Answer: E

Small Hint:

Subtract the apple cost from five dollars.

Big Hint:

Find the total cost of the apples first.

Solution:

The apples cost a total of $3 \cdot 50 = 150$ cents, which equals $\$1.50.$ This means that Margie received $\$5.00-\$1.50=\$3.50$ in change.

Thus, E is the correct answer.

Answer: E

Small Hint:

Compare $20\cdot45$ with $25\cdot40$.

Big Hint:

Compute each rectangular area separately.

Solution:

The area of Karl’s garden is $$20 \text{ft} \cdot 45 \text{ft} = 900 \text{ft}^2.$$ The area of Makenna’s garden is $$25 \text{ft} \cdot 40 \text{ft} = 1000 \text{ft}^2.$$

The difference of these areas is $$1000 \text{ft}^2 - 900 \text{ft}^2 = 100 \text{ft}^2.$$ Therefore, Makenna’s garden is $100 \text{ft}^2$ larger than Karl’s.

Thus, E is the correct answer.

Answer: D

Small Hint:

A border around a $5$ by $5$ square adds the outside ring of tiles.

Big Hint:

The original unshaded tiles stay the same.

Solution:

The original pattern is a $5\times5$ square. Extending it by one tile on every side makes a $7\times7$ square, so the new shaded border contains $7^2-5^2=24$ tiles. Together with the original $8$ shaded tiles, there are $24+8=32$ shaded tiles. The original $17$ unshaded tiles do not change, so the ratio of shaded tiles to unshaded tiles is $32:17.$

Thus, D is the correct answer.

Answer: C

Small Hint:

The mean is the total number of fish divided by $9$.

Big Hint:

Order the nine numbers before finding the median and mode.

Solution:

To find these values more easily, we can get the following ordered list: $$0, 0, 1, 1, 2, 2, 3, 3, 3.$$

From this, we see that the mode is $3,$ the median is $2,$ and the mean is $\frac{15}{9} = \frac{5}{3}.$

Since $\dfrac{5}{3} < 2 < 3,$ we get that:

mean $<$ median $<$ mode.

Thus, C is the correct answer.

Answer: D

Small Hint:

After the first $24$ hours, the remaining time is on January $2.$

Big Hint:

Divide $2011$ minutes by $60$.

Solution:

The remainder when $2011$ is divided by $60$ is $31.$ This means that $2011 = 60 \cdot 33 + 31,$ which means that $2011$ minutes is the same as $33$ hours and $31$ minutes.

$24$ hours takes us to January $2,$ so we get that we are $9$ hours and $31$ minutes into January $2.$

Thus, D is the correct answer.

Answer: D

Small Hint:

Subtract the motorcycle owners from the total adults.

Big Hint:

The adults without motorcycles are exactly the car owners who do not own a motorcycle.

Solution:

We know that $45$ people own motorcycles, so $351 - 45 = 306$ people do not own motorcycles.

Thus, D is the correct answer.

Answer: C

Small Hint:

The four shaded pieces combine to exactly one large square.

Big Hint:

Express each shaded part as a fraction of one large square.

Solution:

The top left and the bottom right shaded regions are both a quarter of each square. The top right is one-eighth, and the bottom left is three-eighths. Their combined area is $$\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{3}{8} = 1.$$

Therefore, the shaded regions combined equal the area of one square, so they are $25 \%$ of the total area.

Thus, C is the correct answer.

Answer: B

Small Hint:

All sums are odd and range from $3$ through $11$.

Big Hint:

List the possible sums from the two bags.

Solution:

The table shows all nine outcomes. The distinct sums are $3,5,7,9,$ and $11$, so there are $5$ possible values.

Thus, B is the correct answer.

Answer: E

Small Hint:

Read the final distance and total hours from the graph.

Big Hint:

Average speed is total distance divided by total time.

Solution:

Carmen travels $35$ miles in $7$ hours, so her average speed is $\frac{35}{7} = 5$ miles per hour.

Thus, E is the correct answer.

