Small Hint:

Line up the decimal points before adding

Big Hint:

Write the numbers as $0.400,$ $0.020,$ and $0.006$

Solution:

Adding place by place, $0.4 + 0.02 + 0.006 = 0.426.$

Thus, the correct answer is E .

Small Hint:

Rewrite the fraction with a denominator of $100$

Big Hint:

Multiply the numerator and denominator by $4$

Solution:

Multiplying top and bottom by $4$ gives $\dfrac{2}{25} = \dfrac{8}{100} = 0.08.$

Thus, the correct answer is B .

Small Hint:

The ten numbers $81,$ $83,$ $\ldots,$ $99$ are evenly spaced

Big Hint:

Their average is $\dfrac{81 + 99}{2} = 90,$ so their sum is $10 \times 90$

Solution:

The ten evenly spaced numbers have average $90,$ so their sum is $10 \times 90 = 900.$

Doubling gives $2 \times 900 = 1800.$

Thus, the correct answer is E .

Small Hint:

A right angle is $\dfrac14$ of a full circle

Big Hint:

Take $\dfrac14$ of $500$ clerts

Solution:

A right angle is a quarter of a full circle, so it contains $\dfrac14 \times 500 = 125$ clerts.

Thus, the correct answer is C .

Small Hint:

The area of a rectangle is its length times its width

Big Hint:

Multiply $0.4$ by $0.22$

Solution:

The area is $0.4 \times 0.22 = 0.088$ square meters.

Thus, the correct answer is A .

Small Hint:

The smallest product is the most negative one, which comes from one negative and one positive factor

Big Hint:

Pair the most negative number with the largest positive number

Solution:

A negative product comes from multiplying a negative number by a positive number, and it is most negative when both factors are largest in size.

The most negative product is $(-7)(3) = -21.$

Thus, the correct answer is B .

Small Hint:

A small cube shows a shaded face only when one of its surface squares is shaded, so count the cubes that stay completely blank, beginning with the single hidden cube at the center of the block

Big Hint:

Besides the hidden center cube, find the cubes whose every exposed square lands on a blank position, then subtract the number of blank cubes from $27$

Solution:

Count the cubes with no shaded face and subtract from $27.$ A small cube is unshaded exactly when every one of its exposed squares is blank.

The three patterns are: the top and bottom faces show only their center square shaded; one pair of opposite side faces shows the four corners and the center shaded; the remaining pair shows the four edge-midpoints shaded.

Exactly $7$ small cubes avoid every shaded square: the one hidden cube at the very center of the block; the two cubes at the centers of the edge-midpoints faces, whose only exposed square is that blank center; and the four cubes at the midpoints of the edges where a corners-and-center face meets the top or bottom face, since there each exposed square is a blank edge cell.

Hence $27 - 7 = 20$ cubes have at least one shaded face.

Thus, the correct answer is C .

Small Hint:

The number $9876$ is already close to $10{,}000,$ so adding anything pushes the sum past it

Big Hint:

Check the largest possible sum too, using $A = B = 9$

Solution:

Since $A$ and $B$ are at least $1,$ the sum is at least $9876 + 132 + 11 = 10{,}019,$ which has $5$ digits.

At the other extreme, with $A = B = 9$ the sum is $9876 + 932 + 91 = 10{,}899,$ still $5$ digits. So the sum always has exactly $5$ digits.

Thus, the correct answer is B .

Small Hint:

The least common denominator is the least common multiple of $2,$ $3,$ $4,$ $5,$ $6,$ $7$

Big Hint:

Use $4$ to cover the factor $4$ (and $2$), then include $3,$ $5,$ and $7$

Solution:

The least common multiple must include $4 = 2^2,$ $3,$ $5,$ and $7$ (the factor $6 = 2 \times 3$ is already covered).

So it is $4 \times 3 \times 5 \times 7 = 420.$

Thus, the correct answer is C .

Small Hint:

Combine the first three terms: $4(299) + 3(299) + 2(299)$ $= 9 \times 299$

Big Hint:

Compute $9 \times 299,$ then add $298$

Solution:

The first three terms combine to $(4 + 3 + 2)(299)$ $= 9 \times 299 = 2691.$

Adding the last term gives $2691 + 298 = 2989.$

Thus, the correct answer is B .

Small Hint:

Add the whole-number parts first: $2 + 3 + 5 = 10$

Big Hint:

The fractional parts are $\dfrac12 + \dfrac17 + \dfrac{1}{19};$ this is more than $\dfrac12$ but less than $1$

Solution:

The whole-number parts sum to $10.$ The fractional parts are $\dfrac12 + \dfrac17 + \dfrac{1}{19}.$

Since $\dfrac17 + \dfrac{1}{19}$ is a small positive amount, the fractional total is more than $\dfrac12$ but well under $1.$ So the sum lies between $10\dfrac12$ and $11.$

Thus, the correct answer is B .

Small Hint:

The horizontal line splits the rectangle into a top half and a bottom half

Big Hint:

The shaded part is the rightmost third of the bottom-right quarter of the rectangle

Solution:

The whole region has area $12 \times 18 = 216.$ The shaded region is $3$ wide and $6$ tall, with area $3 \times 6 = 18.$

So the shaded fraction is $\dfrac{18}{216} = \dfrac{1}{12}.$ Equivalently, it is $\dfrac13$ of $\dfrac14$ of the whole.

Thus, the correct answer is C .

