AMC 8 practice test 2

The remaining $18-11-3=4$ responses are incorrect.

Correct responses earn $11\times 4=44$ points, incorrect responses lose $4$ points, and unanswered stations contribute zero.

The total is $44-4=40$ points.

The score weights apply separately to the three disjoint response categories.

The vertical rectangle's long side equals two short sides, so it is $8\text{cm}$⁠.

Each small rectangle has area $4\times 8=32{\text{cm}}^2$⁠.

The three rectangles therefore have combined area $96{\text{cm}}^2$⁠; the outer dimensions are $12\text{cm}$ by $8\text{cm}$⁠.

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

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

The five-day total is $5\times 18=90$ notebooks.

The four known days account for $12+16+20+22=70$⁠.

The remaining day's sales are $90-70=20$⁠.

The hexagon’s perimeter is $6+7+8+9+10+12=52$ metres.

The square therefore has side length $\frac{52}{4}=13$ metres.

The inside path removes $2$ metres at both ends of each side, leaving a central planting square of side $13-2-2=9$ metres.

Its area is $9^2=81$ square metres.

The path width must be subtracted twice from a side, not once.

If existing tokens have total value $S$ and pay every integer from $0$ through $S$⁠, a token of value $v$ extends that interval through $S+v$ exactly when $v\le S+1$⁠; otherwise $S+1$ is a gap.

In a sorted collection that covers the required range, the first three tokens must be 1-point tokens before a 4-point token can appear.

Choosing the largest next value that avoids a gap gives covered maxima $1,2,3,7,11,23,35,47$ and $59$⁠.

This sequence is an upper bound for any collection with the same number of tokens: a smaller covered maximum cannot allow a larger next denomination.

Thus eight tokens cannot cover $59$⁠.

Three 1-point tokens, two 4-point tokens and four 12-point tokens attain every value through $59$ with nine tokens.

The terms are the $50$ positive multiples of $3$ from $150$ down to $3$⁠.

Pair them in order.

Each pair contributes $3$⁠, and there are $25$ pairs, giving $25\times 3=75$⁠.

The squared radius of $A$ is $30$⁠.

From $2\pi r=11\pi$⁠, $B$ has radius $\frac{11}{2}$ and squared radius $\frac{121}{4}$⁠.

From $\pi r^2=24\pi$⁠, $C$ has squared radius $24$⁠.

In a common denominator these squared radii are $\frac{96}{4}$⁠, $\frac{120}{4}$ and $\frac{121}{4}$⁠, so $24<30<\frac{121}{4}$⁠.

All radii are positive, and squaring preserves their order.

The order is therefore $C,A,B$⁠.

Comparing the raw circumference and area numbers would compare different kinds of measurement.

The allowed offsets are $2,4,5,7,8$⁠.

They give three rankings: $A,D,B,C,E$⁠; $A,B,D,C,E$⁠; $A,B,C,D,E$⁠.

Values $4$ and $5$ share an order, as do $7$ and $8$⁠.

Count: $3$⁠.

Successive factorisation gives $$2772=4\times 693=2^2\times 3^2\times 7\times 11$$

The different prime divisors are $2,3,7$ and $11$⁠, whose sum is $23$⁠.

If $AB$ is wrong, the four other equations imply $C=\frac{11}{2}$⁠, not an integer.

If $BC$ is wrong, they imply $A=\frac{3}{2}$⁠, also invalid.

If $CD$ is wrong, $AB=6$⁠, $BC=9$ and $AC=7$ force $A=2,B=4$ and $C=5$⁠; $DA=8$ then gives $D=6$⁠.

Their true $CD$ sum is $11$⁠, so exactly its report is wrong.

If $DA$ is wrong, the others force $D=7$⁠, outside $1$ through $6$⁠.

If $AC$ is wrong, the other four cycle-edge reports imply $AB+CD=18$ but $BC+DA=17$⁠, although both sums must equal $A+B+C+D$⁠.

The only assignment is $(2,4,5,6)$⁠, so the count is $1$⁠.

There are $4\times 6\times 5\times 3=360$ legal codes.

The first and last symbols are fixed in the required event.

Any unordered pair of distinct middle letters has exactly one reverse alphabetical ordering, so it contributes one favourable code.

There are $6\times \frac{5}{2}=15$ such pairs.

The probability is $$\frac{15}{360}=\frac{1}{24}$$

Each unordered pair of distinct letters has two orderings, which is why the count is halved.

A leading $9$ is feasible because the remaining product is $16$ and three digits can supply it as $8,2$ and $1$⁠.

Since no digit exceeds $9$⁠, the maximum must start with $9$⁠.

Among possible second digits, $9$ cannot divide $16$ but $8$ can, and leaves product $2$⁠.

The last two digits are therefore $2$ and $1$ in descending order.

The number is $9821$⁠.

Each small radius is one-third of the large radius.

Therefore the large circle's area is nine times one small circle's area.

A small circle has area $\frac{27}{3}=9{\text{cm}}^2$⁠, so the large area is $81{\text{cm}}^2$⁠.

