AMC 8-style practice test

Hint 1: Find the adult-ticket cost and the child-ticket cost separately.

Hint 2: The two costs are 2 x $7 and 3 x $4.

Answer: D. $26

The adult tickets cost 2 x $7 = $14, and the child tickets cost 3 x $4 = $12. The total is $14 + $12 = $26.

Hint 1: Apply the discount before calculating the tax.

Hint 2: After the discount, the price is 80% of $50.

Answer: C. $44

The discounted price is 0.80 x $50 = $40. The tax is 0.10 x $40 = $4, so the final price is $44.

Hint 1: The angles of a triangle add to 180 degrees.

Hint 2: There are 2 + 3 + 4 = 9 equal angle-parts.

Answer: C. 40 degrees

Each part is 180/9 = 20 degrees. The smallest angle has 2 parts, so it measures 40 degrees.

Hint 1: A number divisible by 5 must end in 0 or 5.

Hint 2: Here the units digit must be 5. List the possible tens digits.

Answer: B. 2

The units digit must be 5, since 0 is unavailable. The tens digit can be 3 or 7. The numbers are 35 and 75, giving 2 numbers.

Hint 1: Half the perimeter is the length plus the width.

Hint 2: If the width is w, then the length plus the width is 3w.

Answer: D. 50 square cm

The length plus the width is 30/2 = 15 cm. Thus 3w = 15 and w = 5 cm. The length is 10 cm, giving area 10 x 5 = 50 square cm.

Hint 1: Recover the sum of the five numbers from their mean.

Hint 2: Subtract 8 from that sum, then divide by 4.

Answer: D. 13

The original sum is 5 x 12 = 60. The remaining sum is 60 - 8 = 52, so the new mean is 52/4 = 13.

Hint 1: Count multiples of 4 and multiples of 6.

Hint 2: Multiples of 12 appear in both lists, so subtract their count once.

Answer: C. 20

There are 15 multiples of 4 and 10 multiples of 6. The 5 multiples of 12 were counted twice. Therefore the required count is 15 + 10 - 5 = 20.

Hint 1: An odd sum uses one odd result and one even result.

Hint 2: There are 3 odd and 3 even faces; count both possible orders.

Answer: C. 1/2

Of the 36 equally likely ordered results, 3 x 3 = 9 are odd then even, and another 9 are even then odd. The probability is 18/36 = 1/2.

Hint 1: To reach the eighth term, make seven additions.

Hint 2: Pair terms in 2 + 4 + 6 + 8 + 10 + 12 + 14.

Answer: D. 59

The seventh addition is 14. The seven added numbers sum to 7 x (2 + 14)/2 = 56. Thus the eighth term is 3 + 56 = 59.

Hint 1: Count all routes, then subtract the routes that visit (1, 1).

Hint 2: There are 2 routes to (1, 1) and 3 routes from there to (3, 2).

Answer: B. 4

Every route consists of 3 right steps and 2 up steps, so there are 10 ways to choose the positions of the up steps. A route through (1, 1) has 2 choices before that point and 3 after it, giving 6 routes. Therefore 10 - 6 = 4 routes avoid the point.

Hint 1: Remove the points earned in tied games.

Hint 2: The wins account for 22 - 4 = 18 points.

Answer: C. 2

The team has 18/3 = 6 wins. Its 12 games consist of 6 wins, 4 ties, and 12 - 6 - 4 = 2 losses.

Hint 1: The tile side length must divide both 84 and 60.

Hint 2: Their greatest common divisor is 12.

Answer: D. 35

The largest possible tile side is gcd(84, 60) = 12 cm. There are 84/12 = 7 tiles along one side and 60/12 = 5 along the other, so 7 x 5 = 35 tiles are needed.

Hint 1: Find what fraction of the capacity was added.

Hint 2: The increase is 5/6 - 1/3 = 1/2 of the tank.

Answer: C. 24 liters

The added 12 liters fill 1/2 of the tank. The full capacity is therefore 12 divided by 1/2, or 24 liters.

Hint 1: Small cubes with exactly two painted faces lie on edges but not at corners.

Hint 2: Each of the large cube's 12 edges has one noncorner small cube.

Answer: C. 12

A corner small cube has three painted faces. On each edge, the one small cube between the corners has exactly two painted faces. The 12 edges give 12 such cubes; face-interior and interior cubes have fewer painted faces.

