AJHSME · Test Mode

1993 AJHSME

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Problem 1 · 1993 AJHSME Easy
Arithmetic & Operations check-choices
ajhsme-1993-01
Show hint (soft nudge)
Multiply each pair and look for the one that isn't 36.
Show hint (sharpest)
Watch the signs — two negatives make a positive.
Show solution
  1. (−4)(−9) = 36, (−3)(−12) = 36, (1)(36) = 36, (3/2)(24) = 36 — all equal 36.
  2. But (1/2)(−72) = −36, not 36, so the odd pair is {1/2, −72}.
C {1/2, −72}.
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Problem 2 · 1993 AJHSME Easy
Fractions, Decimals & Percents simplify-fraction

When the fraction 4984 is expressed in simplest form, the sum of the numerator and the denominator will be

Show hint (soft nudge)
Both 49 and 84 share a factor of 7.
Show hint (sharpest)
Reduce, then add the two parts.
Show solution
  1. 49/84 = 7/12 after dividing top and bottom by 7.
  2. 7 + 12 = 19.
C 19.
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Problem 3 · 1993 AJHSME Medium
Number Theory prime-factorization

Which of the following numbers has the largest prime factor?

Show hint (soft nudge)
Factor each number into primes.
Show hint (sharpest)
Then compare the biggest prime in each.
Show solution
  1. 39 = 3·13, 51 = 3·17, 77 = 7·11, 91 = 7·13, 121 = 11².
  2. The largest prime factor among these is 17, from 51.
B 51.
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Problem 4 · 1993 AJHSME Medium
Arithmetic & Operations regrouping

1000 × 1993 × 0.1993 × 10 =

Show hint (soft nudge)
Group the powers of ten with the decimal.
Show hint (sharpest)
1000 × 0.1993 × 10 collapses neatly.
Show solution
  1. 1000 × 0.1993 × 10 = 1993, leaving 1993 × 1993.
  2. So the product is (1993)².
E (1993)².
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Problem 5 · 1993 AJHSME Medium
Fractions, Decimals & Percents read-graph proportion
ajhsme-1993-05
Show hint (soft nudge)
Read the three slice sizes: one is half, the other two are quarters.
Show hint (sharpest)
The matching bar graph has one bar twice as tall as the two equal ones.
Show solution
  1. The circle is half white and two equal quarters (black and gray).
  2. So the bars should show one tall bar (white) twice the height of two equal shorter bars — that's graph C.
C Bar graph C.
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Problem 6 · 1993 AJHSME Medium
Ratios, Rates & Proportions unit-rate

A can of soup can feed 3 adults or 5 children. If there are 5 cans of soup and 15 children are fed, then how many adults would the remaining soup feed?

Show hint (soft nudge)
First find how many cans the 15 children use.
Show hint (sharpest)
The leftover cans each feed 3 adults.
Show solution
  1. 15 children need 15 ÷ 5 = 3 cans, leaving 5 − 3 = 2 cans.
  2. Those 2 cans feed 2 × 3 = 6 adults.
B 6 adults.
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Problem 7 · 1993 AJHSME Medium
Number Theory exponent-rules

33 + 33 + 33 =

Show hint (soft nudge)
Three copies of the same thing is 3 times it.
Show hint (sharpest)
3 × 3³ = 3¹ × 3³.
Show solution
  1. 3³ + 3³ + 3³ = 3 × 3³.
  2. Adding exponents, 3 × 3³ = 3⁴.
A 3⁴.
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Problem 8 · 1993 AJHSME Medium
Ratios, Rates & Proportions rate

To control her blood pressure, Jill's grandmother takes one half of a pill every other day. If one supply of medicine contains 60 pills, then the supply would last approximately