Removing the three small circles leaves $81-27=54{\text{cm}}^2$⁠.

There are $6\times \frac{5}{2}=15$ unordered pairs of school teams.

Three games per pair give $45$ internal games.

There are $6\times 4=24$ school-visitor pairs, and two games per such pair give $48$ external games.

The two types are disjoint and exhaust the schedule, so the total is $45+48=93$⁠.

Dividing the external count by two would be wrong because it already identifies each game through its unique school team and visitor.

There are two first moves to $A$⁠.

From either $A$⁠, both permitted moves reach $aT$⁠.

From every $T$⁠, both permitted moves reach an $H$⁠.

Thus there are $2\times 2\times 2=8$ paths.

Paths with the same endpoint still differ if their earlier moves differ.

For the sum to be largest, the thousands digits must be $7$ and $6$⁠, the hundreds digits $5$ and $4$⁠, the tens digits $3$ and $2$⁠, and the units digits $1$ and $0$⁠.

Otherwise exchanging a larger digit into a higher place would increase the sum.

The thousands places therefore create a difference of $1000$⁠.

The other three place differences each have magnitude $100,10$ and $1$⁠, so they can reduce the leading difference by at most $111$⁠.

The difference is at least $889$⁠.

The pair $7420$ and $6531$ attains it while using exactly the required place-value pairs.

Their sum is $13951$⁠, the maximum, and their difference is $889$⁠.

At time $\frac{1}{4}$ hour the first reaches mile $3$⁠.

At time $\frac{3}{8}$ hour it is at mile $6$⁠, while the second has just reached mile $9$⁠.

Both are then in the middle segment, closing the 3-mile gap at $48\text{mph}$⁠.

They meet $\frac{1}{16}$ hour later.

The first travels a further $\frac{24}{16}=\frac{3}{2}$ miles, so the meeting point is $6+\frac{3}{2}=\frac{15}{2}$ miles from $A$⁠.

First choose both vertical sides from $x=0,2$ and $3$⁠.

There are three such pairs, and every pair admits any two of the four horizontal levels, giving $3\times 6=18$ rectangles.

Now use $x=1$ as one vertical side.

Its partner can be any of the other three full vertical lines.

The only valid level pairs are $(0,1)$ and $(2,3)$⁠, because every other pair crosses the missing interval.

This adds $3\times 2=6$ rectangles, so the total is $24$⁠.

Every rectangle has one unique vertical-side pair, so the cases are disjoint.

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 $17$⁠.

$EB$ has equation $y=\frac{9x}{4}$ and $FC$ has equation $y=9-\frac{9(x-2)}{4}$⁠.

They meet at $$x=3,y=\frac{27}{4}$$

Triangle $ECB$ has base $CB=2$ and height $9$⁠, so its area is $9$⁠.

Triangle $XCB$ has the same base and height $9-\frac{27}{4}=\frac{9}{4}$⁠, so its area is $\frac{9}{4}$⁠.

Their difference, triangle $ECX$⁠, has area $\frac{27}{4}$⁠.

Reflection gives the same area for $FBX$⁠; the combined area is $\frac{27}{2}$⁠.

Let initial counts be $2k$ and $3k$⁠.

The exchange leaves $2k+3$ red and $3k-3$ blue.

The ratio gives $4(2k+3)=3(3k-3)$⁠, so $k=21$⁠.

The new counts are $45$ red and $60$ blue, total $105$⁠.

Red is a strict majority once it reaches at least $53$⁠.

Moving $8$ blue counters to red gives $53$ and $52$⁠; moving $7$ gives $52$ and $53$ and still fails.

Thus the minimum is $8$⁠.

Reverse one step at a time.

The positive predecessor of $1$ is $2$⁠; the predecessor of $2$ is $4$⁠.

The predecessors of $4$ are $8$ through the even rule and $1$ through the odd rule.

Reversing once more gives $16$ from $8$ and $2$ from $1$⁠.

No other positive odd predecessors satisfy the rule.

The two forward paths are $16\to 8\to 4\to 2\to 1$ and $2\to 1\to 4\to 2\to 1$⁠.

Thus the starting values have sum $18$⁠.

There are $35$ unrestricted paths; reflection pairs complementary areas summing to $12$⁠, so their total area is $35\times 6=210$⁠.

The $18$ paths through $(2,1)$ split into $3$ prefixes and $6$ suffixes.

The prefix areas sum to $3$⁠, contributing $6\times 3=18$ across all splits.

Each suffix starts at height $1$⁠, contributes a baseline $2$⁠, and has mean relative area $2$ in its 2-by-2 box.

Their total contribution across all splits is $3\times 6\times 4=72$⁠.

Forbidden total area is $90$⁠, leaving $120$ across $17$ allowed paths.

The mean is $\frac{120}{17}$⁠.

All three can be written with exponent $5$⁠:$$3^{30}=729^5,6^{20}=1296^5$$$$10^{15}=1000^5.$$

For positive numbers, raising to the fifth power preserves strict inequalities.

Since $729<1000<1296$⁠, the required order is $3^{30}<10^{15}<6^{20}$⁠.