Pair the terms so each pair sums to $50$: $1 + 49,$ $11 + 39,$ $21 + 29,$ $31 + 19,$ and $41 + 9.$

There are $5$ such pairs, so the total is $5 \times 50 = 250.$

Thus, the correct answer is E .

Each fraction is a decimal in a different place: $\frac{2}{10} = .2,$ $\frac{4}{100} = .04,$ and $\frac{6}{1000} = .006.$

Adding these gives $.2 + .04 + .006 = .246.$

Thus, the correct answer is D .

Write each number with four decimal places: $.9900,$ $.9099,$ $.9000,$ $.9090,$ $.9009.$

Comparing from the left, $.9900$ has the largest hundredths digit, so $.99$ is the largest.

Thus, the correct answer is A .

Round the numerator and denominator: $\frac{401}{.205} \approx \frac{400}{.2}.$

Since $\frac{400}{.2} = 2000,$ the value is closest to $2000.$

Thus, the correct answer is E .

Inside the parentheses, $6 \div 3 = 2.$ Then multiplication comes before addition, so $9 \times 2 = 18.$

Finally, $-15 + 18 = 3.$

Thus, the correct answer is D .

From $0$ to $20$ there are $5$ equal spaces, so each space is $\frac{20}{5} = 4.$

The mark labeled $y$ is $3$ spaces from $0,$ so $y = 3 \times 4 = 12.$

Thus, the correct answer is C .

The value of $20$ quarters and $10$ dimes is $\$5.00 + \$1.00 = \$6.00.$

Ten quarters are worth $\$2.50,$ so the $n$ dimes must supply $\$6.00 - \$2.50 = \$3.50.$ That is $35$ dimes, so $n = 35.$

Thus, the correct answer is D .

Since $2 \times 3 \times 4 = 24,$ distribute it over the sum: $24 \cdot \frac{1}{2} + 24 \cdot \frac{1}{3} + 24 \cdot \frac{1}{4}.$

This equals $12 + 8 + 6 = 26.$

Thus, the correct answer is E .

Boys make up $2$ out of every $2 + 3 = 5$ students, which is $\frac{2}{5}$ of the class. The total of $30$ students is not even needed.

As a percent, $\frac{2}{5} = 40\%.$

Thus, the correct answer is C .

The $12$ numbers divide the clock into $12$ equal sections of $\frac{360^\circ}{12} = 30^\circ$ each.

At seven o'clock the hands point to $12$ and $7,$ which are $5$ sections apart the short way, giving $5 \times 30^\circ = 150^\circ.$

Thus, the correct answer is D .

Two figures are symmetric with respect to the dashed line when folding the paper along that line makes one coincide with the other. Reflecting across the vertical dashed line swaps left and right while leaving top and bottom unchanged.

Under this reflection the small square in the top-left corner moves to the top-right corner, the straight crossbar changes its slant from "/" to "\", and the stem that points down and to the right must instead point down and to the left. Only one choice has all three of these features.

Thus, the correct answer is B .

The numerator is $1 - \frac{1}{3} = \frac{2}{3},$ and the denominator is $1 - \frac{1}{2} = \frac{1}{2}.$

Dividing gives $\frac{2}{3} \div \frac{1}{2} = \frac{2}{3} \times 2 = \frac{4}{3}.$

Thus, the correct answer is E .

To rewrite the numerator $9$ as $.9,$ divide it by $10.$ To keep the fraction equal, divide the denominator by $10$ too, which means dividing one of its factors by $10.$

Dividing the factor $7$ by $10$ gives $\frac{.9}{.7 \times 53},$ which equals the original. Each other choice changes the value by a factor of $10$ or $100.$

Thus, the correct answer is A .

The difference is smallest when the three-digit number (the top) is as small as possible and the two-digit number (subtracted) is as large as possible.

The largest two-digit number is $96,$ using the digits $9$ and $6.$ The smallest three-digit number from the remaining digits $2,$ $4,$ $5$ is $245.$ So the smallest difference is $245 - 96 = 149.$

Thus, the correct answer is C .

