AJHSME

1988 AJHSME

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

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Problem 1 · 1988 AJHSME Easy
Fractions, Decimals & Percents scale-reading
ajhsme-1988-01
Show answer
Answer: C — 10.25.
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Hint 1 of 2
Between the 10 and 11 marks, the arrow points roughly a quarter of the way past 10.
Still stuck? Show hint 2 →
Hint 2 of 2
A quarter of the gap from 10 to 11 is 0.25.
Show solution
Approach: estimate the fraction of the gap
  1. The arrow sits about one-fourth of the way from 10 toward 11.
  2. That's 10 + 0.25 = 10.25.
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Problem 2 · 1988 AJHSME Easy
Fractions, Decimals & Percents pair-decimals

The product 8 × .25 × 2 × .125 =

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Answer: C — 1⁄2.
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Hint 1 of 2
Pair numbers whose product is a round number.
Still stuck? Show hint 2 →
Hint 2 of 2
8 · 0.125 = 1 and 2 · 0.25 = 0.5.
Show solution
Approach: pair friendly factors
  1. Rearrange: (8 × 0.125) × (2 × 0.25) = 1 × 0.5.
  2. Product = 1⁄2.
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Problem 3 · 1988 AJHSME Easy
Fractions, Decimals & Percents simplify-each-fraction

1⁄10 + 2⁄20 + 3⁄30 =

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Answer: D — .3.
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Hint 1 of 2
Each fraction simplifies to the same thing.
Still stuck? Show hint 2 →
Hint 2 of 2
1⁄10 = 2⁄20 = 3⁄30.
Show solution
Approach: simplify each term
  1. 2⁄20 = 1⁄10 and 3⁄30 = 1⁄10, so the sum is 1⁄10 + 1⁄10 + 1⁄10 = 3⁄10.
  2. 3⁄10 = 0.3.
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Problem 4 · 1988 AJHSME Medium
Counting & Probability count-by-rowsymmetry
ajhsme-1988-04
Show answer
Answer: E — 11.
Show hints
Hint 1 of 2
The figure is symmetric top to bottom — count one half and double, watching the shared middle row.
Still stuck? Show hint 2 →
Hint 2 of 2
In each row, count dark squares minus light squares; sum over all rows.
Show solution
Approach: tally dark − light row by row
  1. In each row, dark squares outnumber light squares by exactly one because the row starts and ends with dark.
  2. There are 11 rows in the diamond, so dark exceeds light by 11.
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Problem 5 · 1988 AJHSME Medium
Geometry & Measurement protractor-readingangle-subtraction
ajhsme-1988-05
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Answer: C — 50°.
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Hint 1 of 2
The protractor reads A at 20° and D at 160°.
Still stuck? Show hint 2 →
Hint 2 of 2
If ∠CBD = 90°, then C reads at 160° − 90° = 70°.
Show solution
Approach: subtract protractor readings
  1. D reads 160°; since ∠CBD = 90°, C reads 160° − 90° = 70°.
  2. ∠ABC = 70° − 20° = 50°.
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Problem 6 · 1988 AJHSME Medium
Fractions, Decimals & Percents ratio-of-powers

(.2)³ ⁄ (.02)² =

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Answer: E — 20.
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Hint 1 of 2
Factor out (.2)² from the numerator first.
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Hint 2 of 2
(.2 ⁄ .02)² = 10².
Show solution
Approach: pull out a (.2)² and use .2⁄.02 = 10
  1. (.2)³ ⁄ (.02)² = .2 × (.2⁄.02)² = .2 × 10² = .2 × 100.
  2. = 20.
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Problem 7 · 1988 AJHSME Medium
Arithmetic & Operations round-and-multiply

2.46 × 8.163 × (5.17 + 4.829) is closest to

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Answer: B — 200.
Show hints
Hint 1 of 2
Round each factor to a friendly number.
Still stuck? Show hint 2 →
Hint 2 of 2
2.46 ≈ 2.5, 8.163 ≈ 8, the sum ≈ 10.
Show solution
Approach: round to easy numbers
  1. 2.46 × 8.163 × (5.17 + 4.829) ≈ 2.5 × 8 × 10.
  2. = 200.
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Problem 8 · 1988 AJHSME Medium
Fractions, Decimals & Percents count-decimal-places

