Small Hint:

A digit in the hundreds place counts $100$ times, so put the smallest digits there

Big Hint:

Hundreds digits should be $4$ and $5$, tens digits $6$ and $7$, units digits $8$ and $9$

Solution:

To make the sum small, the two smallest digits go in the hundreds places, the next two in the tens places, and the two largest in the units places.

One such arrangement is $468 + 579 = 1047$, and every arrangement of this type gives the same sum.

Thus, the correct answer is C .

Small Hint:

Changing a digit in a higher place value raises the number by more

Big Hint:

The leftmost digit $1$ sits in the tenths place, the largest place value shown

Solution:

Changing a digit in the tenths place changes the number more than changing any digit farther to the right. The digit in the tenths place is $1$.

Changing it gives $0.92345$, which is the largest possible result.

Thus, the correct answer is A .

Small Hint:

Reflect the figure across the main diagonal that runs corner to corner

Big Hint:

Each shaded piece on one side of the diagonal matches an unshaded piece of equal area on the other side

Solution:

The diagonal from one corner to the opposite corner splits the square into two equal halves. Every shaded piece on one side of the diagonal is the mirror image of an equal unshaded piece on the other side.

So the shaded and unshaded areas are equal, and exactly $\dfrac12$ of the square is shaded.

Thus, the correct answer is E .

Small Hint:

The units digit of a square depends only on the units digit of the number being squared

Big Hint:

Square each of the digits $0$ through $9$ and record only the units digit of each result

Solution:

The units digit of a square is determined by the units digit of the number squared. Squaring $0$ through $9$ gives units digits $0,$ $1,$ $4,$ $9,$ $6,$ $5,$ $6,$ $9,$ $4,$ $1$.

So a square can only end in $0,$ $1,$ $4,$ $5,$ $6,$ or $9$; it can never end in $2,$ $3,$ $7,$ or $8$. Among the choices, only $8$ is impossible.

Thus, the correct answer is E .

Small Hint:

$0.48017$ is very close to $\dfrac12$

Big Hint:

Estimate the product as $\left(\dfrac12\right)^3$

Solution:

Since $0.48017$ is close to $\dfrac12$, the product is close to

$$\left(\dfrac12\right)^3 = \dfrac18 = 0.125.$$

The choice closest to $0.125$ is $0.110$.

Thus, the correct answer is B .

Small Hint:

Dividing by $\dfrac{1}{2468}$ is the same as multiplying by $2468$

Big Hint:

All the other choices stay near $13579$, but one of them becomes thousands of times larger

Solution:

Choices A, B, and E are all very close to $13579$, and choice C multiplies $13579$ by a tiny number, making it much smaller.

Choice D divides by $\dfrac{1}{2468}$, which is the same as multiplying by $2468$. This makes it thousands of times larger than $13579$, so it is the largest.

Thus, the correct answer is D .

Small Hint:

A product of three numbers is positive only when none or two of them are negative

Big Hint:

There are only two positive numbers, so use two negatives and the largest positive

Solution:

For the product of three numbers to be positive, either all three are positive or exactly two are negative. There are only two positive numbers, so we must use two negatives and one positive.

To maximize, take the two most negative numbers and the largest positive: $(-3)(-2)(5) = 30$.

Thus, the correct answer is C .

Small Hint:

A $25\%$ discount leaves $75\%$ of the price

Big Hint:

Adding $10\%$ tax multiplies the sale price by $1.1$

Solution:

The sale price is $\dfrac34 \cdot \$80 = \$60.$

The tax is $10\%$ of $\$60,$ which is $\$6,$ so the total is $\$60 + \$6 = \$66.$

Thus, the correct answer is D .

Small Hint:

Count how many scores fall in the range $75\text{--}84$

Big Hint:

Divide that count by the $15$ students to get the fraction, then convert to a percent

Solution:

A grade of C corresponds to scores from $75$ to $84$. The qualifying scores are $77,$ $75,$ $84,$ $78,$ $80$, which is $5$ students.

