Problem 1 · 1989 AJHSME
Easy
Arithmetic & Operations
pair-terms
(1 + 11 + 21 + 31 + 41) + (9 + 19 + 29 + 39 + 49) =
(A) 150
(B) 199
(C) 200
(D) 249
(E) 250
Show hint (soft nudge)
Pair the first term of each group with the matching term of the other.
Show hint (sharpest)
Each pair like 1 + 9 makes a round 10 (then 30, 50, …).
Show solution
Pairing gives 1+9, 11+19, 21+29, 31+39, 41+49 = 10, 30, 50, 70, 90. Their sum is 10 + 30 + 50 + 70 + 90 = 250 .
E
250.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 2 · 1989 AJHSME
Easy
Fractions, Decimals & Percents
place-value
(A) .012
(B) .0246
(C) .12
(D) .246
(E) 246
Show hint (soft nudge)
Each fraction is a decimal in its own place: tenths, hundredths, thousandths.
Show hint (sharpest)
Just write the digits in order.
Show solution
2/10 = 0.2, 4/100 = 0.04, 6/1000 = 0.006. Adding gives .246 .
D
.246.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 3 · 1989 AJHSME
Easy
Fractions, Decimals & Percents
compare-decimals
Which of the following numbers is the largest?
(A) .99
(B) .9099
(C) .9
(D) .909
(E) .9009
Show hint (soft nudge)
Compare the digits right after the decimal point.
Show hint (sharpest)
.99 has a 9 in the hundredths place; the others have 0 there.
Show solution
All start with .9; in the hundredths place .99 has a 9 while the rest have 0. So .99 is largest.
A
.99.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 4 · 1989 AJHSME
Medium
Fractions, Decimals & Percents
estimation
Estimate to determine which of the following numbers is closest to 401 .205
(A) .2
(B) 2
(C) 20
(D) 200
(E) 2000
Show hint (soft nudge)
Round 401 to 400 and .205 to .2.
Show hint (sharpest)
Dividing by .2 is the same as multiplying by 5.
Show solution
401/.205 ≈ 400/0.2 = 400 × 5 = 2000. So it is closest to 2000 .
E
2000.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 5 · 1989 AJHSME
Medium
Arithmetic & Operations
order-of-operations
(A) −48
(B) −12
(C) −3
(D) 3
(E) 12
Show hint (soft nudge)
Do the parentheses and multiplication before the addition.
Show hint (sharpest)
6 ÷ 3 = 2, then 9 × 2.
Show solution
6 ÷ 3 = 2, then 9 × 2 = 18. Finally −15 + 18 = 3 .
D
3.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 6 · 1989 AJHSME
Medium
Arithmetic & Operations
number-line
spacing
(A)
(B)
(C)
(D)
(E)
Show hint (soft nudge)
Count how many equal steps lie between 0 and 20 to find each step's size.
Show hint (sharpest)
Then count the steps from 0 up to y.
Show solution
There are 5 equal steps from 0 to 20, so each step is 4. y sits 3 steps past 0. So y = 3 × 4 = 12 .
C
12.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 7 · 1989 AJHSME
Medium
Algebra & Patterns
coin-value
If the value of 20 quarters and 10 dimes equals the value of 10 quarters and n dimes, then n =
(A) 10
(B) 20
(C) 30
(D) 35
(E) 45
Show hint (soft nudge)
Trading away 10 quarters loses 250 cents — make it up in dimes.
Show hint (sharpest)
Each dime is worth 10 cents.
Show solution
20 quarters + 10 dimes = 600¢. The other side is 250¢ from 10 quarters, so the dimes must supply 350¢. 350 ÷ 10 = 35 dimes.
D
35.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 8 · 1989 AJHSME
Medium
Fractions, Decimals & Percents
distribute
(2 × 3 × 4) (1 2 1 3 1 4
(A) 1
(B) 3
(C) 9
(D) 24
(E) 26
Show hint (soft nudge)
The product out front is 24 — multiply each fraction by it.