Betty used a calculator to find the product 0.075 × 2.56. She forgot to enter the decimal points. The calculator showed 19200. If Betty had entered the decimal points correctly, the answer would have been

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Answer: B — .192.
Show hints
Hint 1 of 2
Count the total number of digits after the decimal point in the two factors.
Still stuck? Show hint 2 →
Hint 2 of 2
Move the decimal point in 19200 that many places to the left.
Show solution
Approach: count decimal places
  1. 0.075 has 3 decimal places, 2.56 has 2 — total 5. Move the decimal in 19200 five places to the left.
  2. 19200 → 0.19200 = 0.192.
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Problem 9 · 1988 AJHSME Medium
Geometry & Measurement distance-on-gridisosceles-check
ajhsme-1988-09
Show answer
Answer: D — 4.
Show hints
Hint 1 of 2
For each triangle, check whether at least two side lengths match — count squares horizontally, vertically, and diagonally using a²+b².
Still stuck? Show hint 2 →
Hint 2 of 2
Only one of the five fails the test.
Show solution
Approach: compare side lengths on each triangle
  1. Compute the three side lengths of each triangle from the grid (use a² + b² for slanted sides).
  2. Four of the five have two matching sides; only one is scalene, so 4 are isosceles.
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Problem 10 · 1988 AJHSME Medium
Number Theory mod-7

Chris's birthday is on a Thursday this year. What day of the week will it be 60 days after her birthday?

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Answer: A — Monday.
Show hints
Hint 1 of 2
Days of the week repeat every 7 days, so reduce 60 modulo 7.
Still stuck? Show hint 2 →
Hint 2 of 2
60 = 8·7 + 4.
Show solution
Approach: reduce mod 7
  1. 60 days = 8 full weeks plus 4 extra days.
  2. 4 days after Thursday → Friday, Saturday, Sunday, Monday.
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Problem 11 · 1988 AJHSME Hard
Number Theory bound-by-perfect-squares

√164 is

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Answer: E — between 12 and 13.
Show hints
Hint 1 of 2
Bound 164 between two consecutive perfect squares.
Still stuck? Show hint 2 →
Hint 2 of 2
12² = 144 and 13² = 169.
Show solution
Approach: squeeze between perfect squares
  1. 12² = 144 < 164 < 169 = 13².
  2. So √164 lies between 12 and 13.
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Problem 12 · 1988 AJHSME Hard
Fractions, Decimals & Percents cancel-powers-of-ten

Suppose the estimated 20 billion dollar cost to send a person to the planet Mars is shared equally by the 250 million people in the U.S. Then each person's share is

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Answer: C — 80 dollars.
Show hints
Hint 1 of 2
billion = 1000 × million, so 20 billion = 20,000 million.
Still stuck? Show hint 2 →
Hint 2 of 2
Divide 20,000 by 250.
Show solution
Approach: scale both numbers in millions
  1. 20 billion ⁄ 250 million = 20,000 million ⁄ 250 million = 20,000 ⁄ 250.
  2. = 80 dollars.
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Problem 13 · 1988 AJHSME Hard
Geometry & Measurement circumferenceapproximation

If rose bushes are spaced about 1 foot apart, approximately how many bushes are needed to surround a circular patio whose radius is 12 feet?

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Answer: D — 75.
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Hint 1 of 2
The number of bushes ≈ the circumference (in feet).
Still stuck? Show hint 2 →
Hint 2 of 2
Circumference = 2π · 12.
Show solution
Approach: use circumference = 2πr
  1. Circumference = 2π · 12 = 24π ≈ 75.4 feet.
  2. With 1-foot spacing, about 75 bushes fit around.
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Problem 14 · 1988 AJHSME Hard
Number Theory factor-pairsmax-sum

