AMC 8 practice test 1

The carrier amount is $5\times 2=10$ litres.

The base amount is $3\times 10=30$ litres.

Both links are needed, giving the combined factor $3\times 5=15$⁠.

Cleo leaves at $3$⁠.

Counting continues with Dara at $4$⁠, so Faye leaves at $6$⁠.

Ari and Bela say $7$ and $8$⁠, and Dara leaves at $9$⁠.

Evan and Ari say $10$ and $11$⁠, Bela leaves at $12$⁠, and Evan leaves at $13$ because the number contains $3$⁠.

Ari remains.

Take clockwise turns as positive.

The combined turn is $$\frac{11}{6}-\frac{3}{4}=\frac{22}{12}-\frac{9}{12}=\frac{13}{12}$$ revolutions.

Removing one full revolution leaves $\frac{1}{12}$ of a revolution.

Since a full revolution measures $360^{\circ }$⁠, the final angle clockwise from north is $\frac{360^{\circ }}{12}=30^{\circ }$⁠.

The whole revolution changes the travelled angle but does not change the final direction.

The four rings contain $6,12,18$ and $24$ lamps.

Including the central lamp gives $1+6+12+18+24=61$ lamps.

Each container receives $\frac{\frac{7}{8}}{7}=\frac{1}{8}$ of the tank's full capacity.

As a percentage, $$\frac{100}{8}=12.5\%$$

Since $72=2^3\times 3^2$⁠, its factor $2$ has an odd exponent.

Multiplication by $2$ gives $$144=2^4\times 3^2=12^2$$

The only smaller positive multiplier is $1$⁠, which leaves the odd exponent unchanged, so $k=2$⁠.

The initial total is $6+9+5+4=24$ participants.

Moving two participants from $B$ to $D$ changes category counts but not the total or the count choosing $C$⁠.

The six arrivals raise the total to $30$ and the $C$ count to $5+6=11$⁠.

The desired fraction is $\frac{11}{30}$⁠, corresponding to $(\frac{11}{30})\times 100\%=\frac{110}{3}\%$⁠, or $36\frac{2}{3}\%$⁠.

The final denominator includes all participants, not just those who changed or arrived.

The original total is $2\times 6+3\times 8+5\times 10=86$⁠.

A mean of $9$ over ten measurements requires total $90$⁠.

Adding $x$ to each of the two first-group measurements increases the total by $2x$⁠, so $86+2x=90$ and $x=2$⁠.

The new first-group mean is $8$⁠, and the weighted total becomes $2\times 8+3\times 8+5\times 10=90$⁠.

On the four top edges, exclude both three-painted-face corners: there are $4\times 3=12$ pieces.

On each of the four vertical corner columns, every piece below the top has exactly two painted faces, including the bottom piece, giving $4\times 4=16$ more.

These sets are disjoint, so the total is $28$⁠.

Write each number as $10a+b$⁠, where $a$ is $4,5,6$ or $7$ and $b$ is between $1$ and $9$⁠.

Since $b$ divides itself, divisibility is equivalent to $b$ dividing $10a$⁠.

For $b=1$⁠, $2$ or $5$⁠, all four choices of a work, contributing $12$⁠.

For $b=3$⁠, only $63$ works; for $b=4,44$ and $64$ work; for $b=6$⁠, only $66$ works; for $b=7$⁠, only $77$ works; and for $b=8$⁠, only $48$ works.

No $b=9$ case works.

Thus there are $12+1+2+1+1+1=18$ numbers.

Without restrictions there are $7!/(4!3!)=35$ routes.

A route through $(1,1)$ has two choices before that point and $5!/(3!2!)=10$ choices afterward, for $20$ routes.

A route through $(3,2)$ has ten choices before it and two afterward, also $20$⁠.

Routes through both forbidden points have $2\times 3\times 2=12$ choices, since the middle leg uses two east moves and one north move.

Subtracting both forbidden families removes this intersection twice, so restore it once:$$35-20-20+12=7.$$

Monotonicity fixes the order of the two forbidden points on every route that visits both.

The photographs show $A$ sharing an edge with $B,C,D$ and $E$⁠.

A cube face has exactly four edge-neighbours and one opposite face.

The only letter left is $F$⁠, so $F$ is opposite $A$⁠.

Since $$6!/10=\frac{720}{10}=72=8\times 9$$ the equation gives $$N!=8\times 9\times 7!=9!$$

Factorials are strictly increasing for integer indices at least $2$⁠, so $N=9$⁠.

The possible routes are $A-B-D$⁠, $A-C-D$⁠, $A-C-E-D$⁠, $A-B-C-D$ and $A-B-C-E-D$⁠.

For a fixed route, discounting its largest edge gives the smallest cost.

The resulting costs are respectively $11,11,\frac{19}{2},16$ and $13$⁠.

The minimum is $\frac{19}{2}$ on $A-C-E-D$⁠, with the coupon used on $A-C$⁠:$$\frac{5}{2}+4+3=\frac{19}{2}.$$

All possible directed routes and coupon placements have been covered.

The overlap is $(\frac{2}{3})\times 45=30$ people.

Selecting from the swimmers changes the denominator to $60$⁠, so the required probability is $\frac{30}{60}=\frac{1}{2}$⁠.

Every cube needs at least two perpendicular face neighbours.

Thus the contact graph has minimum degree at least two and contains a cycle.

A cubic-grid cycle is even, so fewer than six cubes would require a four-cycle.

Any four-cycle in this grid is a planar 2-by-2 square, which alone does not extend in three directions.

A fifth cube would need two contacts to that square.

