AJHSME

1990 AJHSME

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

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Problem 1 · 1990 AJHSME Medium
Number Theory place-valueminimize

What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in this addition problem?

   
+     
 
Show answer
Answer: C — 1047.
Show hints
Hint 1 of 2
The hundreds digits matter most, so put the two smallest there.
Still stuck? Show hint 2 →
Hint 2 of 2
Then the next-smallest in the tens, and the largest in the units.
Show solution
Approach: small digits into the big place values
  1. Put 4 and 5 in the hundreds (900), 6 and 7 in the tens (130), and 8 and 9 in the units (17).
  2. The smallest sum is 900 + 130 + 17 = 1047.
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Problem 2 · 1990 AJHSME Easy
Fractions, Decimals & Percents place-value

Which digit of .12345, when changed to 9, gives the largest number?

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Answer: A — the 1.
Show hints
Hint 1 of 2
Changing a digit helps most where its place value is largest.
Still stuck? Show hint 2 →
Hint 2 of 2
The leftmost digit is the tenths place.
Show solution
Approach: change the highest-value digit
  1. The 1 sits in the tenths place, the most valuable spot. Changing it to 9 makes .92345.
  2. That beats changing any digit further right, so the answer is the 1.
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Problem 3 · 1990 AJHSME Hard
Geometry & Measurement area-fractionrearrangement
ajhsme-1990-03
Show answer
Answer: E — 1/2.
Show hints
Hint 1 of 2
The square's diagonal splits it into two equal halves.
Still stuck? Show hint 2 →
Hint 2 of 2
Rearrange the shaded pieces — they fill exactly one of those halves.
Show solution
Approach: the shaded pieces reassemble to half the square
  1. The diagonal cuts the square into two equal triangles. The shaded shapes can be slid together to cover exactly one triangle's worth of area.
  2. So the shaded fraction is 1/2.
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Problem 4 · 1990 AJHSME Medium
Number Theory units-digit-of-squares

Which of the following could not be the units digit (one's digit) of the square of a whole number?

Show answer
Answer: E — 8.
Show hint
Hint 1
The last digit of n² depends only on the last digit of n — there are just 10 cases to check.
Show solution
Approach: list the possible last digits of squares
  1. Squares end only in 0, 1, 4, 5, 6, or 9.
  2. 8 is not on that list, so a square can never end in 8.
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Problem 5 · 1990 AJHSME Medium
Fractions, Decimals & Percents estimation

Which of the following is closest to the product (.48017)(.48017)(.48017)?

Show answer
Answer: B — 0.110.
Show hints
Hint 1 of 2
Round .48017 to about 0.5 to estimate.
Still stuck? Show hint 2 →
Hint 2 of 2
Cube the rounded value.
Show solution
Approach: round and cube
  1. .48017 ≈ 0.5, and 0.5³ = 0.125.
  2. That is closest to 0.110.
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Problem 6 · 1990 AJHSME Medium
Fractions, Decimals & Percents compare-operations

Which of these five numbers is the largest?

Show answer
Answer: D — 13579 ÷ (1/2468).
Show hints
Hint 1 of 2
Adding, subtracting, or appending decimals barely changes 13579.
Still stuck? Show hint 2 →
Hint 2 of 2
Dividing by a tiny fraction like 1/2468 multiplies by 2468.
Show solution
Approach: spot the operation that explodes the value
  1. Choices A, B, C, E all stay close to 13579 (or smaller).
  2. But dividing by 1/2468 means multiplying by 2468, giving about 33 million — by far the largest. So 13579 ÷ (1/2468).
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Problem 7 · 1990 AJHSME Medium
Arithmetic & Operations sign-productmaximize

When three different numbers from the set {−3, −2, −1, 4, 5} are multiplied, the largest possible product is

Show answer
Answer: C — 30.
Show hints
Hint 1 of 2
A positive product needs an even number of negatives — here, two.
Still stuck? Show hint 2 →
Hint 2 of 2
Pair the two most-negative numbers with the largest positive.
Show solution
Approach: two negatives times the biggest positive
  1. Using (−3)(−2) = 6 and the largest positive 5 gives a positive product.
  2. 6 × 5 = 30, the largest possible.
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Problem 8 · 1990 AJHSME Medium
Fractions, Decimals & Percents discount-tax

