2003 AMC 8 Solutions

A cube has a fixed number of edges, corners, and faces.

Add the counts for edges, corners, and faces.

A cube has $12$ edges, $8$ corners, and $6$ faces. Adding these together yields $$12 + 8 + 6 = 26.$$

Thus, E is the correct answer.

Look for a number with prime factor $2.$

Among the choices, only one is even.

Note that the smallest prime number is $2.$ This means that any even number would be our answer.

Thus, C is the correct answer.

First find how many grams are not filler.

Compare the non-filler grams to $120.$

We get that $120 - 30 = 90$ grams are not filler. The percentage is therefore $$100 \cdot \dfrac{90}{120} = 100 \cdot \dfrac{3}{4} = 75 \%.$$

Thus, D is the correct answer.

If all $7$ children had bicycles, count the wheels.

Each tricycle adds one wheel compared with a bicycle.

Let $b$ be the number of bicycles and $t$ be the number of tricycles. Then we can set up the following system of equations: $$\begin{gather*} b + t = 7, \\ 2b + 3t = 19. \end{gather*}$$ Multiplying the first equation by $2$ and subtracting from the second equation, we get $t = 5.$

Thus, C is the correct answer.

$20\%$ is one fifth.

Use $10\%$ as a stepping stone from $20\%$ to $30\%.$

Since $20\%$ of the number is $12,$ $10\%$ of the number is $6.$

Therefore $30\%$ of the same number is $3\cdot6=18.$

Thus, B is the correct answer.

The square areas give the side lengths of the three squares.

The side lengths form a $5-12-13$ right triangle.

The side lengths of the squares are $$\sqrt{169} = 13,$$$$\sqrt{144} = 12,$$ and $$\sqrt{25} = 5.$$ These lengths form a Pythagorean triple.

Therefore, the interior triangle is right. Its area is $$\dfrac{1}{2} \cdot 5 \cdot 12 = 30.$$

Thus, B is the correct answer.

Add Jenny’s four score differences from Blake’s.

Divide the total difference by $4$ to compare averages.

The total point difference between the two is $$10 - 10 + 20 \cdot 2 = 40.$$ The average of this difference is $40 \div 4 = 10.$

Thus, A is the correct answer.

Same dough and thickness means larger cookie area gives fewer cookies.

Compare the areas of the four cookie shapes.

Since the cookies all have the same thickness and use the same amount of dough, the largest cookie shape produces the fewest cookies.

Art’s trapezoid has area $\frac12(3+5)\cdot3=12.$ Roger’s rectangle has area $4\cdot2=8.$ Paul’s parallelogram has area $3\cdot2=6,$ and Trisha’s triangle has area $\frac12\cdot3\cdot4=6.$

Art has the largest cookie area, so Art gets the fewest cookies from one batch.

Thus, A is the correct answer.

First find the total money Art earns from a batch.

Use the number of Roger’s cookies from the shared dough amount.

Art makes $12$ cookies that sell for $60$ cents each, so one batch earns $12\cdot60=720$ cents.

The batch has $12\cdot12=144$ square inches of dough area, and each Roger cookie has area $4\cdot2=8.$ Thus Roger makes $144\div8=18$ cookies.

To earn $720$ cents from $18$ cookies, each Roger cookie should cost $720\div18=40$ cents.

Thus, C is the correct answer.

Trisha’s cookie area is half of Art’s cookie area.

Same dough amount means half-size cookies make twice as many.

Art’s cookie area is $12$ square inches, so the whole batch has area $12\cdot12=144$ square inches.

Trisha’s triangular cookie has area $\frac12\cdot3\cdot4=6$ square inches.

Therefore Trisha can make $144\div6=24$ cookies per batch.

Thus, E is the correct answer.

The Monday discount is taken from Friday’s increased price.

Multiply the original price by $1.10,$ then by $0.90.$

On Friday, the shoes are marked up by $10\%,$ so the listed price becomes $40\cdot1.10=44$ dollars.

On Monday, the $10\%$ discount is taken from $\$44,$ giving $44\cdot0.90=39.60$ dollars.

Thus, B is the correct answer.

Check the only case where the face numbered $6$ is hidden.

If $6$ is hidden, the visible faces still include $2$ and $3.$

If the face numbered $6$ is visible, then the visible product is divisible by $6.$

If the face numbered $6$ is on the bottom, then the visible faces include both $2$ and $3,$ so their product is still divisible by $6.$

Every possible toss works, so the probability is $1.$

Thus, E is the correct answer.

A cube has four painted faces exactly when it touches two other cubes.

Separate the top cubes, bottom corner cubes, and remaining cubes.

A cube has exactly four painted faces exactly when it is attached to exactly two other cubes.

The $4$ top cubes touch only one other cube, so they have $5$ painted faces. The $4$ bottom corner cubes touch three other cubes, so they have $3$ painted faces.

The remaining $14-4-4=6$ cubes each touch exactly two other cubes, so they have exactly four painted faces.

Thus, B is the correct answer.

Use the hundreds column after $T = 7.$

The ones column and the fact that $O$ is even force $O.$

Since both $T$’s are $7,$ we get that $O$ is either $4$ or $5.$ Since $O$ is even, we get that $O = 4.$

Then, we get that $R = 4 + 4 = 8.$ We also know that $W + W$ doesn’t carry over, since otherwise $O$ would be $5.$

Therefore, $W$ is less than $5$ and cannot be $4$ or $1.$ If $W = 0,$ then $U = 0,$ which gives two letters the same digit. If $W = 2,$ then $U = 4,$ which is also not allowed.

