AMC 8 Practice Test 8

Answer: A, 40 minutes. First find the total time in seconds; then change the unit to minutes.

1. There are 600 dimples, each requiring 4 seconds: \(600\times4=2400\) seconds.
2. Every minute contains 60 seconds, so \(2400\div60=40\) minutes.

Check: In one minute she paints \(60\div4=15\) dimples. Forty minutes gives \(40\times15=600\).

SOMATH takeaway: Write the units beside each result. A correct number in the wrong unit is not the answer.

Answer: D, 12. We need a pair that passes two tests: multiplication gives 132 and addition gives 23.

1. Try nearby factors around the square root of 132: \(11\times12=132\).
2. Check their sum: \(11+12=23\).
3. The larger number is 12, not 11.

Why factor pairs help: The other positive pairs are \((1,132),(2,66),(3,44),(4,33),(6,22)\); none adds to 23. Don’t stop after checking only the product.

Answer: E, 88 dollars. Follow the people in the correct order: Gary → Emily → Frank.

1. One third of Gary’s money is \(126\div3=42\).
2. Emily has two thirds, so she has \(42\times2=84\) dollars.
3. Frank has 4 dollars more than Emily: \(84+4=88\).

Common mistake: Adding 4 to Gary’s money uses the wrong person as the starting amount.

Answer: E, 2. A number is odd when its last digit is odd. To make the whole number as large as possible, protect the largest digits for the leftmost places.

1. Put 9 first, then 6, then 5. These are the largest available digits, in order.
2. The remaining digits are 2 and 3. The units digit must be odd, so put 3 last and 2 just before it.
3. The greatest possible number is \(96523\). Its tens digit is 2.

Check: Ending in 5 would force the smaller digit 3 into an earlier place, giving at most 96325. Ending in 9 would lose the leading 9 entirely.

Answer: B, 300 feet. Count the spaces between trees, not the trees themselves.

1. From tree 1 to tree 5 there are \(5-1=4\) equal spaces.
2. Each space is \(80\div4=20\) feet.
3. From tree 1 to tree 16 there are \(16-1=15\) spaces.
4. The distance is \(15\times20=300\) feet.

SOMATH takeaway: A straight row of \(n\) objects has \(n-1\) gaps between its first and last objects.

Answer: E, 4%. Use a convenient starting amount of 100 dollars for Yao. Percent relationships do not depend on the size of that starting amount.

1. James has \(100+20=120\) dollars.
2. Bob has 20% less than 120. Since \(0.20\times120=24\), Bob has \(120-24=96\) dollars.
3. Compared with Yao’s 100 dollars, Bob is short by 4 dollars, or 4%.

Algebra check: \(1.20\times0.80=0.96\). The increase and decrease do not cancel because they use different starting amounts.

Answer: C, 26. Segment \(AB\) is the diagonal of a right triangle we can imagine inside the diagram.

1. The horizontal distance from \(A\) to \(B\) is the combined width of the two squares: \(7+17=24\).
2. The vertical rise is the difference of their heights: \(17-7=10\).
3. Apply the Pythagorean theorem: \(AB^2=24^2+10^2=576+100=676\).
4. Take the positive square root: \(AB=26\).

Check: The side lengths \(10,24,26\) are twice the familiar \(5,12,13\) right triangle. The 17-unit square side is not the vertical rise from \(A\).

Answer: E, \(\frac7{12}\). The sample space is the set of positive factors of 72, not the integers from 1 to 72.

1. List the factors in order: \(1,2,3,4,6,8,9,12,18,24,36,72\). There are 12.
2. The factors below 11 are \(1,2,3,4,6,8,9\). There are 7.
3. Probability is favorable choices divided by all equally likely choices: \(\frac7{12}\).

Factor-count check: \(72=2^3\cdot3^2\), so it has \((3+1)(2+1)=12\) positive factors.

Answer: D, 41235. Sort the numbers into blocks according to their first digit.

1. Once the first digit is chosen, the other four digits can be arranged in \(4\times3\times2\times1=24\) ways.
2. Numbers beginning with 1 occupy positions 1–24. Those beginning with 2 occupy 25–48, and those beginning with 3 occupy 49–72.
3. Position 73 is therefore the first number beginning with 4.
4. Put the remaining digits in increasing order to make the smallest number in that block: \(4\,1235=41235\).

Source correction: The supplied solution prints \(3\times24=71\). The correct total is 72; the answer choice remains D.

Answer: C, 51 square units. The two vertical sides are parallel, so this is a trapezoid turned sideways.

1. The right parallel side has length \(5-(-5)=10\).
2. The left parallel side has length \(2-(-5)=7\).
3. The perpendicular distance between them is \(3-(-3)=6\).
4. Area is \(\frac{10+7}{2}\times6=51\).

Another way: Enclose the figure in a \(6\times10\) rectangle. Remove the top triangle of base 6 and height 3: \(60-\frac12(6)(3)=51\).

Answer: D, 60 degrees. First find the number of cherry votes; then convert that fraction of the class into a fraction of a full circle.

