Answer: E

Solution(s):

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.

Answer: D

Solution(s):

We can set up the following proportion: \[ \dfrac{4}{7} = \dfrac{28}{x}. \] Cross multiplying, we get \[ 4x = 7 \cdot 28 \]\[ x = 49 \]

Thus, D is the correct answer.

Answer: C

Solution(s):

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.

Answer: B

Solution(s):

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.

Answer: D

Solution(s):

We can create the following list to find all the numbers up till the eight number \(a_i\) denotes the \(i\)-th number in the sequence.

\(a_1 = 1,\) \(a_2 = 2,\) and \(a_3 = 3\) from the problem statement.

\[a_4 = 1 + 2 + 3 = 6\]

\[ a_5 = 2 + 3 + 6 = 11 \]

\[ a_6 = 3 + 6 + 11 = 20 \]

\[ a_7 = 6 + 11 + 20 = 37 \]

\[ a_8 = 11 + 20 + 37 = 68 \]

Thus, D is the correct answer.

Answer: A

Solution(s):

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.

Answer: C

Solution(s):

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.

Answer: B

Solution(s):

Let \(l\) be the old width of the rectangle and \(w\) be the old width. The new length is then \(1.1l,\) and the new width is \(.9w.\) This means that the new area is \(1.1 \times .9 lw = .99 lw.\)

This shows that the new area is \(99\%\) of the original area.

Thus, B is the correct answer.

Answer: B

Solution(s):

Notice that every polygon in the middle of the chain (every shape except for the triangle and octagon) contribute all but \(1\) of their sides to the overall polygon. The triangle and octagon contribute all but \(1\) of their sides.

Therefore, the total number of sides is \[ (3 - 1) + (4 - 2) + (5 - 2) + \] \[ (6 - 2) + (7 - 2) + (8 - 1) \] \[ = 2 + 2 + 3 + 4 + 5 + 7 \] \[ = 23. \]

Thus, B is the correct answer.

Answer: D

Solution(s):

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.

Answer: D

Solution(s):

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.

Answer: D

Solution(s):

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.

Answer: B

Solution(s):

Then number is equally likely to end in a \(1,\) \(3,\) or \(5.\) It is only divisible if the last digit is a \(5,\) which happens with a \(\dfrac{1}{3}\) probability.

Thus, B is the correct answer.

Answer: B

Solution(s):

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.

Answer: D

Solution(s):

We need to find which of the ingredients is the limiting factor.

With \(5\) squares of chocolate, Jordan can only make \(\dfrac{5}{2}\) times what the recipe requires.

With \(2\) cups of sugar, Jordan can only make \(\dfrac{2}{\dfrac{1}{4}} = 8\) times the required amount.

With \(7\) cups of milk, Jordan can only make \(\dfrac{7}{4}\) times what is needed.

Therefore, the number of servings is limited by the amount of milk. The amount of servings Jordan can make is \[5 \cdot \dfrac{7}{4} = \dfrac{35}{4} = 8 \dfrac{3}{4}.\]

Thus, D is the correct answer.

Answer: D

Solution(s):

The only triples of integers less than \(10\) that multiply to \(24\) are \[ (1, 3, 8), (1, 4, 6), (2, 2, 6), \]\[\text{and } (2, 3, 4). \]

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.

Answer: B

Solution(s):

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.

Answer: C

Solution(s):

Looking at the example given, note that there are \(4^2\) dark tiles. This is because there are \(4\) rows that contain \(4\) dark tiles each.

For a \(15\)-foot-by-\(15\)-foot floor, there are going to be \(8\) rows with \(8\) dark tiles in each row. This will give us \(8^2 = 64\) dark tiles.

Thus, C is the correct answer.

Answer: D

Solution(s):

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.

Answer: D

Solution(s):

We can label to points to find all the unique triangles.

All the possible distinct triangles are \[ AEF, AEG, AEH, AFG, AFH,\]\[ AGH, AFC, \text{ and } AFD. \]

Any other triangle will be congruent to one of the \(8\) listed above.

Thus, D is the correct answer.

Answer: D

Solution(s):

Each person includes every number in the array in their sums exactly once. This means that before they divide in the average, their sums are the exact same. This means that \(40A = 75B.\) Therefore, \(\dfrac{A}{B} = \dfrac{75}{40} = \dfrac{15}{8}.\)

Thus, D is the correct answer.

Answer: D

Solution(s):

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.

Answer: B

Solution(s):

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.

Therefore, \[ \begin{gather*} b^2 + g^2 = 400 - 6 \\ (g + 2)^2 + g^2 = 394 \\ 2g^2 + 4g + 4 = 394 \\ g^2 + 2g - 195 = 0 \\ (g + 15)(g - 13) = 0. \end{gather*} \] Since \(g\) cannot be negative, we get that \(g = 13.\) This means that \(b = 15,\) so \(b + g = 28.\)

Thus, B is the correct answer.

Answer: E

Solution(s):

Since \(A + B = A,\) we get that \(B = 0.\) Using this and the second equation, we also get that \(A = 5.\) This follows from carrying over a \(10\) to \(B,\) resulting in \(10 - A = A.\) Therefore, \(AB = 50.\)

Since \(50 - C5 = 5,\) we also know that \(C = 4.\) Then \(D = C + 5 = 9.\)

Thus, E is the correct answer.

Answer: E

Solution(s):

We can look at this shape from all \(6\) directions to add up the surface area from each view to get the total surface area.

Looking at it from the ends gives us the face of \(A,\) which has an area of \(\dfrac{1}{2} \cdot 1 = \dfrac{1}{2} \text{ ft}^2.\)

Each side has the same area as if we were to stack up all the pieces into a unit cube and take the area of one side. This would yield \(1 \text{ ft}^2.\)

The top and bottom each have \(4\) unit squares, for a total of \(4 \text{ ft}^2\) from each view.

Adding this all up gives us a total surface area of \[ \dfrac{1}{2} \cdot 2 + 1 \cdot 2 + 4 \cdot 2 = 11 \text{ ft}^2. \]

Thus, E is the correct answer.