Answer: C

Small Hint:

The remaining fare buys extra distance at $20$ cents per tenth of a mile.

Big Hint:

Set aside the tip and the first half-mile fare first.

Solution:

There is a guaranteed $\$2$ tip, so we can subtract that from the total, leaving $\$8.$ This is greater than $\$2.40,$ so we can subtract that and add $\dfrac{1}{2}$ miles to the total distance.

We now have $\$5.60$ to use for additional miles. $\$0.20$ per $0.1$ mile is the same as $\$2$ for $1$ mile. That means one can ride for $\frac{5.60}{2} = 2.8$ more miles with this much money. This leaves a total of $2.8 + \dfrac{1}{2} = 3.3$ miles.

Thus, C is the correct answer.

Answer: A

Small Hint:

Average the five daily differences instead of averaging each person separately.

Big Hint:

Compare Sasha and Asha day by day.

Solution:

We can calculate the difference in average minutes by looking at the differences per day.

Starting with Monday, the differences between Sasha and Asha are $10,$ $-10,$ $20,$ $30,$ and $-20.$ This is a total of $30$ minutes. Therefore, the average difference is $30 \div 5 = 6.$

Thus, A is the correct answer.

Answer: B

Small Hint:

Carlos has three possible seats, only one opposite Angie.

Big Hint:

Fix Angie in one seat by symmetry.

Solution:

Consider that Angie’s seat is chosen. Carlos has an equal probability of being in any of the other $3$ seats. Only one of them is opposite Angie, however. Therefore, the probability is $\dfrac{1}{3}.$

Thus, B is the correct answer.

Answer: C

Small Hint:

Compare the shaded rectangle area with the area of the $15$ by $25$ rectangle.

Big Hint:

The overlap is a rectangle whose width is the sum of two side lengths minus $25$.

Solution:

We get that $$\begin{align*} SC &= DC + SR - DR \\ &= 15 + 15 - 25 \\ &= 5. \end{align*}$$

This means that the area of $PBSC$ is $5 \cdot 15 = 75.$ The area of $AQRD$ is $25 \cdot 15 = 375.$

$\dfrac{75}{375} = \dfrac{1}{5},$ which is $20 \%.$

Thus, C is the correct answer.

Answer: C

Small Hint:

Then divide the total number of girls by $270+180$.

Big Hint:

Find the number of girls at each school from its ratio.

Solution:

The total number of girls is $$\dfrac{4}{9} \cdot 270 + \dfrac{5}{9} \cdot 180 =$$ $$120 + 100 = 220.$$

There are $270 + 180 = 450$ students total, so the fraction of girls is $\dfrac{220}{450} = \dfrac{22}{45}.$

Thus, C is the correct answer.

Answer: D

Small Hint:

Pair $2^{10}$ with $5^{10}$ to make $10^{10}$.

Big Hint:

Rewrite $4^5$ as a power of $2$.

Solution:

To find the number of digits, we can try to express this number in terms of powers of $10.$

We get that $$\begin{align*} 4^5 \cdot 5^{10} &= 2^{10} \cdot 5^{10} \\ &= 10^{10}. \end{align*}$$

This shows that the desired number is $1$ followed by $10$ zeros, for a total of $11$ digits.

Thus, D is the correct answer.

Answer: C

Small Hint:

The two right triangles use legs $15,20,25$ in opposite roles.

Big Hint:

Drop an altitude in each isosceles triangle.

Solution:

Since these triangles are isosceles, the altitude shown from the apex of each triangle bisects its base and creates two congruent right triangles.

Using the Pythagorean theorem, we get that the altitude of the triangle with area $A$ equals $$\sqrt{25^2 - 15^2} = 20.$$ Similarly, we get that the altitude of the triangle with area $B$ equals $$\sqrt{25^2 - 20^2} = 15.$$

With these altitudes, we can calculate the areas of the triangles. We get that $$A = \dfrac{1}{2} \cdot 20 \cdot 30 = 300.$$ Similarly, $$B = \dfrac{1}{2} \cdot 15 \cdot 40 = 300.$$

Therefore, $A = B.$

Thus, C is the correct answer.

