Answer: B

Small Hint:

Undo Connie’s mistaken multiplication to find the original number.

Big Hint:

Then divide that original number by $2$.

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

Small Hint:

Find the total price of the five folders.

Big Hint:

The savings would be $20\%$ of that total.

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

Small Hint:

Reflect the shaded squares across diagonal $\overline{BD}$.

Big Hint:

Only shaded squares without a matching reflected square need to be paired.

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

Small Hint:

First find the triangle’s perimeter.

Big Hint:

That perimeter is also the square’s perimeter.

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

Small Hint:

Use as many $24$-packs as possible without going over $90$.

Big Hint:

After three $24$-packs, finish the remaining cans with the fewest packs.

Solution:

Four packs cannot be enough. Four $24$-packs would contain $96$ cans, and replacing any $24$-pack by a smaller pack decreases the total by either $12$ or $18$ cans. There is no way to decrease $96$ by exactly $6$ cans to reach $90$.

Five packs do work: three $24$-packs, one $12$-pack, and one $6$-pack contain $$3\cdot24+12+6=90$$ cans. Therefore, the minimum number is $5$.

Thus, B is the correct answer.

Answer: C

Small Hint:

Compare the thousandths digit of $2.00d5$ with that of $2.005$.

Big Hint:

The digit $d$ must be at least $5$.

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

Small Hint:

Combine the two southward parts into one vertical displacement.

Big Hint:

Use the Pythagorean theorem with legs $1$ and $\frac{3}{4}$.

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

Small Hint:

Test the choices using parity rules for odd numbers.

Big Hint:

The product of odd integers is odd.

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. Modulo $2$, this gives us $$1 + 1 \cdot 1 \equiv 0 \pmod 2.$$ From our above rules, we know that this is even.

B: $$1 \cdot 1 - 1 \equiv 0 \pmod 2.$$ Once again, this is even.

C: $$1 \cdot 1^2 + 1 \cdot 1^2 \equiv 0 \pmod 2.$$ This is also even.

D: $$(1 \cdot 1 + 1)^2 \equiv 0 \pmod 2.$$ Unfortunately, this is also even.

E: $$1 \cdot 1 \cdot 1 \equiv 1 \pmod 2.$$ This is odd.

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

Thus, E is the correct answer.

Answer: D

Small Hint:

Look only at $\triangle ADC$.

Big Hint:

It is isosceles and has a $60^\circ$ angle.

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

Small Hint:

Joe runs the same half-distance he previously walked.

Big Hint:

Running three times as fast takes one third as much time.

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

Small Hint:

Adding tax and taking the discount are both multiplications.

Big Hint:

Multiplication gives the same result in either order.

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

Small Hint:

The five daily amounts form an arithmetic sequence.

Big Hint:

The middle day equals the average of the five days.

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

Small Hint:

Complete the figure to a rectangle.

Big Hint:

Subtract the polygon area from the rectangle area to get the missing rectangle.

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

Small Hint:

Count each team’s games, then correct for double-counting.

Big Hint:

Each game is counted once for each of its two teams.

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

Small Hint:

Let the equal side length be $a$ and the base be $b$.

Big Hint:

Use $2a+b=23$ and the triangle inequality.

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 $a>\frac{23}{4}$. Thus $a\ge6$.

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

Thus, C is the correct answer.

Answer: D

Small Hint:

In the worst case, the Martian gets only $4$ socks of each color first.

Big Hint:

The next sock must make five of some color.

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

Small Hint:

Average speed is distance divided by time.

Big Hint:

On the graph, compare slopes from the origin.

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

Small Hint:

A three-digit multiple of $13$ has form $13k$.

Big Hint:

Find the smallest and largest possible values of $k$.

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

Small Hint:

Drop perpendiculars to split the trapezoid into right triangles and a rectangle.

Big Hint:

Use the $18-24-30$ and $7-24-25$ right triangles.

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

Small Hint:

Track the relative motion of Alice and Bob around the circle.

Big Hint:

Together they close the gap by $14$ points each turn, modulo $12$.

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 $k$ making $2k$ a multiple of $12$ is $6$.

Thus, A is the correct answer.

Answer: C

Small Hint:

There are $6$ dots total.

Big Hint:

Subtract the choices of three collinear dots from all $3$-dot choices.

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

Small Hint:

Choose convenient numbers for the small and large sizes.

Big Hint:

Compare price per ounce for each size.

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 $S: \frac{1.00}{10}=0.100$, $M: \frac{1.50}{16}=0.09375$, and $L: \frac{1.95}{20}=0.0975$.

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

Thus, E is the correct answer.

Answer: B