AJHSME

1992 AJHSME

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

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Problem 1 · 1992 AJHSME Medium
Arithmetic & Operations pair-terms
10 − 9 + 8 − 7 + 6 − 5 + 4 − 3 + 2 − 11 − 2 + 3 − 4 + 5 − 6 + 7 − 8 + 9=
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Answer: B — 1.
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Hint 1 of 2
Pair consecutive terms in the top, and in the bottom, to add them up quickly.
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Hint 2 of 2
Both the numerator and the denominator come out to 5.
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Approach: pair terms in top and bottom
  1. Top: (10−9)+(8−7)+(6−5)+(4−3)+(2−1) = 5. Bottom: (1−2)+(3−4)+(5−6)+(7−8)+9 = −4+9 = 5.
  2. So the value is 5/5 = 1.
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Problem 2 · 1992 AJHSME Easy
Fractions, Decimals & Percents compare-fractions

Which of the following is not equal to 54?

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Answer: D — 1 1/5.
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Hint 1 of 2
5/4 = 1.25 — convert each choice and compare.
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Hint 2 of 2
A mixed number like 1 1/5 equals 1.2.
Show solution
Approach: convert everything to a decimal
  1. 5/4 = 1.25. The choices 10/8, 1¼, 1 3/12, and 1 10/40 all equal 1.25.
  2. But 1 1/5 = 1.2, so 1 1/5 is the one that's not equal.
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Problem 3 · 1992 AJHSME Easy
Arithmetic & Operations max-min

What is the largest difference that can be formed by subtracting two numbers chosen from the set {−16, −4, 0, 2, 4, 12}?

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Answer: D — 28.
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Hint 1 of 2
A difference is largest when you subtract the smallest number from the largest.
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Hint 2 of 2
Subtracting a negative adds.
Show solution
Approach: largest minus smallest
  1. The biggest difference is 12 − (−16).
  2. That equals 12 + 16 = 28.
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Problem 4 · 1992 AJHSME Medium
Fractions, Decimals & Percents percent

During the softball season, Judy had 35 hits. Among her hits were 1 home run, 1 triple, and 5 doubles. The rest of her hits were singles. What percent of her hits were singles?

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Answer: E — 80%.
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Hint 1 of 2
Count the non-single hits first and subtract.
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Hint 2 of 2
Then write the singles as a fraction of 35.
Show solution
Approach: singles over total
  1. Non-singles: 1 + 1 + 5 = 7, so singles = 35 − 7 = 28.
  2. 28/35 = 80%.
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Problem 5 · 1992 AJHSME Medium
Geometry & Measurement area-subtraction
ajhsme-1992-05
Show answer
Answer: E — 5.
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Hint 1 of 2
Find the rectangle's area, then subtract the circle's.
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Hint 2 of 2
The circle has diameter 1, so radius 1/2.
Show solution
Approach: rectangle minus circle
  1. The rectangle is 2 × 3 = 6, and the circle has area π(1/2)² ≈ 0.79.
  2. 6 − 0.79 ≈ 5.2, closest to 5.
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Problem 6 · 1992 AJHSME Medium
Algebra & Patterns custom-operation
ajhsme-1992-06
Show answer
Answer: D — 1.
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Hint 1 of 2
Each triangle means top + bottom-left − bottom-right.
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Hint 2 of 2
Evaluate both triangles, then add.
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Approach: apply the rule to each triangle
  1. First triangle: 1 + 3 − 4 = 0. Second triangle: 2 + 5 − 6 = 1.
  2. Their sum is 0 + 1 = 1.
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Problem 7 · 1992 AJHSME Hard
Counting & Probability digit-sumparity

The digit-sum of 998 is 9 + 9 + 8 = 26. How many 3-digit whole numbers, whose digit-sum is 26, are even?

