Small Hint:

Expand both terms before subtracting

Big Hint:

Most of the $100\cdot100$ terms cancel

Solution:

We have that $$\begin{aligned}100(100 - 3) &= 100 \cdot 97 \\&= 9700\end{aligned}$$ and $$\begin{aligned}100 \cdot 100 - 3 &= 10000 - 3\\ &= 9997.\end{aligned}$$

As such, their difference is: $$9700 - 9997 = - 297.$$

Thus, C is the correct answer.

Small Hint:

Convert the meeting times to the same units

Big Hint:

The meetings last $45+90$ minutes out of $9$ hours

Solution:

Note that $45$ minutes is $$\dfrac{45}{60} = \dfrac{3}{4}$$ hours. The second meeting is then $$2 \cdot \dfrac{3}{4} = \dfrac{3}{2}$$ hours. Both meetings take a total of $$\dfrac{3}{4} + \dfrac{3}{2} = \dfrac{9}{4}$$ hours then. The percent of Makayla’s work day spent attending meetings is $$100 \% \cdot \dfrac{\frac{9}{4}}{9} = 100 \% \cdot \dfrac{1}{4} = 25 \%.$$

Thus, C is the correct answer.

Small Hint:

Use the worst case before a pair appears

Big Hint:

You can draw one sock of each of the four colors first

Solution:

To maximize the number of socks, we want to grab as many single socks as possible before getting a pair.

There are $4$ colors, which means that we can draw one sock of each color before drawing a pair.

This means that it takes at least $4 + 1 = 5$ socks to be drawn before a pair is guaranteed.

Thus, C is the correct answer.

Small Hint:

Evaluate $\heartsuit(1),\heartsuit(2),\heartsuit(3)$ separately

Big Hint:

$\heartsuit(x)=\dfrac{x+x^2}{2}$

Solution:

We have $$\heartsuit(1) = \dfrac{1 + 1^2}{2} = 1,$$ $$\heartsuit(2) = \dfrac{2 + 2^2}{2} = 3,$$ and $$\heartsuit(3) = \dfrac{3 + 3^2}{2} = 6.$$

Then $$1 + 3 + 6 = 10.$$

Thus, C is the correct answer.

Small Hint:

In $31$ days, three weekdays occur one extra time

Big Hint:

Monday and Wednesday must either both be extra days or both not be extra days

Solution:

Note that $31$ days leaves a remainder of $3$ when divided by $7.$

This means that if the month starts on a Saturday, Sunday, Tuesday, or Wednesday, there will be an uneven number of Mondays and Wednesdays.

Then the month can only start on a Monday, Thursday, or Friday.

Thus, B is the correct answer.

Small Hint:

Relate $\angle COB$ to $\angle AOC$

Big Hint:

$OA=OC$, so $\triangle AOC$ is isosceles

Solution:

Consider the following diagram:

We have that $$\angle AOC = 180^{\circ} - 50^{\circ} = 130^{\circ}.$$

Since $\triangle AOC$ is isosceles, we have that $$\angle CAO = \dfrac{180^{\circ} - 130^{\circ}}{2} = 25^{\circ}.$$

Since $\angle CAB = \angle CAO,$ we have that $\angle CAB = 25^{\circ}.$

Thus, B is the correct answer.

Small Hint:

Drop the altitude to the side of length $12$

Big Hint:

The altitude splits the triangle into two $6$-$8$-$10$ triangles

Solution:

To find the area of the triangle, we can drop the altitude to the side of length $12.$

Then we have a right triangle with one leg $12 \div 2 = 6$ and hypotenuse $10.$

The other leg has length $$\sqrt{10^2 - 6^2} = \sqrt{64} = 8.$$

The area of the triangle is then $$\dfrac{8 \cdot 12}{2} = 48.$$

The length of the rectangle is then $48 \div 4 = 12.$ Its perimeter is $$2(4 + 12) = 2 \cdot 16 = 32.$$

Thus, D is the correct answer.

Small Hint:

The ticket price must divide both totals

Big Hint:

Count the positive divisors of $\gcd(48,64)$

Solution:

Note that $x$ must divide both $48$ and $64,$ which means that it must divide their greatest common divisor.

The greatest common divisor of $48$ and $64$ is $16,$ which means $x$ divides $16.$

$16$ has $5$ factors, namely all the powers of $2$ up to $16.$

Thus, E is the correct answer.

