Answer: C

Solution(s):

Converting all the fractions to decimals, we get \[ .1 + .09 + .009 + .0007 = .1997. \]

Thus, C is the correct answer.

Answer: D

Solution(s):

To get the largest number, we would want to subtract the smallest number possible from \(200.\)

The smallest two-digit number is \(10.\) Having Ahn choose this number will give us \[ 2 (200 - 10) = 2 \cdot 190 = 380 \] as the final result.

Thus, D is the correct answer.

Answer: B

Solution(s):

We have that all the tenths place digits are the same, so we then look at the hundredths digit.

The largest hundredths digit is \(7,\) so we can limit the answer choices to the ones with this value.

The largest thousandths digit is \(9,\) which is achieved by \(.979.\)

Thus, B is the correct answer.

Answer: E

Solution(s):

One-half hour is \(30\) minutes, in which Julie can speak \[ 150 \cdot 30 = 4500 \] words. Three-quarters of an hour is \(45\) minutes, in which Julie can speak \[ 150 \cdot 45 = 6750 \] words. The only answer choice in between these two values is \(5650.\)

Thus, E is the correct answer.

Answer: A

Solution(s):

Listing out all the two-digit multiples of \(7,\) we get \[ 14, 21, 28, 35, 42, 49, 56, 63, 70,\] \[ 77, 84, 91, \text{ and } 98. \] We have that \(28\) and \(91\) are the two multiples whose digits add to \(10.\)

The sum of these numbers is \(28 + 91 = 119.\)

Thus, A is the correct answer.

Answer: C

Solution(s):

Note that each digit has ten times the value of the digit to its right.

The place occupied by \(9\) is \(5\) spaces to the right of the place occupied by \(3.\)

This means that it is \[ 10^5 = 100,000 \] times as a great.

Thus, C is the correct answer.

Answer: D

Solution(s):

We can inscribe the circle inside the square so that it is tangent to the midpoints of each side of the square.

This means that the side length of the square is two times the radius of the circle, making it \(2 \cdot 4 = 8.\)

Then the area of the square is \[ 8^2 = 64. \]

Thus, D is the correct answer.

Answer: B

Solution(s):

There are \(8.5\) hours between the time Walter catches the school bus and arrives at home.

This is a total of \[ 60 \cdot 8.5 = 510 \] minutes.

The total time Walter spends at school is \[ 6 \cdot 50 + 30 + 2 \cdot 60 =\]\[ 300 + 30 + 120 = 450 \] minutes.

This means that Walter spends \[ 510 - 450 = 60 \] minutes on the bus.

Thus, B is the correct answer.

Answer: C

Solution(s):

There are \(3\) options for the person in front. Then, there are \(2\) options for the person in the middle.

This leaves \(1\) choice for the person at the end. There are \(3 \cdot 2 \cdot 1 = 6\) ways for the people to line up.

Only one of these lines is correct, and the probability it occurs is \(\dfrac{1}{6}.\)

Thus, C is the correct answer.

Answer: C

Solution(s):

We can split the square into \(6^2 = 36\) unit squares. The number of black squares is \[ 3 + 7 + 11 = 21. \]

The fraction of the square that is shaded is \[ \dfrac{21}{36} = \dfrac{7}{12}. \]

Thus, C is the correct answer.

Answer: A

Solution(s):

We know that \(11\) is prime, which means that it only has \(2\) divisors.

The prime factorization of \(20\) is \[ 2^2 \cdot 5. \] Recall that the number of divisors a number has is the product of all the exponents plus one in the prime factorization.

Here, that product would be \[ (2 + 1) (1 + 1) = 3 \cdot 2 = 6. \]

Then \(2 \cdot 6 = 12.\) We have the prime factorization of \(12\) is \[ 2^2 \cdot 3. \] This also has \[ (2 + 1) (1 + 1) = 3 \cdot 2 = 6 \] divisors.

Thus, A is the correct answer.

Answer: D

Solution(s):

We have that \[ \angle 1 = 180^{\circ} - 70^{\circ} - 40^{\circ} = 70^{\circ} \] using the fact that the interior angles of a triangle add to \(180^{\circ}.\)

This tells us that \[ \angle 2 = 180^{\circ} - \angle 1 = 110^{\circ} \] since \(\angle 1\) and \(\angle 2\) are supplementary.

