Small Hint:

Group the numerator into pairs $(10-9),$ $(8-7),$ $\ldots$, and do the same for the denominator

Big Hint:

Each numerator pair equals $1$; in the denominator group $(1-2),$ $(3-4),$ $\ldots$, where each pair is $-1$, with a $+9$ left over

Solution:

Grouping the numerator in pairs gives $(10-9)$ $+ (8-7)$ $+ (6-5)$ $+ (4-3)$ $+ (2-1) = 5.$

Grouping the denominator as $(1-2)$ $+ (3-4)$ $+ (5-6)$ $+ (7-8)$ $+ 9$ gives four pairs of $-1$ plus $9$, which is $5.$

So the expression is $\dfrac{5}{5} = 1.$

Thus, the correct answer is B .

Small Hint:

Rewrite $\dfrac54$ as the mixed number $1\dfrac14$

Big Hint:

A mixed number $1\dfrac{a}{b}$ equals $\dfrac54$ only when $\dfrac{a}{b} = \dfrac14$; check each choice

Solution:

Note $\dfrac54 = 1\dfrac14.$ Now check each choice: $\dfrac{10}{8} = \dfrac54,$ $1\dfrac14 = \dfrac54,$ $1\dfrac{3}{12} = 1\dfrac14 = \dfrac54,$ and $1\dfrac{10}{40} = 1\dfrac14 = \dfrac54.$

But $1\dfrac15 = \dfrac65 \neq \dfrac54.$

Thus, the correct answer is D .

Small Hint:

A difference is largest when you subtract the smallest number from the largest number

Big Hint:

Subtracting a negative adds; compute $12 - (-16)$

Solution:

The largest difference uses the largest number, $12$, minus the smallest number, $-16.$

This gives $12 - (-16) = 28.$

Thus, the correct answer is D .

Small Hint:

First find how many hits were not singles

Big Hint:

Singles $= 35 - (1+1+5)$; divide by $35$ and convert to a percent

Solution:

The non-single hits number $1 + 1 + 5 = 7,$ so the singles number $35 - 7 = 28.$

The fraction of singles is $\dfrac{28}{35} = \dfrac45 = 80\%.$

Thus, the correct answer is E .

Small Hint:

The rectangle has area $2 \times 3 = 6$; subtract the area of the circle

Big Hint:

The circle has radius $\dfrac12,$ so its area is $\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4},$ a little less than $1$

Solution:

The rectangle has area $6,$ and the removed circle has area $\pi\left(\dfrac12\right)^2 = \dfrac{\pi}{4} \approx 0.79.$

So the shaded region has area $6 - 0.79 \approx 5.2,$ whose closest whole number is $5.$

Thus, the correct answer is E .

Small Hint:

Apply the rule to each triple separately: top plus lower-left minus lower-right

Big Hint:

Compute $(1 + 3 - 4)$ and $(2 + 5 - 6),$ then add the two results

Solution:

The first triple gives $1 + 3 - 4 = 0,$ and the second gives $2 + 5 - 6 = 1.$

Their sum is $0 + 1 = 1.$

Thus, the correct answer is D .

Small Hint:

A digit-sum of $26$ forces the three digits to be very large

Big Hint:

The only digits that work are $9,$ $9,$ and $8$; list the arrangements and keep the even ones

Solution:

A digit-sum of $26$ with three digits requires the digits $9,$ $9,$ and $8.$ The $3$-digit numbers using them are $899,$ $989,$ and $998.$

Of these, only $998$ is even, so there is exactly $1$ such number.

Thus, the correct answer is A .

Small Hint:

First find the total cost of the $1500$ pencils

Big Hint:

The revenue must be the cost plus $\$100;$ divide that needed revenue by the $\$0.25$ price

Solution:

The pencils cost $1500 \times \$0.10 = \$150.$ To make a $\$100$ profit, the revenue must be $\$150 + \$100 = \$250.$

At $\$0.25$ each, he must sell $\dfrac{\$250}{\$0.25} = 1000$ pencils.