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Answer: A — 1.
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Hint 1 of 2
A digit-sum of 26 is just 1 below the maximum 27, so the digits are nearly all 9s.
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Hint 2 of 2
An even number needs an even units digit.
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Approach: list the few digit options, then check parity
  1. Digit-sum 26 means the digits are 9, 9, 8 in some order: 998, 989, 899.
  2. Only 998 is even, so the answer is 1.
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Problem 8 · 1992 AJHSME Medium
Algebra & Patterns cost-revenue-profit

A store owner bought 1500 pencils at $0.10 each. If he sells them for $0.25 each, how many of them must he sell to make a profit of exactly $100.00?

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Answer: C — 1000.
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Hint 1 of 2
First find his total cost, then the revenue needed for a $100 profit.
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Hint 2 of 2
Divide that revenue by the $0.25 selling price.
Show solution
Approach: needed revenue ÷ price
  1. Cost is 1500 × $0.10 = $150, so he needs $150 + $100 = $250 in sales.
  2. At $0.25 each, that's $250 ÷ $0.25 = 1000 pencils.
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Problem 9 · 1992 AJHSME Hard
Fractions, Decimals & Percents read-graphratio
ajhsme-1992-09
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Answer: B — 160.
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Hint 1 of 2
The line across the F bar shows it is twice as tall as the M bar.
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Hint 2 of 2
So females and males split the 480 in a 2 : 1 ratio.
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Approach: read the bar ratio, then split the total
  1. The female bar is twice the male bar, so the town is 2 parts female to 1 part male — 3 parts total.
  2. Each part is 480 ÷ 3 = 160, so there are 160 males.
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Problem 10 · 1992 AJHSME Hard
Geometry & Measurement count-congruent-pieces
ajhsme-1992-10
Show answer
Answer: B — 20.
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Hint 1 of 2
First find the area of one of the 16 small triangles.
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Hint 2 of 2
Then count how many small triangles are shaded.
Show solution
Approach: area of one piece times the shaded count
  1. The big triangle has area ½ · 8 · 8 = 32, so each of the 16 congruent pieces is 2.
  2. Ten of them are shaded, giving 10 × 2 = 20.
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Problem 11 · 1992 AJHSME Medium
Fractions, Decimals & Percents read-graphpercent
ajhsme-1992-11
Show answer
Answer: B — 24%.
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Hint 1 of 2
Read every bar's frequency and add for the total.
Still stuck? Show hint 2 →
Hint 2 of 2
Blue's frequency over the total gives the percent.
Show solution
Approach: blue over the total of all bars
  1. The frequencies are 50, 60, 40, 60, 40, summing to 250.
  2. Blue is 60, so 60/250 = 24%.
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Problem 12 · 1992 AJHSME Medium
Ratios, Rates & Proportions total-divided

The five tires of a car (four road tires and a full-sized spare) were rotated so that each tire was used the same number of miles during the first 30,000 miles the car traveled. For how many miles was each tire used?

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Answer: C — 24,000 miles.
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Hint 1 of 2
Only 4 tires are on the road at any moment, so count total tire-miles.
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Hint 2 of 2
Share those tire-miles equally among all 5 tires.
Show solution
Approach: total tire-miles shared over 5 tires
  1. Four tires are used over 30,000 miles, for 4 × 30,000 = 120,000 tire-miles.
  2. Split among 5 tires: 120,000 ÷ 5 = 24,000 miles each.
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Problem 13 · 1992 AJHSME Hard
Arithmetic & Operations mean-median-mode

Five test scores have a mean of 90, a median of 91, and a mode of 94. The sum of the two lowest test scores is

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Answer: B — 171.
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Hint 1 of 2
The five scores total 5 × 90; the median is the 3rd score.
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Hint 2 of 2
The mode 94 must be the two highest scores.
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Approach: pin the top three, then subtract
  1. The scores total 450. The median is 91 (the 3rd), and the mode 94 (twice) must be the 4th and 5th: so the top three are 91, 94, 94 = 279.
  2. The two lowest sum to 450 − 279 = 171.
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Problem 14 · 1992 AJHSME Medium
Fractions, Decimals & Percents fraction-of