Its two neighbours would be opposite corners of the square, but those corners already have their only two common grid neighbours inside the square.

Therefore a fifth cube cannot have two contacts.

Six cubes work at positions $(0,0,0),(1,0,0),(1,1,0),(1,1,1),(0,1,1),(0,0,1)$⁠.

Consecutive positions, including last to first, share faces.

At each cube those two contacts are perpendicular, so orient its marked faces toward them.

All marks are hidden and all three extents are two cubes.

The minimum is $6$⁠.

With $c$ crates, the first arrangement uses $8(c-3)$ apples.

The second shows the same stock is $5c+6$⁠.

Hence $$8c-24=5c+6$$ giving $c=10$⁠.

The apple count is $5\times 10+6=56$⁠; equivalently, seven filled crates of eight contain $56$⁠.

The leading digit $a$ is nonzero.

Divisibility by $5$ requires $b=0$ or $b=5$⁠.

The digit sum is $3a+3b$⁠, automatically divisible by $3$⁠.

If $b=0$⁠, any $a$ from $1$ through $9$ works, giving $9$ numbers.

If $b=5$⁠, $a$ can be any nonzero digit except $5$⁠, giving $8$⁠.

The total is $17$⁠.

Measure areas in units of $\pi$⁠.

The middle annulus has area $5$ and the outer annulus $16$⁠; the initial gray area is $15$⁠.

Thus the outer gray sector has area $10$⁠, fraction $\frac{5}{8}$ of its annulus, and angle $225^{\circ }$⁠.

Its complement is a 135-degree gap.

After a 180-degree rotation, the two complement gaps are disjoint, so the union of the gray sectors covers the entire outer annulus.

The final gray area is $5+16=21$ of the disk area $25$⁠, giving $\frac{21}{25}$⁠.

Fix the first physical badge.

Among the remaining $23$ badges, $7$ share its shape and $5$ share its code.

The other copy of that exact type belongs to both groups, so subtract that one double count.

There are $7+5-1=11$ winning second badges.

Every first badge has the same counts, so the probability is $\frac{11}{23}$⁠.

Treating the two copies as one object would give the wrong sample space.

The $14$ paths satisfying the neighbour gaps occur in seven reflected pairs.

Reflection exchanges the distinct endpoint labels, so exactly one of each pair meets $A<E$⁠.

Representatives, written from $A$ to $E$⁠, are $(1,3,5,2,4),(1,4,2,5,3),(2,4,1,3,5),(2,4,1,5,3),(2,5,3,1,4),(3,1,4,2,5)$ and $(3,1,5,2,4)$⁠.

A perfect score is $$12\times 6=72$$

Replacing a correct response by an unanswered task loses $4$ points, while replacing it by an incorrect response loses $6$⁠.

Thus every deficit has form $4b+6w$⁠, with nonnegative $b$ and $w$ and at most twelve changed tasks.

In the range from $61$ to $72$⁠, the deficit is between $0$ and $11$⁠.

The only attainable deficits there are $0,4,6,8$ and $10$⁠: they use no change, one blank, one wrong, two blanks, or one blank and one wrong respectively.

Odd deficits are impossible, and deficit $2$ is smaller than either positive loss.

Subtracting these five deficits from $72$ gives attainable scores $72,68,66,64$ and $62$⁠.

All have explicit response patterns using no more than twelve tasks.

Removing every squared prime factor leaves the squarefree kernel.

Two labels have square product exactly when their kernels agree.

Among $1$ through $12$⁠, the nonsingleton kernel classes are $\{1,4,9\},\{2,8\}$ and $\{3,12\}$⁠; every other kernel occurs once.

These classes give $3+1+1=5$ favourable unordered pairs.

There are $12\times \frac{11}{2}=66$ equally likely unordered pairs of different cards, so the probability is $\frac{5}{66}$⁠.

Drawing without replacement excludes a label paired with itself.

The two equilateral triangles $UST$ and $RST$ together have area $2\times (\frac{\sqrt{3}}{4})\times 6^2=18\sqrt{3}$⁠.

Since $V$ and $W$ lie outside $RST$⁠, each minor arc bends into $RST$ and removes one circular segment.

Each segment is a $60^{\circ }$ sector of radius $6$ minus an equilateral triangle, so its area is $(\frac{60}{360})\pi \times 6^2-9\sqrt{3}=6\pi -9\sqrt{3}$⁠.

The cutouts lie on opposite sides of the altitude from $R$ to $ST$ and do not overlap.

Subtracting both gives $$18\sqrt{3}-2(6\pi -9\sqrt{3})=36\sqrt{3}-12\pi$$

A nine-command route is $1\to 2\to 4\to 5\to 10\to 11\to 22\to 44\to 88\to 176$⁠.

Seven or fewer commands cannot exceed $2^7=128$⁠, since doubling is always at least as effective as adding $1$⁠.

Consider eight commands.

The first command has the same effect whichever type is chosen.

Call an increment applied after that first command an effective increment.

With no effective increment the result is $256$⁠.

With exactly one, in position $j$ from $2$ through $8$⁠, the result is $128+2^{8-j}$⁠, giving only $192,160,144,136,132,130$ or $129$⁠.

With at least two effective increments, the largest result comes from placing them as early as possible, because an increment before a later doubling contributes more than the reverse order.

Two early increments give $1\to 2\to 3\to 4$⁠, followed by at most five doublings, reaching at most $128$⁠; replacing further doublings by increments cannot increase it.

Thus eight commands cannot give $176$⁠.

Nine is both attainable and minimal.