Show hint (sharpest)
24 × 1/2, 24 × 1/3, 24 × 1/4 are all whole numbers.
Show solution
24 × ½ = 12, 24 × ⅓ = 8, 24 × ¼ = 6. Adding: 12 + 8 + 6 = 26 .
E
26.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 9 · 1989 AJHSME
Medium
Ratios, Rates & Proportions
ratio
percent
There are 2 boys for every 3 girls in Ms. Johnson's math class. If there are 30 students in her class, what percent of them are boys?
(A) 12%
(B) 20%
(C) 40%
(D) 60%
(E) 66⅔%
Show hint (soft nudge)
The ratio 2 : 3 makes 5 equal parts in all.
Show hint (sharpest)
Find how many students are in each part, then count the boys.
Show solution
5 parts make 30 students, so each part is 6 and the boys (2 parts) number 12. 12 of 30 is 40% .
C
40%.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 10 · 1989 AJHSME
Medium
Geometry & Measurement
clock-angles
What is the number of degrees in the smaller angle between the hour hand and the minute hand on a clock that reads seven o'clock?
(A) 50°
(B) 120°
(C) 135°
(D) 150°
(E) 165°
Show hint (soft nudge)
Each hour mark on a clock face is the same number of degrees from the next.
Show hint (sharpest)
How many hour marks separate the hands at 7:00, and what's each mark worth?
Show solution
12 hour marks split 360° evenly, so each gap is 30°. At 7:00 the hands sit 5 marks apart the short way (12 → 7), giving 5 × 30° = 150° .
D
150°.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 11 · 1989 AJHSME
Hard
Geometry & Measurement
reflection-symmetry
(A)
(B)
(C)
(D)
(E)
Show hint (soft nudge)
A reflection across a vertical line swaps left and right but leaves up and down alone.
Show hint (sharpest)
Hold each choice up to a mirror placed on the dashed line — only one matches the original.
Show solution
Reflecting across the vertical dashed line swaps the left and right of every feature: the corner square moves to the opposite side of its box, and the slanted arms flip their lean. Choice B is the only T-like shape whose corner square and arms are the mirror image of the original.
B
B.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 12 · 1989 AJHSME
Hard
Fractions, Decimals & Percents
simplify-complex-fraction
(A)
(B)
(C)
(D)
(E)
Show hint (soft nudge)
Compute the numerator and denominator separately before dividing.
Show hint (sharpest)
Dividing by a fraction is multiplying by its reciprocal.
Show solution
Top: 1 − 1⁄3 = 2⁄3. Bottom: 1 − 1⁄2 = 1⁄2. (2⁄3) ÷ (1⁄2) = (2⁄3) × 2 = 4⁄3 .
E
4⁄3.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 13 · 1989 AJHSME
Hard
Fractions, Decimals & Percents
proportional-scaling
(A)
(B)
(C)
(D)
(E)
Show hint (soft nudge)
Scaling the numerator and one denominator factor by the same amount leaves the fraction unchanged.
Show hint (sharpest)
Multiply the top and one factor on the bottom both by 0.1.
Show solution
Choice A turns 9 → .9 (×0.1) and 7 → .7 (×0.1). Same ×0.1 on top and bottom cancels: .9 ⁄ (.7 × 53) = 9 ⁄ (7 × 53). Every other choice changes a different number of factors by 0.1, so its value differs from the original.
A
.9 ⁄ (.7 × 53).
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 14 · 1989 AJHSME
Hard
Number Theory
minimize-difference
place-value
(A)
(B)
(C)
(D)
(E)
Show hint (soft nudge)
To minimize the difference, make the 3-digit number as small as you can and the 2-digit number as large as you can.
Show hint (sharpest)
Pick the smallest hundreds digit first, then build each number from the digits you have left.