When four gallons are added to a tank that is one-third full, the tank is then one-half full. The capacity of the tank in gallons is

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Answer: D — 24.
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Hint 1 of 2
The 4 gallons raised the level from 1/3 to 1/2 — what fraction is that?
Still stuck? Show hint 2 →
Hint 2 of 2
Then scale up to the full tank.
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Approach: the added fraction gives the whole
  1. 1/2 − 1/3 = 1/6 of the tank equals 4 gallons.
  2. So the full tank is 6 × 4 = 24 gallons.
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Problem 15 · 1992 AJHSME Hard
Number Theory periodic-sequencemod-arithmetic

What is the 1992nd letter in the sequence ABCDEDCBAABCDEDCBAABCDEDCBA… ?

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Answer: C — C.
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Hint 1 of 2
The block ABCDEDCBA repeats; count its length.
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Hint 2 of 2
Find 1992 modulo that length.
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Approach: use the repeating block
  1. The block ABCDEDCBA has 9 letters and repeats.
  2. 1992 = 9·221 + 3, so the 1992nd letter is the 3rd of the block: C.
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Problem 16 · 1992 AJHSME Hard
Geometry & Measurement volume-cylinder
ajhsme-1992-16
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Answer: B — Cylinder B.
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Hint 1 of 2
Volume of a cylinder is π · radius² · height.
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Hint 2 of 2
Doubling can come from doubling the height (radius squared matters most).
Show solution
Approach: compare volumes to the original
  1. The original has radius 10, height 5: volume π·100·5 = 500π. Twice that is 1000π.
  2. Cylinder B (radius 10, height 10) gives π·100·10 = 1000π — exactly double. So B.
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Problem 17 · 1992 AJHSME Hard
Geometry & Measurement triangle-inequality

The sides of a triangle have lengths 6.5, 10, and s, where s is a whole number. What is the smallest possible value of s?

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Answer: B — 4.
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Hint 1 of 2
The third side must be longer than the difference of the other two.
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Hint 2 of 2
10 − 6.5 = 3.5, and s is a whole number.
Show solution
Approach: triangle inequality on the difference
  1. For a triangle, s must exceed 10 − 6.5 = 3.5.
  2. The smallest whole number greater than 3.5 is 4.
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Problem 18 · 1992 AJHSME Medium
Ratios, Rates & Proportions average-speed

On a trip, a car traveled 80 miles in an hour and a half, then was stopped in traffic for 30 minutes, then traveled 100 miles during the next 2 hours. What was the car's average speed in miles per hour for the 4-hour trip?

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Answer: A — 45.
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Hint 1 of 2
Average speed is total distance divided by total time — include the stop.
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Hint 2 of 2
Add 80 + 100 miles over 1.5 + 0.5 + 2 hours.
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Approach: total distance over total time
  1. Total distance is 80 + 100 = 180 miles over 1.5 + 0.5 + 2 = 4 hours.
  2. Average speed = 180 ÷ 4 = 45 mph.
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Problem 19 · 1992 AJHSME Hard
Algebra & Patterns extremal

The distance between the 5th and 26th exits on an interstate highway is 118 miles. If any two exits are at least 5 miles apart, then what is the largest number of miles there can be between two consecutive exits that are between the 5th and 26th exits?

