Solution:

The coins total $1 + 5 + 10 + 25 = 41$ cents.

Since one dollar is $100$ cents, this is $41\%$ of a dollar.

Thus, the correct answer is D .

Solution:

Zack is $15 + 3 = 18$ years old.

Jose is $4$ years younger, so Jose is $18 - 4 = 14$.

Thus, the correct answer is C .

Solution:

Dividing by $\dfrac35$ is the same as multiplying by $\dfrac53,$ so the two operations together multiply by

$$\dfrac34 \times \dfrac53 = \dfrac54.$$

So the combined effect is multiplying by $\dfrac54$.

Thus, the correct answer is E .

Solution:

Ben computes $(6 + 1) \times 2 = 14$ and gives $14$ to Sue.

Sue computes $(14 - 1) \times 2 = 26$.

Thus, the correct answer is C .

Solution:

The whole-number parts sum to $2 + 3 + 4 + 5 = 14$.

The fractions $\dfrac12 + \dfrac13 + \dfrac14 + \dfrac15$ add to about $1.28,$ which is between $1$ and $2.$ So the total is between $15$ and $16,$ and the smallest whole number larger than it is $16$.

Thus, the correct answer is C .

Solution:

Square $I$ has side $12 \div 4 = 3,$ and square $II$ has side $24 \div 4 = 6.$

From the figure, the side of $III$ is $3 + 6 = 9,$ so its perimeter is $4 \times 9 = 36.$

Thus, the correct answer is C .

Solution:

The students who ride make up

$$\dfrac12 + \dfrac14 + \dfrac{1}{10} = \dfrac{10}{20} + \dfrac{5}{20} + \dfrac{2}{20} = \dfrac{17}{20}.$$

So the fraction who walk is $1 - \dfrac{17}{20} = \dfrac{3}{20}.$

Thus, the correct answer is B .

Solution:

Since $\$1.00$ is $\dfrac{1.00}{1.60} = \dfrac58$ of $\$1.60,$ the traveler gets $\dfrac58$ of $3000$ lire.

That is $\dfrac58 \times 3000 = 1875$ lire.

Thus, the correct answer is D .

Solution:

Each circle has diameter $4.$ The short side of the rectangle equals one diameter, so it is $4.$

Since the circle at $Q$ passes through $P$ and $R,$ all three circles have radius $2,$ and the long side spans two full diameters: $4 + 4 = 8.$ The area is $8 \times 4 = 32.$

Thus, the correct answer is C .

Solution:

The jacket discount saves $40\%$ of $\$80 = \$32,$ and the shirt discount saves $55\%$ of $\$40 = \$22.$ The total saved is $\$32 + \$22 = \$54.$

The original total is $\$80 + \$40 = \$120,$ so the percent saved is $\dfrac{54}{120} = 0.45 = 45\%.$

Thus, the correct answer is A .

Solution:

The perimeter of the region is $18$ blocks, so when Jane and Hector meet they have together walked $18$ blocks. Since Jane walks twice as fast, she covers $12$ blocks and Hector covers $6.$

Starting from the middle of the bottom edge, Hector walks $6$ blocks (to $E,$ then up to $D$), and Jane walks $12$ blocks (to $A,$ up to $B,$ then across the top to $D$). They meet at $D.$

Thus, the correct answer is D .

Solution:

Each of the other years works: $90 = 9 \times 10,$ $91 = 7 \times 13,$ $92 = 4 \times 23,$ and $93 = 3 \times 31,$ each a valid month/day.

For $1994,$ the last two digits factor only as $94 = 2 \times 47,$ and $47$ is too large for a day (and $94$ or $47$ is too large for a month). So $1994$ has no lucky date.

Thus, the correct answer is E .

Solution:

In triangle $BDE,$ the angles at $E$ and $D$ are equal and $\angle B = 90^\circ,$ so $\angle BED = \angle BDE = 45^\circ.$

Then $\angle AED = \angle AEB + \angle BED = 40^\circ + 45^\circ = 85^\circ.$ In quadrilateral $AEDC,$ the angles at $A$ and $C$ are $90^\circ,$ so

$$\angle CDE = 360^\circ - 90^\circ - 90^\circ - 85^\circ = 95^\circ.$$

Thus, the correct answer is E .

