 2
Read Spring, Summer, Fall from the graph and subtract from total.
Show solution
Approach: solve for total, then subtract
  1. Fall ≈ 4 million, and fall is 25%, so total = 4 × 4 = 16 million. Spring ≈ 4.5, Summer ≈ 5.
  2. Winter = 16 − 4.5 − 5 − 4 = 2.5 million.
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Problem 17 · 1986 AJHSME Hard
Number Theory factor-outparity-of-product

Let o be an odd whole number and let n be any whole number. Which of the following statements about the whole number (o² + no) is always true?

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Answer: E — it is odd only if n is even.
Show hints
Hint 1 of 2
Factor: o² + no = o(o + n).
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Hint 2 of 2
o is odd, so the product is odd iff (o + n) is odd, which means n is even.
Show solution
Approach: factor and track parity
  1. o(o + n) is odd ⇔ both factors odd. o is already odd, so the product is odd iff o + n is odd, i.e. iff n is even.
  2. So the expression is odd only if n is even — answer E.
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Problem 18 · 1986 AJHSME Hard
Counting & Probability posts-on-path
ajhsme-1986-18
Show answer
Answer: B — 12.
Show hints
Hint 1 of 2
Place the long 60 m side against the wall so the fence runs 36 + 60 + 36 = 132 m.
Still stuck? Show hint 2 →
Hint 2 of 2
Posts every 12 m on a 132 m path including both ends = 132⁄12 + 1.
Show solution
Approach: lay out the three-sided fence and count posts
  1. With the 60 m side along the wall, the fence has length 36 + 60 + 36 = 132 m. Posts every 12 m including both endpoints give 132⁄12 + 1 = 12 posts; the corners (at 36 m and 96 m) are multiples of 12, so no extras.
  2. Fewest = 12 posts.
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Problem 19 · 1986 AJHSME Hard
Ratios, Rates & Proportions miles-per-gallongas-used-only

At the beginning of a trip, the mileage odometer read 56,200 miles. The driver filled the gas tank with 6 gallons of gasoline. During the trip, the driver filled his tank again with 12 gallons of gasoline when the odometer read 56,560. At the end of the trip, the driver filled his tank again with 20 gallons of gasoline. The odometer read 57,060. To the nearest tenth, what was the car's average miles-per-gallon for the entire trip?

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Answer: D — 26.9.
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Hint 1 of 2
The first 6-gallon fill-up only tops off the tank — gas used during the trip is what was refilled afterwards.
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Hint 2 of 2
Miles ÷ gallons used.
Show solution
Approach: miles ÷ (gas used to refill)
  1. Miles driven = 57,060 − 56,200 = 860. Gas used (the two refills) = 12 + 20 = 32.
  2. MPG = 860 ⁄ 32 ≈ 26.9.
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Problem 20 · 1986 AJHSME Hard
Fractions, Decimals & Percents round-then-estimate

The value of the expression (304)⁵ ⁄ ((29.7)(399)⁴) is closest to

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Answer: D — 3.
Show hints
Hint 1 of 2
Round 304 ≈ 300, 399 ≈ 400, 29.7 ≈ 30.
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Hint 2 of 2
Then (3⁄4)⁴ × 300⁄30 simplifies the powers.
Show solution
Approach: round and pull out a power of 3⁄4
  1. Approximate: 300⁵ ⁄ (30 · 400⁴) = (300⁄400)⁴ · 300⁄30 = (3⁄4)⁴ · 10 = (81⁄256) · 10 ≈ 3.16.
  2. Closest choice: 3.
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Problem 21 · 1986 AJHSME Stretch
Geometry & Measurement net-of-topless-cube
ajhsme-1986-21
Show answer
Answer: E — 6.
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Hint 1 of 2
A topless cube has 5 faces, so the T plus one lettered square gives the right count.
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Hint 2 of 2
Eliminate the lettered squares that would put two faces on the same side of the resulting box.
Show solution
Approach: test each added square
  1. Each added square turns the T into a 5-square net of a topless cube unless it ends up sharing a face position with another square after folding.
  2. Only 2 of the 8 choices create that overlap, so 6 of the 8 fold into a topless cubical box.
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Problem 22 · 1986 AJHSME Stretch
Logic & Word Problems implication-chain

Alan, Beth, Carlos, and Diana were discussing their possible grades in mathematics class this grading period. Alan said, "If I get an A, then Beth will get an A." Beth said, "If I get an A, then Carlos will get an A." Carlos said, "If I get an A, then Diana will get an A." All of these statements were true, but only two of the students received an A. Which two received A's?

Show answer
Answer: C — Carlos, Diana.
Show hints
Hint 1 of 2
Each statement says: A getting an A forces the next person to get one too.
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Hint 2 of 2
If Alan got an A, the chain would force at least three more A's; same for Beth.
Show solution
Approach: chase the chain backwards
  1. Alan's getting an A would force Beth, then Carlos, then Diana — 4 A's, too many. Beth's getting an A would force Carlos and Diana — 3 A's, still too many. So Alan and Beth don't get A's.
  2. If Carlos gets an A, Diana must too — and that's exactly 2 A's: Carlos and Diana.
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Problem 23 · 1986 AJHSME Stretch
Geometry & Measurement area-differencehalf-circles
ajhsme-1986-23
Show answer
Answer: B — 1.
Show hints
Hint 1 of 2
Compare the upper half of the big circle to the upper halves of the two small circles.
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Hint 2 of 2
Big radius = 2, small radius = 1.
Show solution
Approach: shaded = half big disk − two half small disks
  1. Big radius = 2 (since two small circles of radius 1 fit on AC). Shaded = ½(π · 2²) − 2 · ½(π · 1²) = 2π − π = π.
  2. Ratio to one small circle's area (π · 1² = π) is π ⁄ π = 1.
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Problem 24 · 1986 AJHSME Stretch
Counting & Probability fix-one-condition-rest

The 600 students at King Middle School are divided into three groups of equal size for lunch. Each group has lunch at a different time. A computer randomly assigns each student to one of three lunch groups. The probability that three friends, Al, Bob, and Carol, will be assigned to the same lunch group is approximately

Show answer
Answer: B — 1⁄9.
Show hints
Hint 1 of 2
Fix Al's group, then ask the chance Bob and Carol each land in the same one.
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Hint 2 of 2
Each independently lands in any group with chance ≈ 1⁄3.
Show solution
Approach: condition on Al's group
  1. Whatever group Al is in, Bob lands there with chance ≈ 1⁄3 and Carol independently with chance ≈ 1⁄3.
  2. Combined: 1⁄3 × 1⁄3 = 1⁄9.
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Problem 25 · 1986 AJHSME Stretch
Number Theory average-of-arithmetic-progression

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.
Show hints
Hint 1 of 2
For an arithmetic progression, the average is just (first + last)⁄2.
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Hint 2 of 2
Compare each set's first and last terms.
Show solution
Approach: (first + last) ⁄ 2
  1. Averages: multiples of 2 → (2 + 100)⁄2 = 51; of 3 → (3 + 99)⁄2 = 51; of 4 → (4 + 100)⁄2 = 52; of 5 → (5 + 100)⁄2 = 52.5; of 6 → (6 + 96)⁄2 = 51.
  2. Largest is 52.5 — multiples of 5.
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