So the percent is $\dfrac{5}{15} = \dfrac13 = 33\dfrac13\%.$

Thus, the correct answer is D .

Small Hint:

On a calendar, moving one column right adds $1$ to the date and moving one row down adds $7$

Big Hint:

The date behind $C$ is one less than the date behind $A$, so the missing date must be one more than the date behind $B$

Solution:

Moving one column right adds $1$ to the date, and moving one row down adds $7$. Let $C$ have date $n$. Then $A$, one column to the right, is $n+1$. The letter $P$ sits two rows directly below $C$, so $P = n+14$, and $B$, one column left of $P$ in the bottom row, is $n+13$.

We need a letter whose date $d$ satisfies $d + n = (n+1) + (n+13)$, so $d = n + 14$. That date is behind $P$.

Thus, the correct answer is A .

Small Hint:

Five of the faces must be $11,$ $12,$ $13,$ $14,$ $15$; the sixth is either $10$ or $16$

Big Hint:

The three faces $11,$ $14,$ $15$ meet at one corner, so no two of them are opposite; this rules out one of the two possibilities

Solution:

The six consecutive numbers include $11,$ $14,$ $15$, so five of the faces are $11,$ $12,$ $13,$ $14,$ $15$ and the sixth is $10$ or $16$.

If the sixth were $10$, equal opposite sums would force the pairs $(10,15),$ $(11,14),$ $(12,13)$, making $11$ and $14$ opposite. But the figure shows $11,$ $14,$ $15$ meeting at one corner, so no two of them are opposite. Hence the sixth number is $16$, with pairs $(11,16),$ $(12,15),$ $(13,14)$, each summing to $27$.

The total is $3 \cdot 27 = 81$.

Thus, the correct answer is E .

Small Hint:

The $24$ numbers split into four blocks of $6$ by their leading digit

Big Hint:

Positions $1$–$6$ start with $2$, $7$–$12$ start with $4$, $13$–$18$ start with $5$; find the $5$th number starting with $5$

Solution:

Each leading digit accounts for $6$ of the $24$ numbers. Positions $1$–$6$ begin with $2$, positions $7$–$12$ begin with $4$, and positions $13$–$18$ begin with $5$.

So the $17$th number is the $5$th one beginning with $5$: $5247,$ $5274,$ $5427,$ $5472,$ $5724$. The fifth is $5724$.

Thus, the correct answer is B .

Small Hint:

The first ounce costs $30$¢; the remaining $3.5$ ounces are billed as full additional ounces

Big Hint:

A fraction of an ounce counts as a whole additional ounce, so $3.5$ ounces beyond the first counts as $4$

Solution:

The first ounce costs $30$¢. The remaining $3.5$ ounces are charged as $4$ additional ounces, since any fraction rounds up to a full ounce.

That is $4 \cdot 22$¢ $= 88$¢, so the total is $30$¢ $+ 88$¢ $= 118$¢ $= \$1.18.$

Thus, the correct answer is C .

Small Hint:

If the probability of blue is $\dfrac14,$ then blue balls are a quarter of the total

Big Hint:

Use $6$ blue balls to find the total, then subtract to get the green balls

Solution:

Since blue balls make up $\dfrac14$ of the bag and there are $6$ of them, the total is $6 \cdot 4 = 24$ balls.

So the number of green balls is $24 - 6 = 18$.

Thus, the correct answer is B .

Small Hint:

Each of the four squares has area $\dfrac{100}{4} = 25 \text{ cm}^2$

Big Hint:

Find the side length of one square, then count how many side lengths make up the outline

Solution:

Each of the four squares has area $\dfrac{100}{4} = 25 \text{ cm}^2$, so each side is $5 \text{ cm}.$

The outline of this staircase shape is made up of $10$ such sides, so the perimeter is $10 \cdot 5 = 50 \text{ cm}.$

Thus, the correct answer is E .

