Small Hint:

Work backward from the apples Bridget kept and gave to Cassie.

Big Hint:

Those apples were the half not given to Ann.

Solution:

We can work backwards, starting with the $4$ apples that Bridget kept for herself. Adding the $3$ apples that she gave Cassie, she now has $7$ apples.

Finally, we multiply this value by $2$ since she gave half of her initial apples to Ann. $7 \cdot 2 = 14,$ so Bridget started off with $14$ apples.

Thus, E is the correct answer.

Small Hint:

Use the ratio $4$ sports cars to $7$ sedans.

Big Hint:

Going from $4$ to $28$ sports cars multiplies by $7.$

Solution:

Set up the proportion $$\dfrac{4}{7} = \dfrac{28}{x}.$$ Cross-multiplying gives $$4x = 7 \cdot 28$$ and hence $$x = 49.$$

Thus, D is the correct answer.

Small Hint:

Read one clear point from the graph.

Big Hint:

If $15$ minutes gives $3$ miles, then $30$ minutes is twice as long.

Solution:

From the graph, we can see that Suzanna rides $3$ miles in $15$ minutes. This means that in $30$ minutes, she will have ridden $6$ miles.

Thus, C is the correct answer.

Small Hint:

Look for a feature in the five pieces that must appear in the final figure.

Big Hint:

The long $5$-square piece must fit as an unbroken segment somewhere.

Solution:

Note that option B does not have any segments in it that are $5$ blocks long. This means that it is impossible to arrange the $5$ block long piece to fit within the figure.

Thus, B is the correct answer.

Small Hint:

Generate the sequence one term at a time.

Big Hint:

Each new term is the sum of the previous three terms.

Solution:

The sequence begins $1,$ $2,$ $3.$

The next terms are $1+2+3=6,$ $2+3+6=11,$ $3+6+11=20,$ $6+11+20=37,$ and $11+20+37=68.$

Thus, D is the correct answer.

Small Hint:

First find the combined filling rate of the four hoses.

Big Hint:

Convert minutes to hours after finding the total number of minutes.

Solution:

The $4$ hoses together fill the pool with $$2.5 \cdot 4 = 10$$ gallons of water per minute.

To fill $24{,}000$ gallons, it will take the hoses $$24{,}000 \div 10 = 2{,}400$$ minutes to fill the pool.

$2{,}400$ minutes is the same as $$2{,}400 \div 60 = 40$$ hours.

Thus, A is the correct answer.

Small Hint:

Use $CD$ as a base for triangle $ACD.$

Big Hint:

The perpendicular distance from $A$ to the railroad is shown by $AB.$

Solution:

The base of $ADC$ is $CD,$ which is $3.$ The altitude is $3$ as well.

Therefore, the area of $ADC$ is $\dfrac{1}{2} \cdot 3 \cdot 3 = 4.5.$

Thus, C is the correct answer.

Small Hint:

Multiply the scale factors for length and width.

Big Hint:

The new area is $1.10\cdot0.90$ times the old area.

Solution:

Let the old length and width be $l$ and $w,$ so the old area is $lw.$

The new length is $1.1l,$ and the new width is $0.9w.$ Thus the new area is $1.1\cdot0.9lw=0.99lw.$

This shows that the new area is $99\%$ of the old area.

Thus, B is the correct answer.

Small Hint:

Count only the outside sides of each polygon.

Big Hint:

Middle polygons share two sides; the two end polygons share one side each.

Solution:

The triangle and octagon are at the ends of the chain, so each loses one side to a shared edge. The square, pentagon, hexagon, and heptagon are in the middle, so each loses two sides to shared edges.

The resulting polygon has $$\begin{aligned} &(3-1)+(4-2) \\ &\quad {}+(5-2)+(6-2) \\ &\quad {}+(7-2)+(8-1)=23 \end{aligned}$$ sides.

Thus, B is the correct answer.

Small Hint:

Only the squares strictly inside the border do not touch the edge.

Big Hint:

An $8\times8$ board has a $6\times6$ interior.

Solution:

There are $$(8 - 2)^2 = 6^2 = 36$$ squares on the interior.

