Answer: B

Solution:

Since Connie multiplied by $2,$ her original number was $60 \div 2 = 30.$ To get the correct answer, she must divide by $2$ to get $30 \div 2 = 15.$

Thus, B is the correct answer.

Answer: C

Solution:

Karl paid $5\cdot\$2.50=\$12.50$ for the folders.

A $20\%$ discount on $\$12.50$ would save $0.20\cdot12.50=2.50$ dollars.

Thus, C is the correct answer.

Answer: D

Solution:

For diagonal $\overline{BD}$ to be a line of symmetry, every shaded square off the diagonal must have a matching reflected shaded square on the other side of the diagonal.

There are $4$ shaded squares whose reflected partners are not shaded, so $4$ additional small squares must be shaded.

Thus, D is the correct answer.

Answer: C

Solution:

The perimeter of the triangle is $$6.1 + 8.2 + 9.7 = 24 \text{ cm.}$$ This means that the side length of the square is $$24 \div 4 = 6 \text{ cm.}$$ Therefore, the area of the square is $$6^2 = 36 \text{ cm}^2.$$

Thus, C is the correct answer.

Answer: B

Solution:

To minimize the number of packs, we can use the largest packs first.

We can use three $24$-packs to have $$90 - 3 \cdot 24$$$$= 90 - 72$$$$= 18$$ cans left.

From this, we can see that we need one $12$-pack and one $6$-pack.

Therefore, we need $3 + 1 + 1 = 5$ packs.

Thus, B is the correct answer.

Answer: C

Solution:

The numbers $2.00d5$ and $2.005$ first differ in the thousandths place: $d$ is compared with $5$.

Thus $2.00d5>2.005$ exactly when $d\ge 5$. The possible digits are $5,6,7,8,9$, for $5$ values.

Thus, C is the correct answer.

Answer: B

Solution:

Note that what we want is the length of the diagonal. We can use the Pythagorean theorem to find this. $$\sqrt{\left(\dfrac{1}{2} + \dfrac{1}{2}\right)^2 + \dfrac{3}{4}^2}$$ $$= \sqrt{1 + \dfrac{9}{16}} = \sqrt{\dfrac{25}{16}} = \dfrac{5}{4}$$

Thus, B is the correct answer.

Answer: E

Solution:

Recall the four following rules:

• odd plus odd and even plus even is even

• even plus odd is odd

• even times anything is even

• odd times odd is odd

These rules can be easily verified by representing arbitrary odd numbers as $2m+1$ and arbitrary even numbers as $2n$ respectively, for integers $m,n.$

With this in mind, let's examine each answer choice individually:

A:

Note that $3$ is odd. This gives us $$\text{odd} + \text{odd} \cdot \text{odd}.$$ From our above rules, we know that this is even.

B: $$\text{odd} \cdot \text{odd} - \text{odd}.$$ Once again, this is even.

C: $$\text{odd} \cdot \text{odd}^2 + \text{odd} \cdot \text{odd}^2.$$ This is also even.

D: $$(\text{odd} \cdot \text{odd} + \text{odd})^2.$$ Unfortunately, this is also even.

E: $$\text{odd} \cdot \text{odd} \cdot \text{odd}.$$ This is odd.

Therefore, E is the only answer choice that is an odd integer.

Thus, E is the correct answer.

Answer: D

Solution:

Note that $\triangle ADC$ is isosceles. This means that $\angle DAC = \angle DCA.$

We get that $$\begin{gather*} \angle DAC + \angle DCA + \angle ADC = 180^{\circ} \\ \angle DAC + \angle DCA = 120^{\circ} \\ \angle DAC = \angle DCA = 60^{\circ}. \end{gather*}$$

This shows that $\triangle ADC$ is equilateral. This gives us that $$AC = CD = 17.$$

Thus, D is the correct answer.

Answer: D

Solution:

Joe took $6$ minutes to walk the first half of the distance.

He ran the second half, the same distance, at $3$ times his walking speed, so that half took $6\div3=2$ minutes.

His total time was $6+2=8$ minutes.

Thus, D is the correct answer.

Answer: C

Solution:

Jack’s total is $90.00\cdot1.06\cdot0.80$ dollars.

Jill’s total is $90.00\cdot0.80\cdot1.06$ dollars.

These products are equal because multiplication is commutative, so Jack’s total minus Jill’s total is $\$0$.

Thus, C is the correct answer.

Answer: D

Solution:

Let $x$ be the number of bananas Big Al ate on May $3.$ Then Big Al ate the following amounts starting from May $1:$ $$x - 12, x - 6, x, x + 6, x + 12.$$

The sum of this is $5x,$ which equals $100.$ This gives us that $x = 20.$ Then on May $5,$ Big Al ate $20 + 12 = 32$ bananas.

Thus, D is the correct answer.

Answer: C

Solution:

Complete the shape to rectangle $ABCG$, which has area $8\cdot9=72$.

The missing rectangle $FEDG$ has area $72-52=20$. Its height is $DE=BC-FA=9-5=4$.