Thus, the correct answer is C .

Small Hint:

Let the number of males be $M$; then the number of females is $2M$

Big Hint:

Since $M + 2M = 480,$ solve $3M = 480$

Solution:

If there are $M$ males, then there are $2M$ females, and together $M + 2M = 480.$

So $3M = 480,$ giving $M = 160.$

Thus, the correct answer is B .

Small Hint:

The whole triangle has area $\dfrac12 \times 8 \times 8 = 32,$ split into $16$ equal pieces

Big Hint:

Each small triangle has area $2$; count how many are shaded

Solution:

The large triangle has area $\dfrac12 \times 8 \times 8 = 32,$ so each of the $16$ congruent small triangles has area $\dfrac{32}{16} = 2.$

Ten of the small triangles are shaded, so the shaded area is $10 \times 2 = 20.$

Thus, the correct answer is B .

Small Hint:

Add the five counts to get the total number of people surveyed

Big Hint:

The percent preferring blue is (blue count) divided by (total), converted to a percent

Solution:

The total number surveyed is $50 + 60 + 40 + 60 + 40 = 250.$

The percent preferring blue is $\dfrac{60}{250} = 24\%.$

Thus, the correct answer is B .

Small Hint:

At any moment $4$ of the $5$ tires are on the ground, so the total tire-mileage is $4 \times 30{,}000$

Big Hint:

Share that total tire-mileage equally among the $5$ tires

Solution:

During the $30{,}000$ miles, $4$ tires are always in use, so the total tire-mileage is $4 \times 30{,}000 = 120{,}000$ tire-miles.

Split equally among $5$ tires, each tire is used $\dfrac{120{,}000}{5} = 24{,}000$ miles.

Thus, the correct answer is C .

Small Hint:

The five scores add up to $5 \times 90 = 450,$ and the middle score is $91$

Big Hint:

The mode $94$ must appear at least twice, and both copies must be at or above the median, so the three highest scores are $91,$ $94,$ and $94$

Solution:

The five scores sum to $5 \times 90 = 450.$ The median is the third score, $91.$ Since $94$ is the mode, it must appear at least twice, and both copies lie above the median, so the two highest scores are $94$ and $94.$

The three highest scores are $91,$ $94,$ and $94,$ summing to $279.$ So the two lowest sum to $450 - 279 = 171.$

Thus, the correct answer is B .

Small Hint:

Adding $4$ gallons raises the tank from $\dfrac13$ full to $\dfrac12$ full

Big Hint:

So $4$ gallons equals $\dfrac12 - \dfrac13 = \dfrac16$ of the capacity

Solution:

The $4$ gallons account for the change from $\dfrac13$ full to $\dfrac12$ full, which is $\dfrac12 - \dfrac13 = \dfrac16$ of the capacity.

If $\dfrac16$ of the capacity is $4$ gallons, the full capacity is $4 \times 6 = 24$ gallons.

Thus, the correct answer is D .

Small Hint:

The string $ABCDEDCBA$ repeats over and over; count how many letters are in one block

Big Hint:

One block has $9$ letters, so divide $1992$ by $9$ and use the remainder to find the position within a block

Solution:

The sequence is the $9$-letter block $ABCDEDCBA$ repeated again and again.

Since $1992 = 9 \times 221 + 3,$ the $1992$nd letter is the $3$rd letter of a block, which is $C.$

Thus, the correct answer is C .

Small Hint:

The volume of a cylinder is $\pi r^2 h$; first compute the volume of the given cylinder

Big Hint:

Doubling the height doubles the volume, but doubling the radius multiplies the volume by $4$; check the $r$ and $h$ labels on each choice

Solution:

The given cylinder has volume $\pi \times 10^2 \times 5 = 500\pi,$ so a cylinder of twice the volume has volume $1000\pi.$

The choices have volumes $\pi \times 20^2 \times 5 = 2000\pi$ (A), $\pi \times 10^2 \times 10 = 1000\pi$ (B), $\pi \times 5^2 \times 20 = 500\pi$ (C), and $\pi \times 20^2 \times 10 = 4000\pi$ (D). Only (B) equals $1000\pi.$

Thus, the correct answer is B .