This means that the probability of choosing one of these squares is $$\dfrac{36}{64} = \dfrac{9}{16}.$$

Thus, D is the correct answer.

Small Hint:

The pencil price must divide both total costs in cents.

Big Hint:

Use the fact that at most $30$ sixth graders bought pencils to rule out a $1$-cent price.

Solution:

The number of seventh graders that bought a pencil is $143$ divided by the price of a pencil. Similarly, the number of sixth graders that bought a pencil is $195$ divided by the price of a pencil.

This means that the price of a pencil divides both $143$ and $195.$ Prime factorizing, we get $$143 = 11 \cdot 13$$ and $$195 = 3 \cdot 5 \cdot 13.$$ The only numbers that divide both $143$ and $195$ are $1$ and $13.$

If $1$ cent was the price of the pencil, that means $195$ sixth graders bought pencils, which is not possible. Therefore, the price of a pencil is $13$ cents.

This means that $143 \div 13 = 11$ seventh graders bought a pencil, and $195 \div 13 = 15$ sixth graders bought a pencil. Therefore, $4$ more sixth graders than seventh graders bought pencils.

Thus, D is the correct answer.

Small Hint:

Make a $3\times3$ table of possible sums.

Big Hint:

Only sums of $9$ are not prime among the possible outcomes.

Solution:

We can find the sum of the two numbers in every possible outcome.

$$\begin{gather*} 1 + 2 = 3 \\ 1 + 4 = 5 \\ 1 + 6 = 7 \\ 3 + 2 = 5 \\ 3 + 4 = 7 \\ 3 + 6 = 9 \\ 5 + 2 = 7 \\ 5 + 4 = 9 \\ 5 + 6 = 11. \end{gather*}$$

There are only $2$ outcomes where the sum is not prime (the two instances when the sum is $9$). Therefore, the probability that the sum is prime is $\dfrac{7}{9}.$

Thus, D is the correct answer.

Small Hint:

A number is divisible by $5$ exactly when its last digit is $0$ or $5.$

Big Hint:

With digits $1,$ $3,$ and $5,$ each digit is equally likely to be last.

Solution:

The number is equally likely to end in $1,$ $3,$ or $5.$

It is divisible by $5$ only if the last digit is $5,$ which happens with probability $\dfrac{1}{3}.$

Thus, B is the correct answer.

Small Hint:

Average speed is total distance divided by total time.

Big Hint:

The two $50$-mile trips take different amounts of time.

Solution:

The trip from Austin to Temple took $50 \div 60 = \dfrac{5}{6}$ hours. The trip from Temple to Austin took $50 \div 40 = \dfrac{5}{4}$ hours. This means that the total time for the round trip was $\dfrac{5}{6} + \dfrac{5}{4} = \dfrac{25}{12}$ hours.

The total distance of the round trip was $2 \cdot 50 = 100$ miles. Therefore, the average speed for the round trip was $100 \div \dfrac{25}{12} = 48$ miles per hour.

Thus, B is the correct answer.

Small Hint:

Compare how many recipe batches each ingredient can support.

Big Hint:

The ingredient that supports the fewest batches limits the servings.

Solution:

We need to find which ingredient is the limiting factor.

Jordan has enough chocolate for $5\div2=\dfrac{5}{2}$ batches, enough sugar for $2\div\dfrac{1}{4}=8$ batches, and enough milk for $7\div4=\dfrac{7}{4}$ batches.

The milk is limiting, so Jordan can make $5\cdot\dfrac{7}{4}=\dfrac{35}{4}=8\dfrac{3}{4}$ servings.

Thus, D is the correct answer.

Small Hint:

List the unordered triples of digits whose product is $24.$

Big Hint:

Triplets with three distinct digits have $6$ permutations, while $(2,2,6)$ has $3.$

Solution:

The only triples of integers less than $10$ that multiply to $24$ are $$(1, 3, 8), (1, 4, 6), (2, 2, 6),$$$$(2, 3, 4).$$

The triples with $3$ distinct numbers can be rearranged to form $6$ distinct $3$-digit positive integers. The other triple can be arranged to form $3$ distinct $3$-digit positive integers.

This leaves a total of $3 \cdot 6 + 3 = 21$ integers.

