AJHSME · Test Mode

1994 AJHSME

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Problem 1 · 1994 AJHSME Easy
Fractions, Decimals & Percents common-denominator

Which of the following is the largest?

Show hint (soft nudge)
Put them all over a common denominator of 24.
Show hint (sharpest)
Then just compare numerators.
Show solution
  1. Over 24: 8/24, 6/24, 9/24, 10/24, 7/24.
  2. The biggest numerator is 10, so 5/12 is largest.
D 5/12.
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Problem 2 · 1994 AJHSME Easy
Fractions, Decimals & Percents sum-fractions
ajhsme-1994-02
Show hint (soft nudge)
All terms share the denominator 10, so add the numerators first.
Show hint (sharpest)
1 + 2 + … + 9 = 45.
Show solution
  1. The numerators add to (1 + 2 + … + 9) + 55 = 45 + 55 = 100.
  2. So the sum is 100/10 = 10.
D 10.
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Problem 3 · 1994 AJHSME Medium
Arithmetic & Operations time-arithmetic

Each day Maria must work 8 hours. This does not include the 45 minutes she takes for lunch. If she begins working at 7:25 A.M. and takes her lunch break at noon, then her working day will end at

Show hint (soft nudge)
Find how much she works before lunch, then how much is left.
Show hint (sharpest)
Don't forget the 45-minute lunch pushes the afternoon back.
Show solution
  1. From 7:25 to noon she works 4 h 35 min, leaving 8 h − 4 h 35 min = 3 h 25 min.
  2. Lunch ends at 12:45, and 3 h 25 min later is 4:10 P.M.
C 4:10 P.M.
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Problem 4 · 1994 AJHSME Medium
Geometry & Measurement rotation
ajhsme-1994-04
Show hint (soft nudge)
A 120° clockwise turn moves each shape to the next position clockwise.
Show hint (sharpest)
Track where the triangle, circle, and diamond each land.
Show solution
  1. Clockwise 120° sends top → lower-right → lower-left → top.
  2. So the circle moves to the top, the diamond to the lower-left, and the triangle to the lower-right — matching choice B.
B Choice B.
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Problem 5 · 1994 AJHSME Easy
Ratios, Rates & Proportions unit-conversion

Given that 1 mile = 8 furlongs and 1 furlong = 40 rods, the number of rods in one mile is

Show hint (soft nudge)
Convert miles to furlongs, then furlongs to rods.
Show hint (sharpest)
Just multiply the two conversion factors.
Show solution
  1. 1 mile = 8 furlongs, and each furlong = 40 rods.
  2. So 1 mile = 8 × 40 = 320 rods.
B 320.
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Problem 6 · 1994 AJHSME Medium
Number Theory divisibility

The units digit (one's digit) of the product of any six consecutive positive whole numbers is

Show hint (soft nudge)
Among six consecutive numbers, what's guaranteed to appear?
Show hint (sharpest)
A multiple of 5 and an even number together make a multiple of 10.
Show solution
  1. Any six consecutive numbers include a multiple of 5 and at least one even number.
  2. Their product is therefore a multiple of 10, so its units digit is 0.
A 0.
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Problem 7 · 1994 AJHSME Hard
Geometry & Measurement exterior-angle angle-chase
ajhsme-1994-07
Show hint (soft nudge)
First find ∠ABE in triangle ABE from ∠A and ∠E.
Show hint (sharpest)
Since A, B, C are in a line, ∠DBC is the supplement of ∠ABE.
Show solution
  1. In triangle ABE, ∠ABE = 180° − 60° − 40° = 80°.
  2. A, B, C are collinear, so ∠DBC = 180° − 80° = 100°. In triangle BDC, ∠BDC = 180° − 100° − 30° = 50°.
B 50°.
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Problem 8 · 1994 AJHSME Hard
Counting & Probability digit-sum counting

For how many three-digit whole numbers does the sum of the digits equal 25?

Show hint (soft nudge)
The maximum digit sum is 9 + 9 + 9 = 27, so 25 is just 2 short.
Show hint (sharpest)
Spread a shortfall of 2 across the three digits.
Show solution
  1. Digit triples summing to 25 are two 9s and a 7, or a 9 and two 8s.
  2. Arranging gives (9,9,7), (9,7,9), (7,9,9), (9,8,8), (8,9,8), (8,8,9) — 6 numbers.
C 6.
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Problem 9 · 1994 AJHSME Hard
Fractions, Decimals & Percents discount tax

A shopper buys a 100-dollar coat on sale for 20% off. An additional 5 dollars are taken off the sale price by using a discount coupon. A sales tax of 8% is paid on the final selling price. The total amount the shopper pays for the coat is

Show hint (soft nudge)
Apply the discounts in order before the tax.
Show hint (sharpest)
20% off $100, then −$5, then ×1.08.
Show solution
  1. 20% off $100 is $80; the coupon makes it $75.
  2. Adding 8% tax: $75 × 1.08 = $81.00.
A 81.00 dollars.
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Problem 10 · 1994 AJHSME Hard
Number Theory divisors

For how many positive integer values of N is the expression 36N + 2 an integer?