Hint 1: Solve for x in terms of y and require x >= 1.

Hint 2: The possible values of y are the integers from 1 through 7.

Answer: C. 7

We have x = 15 - 2y. Positivity gives 1 <= y <= 7. Each of these seven integer values gives exactly one positive x, so there are 7 ordered pairs.

Hint 1: Triangles with the same altitude have areas in the ratio of their bases.

Hint 2: First replace AB with AM, then replace AC with AN.

Answer: B. 12

Triangle AMC has half the area of ABC because AM = AB/2 and the two triangles share the altitude from C. Its area is 36. Since AN = AC/3, triangle AMN has one-third of the area of AMC, or 12 square units.

Hint 1: List the numbers that are 3 more than a multiple of 7.

Hint 2: After finding the first common solution, add 35 to find the next.

Answer: C. 3

Among 3, 10, 17, 24, and 31, only 17 has remainder 2 on division by 5. Both remainders repeat every 35, giving 17, 52, and 87 in the required range. The next is 122, so there are 3.

Hint 1: Count unordered pairs of counters.

Hint 2: There are 3 pairs of red counters and 1 pair of blue counters.

Answer: C. 2/5

There are 5 x 4 / 2 = 10 equally likely unordered pairs. Of these, 3 x 2 / 2 = 3 are red-red, and 1 is blue-blue. Thus the probability is 4/10 = 2/5.

Hint 1: A palindrome reads the same in either direction.

Hint 2: Write its digits as 2, b, 2 and use the divisibility-by-9 digit-sum test.

Answer: D. 252

The number has the form 2b2, whose digit sum is b + 4. For 0 <= b <= 9, this sum lies from 4 through 13, so the only possible multiple of 9 is 9. Therefore b = 5 and the number is 252.

Hint 1: Start with all selections of two chairs.

Hint 2: There are 5 adjacent pairs to exclude.

Answer: C. 10

There are 6 x 5 / 2 = 15 unordered selections of two chairs. Exactly 5 selections use adjacent chairs, namely consecutive positions 1-2 through 5-6. Hence 15 - 5 = 10 selections are allowed.

Hint 1: The hour hand has moved beyond 3 by 3:20.

Hint 2: The minute hand is 120 degrees clockwise from 12; the hour hand moves 0.5 degree each minute.

Answer: B. 20 degrees

At 3:00 the hour hand is 90 degrees clockwise from 12. In 20 minutes it moves 10 degrees, reaching 100 degrees. The minute hand is at 20 x 6 = 120 degrees. Their smaller separation is 120 - 100 = 20 degrees.

Hint 1: Count separately by side length.

Hint 2: There are 16 of side 1, 9 of side 2, and so on.

Answer: D. 30

The possible side lengths are 1, 2, 3, and 4. A square of side k has (5 - k)^2 possible positions. The total is 16 + 9 + 4 + 1 = 30.

Hint 1: Let the larger number have the larger tens digit.

Hint 2: With consecutive tens digits, the difference is 10 plus the difference of the units digits.

Answer: C. 7

If the tens digits differ by at least 2, the difference is at least 20 - 3 = 17. If they are consecutive, check the three possible pairs of tens digits. Tens 1 and 2 leave units 3 and 4, giving minimum difference 9; tens 2 and 3 leave 1 and 4, giving 10 - 3 = 7; tens 3 and 4 give 9. Thus 31 - 24 = 7 is the minimum.

Hint 1: For a fixed sum, products increase as the two numbers get closer together.

Hint 2: The closest pair is 6 and 6, but its product is divisible by 3. Try the next closest pair.

Answer: D. 35

Since (a - b)^2 >= 0, we have ab <= (a + b)^2/4 = 36, with equality only at a = b = 6. That pair is forbidden. The next integer below 36 is 35, and it is attained by 5 and 7. Since 35 is not divisible by 3, the greatest allowed product is 35.

Hint 1: Look at piles of 4 stones and then 8 stones.

Hint 2: Leave a multiple of 4, then make each pair of moves remove 4 stones.

Answer: B. Remove 2 stones

Remove 2 stones, leaving 28. Whenever the opponent removes r stones, remove 4 - r stones. This is always a legal move and returns a multiple of 4 to the opponent, eventually leaving 0 after your move. A first move of 1 or 3 leaves 29 or 27; the opponent can then leave 28 or 24 and use the same strategy. Thus only removing 2 guarantees a win.