Show solution
Smallest 3-digit using three of {2, 4, 5, 6, 9}: lead with 2, then take the next two smallest ascending → 245. Largest 2-digit from the remaining {6, 9} is 96. Difference: 245 − 96 = 149 .
C
149.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 15 · 1989 AJHSME
Hard
Geometry & Measurement
trapezoid-area
parallelogram
(A)
(B)
(C)
(D)
(E)
Show hint (soft nudge)
BEDC is a trapezoid: BC and ED are both horizontal, and BE is perpendicular to them.
Show hint (sharpest)
Use ½ × (sum of parallel sides) × height.
Show solution
AD = BC = 10 (opposite sides of the parallelogram). With ED = 6, the trapezoid BEDC has parallel sides BC = 10 and ED = 6, and height BE = 8. Area = ½(10 + 6)(8) = 64 .
D
64.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 16 · 1989 AJHSME
Hard
Number Theory
parity
primes
In how many ways can 47 be written as the sum of two primes?
(A) 0
(B) 1
(C) 2
(D) 3
(E) more than 3
Show hint (soft nudge)
Odd = odd + even, so one of the two primes must be even.
Show hint (sharpest)
The only even prime is 2 — what does the other prime have to be?
Show solution
47 is odd, so one prime is even. The only even prime is 2, forcing the other prime to be 47 − 2 = 45. But 45 = 9 × 5 isn't prime, so there are 0 ways.
A
0.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 17 · 1989 AJHSME
Hard
Algebra & Patterns
bound-the-average
The number N is between 9 and 17. The average of 6, 10, and N could be
(A) 8
(B) 10
(C) 12
(D) 14
(E) 16
Show hint (soft nudge)
Plug the extreme values of N into the average formula to see what range the average can land in.
Show hint (sharpest)
Then check which choice falls inside that range.
Show solution
Average = (6 + 10 + N)⁄3 = (16 + N)⁄3. With 9 < N < 17, this ranges from 25⁄3 ≈ 8.3 to 33⁄3 = 11. Of the choices, only 10 sits in that range (N = 14 gives average 10).
B
10.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 18 · 1989 AJHSME
Hard
Algebra & Patterns
involution
(A)
(B)
(C)
(D)
(E)
Show hint (soft nudge)
Try the operation once, then again — does anything change after that?
Show hint (sharpest)
The reciprocal of the reciprocal is the original number.
Show solution
Press once: 32 → 1⁄32. Press again: 1⁄32 → 32. So the display reads 32 again after exactly 2 presses.
B
2.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 19 · 1989 AJHSME
Hard
Ratios, Rates & Proportions
read-cumulative-graph
differences
(A)
(B)
(C)
(D)
(E)
Show hint (soft nudge)
The graph is cumulative, so spending in any time window is the rise of the curve over that window.
Show hint (sharpest)
Read the total at the end of August and subtract the total at the end of May.
Show solution
Summer covers June, July, and August. Read the curve at end-of-May (≈ 2.2) and end-of-August (≈ 4.7). Amount spent over the summer ≈ 4.7 − 2.2 = 2.5 million.
B
2.5.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 20 · 1989 AJHSME
Hard
Geometry & Measurement
cube-net
opposite-faces
(A)
(B)
(C)
(D)
(E)
Show hint (soft nudge)
Pair up the faces that end up opposite each other after folding — opposite faces never share a corner.
Show hint (sharpest)
At every corner, the three meeting faces are one from each opposite pair, so pick the larger number from each pair.
Show solution
Fold the net with 2 as the front: 1 goes on top and 3 on the bottom (1↔3), 6 to the left and 4 to the right (6↔4), and 5 wraps around to the back opposite 2 (2↔5). Each corner has one face from each pair, so the biggest corner sum is 5 + 3 + 6 = 14 .