Finally, \[ \angle 3 + \angle 4 + 110^{\circ} = 180^{\circ} \Rightarrow\]\[ 2 \angle 4 = 70^{\circ} \Rightarrow \angle 4 = 35^{\circ}. \]

Thus, D is the correct answer.

Answer: A

Solution(s):

There are \[ 26 \cdot .5 = 26 \div 2 = 13 \] yellow jelly beans in the first bag, \[ 28 \cdot .25 = 28 \div 4 = 7 \] yellow jelly beans in the second bag, and \[ 30 \cdot .2 = 30 \div 5 = 6 \] yellow jelly beans in the third bag.

The total number of yellow jelly beans is \[ 13 + 7 + 6 = 26 \] and the total number of jelly beans is \[ 26 + 28 + 30 = 84. \] The ratio of yellow jelly beans to all the beans is \[ \dfrac{26}{84} \cdot 100 \% = \dfrac{13}{42} \cdot 100 \%\]\[ \approx 30.9 \%. \]

Thus, A is the correct answer.

Answer: D

Solution(s):

The sum of all the numbers in the list is \(5 \cdot 5 = 25.\)

The only mode is \(8,\) which means that there are guaranteed two \(8\) s.

The sum of the numbers is up to \[ 2 \cdot 8 + 5 = 16 + 5 = 21. \]

The other two numbers must then add to \(4.\) They are both positive and not equal, leaving the only possible two numbers as \[ 1 \text{ and } 3. \]

The desired difference is then \[ 8 - 1 = 7. \]

Thus, D is the correct answer.

Answer: B

Solution(s):

Let \(3x\) be the side length of the large square. Then we can find the side length of the inner square via \[ \sqrt{(2x)^2 + x^2} = \sqrt{5x^2} = x\sqrt{5} \] from the Pythagorean Theorem.

The area of the larger square is \[ (3x)^2 = 9x^2 \] and that of the inner square is \[ (x\sqrt{5})^2 = 5x^2. \]

The ratio of the areas is then \(\dfrac{5}{9}.\)

Thus, B is the correct answer.

Answer: E

Solution(s):

After the first year, AA's stock's worth goes up to \[ $ 100 \cdot 1.2 = $ 120 \] and BB's goes down to \[ $ 100 \cdot .75 = $ 75. \]

After the second year, AA's stock is worth \[ $ 120 \cdot .8 = $ 96 \] and BB's is worth \[ $ 75 \cdot 1.25 = $ 93.75. \]

CC's stock worth remains the same, so the ordering of the stock worths is now \[ B \lt A \lt C. \]

Thus, E is the correct answer.

Answer: E

Solution(s):

Each face has two diagonals connecting each of the two pairs of opposite vertices.

Also, for each vertex, there is one corresponding vertex that lies opposite it on the cube.

There are then \(8 \div 2 = 4\) interior space diagonals in the cube.

The total number of diagonals is then \[ 6 \cdot 2 + 4 = 12 + 4 = 16. \]

Thus, E is the correct answer.

Answer: B

Solution(s):

Originally, each box is worth \[ $ 5 \div 4 = $ 1.25. \]

Now, each box is worth \[ $ 4 \div 5 = $ .8. \]

The percent decrease is then \[ \dfrac{1.25 - .8}{1.25} \cdot 100 \% = \dfrac{45}{125} \cdot 100 \% \]\[= \dfrac{9}{25} \cdot 100 \% = 36 \%. \]

Thus, B is the correct answer.

Answer: D

Solution(s):

Note that the numerator of each fraction cancels with the denominator of the fraction to its right.

We can then cancel out all these terms to get a final equation of \[ \dfrac{a}{2} = 9 \Rightarrow a = 18 \Rightarrow b = 17. \]

The desired sum is then \[ 18 + 17 = 35. \]

Thus, D is the correct answer.

Answer: A

Solution(s):

We can case on the value of the first die. If its value is \(1 - 4,\) then it is impossible for the product be greater than \(36.\)

If it is a \(5,\) then the other dice has to roll an \(8,\) otherwise the product is less than \(36.\)

If the first die is a \(6\) or \(7,\) then other die has to be at least a \(6,\) giving \(2 \cdot 3 = 6\) possibilities.