Show hint (soft nudge)
How many doses are in 60 pills if each dose is half a pill?
Show hint (sharpest)
Each dose covers two days.
Show solution
  1. 60 pills ÷ ½ per dose = 120 doses, and each dose lasts 2 days, so 240 days.
  2. 240 days is about 8 months.
D 8 months.
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Problem 9 · 1993 AJHSME Hard
Algebra & Patterns read-table
ajhsme-1993-09
Show hint (soft nudge)
Read each starred product straight off the table.
Show hint (sharpest)
Do the two inner operations first, then combine.
Show solution
  1. From the table, 2 ∗ 4 = 3 and 1 ∗ 3 = 3.
  2. Then 3 ∗ 3 = 4.
D 4.
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Problem 10 · 1993 AJHSME Medium
Arithmetic & Operations read-graph
ajhsme-1993-10
Show hint (soft nudge)
A 'drop' is a downward segment of the line.
Show hint (sharpest)
Compare the sizes of the downward steps and pick the steepest one.
Show solution
  1. Look only at the segments where the price falls and measure how far each drops.
  2. The steepest downward step starts at the March price, so the greatest monthly drop occurred during March.
B March.
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Problem 11 · 1993 AJHSME Hard
Arithmetic & Operations median cumulative-count
ajhsme-1993-11
Show hint (soft nudge)
With 81 students, the median is the 41st score from the bottom.
Show hint (sharpest)
Add bar heights from the left until you pass 41.
Show solution
  1. The median is the 41st of 81 students. Adding the bars from the low end: 1, 3, 7, 12, 18, 28, 42 — the running total passes 41 at the 70 bar.
  2. So the median lies in the interval labeled 70.
C 70.
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Problem 12 · 1993 AJHSME Hard
Arithmetic & Operations order-of-operations trial

If each of the three operation signs +, −, × is used exactly once in one of the blanks in the expression 5 __ 4 __ 6 __ 3, then the value of the result could equal

Show hint (soft nudge)
Remember multiplication happens before addition and subtraction.
Show hint (sharpest)
Try placing × between 6 and 3.
Show solution
  1. Take 5 − 4 + 6 × 3: the multiplication gives 18 first.
  2. Then 5 − 4 + 18 = 19, using each sign once.
E 19.
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Problem 13 · 1993 AJHSME Hard
Geometry & Measurement area-subtraction
ajhsme-1993-13
Show hint (soft nudge)
White area = whole sign − black letters.
Show hint (sharpest)
Count the unit squares each block letter covers.
Show solution
  1. The sign is 5 × 15 = 75 square units, and the four block letters (1-unit strokes) cover 39 squares in total.
  2. So the white area is 75 − 39 = 36.
D 36.
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Problem 14 · 1993 AJHSME Hard
Logic & Word Problems latin-square deduction
ajhsme-1993-14
Show hint (soft nudge)
Every row and every column must contain 1, 2, 3 exactly once — like a mini Sudoku.
Show hint (sharpest)
Fill in forced cells one at a time.
Show solution
  1. The top row already has 1, so its other cells are 2 and 3; the column with 2 and the diagonal force the middle column to read 3, 2, 1.
  2. Working through, A = 1 and B = 3, so A + B = 4.
C 4.
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Problem 15 · 1993 AJHSME Medium
Arithmetic & Operations average-sum

The arithmetic mean (average) of four numbers is 85. If the largest of these numbers is 97, then the mean of the remaining three numbers is

Show hint (soft nudge)
Find the total of all four numbers from the mean.
Show hint (sharpest)
Remove the largest, then average the other three.
Show solution
  1. The four numbers total 4 × 85 = 340; removing 97 leaves 243.
  2. Their mean is 243 ÷ 3 = 81.0.
A 81.0.
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Problem 16 · 1993 AJHSME Hard
Fractions, Decimals & Percents continued-fraction
ajhsme-1993-16
Show hint (soft nudge)
Work from the bottom up.
Show hint (sharpest)
Simplify 2 + 1/3 first, then move outward.
Show solution
  1. 2 + 1/3 = 7/3, so 1 ÷ (7/3) = 3/7, and 1 + 3/7 = 10/7.
  2. Finally 1 ÷ (10/7) = 7/10.
C 7/10.
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Problem 17 · 1993 AJHSME Hard
Geometry & Measurement surface-area net
ajhsme-1993-17
Show hint (soft nudge)
After cutting 5×5 corners, the base is (20−10) by (30−10) and the height is 5.
Show hint (sharpest)
The open box has a bottom and four inner walls — no top.
Show solution
  1. The base is 10 × 20 = 200 and the height is 5, so the four walls add 2(10·5) + 2(20·5) = 300.
  2. The interior surface (no top) is 200 + 300 = 500.
B 500.
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Problem 18 · 1993 AJHSME Hard
Geometry & Measurement coordinates shoelace
ajhsme-1993-18
Show hint (soft nudge)
Place the rectangle on coordinates: A(0,20), the midpoints give B(16,20) and F(0,10), and D(32,0).
Show hint (sharpest)
Use the shoelace formula on A, B, D, F.
Show solution
  1. With A(0,20), B(16,20), D(32,0), F(0,10), the shoelace formula gives area ½|0·(20−10) + 16·(0−20) + 32·(10−20) + 0·(20−0)|.
  2. That is ½(320 + 320) = 320.
A 320.
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Problem 19 · 1993 AJHSME Hard
Algebra & Patterns pair-terms