The parallelogram has base $10$ and height $8,$ so its area is $10 \times 8 = 80.$

The unshaded triangle $ABE$ has base $AE = AD - ED = 10 - 6 = 4$ and height $8,$ so its area is $\frac{1}{2} \times 4 \times 8 = 16.$ The shaded area is $80 - 16 = 64.$

Thus, the correct answer is D .

Since $47$ is odd, a sum of two primes equal to $47$ needs one even prime and one odd prime. The only even prime is $2.$

That would require the other number to be $47 - 2 = 45,$ but $45 = 9 \times 5$ is not prime. So there is no way.

Thus, the correct answer is A .

The average is $\frac{6 + 10 + N}{3} = \frac{16 + N}{3}.$ When $N = 9$ this is $\frac{25}{3} \approx 8.3,$ and when $N = 17$ it is $11.$

So the average lies strictly between about $8.3$ and $11.$ Among the choices, only $10$ falls in this range.

Thus, the correct answer is B .

Pressing the key once changes $32$ to its reciprocal $\frac{1}{32}.$

Pressing it a second time takes the reciprocal again, returning to $\frac{1}{1/32} = 32.$ So $2$ presses are enough.

Thus, the correct answer is B .

The graph gives the total accumulated spending, so the amount spent during June, July, and August equals the accumulated total at the end of August minus the accumulated total at the beginning of June.

The curve is at a bit more than $2$ million at the beginning of June and a bit more than $4.5$ million by the end of August. The difference is about $4.5 - 2 = 2.5$ million.

Thus, the correct answer is B .

When the net is folded, the pairs of opposite faces are $1$ and $3,$ $2$ and $5,$ and $4$ and $6.$ Three faces meeting at a corner must come one from each opposite pair.

To maximize the sum, take the larger number from each pair: $3,$ $5,$ and $6.$ These three faces do meet at a corner, and their sum is $3 + 5 + 6 = 14.$

Thus, the correct answer is D .

After selling $25\%$ to Jill, Jack keeps $\frac{3}{4} \times 128 = 96$ apples. After selling $25\%$ of those to June, he keeps $\frac{3}{4} \times 96 = 72$ apples.

He then gives $1$ to his teacher, leaving $72 - 1 = 71.$

Thus, the correct answer is D .

The six letters cycle back to their original order AJHSME every $6$ lines, and the four digits cycle back to $1989$ every $4$ lines.

Both happen on the same line at the least common multiple of $6$ and $4,$ which is $12.$ So AJHSME 1989 first reappears on line $12.$

Thus, the correct answer is C .

Because each higher layer is flush into one corner, each of the four vertical sides shows a stepped profile of $3 + 2 + 1 = 6$ exposed square faces, for $4 \times 6 = 24$ side faces.

Viewed from directly above, the top faces cover the full $3 \times 3$ footprint, adding $9$ more faces. The bottom rests on the ground and is not painted, so the total is $24 + 9 = 33$ square meters.

Thus, the correct answer is C .

Let the square have side $4.$ Folding in half makes a $2 \times 4$ stack of two layers. Cutting parallel to the fold splits it into a strip containing the fold and a strip that does not.

The strip with the fold unfolds into the large rectangle, $2 \times 4,$ with perimeter $2(2 + 4) = 12.$ The other strip is two separate small rectangles, each $1 \times 4,$ with perimeter $2(1 + 4) = 10.$ The ratio is $\frac{10}{12} = \frac{5}{6}.$

Thus, the correct answer is E .

The sum is even when both numbers are even or both are odd. The first wheel has evens $\{4, 8\}$ and odds $\{3, 5\},$ each with probability $\frac{2}{4} = \frac{1}{2}.$ The second wheel has even $\{6\}$ with probability $\frac{1}{3}$ and odds $\{7, 9\}$ with probability $\frac{2}{3}.$

So the probability of an even sum is $\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}.$

Thus, the correct answer is C .