1. The three named groups contain \(14+18+8=40\) students.
2. That leaves \(60-40=20\), half of whom choose cherry: \(20\div2=10\).
3. Cherry represents \(\frac{10}{60}=\frac16\) of the class.
4. Its angle is \(\frac16\times360^\circ=60^\circ\).

Check: Each of the 60 students represents \(360\div60=6\) degrees, and \(10\times6=60\).

Answer: C, 1000. Treat each category as a selection of distinct menu items. Order does not matter, so beef–chicken and chicken–beef are the same meat choice.

1. Choose 2 of 5 meats: \(\frac{5\times4}{2}=10\) selections.
2. Choose 3 of 6 vegetables: \(\frac{6\times5\times4}{3\times2\times1}=20\) selections.
3. Choose 4 of 5 desserts: there are 5 choices, because choosing the one dessert to leave out determines the other four.
4. Combine independent category choices: \(10\times20\times5=1000\).

Why multiply? Every meat selection can be paired with every vegetable selection and every dessert selection. Adding would count individual choices, not complete meals.

Answer: A, 70 potatoes. Split the work into two periods: Helen alone, then both together.

1. During her five-minute head start, Helen peels \(5\times5=25\) potatoes.
2. There are \(145-25=120\) potatoes left.
3. Together they peel \(5+7=12\) per minute, so the remaining work takes \(120\div12=10\) minutes.
4. Charles works only those 10 minutes: \(7\times10=70\) potatoes.

Check: Helen works 15 minutes and peels 75. Charles peels 70, and \(75+70=145\).

Answer: A, 1141 circles. See the figures as a center circle surrounded by larger hexagonal rings.

1. Figure 1 has 1 circle. To make Figure 2, add a ring of 6 circles.
2. The next ring adds 12, then the next would add 18. Ring \(k\) adds \(6k\) circles.
3. Figure 20 has the center plus 19 rings: \(1+6(1+2+\cdots+19)\).
4. Pair the sum or use its formula: \(1+2+\cdots+19=\frac{19\times20}{2}=190\).
5. The total is \(1+6(190)=1141\).

Check the rule first: It gives 7 circles for Figure 2 and 19 for Figure 3, matching the drawing. The general total is \(1+3n(n-1)\) for Figure \(n\).

Answer: D, 2014. Avoid squaring 2013 by long multiplication. Make the expression fit \((n+1)^2=n^2+2n+1\).

1. Write \(2014=2013+1\).
2. The expression under the square root becomes \(2013^2+2(2013)+1\).
3. This is \((2013+1)^2=2014^2\).
4. The positive square root is \(a=2014\).

SOMATH takeaway: Before calculating a large expression, look for a familiar structure. Here the two final terms complete the square.

Answer: A, \(\frac{91}{216}\). For ordinary fair six-sided dice, only a face showing 5 contributes a factor of 5.

1. The product is a multiple of 5 exactly when at least one die shows 5.
2. The opposite event is that no die shows 5. Each die has probability \(\frac56\) of avoiding 5.
3. For three independent dice, the probability of avoiding 5 every time is \((\frac56)^3=\frac{125}{216}\).
4. Subtract from 1: \(1-\frac{125}{216}=\frac{91}{216}\).

Common mistake: \(3\times\frac16\) double-counts some outcomes with two or three fives. “At least one” often suggests using the complement.

Answer: D, 36. Imagine ten counters in a row. Two dividers split that row into three nonempty groups.

1. There are 9 gaps between the 10 counters.
2. Choose 2 different gaps for the dividers. The resulting group sizes are the first, second, and third numbers.
3. The number of choices is \(\binom92=\frac{9\times8}{2}=36\).

Why this matches the wording: Choosing internal gaps keeps every group positive; reading the groups left to right preserves order.

Alternative check: If the first number is 1, there are 8 choices for the second; if it is 2, there are 7, and so on. Thus \(8+7+\cdots+1=36\).

Answer: A, 12 meetings. Their opposite directions mean their speeds add when measuring how quickly they close a gap.

1. The initial gap is half the track: \(15\div2=7.5\) km.
2. The relative speed is \(35+25=60\) km/h, so the first meeting occurs after \(7.5\div60=\frac18\) hour.
3. Each later meeting requires another full track of relative travel: \(15\div60=\frac14\) hour.
4. Meeting times are \(\frac18,\frac38,\frac58,\ldots,\frac{23}{8}\) hours. These are 12 times, ending at \(2\frac78\) hours.
5. The next would be \(\frac{25}{8}=3\frac18\) hours, after they stop.

Important detail: They do not meet at time zero or exactly at 3 hours. Accounting for the initial half-lap avoids an off-by-one error.

Answer: A, \(8\sqrt{34}\) square centimeters. A triangular face needs its own perpendicular height along the sloping surface, not the pyramid’s vertical height.