A dress originally priced at 80 dollars was put on sale for 25% off. If 10% tax was added to the sale price, then the total selling price (in dollars) of the dress was

Show answer
Answer: D — 66 dollars.
Show hints
Hint 1 of 2
Take 25% off first, then add the tax on the reduced price.
Still stuck? Show hint 2 →
Hint 2 of 2
25% off $80 leaves $60.
Show solution
Approach: discount, then tax
  1. 25% off $80 is $60.
  2. Adding 10% tax: $60 × 1.10 = $66.
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Problem 9 · 1990 AJHSME Hard
Fractions, Decimals & Percents count-in-rangepercent
ajhsme-1990-09
Show answer
Answer: D — 33⅓%.
Show hints
Hint 1 of 2
A C is a score from 75 to 84 — count how many of the 15 scores fall there.
Still stuck? Show hint 2 →
Hint 2 of 2
Then divide by 15.
Show solution
Approach: count C scores, then take the percent
  1. Scores in 75–84 are 77, 75, 84, 78, 80 — that's 5 of the 15.
  2. 5/15 = 33⅓%.
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Problem 10 · 1990 AJHSME Hard
Algebra & Patterns calendar-arithmeticsubstitution
ajhsme-1990-10
Show answer
Answer: A — P.
Show hints
Hint 1 of 2
Dates one row apart on a calendar differ by 7; A is one day after C.
Still stuck? Show hint 2 →
Hint 2 of 2
Set up the equation (date) + C = A + B and solve for the letter.
Show solution
Approach: use the +7 (down a row) and +1 (next day) structure
  1. Let C be its date. Then A = C + 1, B = C + 13, and P = C + 14 (two rows below C).
  2. Since A + B = 2C + 14 and P + C = 2C + 14, the matching letter is P.
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Problem 11 · 1990 AJHSME Stretch
Number Theory consecutiveopposite-pairs
ajhsme-1990-11
Show answer
Answer: E — 81.
Show hints
Hint 1 of 2
The six faces are six consecutive numbers including the visible 11, 14, 15.
Still stuck? Show hint 2 →
Hint 2 of 2
The visible faces meet at a corner, so they can't be opposite each other — that pins which six numbers.
Show solution
Approach: use the equal-opposite-sums to fix the six numbers
  1. The visible 11, 14, 15 are mutually adjacent, so they aren't opposite pairs. The six consecutive numbers must be 11–16, pairing as 11+16, 12+15, 13+14 (each summing to 27).
  2. Their total is 11 + 12 + 13 + 14 + 15 + 16 = 81.
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Problem 12 · 1990 AJHSME Hard
Counting & Probability ordering-permutations

There are twenty-four 4-digit numbers that use each of the four digits 2, 4, 5, and 7 exactly once. Listed in numerical order from smallest to largest, the number in the 17th position in the list is

Show answer
Answer: B — 5724.
Show hints
Hint 1 of 2
Each choice of leading digit fixes 3! = 6 numbers.
Still stuck? Show hint 2 →
Hint 2 of 2
Find which leading-digit block holds the 17th, then list within it.
Show solution
Approach: group by leading digit, then order within
  1. Leading 2: positions 1–6, leading 4: 7–12, leading 5: 13–18. So the 17th is the 5th number starting with 5.
  2. Those are 5247, 5274, 5427, 5472, 5724 — the 5th is 5724.
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Problem 13 · 1990 AJHSME Hard
Arithmetic & Operations round-uprate

One proposal for new postage rates for a letter was 30 cents for the first ounce and 22 cents for each additional ounce (or fraction of an ounce). The postage for a letter weighing 4.5 ounces was

Show answer
Answer: C — 1.18 dollars.
Show hints
Hint 1 of 2
After the first ounce, 3.5 ounces remain — each fraction counts as a whole charge.
Still stuck? Show hint 2 →
Hint 2 of 2
Round 3.5 up to 4 additional charges.
Show solution
Approach: first ounce plus rounded-up additional ounces
  1. After the first ounce (30¢), 3.5 ounces remain, charged as 4 additional ounces.
  2. Total = 30 + 4 × 22 = 118¢ = $1.18.
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Problem 14 · 1990 AJHSME Medium
Fractions, Decimals & Percents probabilityfraction-to-count