◇ and △ are whole numbers and ◇ × △ = 36. The largest possible value of ◇ + △ is

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Answer: E — 37.
Show hints
Hint 1 of 2
List all factor pairs of 36 and add each pair.
Still stuck? Show hint 2 →
Hint 2 of 2
Sums are biggest when the factors are most spread out.
Show solution
Approach: list factor pairs
  1. Factor pairs of 36: (1,36), (2,18), (3,12), (4,9), (6,6). Their sums: 37, 20, 15, 13, 12.
  2. The largest is 1 + 36 = 37.
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Problem 15 · 1988 AJHSME Hard
Fractions, Decimals & Percents add-fractions-then-flip

The reciprocal of (1⁄2 + 1⁄3) is

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Answer: C — 6⁄5.
Show hints
Hint 1 of 2
Add the fractions first, then flip.
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Hint 2 of 2
1⁄2 + 1⁄3 = 5⁄6.
Show solution
Approach: add then take reciprocal
  1. 1⁄2 + 1⁄3 = 3⁄6 + 2⁄6 = 5⁄6.
  2. Reciprocal of 5⁄6 is 6⁄5.
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Problem 16 · 1988 AJHSME Hard
Counting & Probability block-each-line
ajhsme-1988-16
Show answer
Answer: E — 6.
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Hint 1 of 2
There are 8 lines (3 rows, 3 columns, 2 diagonals); each must miss at least one square.
Still stuck? Show hint 2 →
Hint 2 of 2
Leaving the center plus two opposite corners empty kills every line.
Show solution
Approach: find a smallest set of empty squares that breaks every line
  1. Each of the 8 lines (3 rows, 3 cols, 2 diagonals) must contain at least one empty square. A single empty square covers at most 4 of those lines (only the center does that much), so 2 empties cover at most 4 + 3 = 7 lines — not enough. So at least 3 squares must be empty.
  2. Three is achievable: leave the center and two opposite corners empty. Center kills the middle row, middle column, and both diagonals; the two opposite corners kill the four remaining edge lines. So 9 − 3 = 6 X's can be placed.
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Problem 17 · 1988 AJHSME Hard
Geometry & Measurement inclusion-exclusion-area
ajhsme-1988-17
Show answer
Answer: B — 38.
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Hint 1 of 2
Add the two rectangles' areas, then subtract the overlap so it isn't counted twice.
Still stuck? Show hint 2 →
Hint 2 of 2
The overlap is a 3 × 2 rectangle.
Show solution
Approach: inclusion–exclusion on overlapping rectangles
  1. Horizontal rectangle: 10 × 2 = 20. Vertical rectangle: 3 × 8 = 24. Overlap: 3 × 2 = 6.
  2. Shaded area = 20 + 24 − 6 = 38.
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Problem 18 · 1988 AJHSME Hard
Ratios, Rates & Proportions weighted-average

The average weight of 6 boys is 150 pounds and the average weight of 4 girls is 120 pounds. The average weight of the 10 children is

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Answer: C — 138 pounds.
Show hints
Hint 1 of 2
Average isn't just the average of the two averages — weight by group size.
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Hint 2 of 2
Total weight ÷ total count.
Show solution
Approach: total weight ÷ total count
  1. Total weight = 6 × 150 + 4 × 120 = 900 + 480 = 1380.
  2. Average = 1380 ⁄ 10 = 138 pounds.
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Problem 19 · 1988 AJHSME Hard
Algebra & Patterns arithmetic-sequence-nth-term

What is the 100th number in the arithmetic sequence: 1, 5, 9, 13, 17, 21, 25, …?

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Answer: A — 397.
Show hints
Hint 1 of 2
The common difference is 4; the 100th term sits 99 steps after the first.
Still stuck? Show hint 2 →
Hint 2 of 2
100th term = 1 + 99 × 4.
Show solution
Approach: a + (n−1)d
  1. First term 1, common difference 4. The 100th term = 1 + 99 × 4 = 1 + 396.
  2. = 397.
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Problem 20 · 1988 AJHSME Hard
Fractions, Decimals & Percents percent-of-whole
ajhsme-1988-20
Show answer
Answer: C — 125.
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Hint 1 of 2
45 cups is 36% of the full capacity.
Still stuck? Show hint 2 →
Hint 2 of 2
Divide 45 by 0.36.
Show solution
Approach: divide by the percent
  1. If 45 = 0.36 · Full, then Full = 45 ⁄ 0.36 = 4500 ⁄ 36.
  2. = 125 cups.
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Problem 21 · 1988 AJHSME Stretch
Algebra & Patterns case-on-where-n-falls

