Small Hint:

Look for pairs of terms that combine into round numbers

Big Hint:

Pairing $1$ with $49,$ $11$ with $39,$ and so on, each pair sums to $50$

Solution:

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 .

Small Hint:

Write each fraction as a decimal using its place value

Big Hint:

$\frac{2}{10} = 0.2,$ $\frac{4}{100} = 0.04,$ and $\frac{6}{1000} = 0.006$

Solution:

Each fraction is a decimal in a different place: $\frac{2}{10} = 0.2,$ $\frac{4}{100} = 0.04,$ and $\frac{6}{1000} = 0.006.$

Adding these gives $0.2 + 0.04 + 0.006 = 0.246.$

Thus, the correct answer is D .

Small Hint:

Give every number the same number of decimal places by appending zeros

Big Hint:

Compare digit by digit from the left; the hundredths place decides between the leading $0.9$ numbers

Solution:

Write each number with four decimal places: $0.9900,$ $0.9099,$ $0.9000,$ $0.9090,$ $0.9009.$

Comparing from the left, $0.9900$ has the largest hundredths digit, so $0.99$ is the largest.

Thus, the correct answer is A .

Small Hint:

Round $401$ and $0.205$ to simple numbers

Big Hint:

$\frac{401}{0.205} \approx \frac{400}{0.2}$

Solution:

Round the numerator and denominator: $\frac{401}{0.205} \approx \frac{400}{0.2}.$

Since $\frac{400}{0.2} = 2000,$ the value is closest to $2000.$

Thus, the correct answer is E .

Small Hint:

Follow the order of operations: parentheses first, then multiplication, then addition

Big Hint:

$6 \div 3 = 2,$ so the expression becomes $-15 + 9 \times 2$

Solution:

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 .

Small Hint:

Count how many equal spaces lie between $0$ and $20$

Big Hint:

There are $5$ spaces from $0$ to $20,$ so each space is $\frac{20}{5} = 4;$ then count the spaces up to $y$

Solution:

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 .

Small Hint:

First find the total value of $20$ quarters and $10$ dimes

Big Hint:

After using $10$ quarters, find how much value the remaining dimes must supply, then divide by $10$¢

Solution:

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 .

Small Hint:

$2 \times 3 \times 4 = 24;$ distribute it across the three fractions

Big Hint:

$24 \cdot \frac{1}{2} + 24 \cdot \frac{1}{3} + 24 \cdot \frac{1}{4}$

Solution:

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 .

Small Hint:

For every $2$ boys there are $3$ girls, so boys are $2$ out of every $5$ students

Big Hint:

Convert the fraction $\frac{2}{5}$ to a percent

Solution:

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 .

Small Hint:

The $12$ hour marks split the clock face into $12$ equal angles

Big Hint:

Each hour mark is $30^\circ;$ at seven o’clock the hands are $5$ marks apart the short way

Solution:

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 .

Small Hint:

Two figures are symmetric across the dashed line when folding the paper along that line makes one land exactly on the other

Big Hint:

A reflection across a vertical line keeps top and bottom fixed but swaps left and right; track where the small corner square ends up and how the slanted strokes turn

Solution:

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 reverses its slant, 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 .

Small Hint:

Simplify the numerator and denominator separately

Big Hint:

$1 - \frac{1}{3} = \frac{2}{3}$ and $1 - \frac{1}{2} = \frac{1}{2};$ then divide

Solution:

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 .

Small Hint:

Dividing the numerator and one factor of the denominator by the same number keeps the fraction unchanged

Big Hint:

To turn $9$ into $0.9,$ divide by $10;$ to keep the value equal, divide one factor of the denominator by $10$ as well

Solution:

To rewrite the numerator $9$ as $0.9,$ 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 $\frac{0.9}{0.7 \times 53},$ which equals the original. Each other choice changes the value by a factor of $10$ or $100.$

Thus, the correct answer is A .

Small Hint:

Make the three-digit number as small as possible and the two-digit number as large as possible

Big Hint:

The largest two-digit number from these digits uses $9$ and $6;$ build the smallest three-digit number from the digits that remain

Solution:

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 .

Small Hint:

The shaded region is the whole parallelogram with the triangle $ABE$ removed

Big Hint:

The parallelogram area is $10 \times 8;$ triangle $ABE$ has base $AE = 10 - 6$ and height $8$

Solution:

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 .

Small Hint:

$47$ is odd, so one of the two primes must be even

Big Hint:

The only even prime is $2,$ which would force the other prime to be $45$

Solution:

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 .

Small Hint:

The average is $\frac{6 + 10 + N}{3};$ find its value at each end of the range for $N$

Big Hint:

As $N$ runs from $9$ to $17,$ the average runs from $\frac{25}{3}$ to $11$

Solution:

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 .

Small Hint:

Pressing the key once turns $32$ into $\frac{1}{32}$

Big Hint:

The reciprocal of the reciprocal of a number is the number itself

Solution:

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

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

Thus, the correct answer is B .

Small Hint:

Because the graph shows an accumulated total, the amount spent over a span of months is the increase in that total across the span

Big Hint:

Read the height of the curve at the beginning of June and at the end of August, then subtract the two values

Solution:

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 .

Small Hint:

A corner uses three faces that are mutually adjacent, so no two of them can be opposite faces

Big Hint:

Folding the net, the opposite face pairs are $1$ & $3,$ $2$ & $5,$ and $4$ & $6;$ take the larger of each pair

Solution:

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 .

Small Hint:

Selling $25\%$ leaves $75\% = \frac{3}{4}$ of the apples

Big Hint:

Multiply $128$ by $\frac{3}{4}$ twice, then subtract the one given to the teacher

Solution:

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 .

Small Hint:

The six letters return to AJHSME every $6$ lines; the four digits return to $1989$ every $4$ lines

Big Hint:

Both return together on the least common multiple of $6$ and $4$

Solution:

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 .

Small Hint:

Count the exposed faces on the four vertical sides, then the faces seen from directly above

Big Hint:

Each of the four sides shows $6$ exposed faces, and looking straight down covers the full $3 \times 3$ footprint of top faces

Solution:

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 .