A bag contains only blue balls and green balls. There are 6 blue balls. If the probability of drawing a blue ball at random from this bag is 14, then the number of green balls in the bag is

Show answer
Answer: B — 18.
Show hints
Hint 1 of 2
If blue is 1/4 of the balls, the total is 4 times the blue count.
Still stuck? Show hint 2 →
Hint 2 of 2
Subtract the blue balls to get the green ones.
Show solution
Approach: find the total, then subtract blue
  1. Blue = 1/4 of all, so the total is 4 × 6 = 24 balls.
  2. Green = 24 − 6 = 18.
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Problem 15 · 1990 AJHSME Hard
Geometry & Measurement perimetertetromino
ajhsme-1990-15
Show answer
Answer: E — 50 cm.
Show hints
Hint 1 of 2
Four equal squares with area 100 means each square has area 25 and side 5.
Still stuck? Show hint 2 →
Hint 2 of 2
Count the outside edges of the S-shaped figure — there are 10.
Show solution
Approach: find the side, then count outer edges
  1. Each of the four squares has area 25, so side 5. The offset (S-shaped) figure has 10 unit-side edges on its boundary.
  2. Perimeter = 10 × 5 = 50 cm.
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Problem 16 · 1990 AJHSME Hard
Algebra & Patterns pair-terms

1990 − 1980 + 1970 − 1960 + … − 20 + 10 =

Show answer
Answer: D — 1000.
Show hints
Hint 1 of 2
Pair the terms: (1990 − 1980), (1970 − 1960), … each equals 10.
Still stuck? Show hint 2 →
Hint 2 of 2
Don't forget the leftover +10 at the end.
Show solution
Approach: pair the terms, then add the leftover
  1. Pairs (1990 − 1980), (1970 − 1960), …, (30 − 20) each give 10, and there are 99 such pairs: 990.
  2. Adding the final +10 gives 990 + 10 = 1000.
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Problem 17 · 1990 AJHSME Hard
Geometry & Measurement volumeunit-conversionround-up

A straight concrete sidewalk is to be 3 feet wide, 60 feet long, and 3 inches thick. How many cubic yards of concrete must a contractor order for the sidewalk if concrete must be ordered in a whole number of cubic yards?

Show answer
Answer: A — 2.
Show hints
Hint 1 of 2
Convert 3 inches to feet (1/4 ft) and find the volume in cubic feet.
Still stuck? Show hint 2 →
Hint 2 of 2
There are 27 cubic feet in a cubic yard; round up.
Show solution
Approach: volume in cubic feet, convert, round up
  1. Volume = 3 × 60 × ¼ = 45 cubic feet, and 45 ÷ 27 ≈ 1.67 cubic yards.
  2. Ordering a whole number rounds up to 2 cubic yards.
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Problem 18 · 1990 AJHSME Stretch
Geometry & Measurement truncationedge-counting
ajhsme-1990-18
Show answer
Answer: C — 36.
Show hints
Hint 1 of 2
Each corner cut adds a new triangular face with 3 new edges.
Still stuck? Show hint 2 →
Hint 2 of 2
The original 12 edges of the prism survive (just shortened).
Show solution
Approach: original edges plus new edges from each corner
  1. Cutting all 8 corners adds 8 triangular faces, each with 3 new edges: 8 × 3 = 24 new edges.
  2. Adding the 12 original edges gives 12 + 24 = 36.
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Problem 19 · 1990 AJHSME Hard
Logic & Word Problems optimizationspacingcovering

There are 120 seats in a row. What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?

Show answer
Answer: B — 40.
Show hints
Hint 1 of 2
You want every empty seat to touch an occupied one.
Still stuck? Show hint 2 →
Hint 2 of 2
Each occupied seat can 'cover' itself plus its two neighbors — three seats per person.
Show solution
Approach: each occupied seat covers a block of 3
  1. Each occupied seat covers itself and its two neighbors — at most 3 seats. So 120 ÷ 3 = 40 people are necessary, and 39 leaves at least one empty seat with two empty neighbors.
  2. 40 is also enough: seat one person in every block of (occupied, empty, empty). Every empty seat touches an occupied one. So the answer is 40.
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Problem 20 · 1990 AJHSME Hard
Arithmetic & Operations mean-error