Finally, if the first roll is an \(8,\) the other die must roll at least a \(5,\) giving us \(3\) more possibilities.

The total number of working pairs is \[ 1 + 6 + 3 = 10, \] and the total number of pairs is \(8^2 = 64.\) The desired probability is then \[ \dfrac{10}{64} = \dfrac{5}{32}. \]

Thus, A is the correct answer.

Answer: D

Solution(s):

Note that there is one unit cube between any pair of corner cubes, so the removal of each does not affect the others.

When we remove a corner, we are losing three unit squares. We, however, gain these back from the three faces that get uncovered.

This means that removing a corner cube does not change the surface area. The surface area of the original cube is \[ 6 \cdot 3^2 = 6 \cdot 9 = 54 \] square centimeters.

Thus, D is the correct answer.

Answer: E

Solution(s):

The two-inch cube consists of \(2^3 = 8\) unit cubes. Each of these unit cubes is worth \[ 200 \div 8 = 25 \] dollars. To form a three-inch cube, you need \(3^3 = 27\) unit cubes. This means that it is worth \[ 25 \cdot 27 = 675 \] dollars.

Thus, E is the correct answer.

Answer: C

Solution(s):

Note that if the number has \(5\) digits, then the sum of the squares of the digits is at least \[ 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55, \] which violates the first condition.

This means that we should aim to find the largest four digit number that satisfies the properties.

Let the four digit number be \(abcd\) where \[ 0 \lt a \lt b \lt c \lt d. \] We must have that \(d \lt 8\) since otherwise \[ d \geq 8 \Rightarrow d^2 \geq 64 \gt 50. \]

If \(d = 7,\) then \[ a^2 + b^2 + c^2 \geq 1^2 + 2^2 + 3^2 = 14, \] which means that the sum of the digits is greater than \(50.\)

Then if \(d = 6,\) we have that \(a = 1,\) \(b = 2,\) and \(c = 3\) work. If \(d = 5,\) then \[ a^2 + b^2 + c^2 = 25. \] The only option is \((0, 4, 5)\) which causes us to end up with a \(3\)-digit number.

If \(d = 4,\) then \[ a^2 + b^2 + c^2 + d^2 = 14, \] which does not satisfy the first condition. The only viable option is then \(1236.\) The product of the digits is then \[ 1 \cdot 2 \cdot 3 \cdot 6 = 36. \]

Thus, C is the correct answer.

Answer: C

Solution(s):

WLOG, let \(AE = 10.\) Then \[ AC = \dfrac{2}{5} \cdot 10 = 4 \] and \[ CE = \dfrac{3}{5} \cdot 10 = 6. \] Then the area of the semicircle \(ABC\) is \[ \dfrac{1}{2} \cdot \pi (4 \div 2)^2 = 2 \pi. \]

We also have that the area of semicircle \(CDE\) is \[ \dfrac{1}{2} \cdot \pi (6 \div 2)^2 = \dfrac{9}{2}\pi. \]

Then the area of the upper shaded region is \[ \dfrac{1}{2} \cdot (10 \div 2)^2 - 2 \pi + \dfrac{9}{2}\pi = 15 \pi. \]

Subtracting this from the area of the total circle gives us that the area of the lower region is \[ \pi 5^2 - 15 \pi = 10\pi. \]

The desired ratio is then \[ \dfrac{15\pi}{10\pi} = \dfrac{3}{2} = 3:2. \]

Thus, C is the correct answer.

Answer: D

Solution(s):

We only care about the units digit, which means that the tens digits don't matter.

Then we have \(10\) groups of \[ 2 \cdot 4 \cdot 6 \cdot 8 = 384. \]

The units digit of each group is \(4.\) We now need to find the units digit of \(4^{10}.\)

The units digit of \(4^2 = 16\) is \(6.\) This means we only need to find the units digit of \(6^5.\)

Note that every power of \(6\) always ends in a \(6\) (e.g. \(6, 36, 216, \cdots\)).

Thus, D is the correct answer.