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Answer: C — 18 miles.
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Hint 1 of 2
There are 21 gaps between the 5th and 26th exits.
Still stuck? Show hint 2 →
Hint 2 of 2
To stretch one gap as far as possible, make all the others as small as allowed (5 miles).
Show solution
Approach: shrink the other gaps to free up room
  1. There are 26 − 5 = 21 gaps. Making 20 of them the minimum 5 miles uses 100 miles.
  2. The last gap can be 118 − 100 = 18 miles.
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Problem 20 · 1992 AJHSME Hard
Geometry & Measurement net-folding
ajhsme-1992-20
Show answer
Answer: D — Pattern D.
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Hint 1 of 2
A cube net needs all six squares to fold without two landing on the same face.
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Hint 2 of 2
Mentally fold each; the bad one forces two squares onto the same face.
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Approach: fold each net and watch for overlaps
  1. Four of the patterns fold neatly into a cube, each square becoming a different face.
  2. Pattern D forces two squares onto the same face (leaving another open), so it cannot form a cube.
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Problem 21 · 1992 AJHSME Hard
Fractions, Decimals & Percents read-graphpercent-comparison
ajhsme-1992-21
Show answer
Answer: B — February.
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Hint 1 of 2
Greatest percent, not greatest amount — a small bar makes a small lead look big.
Still stuck? Show hint 2 →
Hint 2 of 2
Check the month with the lowest sales, where one bar is a large fraction above the other.
Show solution
Approach: compare the relative (percent) gap, not the absolute one
  1. In February the bars are smallest, so the one-unit lead of drums over bugles is a 50% difference.
  2. No other month's bars give that large a percentage gap, so the answer is February.
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Problem 22 · 1992 AJHSME Stretch
Geometry & Measurement perimeter-changeparity
ajhsme-1992-22
Show answer
Answer: C — 18.
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Hint 1 of 2
Adding a tile that shares exactly one edge raises the perimeter by 2; sharing more raises it less.
Still stuck? Show hint 2 →
Hint 2 of 2
Two tiles can add at most 2 + 2 = 4.
Show solution
Approach: track how each added tile changes the perimeter
  1. Each new tile shares at least one edge with the existing shape, so it changes the perimeter by an even amount, and at most by +2 (3 new edges, 1 hidden).
  2. Starting at 14, two tiles add at most 4, capping the perimeter at 18. Placing each tile so it touches only one edge of the figure attains that — answer 18.
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Problem 23 · 1992 AJHSME Stretch
Counting & Probability casework

If two dice are tossed, the probability that the product of the numbers showing on the tops of the dice is greater than 10 is

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Answer: B — 17/36.
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Hint 1 of 2
Case on the larger die value and count how many partners push the product over 10.
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Hint 2 of 2
There are 36 equally likely ordered outcomes.
Show solution
Approach: count winning ordered pairs
  1. Counting products over 10: a 2 works with 6 (1 way), a 3 with 4–6 (3), a 4 with 3–6 (4), a 5 with 3–6 (4), a 6 with 2–6 (5).
  2. That's 1 + 3 + 4 + 4 + 5 = 17 of 36, a probability of 17/36.
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Problem 24 · 1992 AJHSME Stretch
Geometry & Measurement area-subtraction
ajhsme-1992-24
Show answer
Answer: A — 7.7.
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Hint 1 of 2
The shaded region is the square minus the four quarter-circle corners.
Still stuck? Show hint 2 →
Hint 2 of 2
Four quarter-circles make one whole circle.
Show solution
Approach: square minus one full circle's worth of corners
  1. The centers form a square of side 2·3 = 6 (area 36), and the four quarter-circles total one circle of area π·3² ≈ 28.3.
  2. So the shaded area is 36 − 28.3 ≈ 7.7.
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Problem 25 · 1992 AJHSME Stretch
Fractions, Decimals & Percents telescoping-product

One half of the water is poured out of a full container. Then one third of the remainder is poured out. Continue the process: one fourth of the remainder for the third pouring, one fifth of the remainder for the fourth pouring, and so on. After how many pourings does exactly one tenth of the original water remain?

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Answer: D — 9.
Show hints
Hint 1 of 2
After each pouring, multiply by what's left: 1/2, then 2/3, then 3/4, …
Still stuck? Show hint 2 →
Hint 2 of 2
These products telescope to a simple fraction.
Show solution
Approach: multiply the surviving fractions (they telescope)
  1. After k pourings the remaining fraction is ½ · ⅔ · ¾ · … · k/(k+1), which telescopes to 1/(k+1).
  2. Setting 1/(k+1) = 1/10 gives k = 9.
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