Therefore $EF=20\div4=5$, so $DE+EF=4+5=9$.

Thus, C is the correct answer.

Answer: B

Solution:

Let us focus on one team. This team plays against $5$ other teams twice in its own division, for a total of $2 \cdot 5 = 10$ games.

This team also plays $6$ games with teams from the other division. Therefore, they play a total of $10 + 6 = 16$ games.

There are $12$ teams total, so there are $12 \cdot 16 = 192$ games conducted. Each game, however, involves $2$ teams, so we have to divide by $2.$

This gives us the actual total number of games to be $192 \div 2 = 96.$

Thus, B is the correct answer.

Answer: C

Solution:

Let the equal side length be $a$ and the base be $b$. Then $2a+b=23$, so $b$ must be odd.

The triangle inequality requires $b<2a$. Since $b=23-2a$, this gives $23-2a<2a$, so $a>23/4$. Thus $a\ge6$.

Also $b\ge1$, so $2a\le22$ and $a\le11$. The possible values $a=6,7,8,9,10,11$ all work, giving $6$ triangles.

Thus, C is the correct answer.

Answer: D

Solution:

Note that the Martian can pull out $4$ socks of each color without drawing $5$ of the same color. The $13$th sock, however, must be the $5$th sock of some color.

Thus, D is the correct answer.

Answer: E

Solution:

Average speed is distance over time, which is given by the slope of the line through the point and the origin.

Evelyn has the steepest slope, telling us that she had the greatest average speed.

Thus, E is the correct answer.

Answer: C

Solution:

We want to find how many $k$ exist such that $13k$ is a three-digit number.

The smallest possible $k$ such that $13k \gt 99$ is $k = 8.$

The largest possible $k$ such that $13k \lt 1000$ is $k = 76.$

This tells us that $k$ can range from $8$ to $76,$ which gives us $69$ values for $k.$

Thus, C is the correct answer.

Answer: A

Solution:

Drop the altitude from $C$ to meet $\overline{AD}$ at $F$. The existing altitude from $B$ meets $\overline{AD}$ at $E$.

By the Pythagorean theorem, $AE=\sqrt{30^2-24^2}=18$ and $FD=\sqrt{25^2-24^2}=7$. Also $EF=BC=50$.

Thus $AD=18+50+7=75$, and the trapezoid perimeter is $30+50+25+75=180$.

Thus, A is the correct answer.

Answer: A

Solution:

After one turn, Alice has moved $5$ points clockwise and Bob has moved $9$ points counterclockwise. Their relative movement is $5+9=14$ points per turn.

They meet when $14k$ is a multiple of $12$, where $k$ is the number of turns.

Since $14k\equiv2k\pmod{12}$, the smallest positive $k$ with $12\mid2k$ is $6$.

Thus, A is the correct answer.

Answer: C

Solution:

Notice that choosing any of these $3$ dots forms a triangle, except for the $2$ triples that form a straight line.

There are $\binom{6}{3} = 20$ ways to choose the points, and then we subtract $2$ to get $20 - 2 = 18.$

Thus, C is the correct answer.

Answer: E

Solution:

Choose convenient values: let the small box cost $\$1.00$ and contain $10$ ounces. Then the large box contains $20$ ounces.

The medium box costs $\$1.50$ and contains $20\%$ less than the large box, or $16$ ounces. The large box costs $30\%$ more than the medium, or $\$1.95$.

The costs per ounce are $S: 1.00/10=0.100$, $M: 1.50/16=0.09375$, and $L: 1.95/20=0.0975$.

From best to worst buy, the order is $M,L,S$.

Thus, E is the correct answer.

Answer: B

Solution:

Reflect the triangle and semicircle across hypotenuse $\overline{AB}$. This forms a full circle inscribed in a square.

The semicircle has area $2\pi$, so the full circle has area $4\pi$ and radius $2$. The square side length is therefore $4$.

The square area is $16$, so the original triangle has half that area, $8$.

Thus, B is the correct answer.

Answer: B

Solution:

Work backward from $200$. When the number is even, undo $\times2$ by dividing by $2$; when it is odd, undo $+1$ by subtracting $1$.

$$200,100,50,25,24,12,6,3,2,1$$ uses $9$ reverse steps, so $9$ keystrokes are enough.

Eight keystrokes are not enough: the largest value below $200$ reachable in $8$ keystrokes is obtained by $\times2,+1$ followed by six doublings, giving $3\cdot2^6=192$. The next larger possibilities overshoot to at least $256$, so $200$ cannot be reached in $8$ keystrokes.

Thus, B is the correct answer.

Answer: A

Solution:

Let $S$ be the common area inside both the square and circle. The problem says the circle-only area equals the square-only area.

Adding $S$ to both equal areas shows that the total area of the circle equals the total area of the square.

The square area is $2^2=4$, so $\pi r^2=4$. Hence $r=\dfrac{2}{\sqrt{\pi}}$.

Thus, A is the correct answer.