Small Hint:

Compare Larry’s ignored-parentheses value to the correct value

Big Hint:

Correctly evaluated, the expression is $-2-e$

Solution:

Ignoring the parentheses, Larry would get $$1 - 2 - 3 - 4 + e = e - 8.$$

Evaluating with the parentheses, one would get $$\begin{aligned} &1 - (2 -(3 - (4 + e)))\\ & = 1 - (2 - (-1 - e))\\ &= 1 - (3 + e) \\ &=-2 - e.\end{aligned}$$

Both of these values are the same, so $$e - 8 = -2 - e$$$$e = 3.$$

Thus, D is the correct answer.

Small Hint:

Let $t$ be the rain time in minutes

Big Hint:

Use $20\cdot\dfrac{t}{60}+30\cdot\dfrac{40-t}{60}=16$

Solution:

Note that $40$ minutes is $\frac{2}{3}$ hours. Let Shelby drive $h$ hours in the sun.

Then she drives $\frac{2}{3} - h$ hours in the rain, which means she travels a total of $$30h + 20\left(\dfrac{2}{3} - h\right) = 10h + \dfrac{40}{3}$$ miles. We have this equals $16,$ so $$10h + \dfrac{40}{3} = 16$$$$h = \dfrac{4}{15}.$$

Then Shelby drives $$60\left(\dfrac{2}{3} - \dfrac{4}{15}\right) = 60 \cdot \dfrac{2}{5} = 24$$ minutes in the rain.

Thus, C is the correct answer.

Small Hint:

Write each coupon’s savings in terms of the listed price

Big Hint:

Coupon A must beat both $\$30$ and $0.25(p-100)$

Solution:

Let $p$ be the price of the item. Then coupon A saves $0.15p.$ Coupon B saves $\$ 30.$

Coupon C will save $$0.25(p - 100) = 0.25p - 25.$$

We must have that $$0.15p \ge 30$$$$p \geq 200$$ and $$0.15p \ge 0.25p - 25$$$$250 \geq p.$$

This shows that $x = 200$ and $y = 250.$ Therefore $y - x = 50.$

Thus, A is the correct answer.

Small Hint:

Think of $100$ students

Big Hint:

Minimize or maximize the number who switch answers

Solution:

To minimize $x,$ we want to have as many kids as possible maintain their answer.

We then need at least $70 - 50 = 20$ percent of the students to change their answer.

To maximize $x,$ we can have everybody that answered no change their answer, but some who answered yes must stay yes.

We need $70 - 50 = 20$ percent of the people who said yes to stay yes, which means only $50 - 20 = 30$ percent can switch.

This makes the maximum percent of students that can switch $50 + 30 = 80$ percent. The difference is then $80 - 20 = 60.$

Thus, D is the correct answer.

Small Hint:

Split cases based on the sign of $60-2x$

Big Hint:

After resolving the inner absolute value, solve the resulting absolute-value equations

Solution:

We first take care of the outer absolute value. We have either $$x = 2x - |60 - 2x|$$ or $$x = -2x + |60 - 2x|.$$

These simplify to $$x = |60 - 2x|$$ and $$3x = |60 - 2x|.$$

We have $2$ cases for each equation. For the first one, we have $$x = 60 - 2x$$ and $$x = 2x - 60.$$

Solving both gives us $$3x = 60$$$$x = 20$$ and $$-x = -60$$$$x = 60.$$

For the other equation, we have $$3x = 60 - 2x$$ and $$3x = 2x - 60.$$

Again solving both, we have $$5x = 60$$$$x = 12$$ and $x = -60.$ Note that $x$ cannot be negative since the right side of the original equation is nonnegative.

Adding up all the solutions gives us $$20 + 60 + 12 = 92.$$

Thus, C is the correct answer.

Small Hint:

Use the sum $1+2+\cdots+99$

Big Hint:

Set $\dfrac{99\cdot50+x}{100}=100x$

Solution:

Recall that the sum of the first $n$ integers is $\dfrac{n(n + 1)}{2}.$

Then, we have that $$\dfrac{\frac{99 \cdot 100}{2} + x}{100} = 100x,$$ which simplifies to $$99 \cdot 50 = (100^2 - 1)x$$$$= 101 \cdot 99x,$$ by difference of squares. Dividing gives us $x = \dfrac{50}{101}.$

Thus, B is the correct answer.

Small Hint:

Let $R$ be correct answers and $W$ wrong answers

Big Hint:

Use $4R-W=99$ and $R+W\le50$

Solution:

Let $x$ be the number of questions Jesse answered correctly and $y$ be the number he answered incorrectly.