Small Hint:

Group the terms in pairs: $(1990 - 1980), (1970 - 1960), \ldots$

Big Hint:

Each pair equals $10$; count the pairs and remember the leftover $+10$ at the end

Solution:

Group the terms as $(1990 - 1980)$ $+ (1970 - 1960)$ $+ \cdots$ $+ (30 - 20)$ $+ 10.$ Each parenthesized pair equals $10$.

The first elements $1990, 1970, \ldots, 30$ number $99$, so there are $99$ pairs, giving $990$, plus the leftover $+10$ at the end for a total of $1000$.

Thus, the correct answer is D .

Small Hint:

First convert $3$ inches to feet: $3 \text{ in} = \dfrac14 \text{ ft}$

Big Hint:

Find the volume in cubic feet, then divide by $27$ since $1$ cubic yard $= 27$ cubic feet

Solution:

The thickness is $3$ inches $= \dfrac14$ foot, so the volume is $3 \cdot 60 \cdot \dfrac14 = 45$ cubic feet.

Since $1$ cubic yard $= 27$ cubic feet, this is $\dfrac{45}{27} = 1\dfrac23$ cubic yards. Rounding up to a whole number, the contractor must order $2$ cubic yards.

Thus, the correct answer is A .

Small Hint:

The original rectangular prism has $12$ edges, and none of them disappear

Big Hint:

Each corner cut creates a new triangular face, adding $3$ new edges

Solution:

A rectangular prism starts with $12$ edges, and cutting corners only shortens them without removing any.

Each of the $8$ corner cuts creates a small triangular face with $3$ new edges, adding $8 \cdot 3 = 24$ edges. The total is $12 + 24 = 36$.

Thus, the correct answer is C .

Small Hint:

To force the next person to sit adjacent to someone, no empty seat may have both neighbors empty

Big Hint:

Seat people so there is exactly one empty seat between occupied ones: occupy every third seat

Solution:

To force the next person next to someone, every empty seat must be adjacent to an occupied one. Each occupied seat can account for at most itself and its two neighboring seats, so at least $\frac{120}{3} = 40$ seats must be occupied.

This bound is attainable: use a repeating pattern of empty-occupied-empty, filling the middle seat of every group of three. This uses exactly $40$ occupied seats.

Thus, the correct answer is B .

Small Hint:

Only one value changed, so only the total sum changes; the count stays at $1000$

Big Hint:

Find how much the one wrong entry inflated the total, then divide that by $1000$

Solution:

Only one entry changed. The incorrect total exceeds the actual total by $\$980{,}000 - \$98{,}000 = \$882{,}000.$

Since this extra amount is spread over $1000$ families, the means differ by $\dfrac{\$882{,}000}{1000} = \$882.$

Thus, the correct answer is A .

Small Hint:

If a term is the product of the two before it, then dividing a term by the one just before it recovers the term two positions back

Big Hint:

Work backwards from the end, filling one earlier term at a time by dividing

Solution:

Since each term is the product of the two before it, dividing any term by the term just before it recovers the term two positions earlier. Working backwards from the last three terms $16,$ $64,$ $1024$:

$$\begin{gathered} 64 \div 16 = 4, \quad 16 \div 4 = 4, \\ 4 \div 4 = 1, \quad 4 \div 1 = 4, \\ 1 \div 4 = \dfrac14. \end{gathered}$$

These fill in the earlier terms, giving the full list $\dfrac14,$ $4,$ $1,$ $4,$ $4,$ $16,$ $64,$ $1024$. The first number is $\dfrac14$.

Thus, the correct answer is B .

Small Hint:

Chris takes piece $1$ and piece $100$, so after Chris’s first piece, the remaining $99$ pieces bring the bag exactly back to Chris

Big Hint:

The number of students must divide $99$ evenly

Solution:

Chris takes the first piece, and then the bag goes around until Chris takes the last ($100$th) piece. So the $99$ pieces after Chris’s first must be exactly a whole number of trips around the table back to Chris.