Small Hint:

For a triangle, the two shorter sides must add to more than the longest side

Big Hint:

You need $6.5 + s \gt 10$; find the smallest whole number $s$

Solution:

By the triangle inequality, $6.5 + s$ must exceed the longest side $10,$ so $s \gt 3.5.$

The smallest whole number greater than $3.5$ is $4,$ and it does form a valid triangle.

Thus, the correct answer is B .

Small Hint:

Average speed is total distance divided by total time

Big Hint:

Total distance is $80 + 100$ miles; total time is $1.5 + 0.5 + 2 = 4$ hours, since the stop still counts

Solution:

The car covered $80 + 100 = 180$ miles in total, over $1.5 + 0.5 + 2 = 4$ hours (the traffic stop still counts as time).

The average speed is $\dfrac{180}{4} = 45$ miles per hour.

Thus, the correct answer is A .

Small Hint:

Between the $5$th and $26$th exits there are $21$ gaps

Big Hint:

To make one gap as large as possible, make the other $20$ gaps as small as allowed, $5$ miles each

Solution:

From the $5$th exit to the $26$th exit there are $26 - 5 = 21$ gaps between consecutive exits, each at least $5$ miles.

To maximize one gap, make the other $20$ gaps exactly $5$ miles, using $20 \times 5 = 100$ miles. The remaining gap is $118 - 100 = 18$ miles.

Thus, the correct answer is C .

Small Hint:

A cube has $6$ faces, so each pattern has $6$ squares; try folding each one up in your mind

Big Hint:

Track where each square lands as the pattern folds; a pattern fails when two squares would cover the same face

Solution:

Each of the five patterns has $6$ squares. Folding patterns (A), (B), (C), and (E) wraps the squares neatly onto the six faces of a cube.

For pattern (D), any attempt to fold forces two of the squares to land on the same face, so they overlap and no cube can be formed.

Thus, the correct answer is D .

Small Hint:

For each month, read off the two bar heights and find the gap between them

Big Hint:

Divide the difference by the smaller value to find the percent by which the larger exceeds it; compare this ratio across the months

Solution:

Reading the graph gives (drums, bugles): January $(7, 9),$ February $(5, 3),$ March $(9, 6),$ April $(9, 12),$ and May $(8, 10).$

The percent excess of the larger over the smaller is $\tfrac{2}{7} \approx 29\%$ in January, $\tfrac{2}{3} \approx 67\%$ in February, $\tfrac{3}{6} = 50\%$ in March, $\tfrac{3}{9} \approx 33\%$ in April, and $\tfrac{2}{8} = 25\%$ in May. The greatest is February, where $5$ exceeds $3$ by about $67\%.$

Thus, the correct answer is B .

Small Hint:

Adding a tile that shares exactly one side changes the perimeter by $+2$; sharing two sides changes it by $0$

Big Hint:

With two added tiles, the total change can be $0,$ $2,$ or $4$; add these to $14$

Solution:

A tile that shares exactly one side adds $2$ to the perimeter (four new edges minus two hidden), while a tile sharing two sides adds $0.$

With two added tiles the perimeter can change by $0,$ $2,$ or $4,$ giving new perimeters of $14,$ $16,$ or $18.$ Of the choices, only $18$ is possible.

Thus, the correct answer is C .

Small Hint:

There are $6 \times 6 = 36$ equally likely outcomes; count those whose product exceeds $10$

Big Hint:

Make a $6 \times 6$ multiplication table and mark every product greater than $10$

Solution:

There are $36$ equally likely outcomes. Counting the ordered pairs whose product exceeds $10$: with a first die of $2$ there is $1$ (namely $2 \times 6$); of $3$, there are $3$; of $4$, there are $4$; of $5$, there are $4$; of $6$, there are $5.$