D
14.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 21 · 1989 AJHSME
Stretch
Fractions, Decimals & Percents
keep-fraction
percent
Jack had a bag of 128 apples. He sold 25% of them to Jill. Next he sold 25% of those remaining to June. Of those apples still in his bag, he gave the shiniest one to his teacher. How many apples did Jack have then?
(A) 7
(B) 63
(C) 65
(D) 71
(E) 111
Show hint (soft nudge)
Selling 25% means keeping 75% = 3⁄4.
Show hint (sharpest)
Multiply by 3⁄4 twice, then subtract 1 for the apple given to the teacher.
Show solution
After Jill: 128 × 3⁄4 = 96. After June: 96 × 3⁄4 = 72. Give 1 to the teacher: 72 − 1 = 71 .
D
71.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 22 · 1989 AJHSME
Stretch
Number Theory
lcm-of-cycle-lengths
The letters A, J, H, S, M, E and the digits 1, 9, 8, 9 are "cycled" separately as follows and put together in a numbered list:
AJHSME 1989
1. JHSMEA 9891
2. HSMEAJ 8919
3. SMEAJH 9198
......... What is the number of the line on which AJHSME 1989 will appear for the first time?
(A) 6
(B) 10
(C) 12
(D) 18
(E) 24
Show hint (soft nudge)
The letters return to AJHSME every 6 lines; the digits return to 1989 every 4 lines.
Show hint (sharpest)
The two parts line up again the first time both cycles finish together.
Show solution
The 6 letters come back after 6 cycles; the 4 digits come back after 4 cycles. Both line up together at the smallest common multiple: LCM(6, 4) = 12 .
C
12.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 23 · 1989 AJHSME
Stretch
Geometry & Measurement
surface-area
exposed-faces
(A)
(B)
(C)
(D)
(E)
Show hint (soft nudge)
Each cube face is 1 m². Count only the faces that are visible — none of those touching the ground or another cube.
Show hint (sharpest)
Sum the exposed faces from the top, front, back, left, right; faces hidden by neighbors don't count.
Show solution
Top faces (one per cube whose top isn't covered) total 10. Front-facing exposed faces total 6, back 6, and the two side walls together account for the remaining 11. Adding gives 10 + 6 + 6 + 11 = 33 square meters of paint.
C
33.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 24 · 1989 AJHSME
Stretch
Geometry & Measurement
fold-and-cut
perimeter-ratio
(A)
(B)
(C)
(D)
(E)
Show hint (soft nudge)
Pick a convenient side length for the square (say 4) so the pieces have whole-number sides.
Show hint (sharpest)
The cut parallel to the fold leaves the fold-side as one rectangle; the open side falls apart into two.
Show solution
Take the square 4 × 4. Folding halves the width to a 2 × 4 stack; cutting it in half parallel to the fold means the cut sits at 1 from the fold. The fold-side unfolds back into one 2 × 4 large rectangle; the open side becomes two separate 1 × 4 small rectangles. Small perimeter 2(1 + 4) = 10, large perimeter 2(2 + 4) = 12, ratio 10⁄12 = 5⁄6 .
E
5⁄6.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
Problem 25 · 1989 AJHSME
Stretch
Counting & Probability
parity
independent-events
(A)
(B)
(C)
(D)
(E)
Show hint (soft nudge)
The sum is even exactly when both wheels land on numbers of the same parity (both even or both odd).
Show hint (sharpest)
Compute the even/odd probabilities of each wheel separately, then combine.
Show solution
Wheel 1 has equal sectors {5, 3, 8, 4}: P(even) = 2⁄4 = ½, P(odd) = ½. Wheel 2 has equal sectors {6, 9, 7}: P(even) = 1⁄3, P(odd) = 2⁄3. P(both even) + P(both odd) = ½·⅓ + ½·⅔ = 1⁄6 + 2⁄6 = 1⁄2 .
C
1⁄2.
Mark:
easy
got-it
tough
follow-up
clear
· log in to save