Then $$4x-y=99$$ and $$x+y\leq50.$$

We have that $$y = 4x - 99$$$$5x - 99 \leq 50.$$

Rearranging and simplifying tells us that $x \leq 29.8.$ Since $x$ is an integer, its maximum value is $29.$

Thus, C is the correct answer.

Small Hint:

Find where the circle cuts one side of the square

Big Hint:

One side contributes a circular segment outside the square

Solution:

Drop the altitude from $O$ to $\overline{AB}$ at $X.$

Looking at the side lengths, we see that $\triangle OBX$ is a $30-60-90$ triangle.

This means that the area of sector $AOB$ is $$\dfrac{1}{6} \cdot \pi \cdot \left(\dfrac{\sqrt3}{3}\right)^2 = \dfrac{\pi}{18}.$$

We also have that the area of $\triangle AOB$ is $$\dfrac{1}{2} \cdot 2 \cdot \dfrac{\sqrt3}{6} \cdot \dfrac{1}{2} = \dfrac{\sqrt3}{12}.$$

The area of the sector outside the square and inside the circle is then $$\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}.$$

There are $4$ of these regions, which gives us a total area of $$4\left(\dfrac{\pi}{18} - \dfrac{\sqrt3}{12}\right) = \dfrac{2\pi}{9} - \dfrac{\sqrt3}{3}.$$

Thus, B is the correct answer.

Small Hint:

If there are $n$ schools, there are $3n$ contestants

Big Hint:

Andrea’s median place is $\dfrac{3n+1}{2}$

Solution:

Let there be $n$ schools. Then there are $3n$ participants in the contest.

Because the median is one student’s score, $3n$ is odd, so $n$ is odd. Carla’s $64$th place requires $3n\ge64,$ and hence the odd integer $n$ is at least $23.$

Andrea’s median place is $\dfrac{3n+1}{2}.$ Because she had the highest score on her team, she finished ahead of Beth, so $\dfrac{3n+1}{2}\lt37.$ This gives $3n\lt73,$ and therefore the odd integer $n$ is at most $23.$

Thus there are $23$ schools.

Thus, B is the correct answer.

Small Hint:

Factor the expression as $a(bc+b+1)$

Big Hint:

If $a$ is not divisible by $3$, work modulo $3$ with $b(c+1)$

Solution:

Note that $$abc + ab + a = a(bc + b + 1).$$ This means that if $a$ is divisible by $3,$ the whole expression is as well.

Since $2010$ is divisible by $3,$ we have that $a$ is divisible by $3$ with probability $\frac{1}{3}.$

Now consider $a$ not divisible by $3.$ For the expression to be divisible by $3,$ we must have that $bc + b + 1$ is divisible by $3.$

This means that $$bc + b = b(c + 1) \equiv 2 \pmod{3}.$$

The only possibility for this is that one of the factors is $2$ mod $3$ and the other is $1$ mod $3.$

For each of the two cases, there is a $\frac{1}{3}$ chance that each of the factors is the desired modulus, for a probability of $$\dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9}.$$

There are two cases, which means that this happens with a $\dfrac{2}{9}$ probability.

The total probability is then $$\dfrac{1}{3} \cdot 1 + \dfrac{2}{3} \cdot \dfrac{2}{9} = \dfrac{13}{27}.$$

Thus, E is the correct answer.

Small Hint:

Let $s$ be the side length and drop the altitude from $A$

Big Hint:

Use the circle radius $\sqrt{156}$ and the equilateral triangle altitude

Solution:

Consider the following diagram:

Using the formula for the area of a circle, we have that $BO = \sqrt{156}$ since it is a radius.

Extend $\overline{AO}$ to intersect $\overline{BC}$ at $X.$ Let $s$ be the side length of $\triangle ABC.$

Then we have that $$BX = \dfrac{s}{2}$$ and $$AX = \dfrac{s\sqrt3}{2}.$$

We have that $\triangle OXB$ is right, which means that we can apply the Pythagorean Theorem. This gives us $$(\sqrt{156})^2 = \left(\dfrac{s}{2}\right)^2$$$$+ \left(\dfrac{s\sqrt3}{2} + 4\sqrt3\right)^2.$$

Simplifying, we get $$156 = s^2 + 12s + 48$$ $$s^2 + 12s - 108 = 0$$ $$(s - 6)(s + 18) = 0.$$