Answer: A

Small Hint:

Match the exponents of $2$, $3$, $5$, and $7$.

Big Hint:

Prime-factorize $588$.

Solution:

To find the desired exponents, note that all the bases are prime numbers. This means that finding the prime factorization will be helpful.

We get that $$588 = 2^2 \cdot 3^1 \cdot 7^2.$$

From this, it is clear that $w = 2,$ $x = 1,$ $y = 0,$ and $z = 2$ ($y = 0$ since that makes the $5^y$ term equal $1$).

Therefore, $$\begin{gather*} 2w + 3x + 5y + 7z \\ = 2 \cdot 2 + 3 \cdot 1 + 5 \cdot 0 + 7 \cdot 2 \\ = 21. \end{gather*}$$

Thus, A is the correct answer.

Answer: D

Small Hint:

By symmetry, half of the unequal rolls have the first roll larger.

Big Hint:

Separate the equal rolls from the unequal rolls.

Solution:

There are $3$ possible outcomes when rolling a die twice: the first number is greater than the second, both numbers are equal, or the first number is less than the second number. The first and third outcomes have the same probability since they are symmetric.

The second outcome has a $\dfrac{1}{6}$ chance of happening, since the first number can be anything, and the second number must equal the first number. The other two outcomes have a combined probability of $1 - \dfrac{1}{6} = \dfrac{5}{6}.$ This means that each outcome has a $\dfrac{5}{6} \div 2 = \dfrac{5}{12}$ chance of happening.

The desired probability is the first outcome plus the second outcome, for a total probability of $\dfrac{5}{12} + \dfrac{1}{6} = \dfrac{7}{12}.$

Thus, D is the correct answer.

Answer: D

Small Hint:

Count all single-region and multi-region rectangles systematically.

Big Hint:

Break the figure into labeled small regions.

Solution:

We can split the figure into these regions to make it easier to count the rectangles.

The rectangles in this figure are $b,$ $c,$ $d,$ $ab,$ $bc,$ $cd,$ $cf,$ $de,$ $abcd,$ $cdef,$ and $bcfg.$ These form $11$ rectangles.

Thus, D is the correct answer.

Answer: D

Small Hint:

Use $9$-$12$-$15$ and $12$-$16$-$20$ right triangles.

Big Hint:

Drop perpendiculars from the top vertices to the base.

Solution:

We can drop the following altitudes to more easily find the area.

We can use the Pythagorean Theorem to get that $$DE = \sqrt{15^2 - 12^2} = 9$$ and $$FC = \sqrt{20^2 - 12^2} = 16.$$

We also know that $$EF = AB = 50,$$ so $$DC = DE + EF + FC = 75.$$

Then the area of $ABCD$ is $$\dfrac{1}{2} \cdot (DC + 50) \cdot 12 = 6 \cdot 125$$ $$= 750.$$

Thus, D is the correct answer.

Answer: C

Small Hint:

The two guesses off by one force Norb near either $37$ or $48$.

Big Hint:

At least half too low puts Norb above the fifth listed guess.

Solution:

The first part of the statement means that Norb’s age is greater than $36.$

The second part means that Norb’s age is either between $36$ and $38$ or between $47$ and $49.$

Since $37$ is prime and $48$ is not, Norb’s age is $37.$

Thus, C is the correct answer.

Answer: D

Small Hint:

The last two digits of powers of $7$ repeat every $4$ powers.

Big Hint:

Work modulo $100$, since the tens digit depends on the last two digits.

Solution:

To find the tens digit, it is enough to track powers of $7$ modulo $100$. Since $7^4=2401\equiv1\pmod{100}$, the last two digits repeat every four powers.

Because $2011=4\cdot502+3$, we have $$\begin{aligned}7^{2011}&\equiv7^3=343\\&\equiv43\pmod{100}\end{aligned}$$ Thus the tens digit is $4$.

Thus, D is the correct answer.

Answer: D