Solution:

The season has $50 + 40 = 90$ games, and $70\%$ of $90$ is $63$ wins.

The team already has $40$ wins, so it needs $63 - 40 = 23$ more.

Thus, the correct answer is B .

Solution:

$\dfrac{4}{37} = 0.\overline{108},$ repeating with block length $3.$ The digits in positions $3, 6, 9, \ldots$ (multiples of $3$) are $8.$

Since $99$ is a multiple of $3,$ the $99$th digit is $8,$ so the $100$th digit starts the next block: it is $1.$

Thus, the correct answer is B .

Solution:

The total student-days are $7 \times 3 + 4 \times 5 + 5 \times 9 = 21 + 20 + 45 = 86.$

So each student-day pays $\$774 \div 86 = \$9.$ Balboa worked $20$ student-days, earning $20 \times \$9 = \$180.$

Thus, the correct answer is C .

Solution:

Annville has $11\%$ of $100 = 11$ sixth graders, and Cleona has $17\%$ of $200 = 34$ sixth graders.

Combined, that is $11 + 34 = 45$ out of $300$ students, which is $\dfrac{45}{300} = 15\%.$

Thus, the correct answer is D .

Solution:

The four L-shaped regions cover $4 \times \dfrac{3}{16} = \dfrac34$ of the square, so the center square is the remaining $\dfrac14$ of the total area.

The total area is $100 \times 100 = 10000$ square inches, so the center square has area $\dfrac14 \times 10000 = 2500,$ and its side is $\sqrt{2500} = 50$ inches.

Thus, the correct answer is C .

Solution:

The graph gives $2$ families with $1$ child, $1$ with $2,$ $2$ with $3,$ $2$ with $4,$ and $6$ with $5,$ for $2 + 1 + 2 + 2 + 6 = 13$ families.

The median is the $7$th value in order. Listing the family sizes, the $7$th value is $4.$

Thus, the correct answer is D .

Solution:

There are $6 \times 6 = 36$ equally likely outcomes, of which $6$ are ties, leaving $30$ outcomes with different numbers.

By symmetry, Diana is larger in exactly half of those, or $15,$ so the probability is $\dfrac{15}{36} = \dfrac{5}{12}.$

Thus, the correct answer is B .

Solution:

Every cube's single snap must be plugged into another cube's hole to be hidden. With one, two, or three cubes, at least one snap is always left exposed.

Four cubes can be arranged in a square ring, each snap fitting into the neighbor's hole, so only receptacle holes show. The smallest number is $4.$

Thus, the correct answer is B .

Solution:

The prime factorization is $6545 = 5 \times 7 \times 11 \times 17.$ To split into two two-digit factors, pair the primes: $5 \times 17 = 85$ and $7 \times 11 = 77.$

These are the only two-digit pair, and their sum is $85 + 77 = 162.$

Thus, the correct answer is A .

Solution:

The first digit is odd: $5$ choices. The second is even: $5$ choices (none of which repeats the odd first digit).

The third digit is any of the $8$ unused digits, and the fourth is any of the $7$ remaining. In total, $5 \times 5 \times 8 \times 7 = 1400.$

Thus, the correct answer is B .

Solution:

Since $AB = DC = 12,$ we get $AE = 12 - 4 = 8.$ In right triangle $ADE,$ $AD = \sqrt{8^2 + 6^2} = 10,$ so $BC = AD = 10.$

The area is $AB \cdot DE = 12 \cdot 6 = 72,$ and also $BC \cdot DF = 10 \cdot DF.$ So $DF = \dfrac{72}{10} = 7.2.$

Thus, the correct answer is C .

Solution:

Consider a bus leaving Dallas at $6\!:\!00,$ arriving in Houston at $11\!:\!00.$ It meets every Dallas-bound bus that is on the highway during that window.

Dallas-bound buses leave Houston on the half hour and take $5$ hours. The ones sharing the road (meeting away from a station) are those that left Houston at $1\!:\!30, 2\!:\!30, \ldots, 10\!:\!30,$ which is $10$ buses.

Thus, the correct answer is D .