(1901 + 1902 + 1903 + … + 1993) − (101 + 102 + 103 + … + 193) =

Show hint (soft nudge)
Line up the two sums term by term.
Show hint (sharpest)
Each top term is exactly 1800 more than the matching bottom term.
Show solution
  1. Both sums have 93 terms, and each top term beats its partner by 1901 − 101 = 1800.
  2. So the difference is 93 × 1800 = 167,400.
A 167,400.
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Problem 20 · 1993 AJHSME Stretch
Number Theory borrowing-pattern digit-sum

When 1093 − 93 is expressed as a single whole number, the sum of the digits is

Show hint (soft nudge)
Try a small case like 10⁴ − 93 = 9907 to see the shape.
Show hint (sharpest)
10ⁿ − 93 is a string of nines ending in 07.
Show solution
  1. 10⁴ − 93 = 9907, 10⁵ − 93 = 99907 — so 10ⁿ − 93 is (n − 2) nines followed by 07.
  2. For n = 93 that's 91 nines and a 0 and 7: 91 × 9 + 7 = 826.
D 826.
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Problem 21 · 1993 AJHSME Hard
Fractions, Decimals & Percents percent-area

If the length of a rectangle is increased by 20% and its width is increased by 50%, then the area is increased by

Show hint (soft nudge)
Area multiplies by the two growth factors.
Show hint (sharpest)
1.2 × 1.5 — how much more than 1 is that?
Show solution
  1. The new area is 1.2 × 1.5 = 1.8 times the old area.
  2. That's an increase of 80%.
D 80%.
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Problem 22 · 1993 AJHSME Stretch
Counting & Probability digit-counting

Pat Peano has plenty of 0's, 1's, 3's, 4's, 5's, 6's, 7's, 8's and 9's, but he has only twenty-two 2's. How far can he number the pages of his scrapbook with these digits?

Show hint (soft nudge)
The only limit is the supply of 2's — count how many 2's the page numbers use.
Show hint (sharpest)
Pages 1–99 use 20 twos; then keep going until the 2's run out.
Show solution
  1. Pages 1–99 use 20 twos (ten in the units place, ten in the tens place). That leaves 2 twos.
  2. Pages 102 and 112 use one 2 each, exhausting the supply; pages 113–119 need no 2, but 120 would, so he can reach 119.
D 119.
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Problem 23 · 1993 AJHSME Stretch
Logic & Word Problems ordering constraints

Five runners, P, Q, R, S, T, have a race. P beats Q, P beats R, Q beats S, and T finishes after P and before Q. Who could NOT have finished third in the race?

Show hint (soft nudge)
Combine the clues into chains: P is ahead of Q, R, and T, and S is behind Q.
Show hint (sharpest)
Count how many runners must be ahead of each person.
Show solution
  1. P beats Q, R, and T, and P < T < Q < S, so P is always first — it can't be third.
  2. S comes after Q, which comes after both P and T, so at least three runners beat S, putting it 4th or later — also never third. So the answer is P and S.
C P and S.
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Problem 24 · 1993 AJHSME Stretch
Number Theory pattern rows
ajhsme-1993-24
Show hint (soft nudge)
Row k ends at the perfect square k², and holds 2k − 1 numbers.
Show hint (sharpest)
Find which row holds 142, then the number sitting one row up and aligned with it.
Show solution
  1. Rows end at 1, 4, 9, 16, …, k², so 142 is in row 12 (122–144), as its 21st of 23 entries.
  2. Row 11 (101–121) sits centered above, so directly above the 21st entry is the 20th entry of row 11: 101 + 19 = 120.
C 120.
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Problem 25 · 1993 AJHSME Stretch
Geometry & Measurement covering tilting

A checkerboard consists of one-inch squares. A square card, 1.5 inches on a side, is placed on the board so that it covers part or all of the area of each of n squares. The maximum possible value of n is

Show hint (soft nudge)
Don't keep the card lined up with the grid — tilt it.
Show hint (sharpest)
A tilted card pokes its corners into many extra squares.
Show solution
  1. Lined up, the 1.5-inch card touches only up to a 3 × 3 block (9 squares).
  2. But tilting it lets its corners reach into still more squares, so it can cover 12 or more.
E 12 or more.
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