The annual incomes of 1,000 families range from 8200 dollars to 98,000 dollars. In error, the largest income was entered on the computer as 980,000 dollars. The difference between the mean of the incorrect data and the mean of the actual data is

Show answer
Answer: A — 882 dollars.
Show hints
Hint 1 of 2
Only one entry changed, so the totals differ by that one error.
Still stuck? Show hint 2 →
Hint 2 of 2
Divide the size of the error by the number of families.
Show solution
Approach: spread the single error over all families
  1. The wrong entry overstates the total by 980,000 − 98,000 = 882,000.
  2. Over 1,000 families, the mean is off by 882,000 ÷ 1000 = $882.
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Problem 21 · 1990 AJHSME Stretch
Algebra & Patterns work-backward

A list of 8 numbers is formed by beginning with two given numbers. Each new number in the list is the product of the two previous numbers. Find the first number if the last three numbers are 16, 64, 1024.

Show answer
Answer: B — 1/4.
Show hints
Hint 1 of 2
Each term is the product of the two before it, so divide to step backward.
Still stuck? Show hint 2 →
Hint 2 of 2
From 16, 64, 1024 work back: the term before 16 is 64 ÷ 16, and so on.
Show solution
Approach: divide to walk back to the start
  1. Since 1024 = 16 · 64, the term before 16 is 64 ÷ 16 = 4, then 16 ÷ 4 = 4, then 4 ÷ 4 = 1, then 4 ÷ 1 = 4, and finally 1 ÷ 4 = 1/4.
  2. So the first number is 1/4.
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Problem 22 · 1990 AJHSME Stretch
Number Theory mod-arithmeticdivisors

Several students are seated at a large circular table. They pass around a bag of 100 pieces of candy. Each person takes one piece and passes the bag to the next person. If Chris takes the first and the last piece of candy, then the number of students at the table could be

Show answer
Answer: B — 11.
Show hints
Hint 1 of 2
Chris takes pieces 1, n+1, 2n+1, … where n is the number of students.
Still stuck? Show hint 2 →
Hint 2 of 2
Taking the 100th piece too means 99 is a multiple of n.
Show solution
Approach: the gap between Chris's pieces must divide 99
  1. Chris takes the 1st and every nth piece after, so for the 100th to be his, n must divide 100 − 1 = 99.
  2. Of the choices, only 11 divides 99, so there could be 11 students.
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Problem 23 · 1990 AJHSME Hard
Ratios, Rates & Proportions read-graphslope
ajhsme-1990-23
Show answer
Answer: B — The second hour (1-2).
Show hints
Hint 1 of 2
Average speed in an hour is how much the distance rose that hour — the steepness.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the steepest one-hour rise on the graph.
Show solution
Approach: steepest one-hour climb on the graph
  1. The distance jumps the most between hours 1 and 2 (about 500 miles), steeper than any other hour.
  2. So the largest average speed is during the second hour (1-2).
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Problem 24 · 1990 AJHSME Stretch
Algebra & Patterns substitutionbalance
ajhsme-1990-24
Show answer
Answer: C — 3.
Show hints
Hint 1 of 2
Write the two balances as equations: 3 triangles + 1 diamond = 9 circles, and 1 triangle = 1 diamond + 1 circle.
Still stuck? Show hint 2 →
Hint 2 of 2
Substitute to eliminate the triangles.
Show solution
Approach: substitute one balance into the other
  1. From 3T + D = 9C and T = D + C: 3(D + C) + D = 9C gives 4D = 6C, so 2D = 3C.
  2. Thus two diamonds balance 3 circles.
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Problem 25 · 1990 AJHSME Stretch
Counting & Probability counting-up-to-symmetry
ajhsme-1990-25
Show answer
Answer: C — 8.
Show hints
Hint 1 of 2
The nine cells come in three kinds: center, edges, and corners.
Still stuck? Show hint 2 →
Hint 2 of 2
Group the chosen pairs by the kinds of the two cells and their relative position.
Show solution
Approach: classify the shaded pairs up to flips and turns
  1. Sort the two-cell choices by type: center+edge, center+corner, two adjacent edges, two opposite edges, two adjacent corners, two diagonal corners, corner+touching edge, corner+far edge.
  2. These give 8 patterns that can't be matched by any flip or turn, so the answer is 8.
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