This makes $W = 3.$

Thus, D is the correct answer.

Try to satisfy both views with only three cubes.

A four-cube construction is shown; you only need to rule out three.

The front view requires an L-shape with three visible positions, so at least three cubes are needed.

Suppose there were exactly three cubes. The two cubes that appear one above the other must share a face, so they have the same depth. The third cube appears beside the lower one. To share a face with either of the first two cubes, it must share the lower cube’s depth as well; otherwise it would be disconnected from both. Thus all three cubes would occupy one depth column, giving a one-column side view instead of the required L-shape.

The shown four-cube construction has both required views, so the minimum is $4.$

Thus, B is the correct answer.

Choose the driver first.

After the driver is chosen, arrange the other three people in the remaining seats.

There are $2$ options for who sits in the driver’s seat. There are $3$ options for the other front seat, and $2$ options for the first back seat.

The last person has to sit in the last seat, for a total of $$2 \cdot 3 \cdot 2 = 12$$ possible seating arrangements.

Thus, D is the correct answer.

Jim’s siblings must each share eye color or hair color with Jim.

The two siblings must also share at least one characteristic with each other.

Note that Nadeen, Austin, and Sue are the only individuals who share a characteristic with Jim. We need to find which of the $3$ are completely different from the others.

Austin and Sue both have blue eyes, which makes Nadeen the odd one out. Therefore, Austin and Sue are Jim’s siblings.

Thus, E is the correct answer.

Sarah invites vertices at distance $1$ or $2$ from her.

Count the dots not within two line segments of Sarah.

Sarah invites her friends and the classmates who are friends with at least one of her friends. In graph terms, she invites dots that are $1$ or $2$ line segments away from Sarah.

From the graph, $4$ dots are disconnected from Sarah’s component, and $2$ more dots in Sarah’s component are $3$ segments away.

Those $4+2=6$ classmates will not be invited.

Thus, D is the correct answer.

Use the least common multiple of $15,$ $20,$ and $25.$

Count the multiples of that LCM between $1000$ and $2000.$

If a number $x$ has these three numbers as factors, then their least common multiple must also divide $x.$

These numbers have the following prime factorizations: $$15 = 3 \cdot 5,$$$$20 = 2^2 \cdot 5,$$$$25 = 5^2.$$

From these values, we get that the least common multiple is $$2^2 \cdot 3 \cdot 5^2 = 300.$$

Therefore, the multiples of $300$ between $1000$ and $2000$ are $1200,$ $1500,$ and $1800.$

Thus, C is the correct answer.

At $4{:}20,$ the minute hand points at $4.$

The hour hand has moved one third of the way from $4$ to $5.$

At $4{:}20,$ the hour hand will be $\frac{1}{3}$ of the way between $4$ and $5.$

Each hour represents $360 \div 12 = 30^{\circ}.$ This means the hour hand will be $30 \div 3 = 10^{\circ}$ past $4.$

At $20$ minutes, the minute hand points to $4,$ so the acute angle between the hands is $10^{\circ}.$

Thus, D is the correct answer.

Drop perpendiculars from $B$ and $C$ to the base.

Use the $6-8-10$ and $8-15-17$ right triangles.

Drop perpendiculars from $B$ and $C$ to $\overline{AD},$ meeting it at $E$ and $F.$

In the left and right right triangles, $AE=\sqrt{10^2-8^2}=6$ and $FD=\sqrt{17^2-8^2}=15.$

The two side triangles have areas $\frac12\cdot6\cdot8=24$ and $\frac12\cdot15\cdot8=60.$ The middle rectangle has area $8\cdot BC.$

Thus $164=24+60+8BC,$ so $8BC=80$ and $BC=10.$

Thus, B is the correct answer.

Figures $A$ and $B$ leave the same shaded area.

For $C,$ compare $\pi-2$ with $4-\pi.$

In figure $A,$ the shaded area is the area of a $2$ by $2$ square minus a circle of radius $1,$ so it is $4-\pi.$

Figure $B$ is made of four copies of figure $A$ with half the side length, so each copy has one-fourth the area and the total shaded area is also $4-\pi.$

In figure $C,$ the circle has radius $1,$ and the inscribed square has diagonal $2,$ so its area is $2.$ The shaded area is $\pi-2.$

Since $\pi>3,$ we have $\pi-2>4-\pi,$ so figure $C$ has the largest shaded area.

Thus, C is the correct answer.

The cat repeats every $4$ moves.

The mouse repeats every $8$ moves; use the remainders of $247.$

The cat’s position repeats every $4$ moves, and the mouse’s position repeats every $8$ moves.

Since $247\equiv3\pmod{4},$ the cat is in the same position as after the $3$rd move: the lower right square.

Since $247\equiv7\pmod{8},$ the mouse is in the same position as after the $7$th move: the left side of the lower left square.

Thus, A is the correct answer.

Along the semicircle, the distance from $X$ is constant.

On the straight segment, the ship first gets closer to $X,$ then farther away.

Every point on the semicircular path from $A$ to $B$ is the same distance from the center $X,$ so the graph starts horizontally.

On the straight path from $B$ to $C,$ the ship first gets closer to $X$ and then farther away from $X.$

Only graph $B$ starts flat and then decreases before increasing.

Thus, B is the correct answer.