Show hint (soft nudge)
N + 2 must be a divisor of 36.
Show hint (sharpest)
Since N ≥ 1, only divisors that are at least 3 count.
Show solution
  1. N + 2 must divide 36, and N ≥ 1 means N + 2 ≥ 3.
  2. The divisors of 36 that are ≥ 3 are 3, 4, 6, 9, 12, 18, 36 — 7 of them.
A 7.
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Problem 11 · 1994 AJHSME Hard
Logic & Word Problems two-way-table

Last summer 100 students attended basketball camp. Of those, 52 were boys and 48 were girls. Also, 40 students were from Jonas Middle School and 60 were from Clay Middle School. Twenty of the girls were from Jonas Middle School. How many of the boys were from Clay Middle School?

Show hint (soft nudge)
Make a boys/girls by Jonas/Clay table and fill it in.
Show hint (sharpest)
Find the Clay girls first, then the Clay boys.
Show solution
  1. Girls from Clay = 48 − 20 = 28.
  2. Clay has 60 students, so boys from Clay = 60 − 28 = 32.
B 32.
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Problem 12 · 1994 AJHSME Hard
Geometry & Measurement dissection area-comparison
ajhsme-1994-12
Show hint (soft nudge)
Don't compute each area separately — rearrange the shaded pieces.
Show hint (sharpest)
In each square the shaded parts cut and reassemble into the same region.
Show solution
  1. In each square the shaded pieces can be slid together to cover the same fraction of the square.
  2. So the three shaded areas are all equal.
A All three are equal.
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Problem 13 · 1994 AJHSME Medium
Fractions, Decimals & Percents average

The number halfway between 16 and 14 is

Show hint (soft nudge)
Halfway between two numbers is their average.
Show hint (sharpest)
Add the fractions and divide by 2.
Show solution
  1. 1/6 + 1/4 = 2/12 + 3/12 = 5/12.
  2. Half of 5/12 is 5/24.
C 5/24.
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Problem 14 · 1994 AJHSME Medium
Ratios, Rates & Proportions total-divided

Two children at a time can play pairball. For 90 minutes, with only two children playing at a time, five children take turns so that each one plays the same amount of time. The number of minutes each child plays is

Show hint (soft nudge)
Two children play at once, so there are 2 × 90 child-minutes of play available.
Show hint (sharpest)
Share those equally among the five children.
Show solution
  1. Two play at a time for 90 minutes: 2 × 90 = 180 child-minutes.
  2. Split among 5 children: 180 ÷ 5 = 36 minutes each.
E 36.
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Problem 15 · 1994 AJHSME Hard
Number Theory pattern mod-arithmetic
ajhsme-1994-15
Show hint (soft nudge)
The arrow pattern repeats every 4 steps.
Show hint (sharpest)
Find 425 and 426 modulo 4 to see which arrows apply.
Show solution
  1. The arrows repeat with period 4: from a number ≡ 0 (right), ≡ 1 (up), ≡ 2 (right), ≡ 3 (down).
  2. Since 425 ≡ 1 and 426 ≡ 2 (mod 4), the moves are up then right — choice A.
A Up arrow, then right arrow.
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Problem 16 · 1994 AJHSME Medium
Geometry & Measurement scaling

The perimeter of one square is 3 times the perimeter of another square. The area of the larger square is how many times the area of the smaller square?

Show hint (soft nudge)
Perimeter scales with the side, so the sides are in the same 3 : 1 ratio.
Show hint (sharpest)
Area scales with the side squared.
Show solution
  1. The side ratio equals the perimeter ratio, 3.
  2. So the area ratio is 3² = 9.
E 9.
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Problem 17 · 1994 AJHSME Hard
Algebra & Patterns partial-sums

Pauline can shovel snow at the rate of 20 cubic yards for the first hour, 19 cubic yards for the second, 18 for the third, and so on, always shoveling one cubic yard less per hour than the previous hour. If her driveway is 4 yards wide, 10 yards long, and covered with snow 3 yards deep, then the number of hours it will take her to shovel it clean is closest to