Small Hint:

Compare each fraction to $\dfrac12$

Big Hint:

A fraction is more than $\dfrac12$ exactly when its numerator is more than half its denominator

Solution:

For $\dfrac37,$ $\dfrac49,$ $\dfrac{17}{35},$ and $\dfrac{100}{201},$ each numerator is less than half its denominator, so each is less than $\dfrac12.$

For $\dfrac{151}{301},$ the numerator $151$ is more than half of $301,$ so this fraction exceeds $\dfrac12$ and is the largest.

Thus, the correct answer is E .

Small Hint:

There are $60 \times 60 = 3600$ seconds in an hour

Big Hint:

Multiply the rate $10{,}000$ by the number of seconds in an hour

Solution:

One hour has $3600$ seconds, so the computer does $10{,}000 \times 3600 = 36{,}000{,}000$ additions.

That is $36$ million.

Thus, the correct answer is B .

Small Hint:

The $\$240$ covers three tires at the regular price plus $\$3$ for the fourth

Big Hint:

Subtract the $\$3,$ then divide the rest among the three full-price tires

Solution:

The three regular-price tires cost $\$240 - \$3 = \$237.$

So one tire costs $\dfrac{\$237}{3} = \$79.$

Thus, the correct answer is D .

Small Hint:

After the next game she has attempted $30 + 10 = 40$ shots

Big Hint:

A $50\%$ average on $40$ shots means $20$ made in all; she already had $12$

Solution:

After the fourth game she has taken $40$ shots. A $50\%$ average means she made $\tfrac12 \times 40 = 20$ shots in total.

She had already made $12,$ so she made $20 - 12 = 8$ of the last $10.$

Thus, the correct answer is E .

Small Hint:

“Bret is next to Carl” is false, so Carl is not in seat $2$ or seat $4;$ the only remaining seat for Carl is seat $1$

Big Hint:

Once Carl is in seat $1,$ “Abby is between Bret and Carl” is false, so Abby cannot be in seat $2;$ place Abby in the last open seat and see who is left

Solution:

Bret is in seat $3.$ Since “Bret is next to Carl” is false, Carl is not in seat $2$ or seat $4,$ so Carl must be in seat $1.$

The seat between Bret (seat $3$) and Carl (seat $1$) is seat $2.$ Since “Abby is between Bret and Carl” is false, Abby is not in seat $2,$ so Abby must be in seat $4.$

That leaves Dana in seat $2.$

Thus, the correct answer is D .

Small Hint:

After half leave, the number remaining is what the dancing fraction applies to

Big Hint:

The $12$ non-dancers are $\dfrac23$ of those remaining

Solution:

Of those remaining, $\dfrac13$ danced, so $\dfrac23$ did not. Thus $\dfrac23$ of the remaining people equals $12,$ giving $18$ remaining.

Those $18$ are half the original group, so the room started with $2 \times 18 = 36$ people.

Thus, the correct answer is C .

Small Hint:

Starting from $2,$ each press squares the number: $2,$ $4,$ $16,$ $256,$ $\ldots$

Big Hint:

Find the first term in that list that exceeds $500$

Solution:

Pressing the key repeatedly gives $2 \to 4 \to 16 \to 256 \to 65536.$

Since $256 \lt 500 \lt 65536,$ the display first exceeds $500$ on the fourth press.

Thus, the correct answer is A .

Small Hint:

To disprove an “if ... then” statement, find a case where the “if” part holds but the “then” part fails

Big Hint:

Look for $n$ that is not prime while $n - 2$ is prime

Solution:

A counterexample needs $n$ not prime but $n - 2$ prime. For $n = 9,$ the number $9$ is not prime while $9 - 2 = 7$ is prime, so it breaks the statement.

The others fail to be counterexamples: $12 - 2 = 10$ and $16 - 2 = 14$ are not prime, while $13$ and $23$ are themselves prime.

Thus, the correct answer is A .

Small Hint:

Replace each $n^\ast$ with $\dfrac1n$ and test the two sides of each statement

Big Hint:

For $\mathrm{(iv)},$ $10^\ast \div 2^\ast = \dfrac{1}{10} \div \dfrac12 = \dfrac{1}{10} \cdot 2$

Solution:

$\mathrm{(i)}$ $\dfrac13 + \dfrac16 = \dfrac12,$ but $9^\ast = \dfrac19,$ so false. $\mathrm{(ii)}$ $\dfrac16 - \dfrac14 = -\dfrac{1}{12},$ but $2^\ast = \dfrac12,$ so false.

$\mathrm{(iii)}$ $\dfrac12 \cdot \dfrac16 = \dfrac{1}{12} = 12^\ast,$ true. $\mathrm{(iv)}$ $\dfrac{1}{10} \div \dfrac12 = \dfrac15 = 5^\ast,$ true.

So exactly $2$ of the statements are true.

Thus, the correct answer is C .

Small Hint:

The diagonal $DB$ is the radius; find it with the Pythagorean theorem

Big Hint:

The shaded region is a quarter of the circle with the rectangle removed: $\dfrac14 \pi (5)^2 - 3 \cdot 4$

Solution:

Since $AD = 4$ and $CD = 3,$ the diagonal $DB = \sqrt{3^2 + 4^2} = 5,$ which is the radius.

The shaded region is the quarter circle at $D$ minus the rectangle: $$\begin{gathered} \dfrac14 \pi (5)^2 - (3)(4) = \dfrac{25\pi}{4} - 12 \\ \approx 19.6 - 12 = 7.6. \end{gathered}$$

This lies between $7$ and $8.$

Thus, the correct answer is D .