Group the terms in threes

$(1+2-3), (4+5-6), \ldots$

Group the expression as $(1+2-3)$ $+(4+5-6)$ $+(7+8-9)$ $+(10+11-12)$. These group sums are $0,3,6,9$, respectively, so the total is $0+3+6+9=18$.

Count how many entries are $1$, $2$, and $3$

Use a weighted sum: $1\cdot\#1+2\cdot\#2+3\cdot\#3$

There are $20$ border entries equal to $1$, $12$ entries equal to $2$, and $3$ entries equal to $3$. The sum is $20\cdot1+12\cdot2+3\cdot3$ $=20+24+9$ $=53$.

Compare each shape’s perimeter with $24$

The right triangle has hypotenuse $10$

The regular hexagon would have perimeter $6\cdot5=30$, too long for the wire. The square has side length $6$, so its perimeter is $24$. The right triangle has hypotenuse $10$, so its perimeter is $6+8+10=24$. Haruki can make the square and the triangle only.

Use $100$ as the original amount

After July, multiply by $0.8$; after August, multiply by $1.5$

If the original amount is $100$, then after a $20\%$ decrease Brynn has $80$. Increasing by $50\%$ gives $80\cdot1.5=120$, so the savings are $120\%$ of the original amount.

Find the driving time from distance and speed

The lunch break is the total time minus the driving time

Driving $100$ miles at $40$ miles per hour takes $\frac{100}{40}=2.5$ hours. The whole trip took $3$ hours, so the lunch break lasted $0.5$ hours, which is $30$ minutes.

Find the unreachable inner rectangle

Reachable area = field area - inner area

The whole field has area $10\cdot8=80$. The unreachable inner rectangle is $8$ meters by $6$ meters, with area $48$. Thus the reachable area is $80-48=32$, and the fraction is $\frac{32}{80}=\frac{2}{5}$.

Add the fractions of battery used on the two trips

$40$ miles is $\frac45$ of a full charge

The two trips used $\frac12+\frac3{10}=\frac45$ of a full battery. Since $\frac45$ of the range is $40$ miles, a full charge gives $40\div\frac45=50$ miles.

Write $74\%$ as a fraction of the total

Reduce the percentage fraction and use its denominator

Exactly $74\%=\frac{74}{100}=\frac{37}{50}$ of the people answered yes. The total number of people must therefore be a multiple of $50$, and $50$ people is possible.

Evaluate the inner square roots first

The numerator is $\sqrt{16\cdot9}$, and the denominator is $\sqrt{81\cdot4}$

The numerator is $\sqrt{16\sqrt{81}}$ $=\sqrt{16\cdot9}$ $=\sqrt{144}$ $=12$. The denominator is $\sqrt{81\sqrt{16}}$ $=\sqrt{81\cdot4}$ $=\sqrt{324}$ $=18$. The value is $\frac{12}{18}=\frac{2}{3}$.

Measure all finish times relative to Pedro

Olympia is $3$ minutes behind Pedro; Melina and Luke are after Olympia

Measure finish times in minutes after Pedro. Then Pedro is at $0$, Olympia is at $3$, Melina is at $5$, Luke is at $9$, and Nico is at $11$. The fourth finisher is Luke.

Each arc is one quarter of a circle

Add the radii $1+1+2+3+5$, then multiply by $\frac{\pi}{2}$

Each piece of the curve is a quarter circle. The radii are $1,1,2,3,5$, so the total length is $\frac14\cdot2\pi(1+1+2+3+5)$ $=\frac{\pi}{2}\cdot12$ $=6\pi$.

Let the top digit be $x$

The two adjacent upper digits are $9-x$ and $6-x$

Let the top digit be $x$. Moving clockwise from the top, the other digits are $9-x$, $x-2$, $7-x$, $x+1$, and $6-x$. For these six values to be exactly $1,2,3,4,5,6$, the value $x=5$ works and gives the set $5,4,3,2,6,1$. Thus the top digit is $5$.

Look at one side of the shaded square

That side moves $3$ units horizontally and $1$ unit vertically

One side of the shaded square goes $3$ unit widths horizontally and $1$ unit vertically in the grid. Its length squared is therefore $3^2+1^2=10$. The area of a square equals its side length squared, so the shaded area is $10$.

Let the three numbers be $a-d, a, a+d$

The two given sums differ by $20$, so $2d=20$

Let the three equally spaced numbers be $a-d$, $a$, and $a+d$. The given sums are $2a-d=40$ and $2a+d=60$. Adding gives $4a=100$, so $a=25$. The total of the three numbers is $3a=75$.

Every shaded face must be glued to another cube

With only $3$ cubes, some cube has at most one face-neighbor

For no shaded face to be visible, each cube must have both of its shaded faces glued to neighboring cubes, so each cube needs two face-neighbors. With only three cubes, the face-adjacency graph is a path, so an end cube has only one face-neighbor. Four cubes arranged as a $2\times2$ square can each be oriented with its two shaded faces pointing inward, so $4$ cubes suffice.

Divisibility by $4$ depends only on the last two digits

For each tens digit, the ones digit must be $0$, $4$, or $8$

There are $4$ choices for the thousands digit and $5$ choices for each other digit. Divisibility by $4$ depends only on the last two digits. Among the $25$ even digit endings, for each tens digit $0,2,4,6,8$, the ones digit can be $0,4,8$, so $15$ endings work. The desired fraction is $\frac{15}{25}=\frac{3}{5}$.

Name the original order $A,B,C,D$

The allowed neighboring pairs are $AC$, $AD$, and $BD$

Name the original order $A,B,C,D$. The forbidden neighboring pairs are $AB$, $BC$, and $CD$, so the only allowed pairs are $AC$, $AD$, and $BD$. To seat all four students in a row, the three row-adjacencies must be exactly these three edges, giving $C,A,D,B$ or its reverse. There are $2$ rearrangements.