Show hint (soft nudge)
First find the total volume of snow: 4 × 10 × 3.
Show hint (sharpest)
Add 20 + 19 + 18 + … until you reach that volume.
Show solution
  1. The snow is 4 × 10 × 3 = 120 cubic yards.
  2. Running totals: 20, 39, 57, 74, 90, 105, 119 — after 7 hours she's at 119, just shy of 120, so the time is closest to 7 hours.
D 7.
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Problem 18 · 1994 AJHSME Hard
Algebra & Patterns interpret-graph
ajhsme-1994-18
Show hint (soft nudge)
Distance from home rises while going out, stays flat while shopping, then falls coming back.
Show hint (sharpest)
Because the speed changes (gentle in city, steep on highway), each side of the graph bends rather than staying a single straight line.
Show solution
  1. Going out, distance rises gently (city) then steeply (highway), so the climb curves and gets steeper; it stays flat for the hour at the mall; coming home it falls steeply (highway) then gently (city).
  2. Graph B shows these two different slopes on each side — the straight-sided trapezoid (A) would mean a single constant speed each way.
B Graph B.
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Problem 19 · 1994 AJHSME Hard
Geometry & Measurement tangent dimensions
ajhsme-1994-19
Show hint (soft nudge)
Each semicircle has diameter 4, so it bulges out 2 (its radius) past each side.
Show hint (sharpest)
The outer square's side is the inner side plus those two bulges.
Show solution
  1. Each semicircle has radius 2 and sticks out 2 beyond its side, so square ABCD has side 4 + 2 + 2 = 8.
  2. Its area is 8² = 64.
E 64.
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Problem 20 · 1994 AJHSME Stretch
Fractions, Decimals & Percents minimize-fractions
ajhsme-1994-20
Show hint (soft nudge)
To make two fractions small, use the smallest numerators and the largest denominators.
Show hint (sharpest)
Test how to pair 1 and 2 over 8 and 9.
Show solution
  1. Use numerators 1, 2 and denominators 8, 9. Pairing as 1/8 + 2/9 = 9/72 + 16/72 = 25/72 beats 1/9 + 2/8 = 26/72.
  2. So the smallest sum is 25/72.
D 25/72.
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Problem 21 · 1994 AJHSME Hard
Counting & Probability pigeonhole

A gumball machine contains 9 red, 7 white, and 8 blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is

Show hint (soft nudge)
Imagine the worst luck: as many gumballs as possible without four of any color.
Show hint (sharpest)
That's three of each color; the next one must make a fourth.
Show solution
  1. You could draw 3 red, 3 white, 3 blue — 9 gumballs — with no color yet reaching four.
  2. The 10th gumball must complete a set of four, so the answer is 10.
C 10.
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Problem 22 · 1994 AJHSME Stretch
Counting & Probability parity probability
ajhsme-1994-22
Show hint (soft nudge)
A sum is even when both numbers are even or both are odd.
Show hint (sharpest)
Read each wheel's chance of landing odd vs. even from the region sizes.
Show solution
  1. Wheel 1 lands even (the 2) with probability 1/4 and odd with 3/4; wheel 2 lands even (6 or 4) with probability 2/3 and odd (5) with 1/3.
  2. Even sum = both even or both odd: (1/4)(2/3) + (3/4)(1/3) = 2/12 + 3/12 = 5/12.
D 5/12.
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Problem 23 · 1994 AJHSME Stretch
Number Theory place-value maximize
ajhsme-1994-23
Show hint (soft nudge)
The sum equals 113·X + 10·Y; keep it to three digits while making it large.
Show hint (sharpest)
Find the digits of that largest sum and match them to X, Y, Z.
Show solution
  1. Adding XXX + YX + X gives 113·X + 10·Y. To stay three digits, X ≤ 8; taking X = 8, Y = 9 gives 904 + 90 = 994.
  2. 994 reads as 9, 9, 4 = Y, Y, (a new digit Z), so the form is YYZ.
D Form YYZ.
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Problem 24 · 1994 AJHSME Stretch
Counting & Probability casework order-ideal

A 2 by 2 square is divided into four 1 by 1 squares. Each small square is painted green or red. In how many ways can this be done so that no green square shares its top or right side with a red square? (There may be from zero to four green squares.)

Show hint (soft nudge)
The rule means: anything directly above or to the right of a green square must also be green.
Show hint (sharpest)
So the green squares must form an 'up-and-right' staircase region.
Show solution
  1. If a square is green, the squares above it and to its right can't be red, so they're green too.
  2. The green sets that satisfy this are: none, just the top-right, top-right + top-left, top-right + bottom-right, those three together, and all four — 6 ways.
B 6.
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Problem 25 · 1994 AJHSME Stretch
Number Theory small-cases pattern
ajhsme-1994-25
Show hint (soft nudge)
Try small versions: 9 × 4, 99 × 44, 999 × 444, and find the digit-sum pattern.
Show hint (sharpest)
The digit sum turns out to be 9 times the number of nines.
Show solution
  1. 9·4 = 36 (digit sum 9), 99·44 = 4356 (18), 999·444 = 443556 (27): the digit sum is 9 × (number of nines).
  2. With 94 nines, the digit sum is 9 × 94 = 846.
A 846.
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