1. The apex is above the square’s center. The horizontal distance from that center to the midpoint of a side is half the side length: \(4\sqrt2\).
2. The face’s slant height \(\ell\) is the hypotenuse of a right triangle with legs 6 and \(4\sqrt2\).
3. Thus \(\ell^2=6^2+(4\sqrt2)^2=36+32=68\), giving \(\ell=2\sqrt{17}\).
4. One face has area \(\frac12(8\sqrt2)(2\sqrt{17})=8\sqrt{34}\).

Common mistake: Using 6 as the height of the triangular face gives the wrong area. Six measures straight up from the base, not along the face.

Answer: A, at least 3 students. It is easier to count students who cannot do each sport.

1. There are \(52-30=22\) who cannot swim, \(52-35=17\) who cannot ride, and \(52-42=10\) who cannot play table tennis.
2. Even if these groups do not overlap, they cover only \(22+17+10=49\) students.
3. At least \(52-49=3\) students are outside all three “cannot” groups. Those students can do all three sports.

Why the minimum is achievable: Let 22 students miss only swimming, 17 miss only cycling, 10 miss only table tennis, and 3 miss none. This produces exactly 30 swimmers, 35 cyclists, and 42 table-tennis players, so the lower bound really can occur.

Answer: D, 25 triangles. There are five marked points on the diameter and two above it. A triangle cannot have all three vertices on the same straight line.

1. Choose two diameter points and one upper point: \(\binom52\times2=10\times2=20\) triangles.
2. Choose one diameter point and both upper points: \(5\times1=5\) triangles.
3. These cases cannot overlap, so add them: \(20+5=25\).

Alternative check: Choose any three of the seven points, then remove the triples entirely on the diameter: \(\binom73-\binom53=35-10=25\).

Answer: D, 17 arrangements. Because the stamps are placed in a row, changing the order can create a new arrangement.

1. With no 4¢ stamp, the possibilities are five 2¢ stamps (1 order) or two 2¢ and two 3¢ stamps (\(\frac{4!}{2!2!}=6\) orders). Total: 7.
2. With one 4¢ stamp, the remaining 6¢ is either three 2¢ stamps (4 positions for the 4¢ stamp) or two 3¢ stamps (3 positions for the 4¢ stamp). Total: 7.
3. With two 4¢ stamps, the remaining 2¢ is one stamp, which can occupy any of 3 positions. Total: 3.
4. Three 4¢ stamps would exceed 10¢. Therefore the full count is \(7+7+3=17\).

SOMATH takeaway: Organize by the number of largest stamps. This keeps the cases separate and makes it easy to see that nothing is missing.

Answer: D, 6 revolutions. The diagram shows external rolling, with circle \(A\) fixed. Count the turns of \(B\) relative to a fixed observer, not relative to the moving line between the centers.

1. Let the fixed radius be \(R=10\) and the rolling radius be \(r=2\). The moving center travels on a circle of radius \(R+r=12\).
2. If that center moves at speed \(v\), no slipping means the contact point is instantaneously at rest. The rotational rim speed must cancel the center’s speed, so \(r\omega=v\).
3. Over the full orbit, the center travels \(2\pi(R+r)\). The total rotation angle is therefore \(\frac{2\pi(R+r)}r\).
4. Divide by \(2\pi\) radians per revolution: \(\frac{R+r}{r}=\frac{12}{2}=6\).

Why not 5? The ratio \(R/r=5\) misses the extra turn caused by going all the way around the curved path. For external rolling, the rule is \(R/r+1\).

Answer: D, \(\frac47\). “Last marble drawn” means the marble at the stopping moment, not necessarily the seventh marble.

1. Imagine secretly arranging all seven marbles in a random order before the drawing begins.
2. If the seventh marble in that full order is red, all three whites must be exhausted before all four reds. The actual drawing stops on a white marble.
3. If the seventh marble is white, the opposite happens: all reds are exhausted first, so the actual drawing stops on red.
4. Thus stopping on white is exactly the event that the seventh marble in the full order is red. Four of the seven marbles are red, so its probability is \(\frac47\).

Counting check: There are \(\binom73=35\) equally likely choices of the white positions. To leave the final position red, choose all three white positions among the first six: \(\binom63=20\). Hence \(20/35=4/7\).

Source correction: The supplied solution labels D correctly but prints \(20/35=5/7\). The correct simplification is \(4/7\).

Answer: A, \(\frac23\). The selection is made from the multiples of at least one of 3, 4, or 5. It is not made from all 300 integers.

1. Count the multiples separately: \(300/3=100\), \(300/4=75\), and \(300/5=60\).
2. Subtract the double-counted overlaps: multiples of 12 give 25, multiples of 15 give 20, and multiples of 20 give 15.
3. Add back the triple overlap, the 5 multiples of 60. The separate counts sum to 235 and the pairwise overlaps sum to 60. The eligible total is \(235-60+5=180\).
4. All 60 multiples of 5 are in this eligible set. Remove them to get \(180-60=120\) eligible numbers not divisible by 5.
5. The desired probability is \(\frac{120}{180}=\frac23\).

Why add the triple overlap back? A number such as 60 was added three times and subtracted three times. It should be counted once, so add it back once.

Check: The complement is \(60/180=1/3\), and \(1-1/3=2/3\).