Thus, D is the correct answer.

Small Hint:

Factor $360$ into primes.

Big Hint:

Make exponents even for a square and multiples of $3$ for a cube.

Solution:

For a number to be a perfect square, every exponent in the prime factorization must be even. For it to be a cube, the exponents must be divisible by $3.$

We can factor $360$ to get $$360 = 2^3 \cdot 3^2 \cdot 5.$$ For $360x$ to be a perfect square and $x$ to be minimized, $x$ must have one factor of $2$ and one factor of $5.$ Therefore, we can let $x = 10.$

For $360y$ to be a cube, $y$ must have one factor of $3$ and two factors of $5.$ Therefore, we can let $y = 75,$ suggesting $x + y = 85.$

Thus, B is the correct answer.

Small Hint:

Look for the pattern in the unshaded tiles of the $7\times7$ floor.

Big Hint:

For an odd side length $2k-1,$ the unshaded tile count follows $k^2.$

Solution:

In the $7\times7$ example, there are $4$ rows that contain $4$ unshaded tiles, for $4^2=16$ unshaded tiles.

For a $15\times15$ floor with the same pattern, there will be $8$ such rows with $8$ unshaded tiles each.

Thus the number of unshaded tiles is $8^2=64.$

Thus, C is the correct answer.

Small Hint:

There are three ways the known $70^\circ$ angle and the $x^\circ$ angle can sit in an isosceles triangle.

Big Hint:

Consider whether they are the equal angles, or whether one is the vertex angle.

Solution:

All the following possibilities are shown below.

In the first scenario, we get $x = 70$ by the properties of the isosceles triangle.

In the second scenario, we get that $$70 \cdot 2 + x = 180,$$ from which we get that $x = 40.$

From the third scenario, we get that $$2x + 70 = 180,$$ from which we get that $x = 55.$

The sum of these values yields $$70 + 40 + 55 = 165.$$

Thus, D is the correct answer.

Small Hint:

Use symmetry to avoid counting every triangle separately.

Big Hint:

Classify triangles by their side lengths after choosing vertices from the two-row array.

Solution:

By symmetry, it is enough to list one representative of each possible triangle shape.

One complete list of non-congruent possibilities is $ABE,$ $ABG,$ $ACE,$ $ABH,$ $AEH,$ $ACF,$ $ACH,$ and $ADF.$

Every other triangle formed from the eight points is congruent to one of these $8$ triangles.

Thus, D is the correct answer.

Small Hint:

Let $S$ be the sum of all entries in the array.

Big Hint:

Andy has average row sum $\frac{S}{40},$ while Bethany has average column sum $\frac{S}{75}.$

Solution:

Let $S$ be the sum of all numbers in the array.

Andy’s $40$ row sums also add to $S,$ so their average is $A=\dfrac{S}{40}.$ Bethany’s $75$ column sums also add to $S,$ so their average is $B=\dfrac{S}{75}.$

Therefore $\dfrac{A}{B}=\dfrac{\frac{S}{40}}{\frac{S}{75}}=\dfrac{75}{40}=\dfrac{15}{8}.$

Thus, D is the correct answer.

Small Hint:

Count by number of digits, or pad numbers with leading zeroes.

Big Hint:

For three padded digits, each position has $9$ choices if digit $1$ is forbidden, but exclude $000.$

Solution:

We can case on the number of digits.

There are $8$ one digit numbers excluding $1.$

There are $8 \cdot 9 = 72$ two digit numbers that lack the digit $1.$

There are $8 \cdot 9 \cdot 9 = 648$ three digit numbers that do not include $1.$

This yields a total of $$8 + 72 + 648 = 728$$ numbers that do not contain the digit $1.$

Thus, D is the correct answer.

Small Hint:

Let the number of girls be $g,$ so the number of boys is $g+2.$

Big Hint:

The total jelly beans given out is $400-6.$

Solution:

Let $b$ be the number of boys in the class and $g$ be the number of girls. From the problem, we get that $b = g + 2.$

If each boy gets $b$ jelly beans, then Mrs. Wonderful will give out a total of $b^2$ jelly beans to all the boys. Similarly, she will give out $g^2$ jelly beans to all the girls.