A fifth number, n, is added to the set {3, 6, 9, 10} to make the mean of the set of five numbers equal to its median. The number of possible values of n is

Show answer
Answer: C — 3.
Show hints
Hint 1 of 2
The median depends on where n falls in the sorted list; split into three cases.
Still stuck? Show hint 2 →
Hint 2 of 2
In each case, set mean = median and solve for n.
Show solution
Approach: case on where n sits in the sorted list
  1. Mean = (28 + n)⁄5. Case n ≤ 6: median = 6 → n = 2. Case 6 ≤ n ≤ 9: median = n → 28 + n = 5n → n = 7. Case n ≥ 9: median = 9 → n = 17.
  2. Each value of n is consistent with its case, so the answers are 2, 7, 17 — 3 values.
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Problem 22 · 1988 AJHSME Stretch
Fractions, Decimals & Percents compound-percent

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.
Show hints
Hint 1 of 2
Apply each percent change as a multiplier and multiply them.
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Hint 2 of 2
1.25 × 0.80 — what does that equal?
Show solution
Approach: multiply the two factors
  1. +25% means ×1.25; −20% off means ×0.80. Combined: 1.25 × 0.80 = 1.00.
  2. So the sale price equals the original price.
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Problem 23 · 1988 AJHSME Stretch
Ratios, Rates & Proportions profit-per-item

Maria buys computer disks at a price of 4 for $5 and sells them at a price of 3 for $5. How many computer disks must she sell in order to make a profit of $100?

Show answer
Answer: D — 240.
Show hints
Hint 1 of 2
Cost per disk = $5⁄4; sale price per disk = $5⁄3.
Still stuck? Show hint 2 →
Hint 2 of 2
Profit per disk = $5⁄3 − $5⁄4.
Show solution
Approach: profit per disk × number = total profit
  1. Profit per disk = 5⁄3 − 5⁄4 = 20⁄12 − 15⁄12 = 5⁄12 dollar.
  2. To make $100: 100 ÷ (5⁄12) = 100 × 12⁄5 = 240 disks.
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Problem 24 · 1988 AJHSME Stretch
Geometry & Measurement roll-around-polygonrotation
ajhsme-1988-24
Show answer
Answer: A — A.
Show hints
Hint 1 of 2
When a square rolls around a hexagon's corner, it pivots through the hexagon's exterior angle.
Still stuck? Show hint 2 →
Hint 2 of 2
Total rotation from position 1 to position 4 = 3 × 60° = 180°.
Show solution
Approach: count pivots × exterior-angle rotation
  1. Each pivot at a hexagon corner rotates the square by the exterior angle 60°. Three pivots take it from the top to the bottom, so the square has rotated 3 × 60° = 180° clockwise.
  2. Rotating the original triangle by 180° gives the orientation in choice A.
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Problem 25 · 1988 AJHSME Stretch
Counting & Probability count-palindromescase-by-digit-count

A palindrome is a whole number that reads the same forwards and backwards. If one neglects the colon, certain times displayed on a digital watch are palindromes. Three examples are: 1:01, 4:44, and 12:21. How many times during a 12-hour period will be palindromes?

Show answer
Answer: A — 57.
Show hints
Hint 1 of 2
Split into 1-digit hours (1–9) and 2-digit hours (10–12) and count each.
Still stuck? Show hint 2 →
Hint 2 of 2
For h:mm to be a palindrome, the last digit of mm must equal h, and the middle digit can be 0–5.
Show solution
Approach: split by hour digit count
  1. 1-digit hours h:mm (h = 1–9): need mm's ones digit = h with mm's tens digit 0–5; that's 6 minutes per hour, so 9 × 6 = 54 palindromes.
  2. 2-digit hours hh:mm (10, 11, 12): only 10:01, 11:11, 12:21 work — 3 more. Total = 54 + 3 = 57.
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