Small Hint:

Multiply the numerators together and the denominators together before simplifying

Big Hint:

The same factors $3,$ $5,$ $7,$ $9,$ $11$ appear in both the numerator and the denominator

Solution:

Combining the two fractions, the numerator is $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11$ and the denominator is $9 \cdot 11 \cdot 3 \cdot 5 \cdot 7.$ These are the same product.

So the value is $1.$

Thus, the correct answer is A .

Small Hint:

There are $10$ terms, from $90$ up to $99$

Big Hint:

The sum equals the average of the first and last term times the number of terms

Solution:

There are $10$ terms. The average of the first and last is $\dfrac{90 + 99}{2} = 94.5.$

So the sum is $10 \cdot 94.5 = 945.$

Thus, the correct answer is B .

Small Hint:

$\dfrac{10^7}{10^4} = 10^3$

Big Hint:

Then divide $10^3 = 1000$ by $5$

Solution:

Since $\dfrac{10^7}{10^4} = 10^3 = 1000,$ the expression is $\dfrac{1000}{5} = 200.$

Thus, the correct answer is D .

Small Hint:

Extend sides $AF$ and $DC$ to complete a full $6 \times 9$ rectangle

Big Hint:

Subtract the area of the small rectangular notch that was cut from the lower-left corner

Solution:

Completing the figure to the full $6 \times 9$ rectangle gives an area of $54.$ The piece removed from the lower-left corner is a rectangle measuring $2$ by $4,$ with area $8.$

So the polygon’s area is $54 - 8 = 46.$

Thus, the correct answer is C .

Small Hint:

Add the counts for the satisfactory grades A, B, C, and D

Big Hint:

Divide the satisfactory count by the total number of grades

Solution:

The satisfactory grades number $5 + 4 + 3 + 3 = 15,$ and the total number of grades is $15 + 5 = 20.$

So the fraction that is satisfactory is $\dfrac{15}{20} = \dfrac{3}{4}.$

Thus, the correct answer is C .

Small Hint:

The number of sheets is proportional to the height of the stack

Big Hint:

A $7.5$ cm stack is $1.5$ times as tall as a $5$ cm stack

Solution:

Since $7.5$ cm is $1.5$ times $5$ cm, the stack holds $1.5$ times as many sheets.

That is $1.5 \times 500 = 750$ sheets.

Thus, the correct answer is D .

Small Hint:

Row $n$ has $2n - 1$ squares, and they alternate starting and ending with an unshaded square

Big Hint:

Since the ends are unshaded and the squares alternate, there is one more unshaded square than shaded square in each row

Solution:

Row $n$ contains $2n - 1$ squares. Because both ends are unshaded and the squares alternate, each row has one more unshaded square than shaded square, so the number of shaded squares is $n - 1.$

For the $37$th row, this is $37 - 1 = 36.$

Thus, the correct answer is C .

Small Hint:

Substitute $a = -2$ into each expression

Big Hint:

Check signs first: $4a$ and $\frac{24}{a}$ are negative, while $-3a$ and $a^2$ are positive

Solution:

Substituting $a = -2$ gives the values $-3a = 6,$ $4a = -8,$ $\dfrac{24}{a} = -12,$ $a^2 = 4,$ and $1.$

The largest of these is $6,$ which is $-3a.$

Thus, the correct answer is A .

Small Hint:

Rewrite each factor: $1 - \dfrac1k = \dfrac{k-1}{k}$

Big Hint:

Write the product as $\dfrac12 \cdot \dfrac23 \cdot \dfrac34 \cdots \dfrac{9}{10}$ and watch consecutive terms cancel

Solution:

Each factor $1 - \dfrac1k$ equals $\dfrac{k-1}{k},$ so the product is

$$\dfrac12 \cdot \dfrac23 \cdot \dfrac34 \cdots \dfrac{9}{10}.$$

Every numerator cancels the previous denominator, leaving $\dfrac{1}{10}.$

Thus, the correct answer is A .

Small Hint:

The point halfway between two numbers is their average

Big Hint:

$\dfrac15 + \dfrac13 = \dfrac{8}{15};$ then divide by $2$

Solution:

The midpoint of $\dfrac15$ and $\dfrac13$ is their average.

$$\dfrac{\frac15 + \frac13}{2} = \dfrac{\frac{8}{15}}{2} = \dfrac{4}{15}.$$

Thus, the correct answer is C .

Small Hint:

Pick one face to be the front and fold the rest of the net around it

Big Hint:

If $X$ becomes the bottom face, track which square ends up on top

Solution:

Fold the net so that $X$ is the bottom face. Then $U,$ $V,$ $W,$ and $Z$ wrap around to become the four side faces.

The only remaining square, $Y,$ becomes the top face, which is opposite the bottom face $X.$

Thus, the correct answer is E .

Small Hint:

Add the triangle’s three sides to get the common perimeter

Big Hint:

The square’s side is one fourth of its perimeter

Solution:

The triangle’s perimeter is $6.2 + 8.3 + 9.5 = 24$ cm, so the square also has perimeter $24$ cm.

The square’s side is $\dfrac{24}{4} = 6$ cm, so its area is $6^2 = 36 \text{ cm}^2.$

Thus, the correct answer is B .

Small Hint:

Convert each time to hours: $45$ minutes $= \dfrac34$ hour and $30$ minutes $= \dfrac12$ hour

Big Hint:

Distance $=$ rate $\times$ time for each part, then add the two distances

Solution:

Walking: $4 \times \dfrac34 = 3$ miles. Running: $10 \times \dfrac12 = 5$ miles.

The total distance is $3 + 5 = 8$ miles.

Thus, the correct answer is B .

Small Hint:

The difference in tax rates is $6.5\% - 6\% = 0.5\%$

Big Hint:

Find $0.5\%$ of $\$20$

Solution:

The two taxes differ by $0.5\%$ of the price.

That is $0.005 \times \$20 = \$0.10.$

Thus, the correct answer is B .

Small Hint:

Every number from $200$ to $299$ already contains a $2$

Big Hint:

For $100$–$199$ and $300$–$399,$ count those with a $2$ in the tens or units place, subtracting the ones counted twice

Solution:

All $100$ numbers from $200$ to $299$ contain a $2.$

Among $100$ to $199,$ there are $10$ with a $2$ in the tens place and $10$ with a $2$ in the units place, but $122$ is counted twice, giving $10 + 10 - 1 = 19.$ The range $300$ to $399$ similarly contributes $19.$

The total is $100 + 19 + 19 = 138.$

Thus, the correct answer is C .

Small Hint:

The class splits into $2 + 3 = 5$ equal parts

Big Hint:

Girls exceed boys by exactly one part, so find the size of one part

Solution:

The $30$ students split into $5$ equal parts of $6.$ Boys make up $2$ parts ($12$) and girls $3$ parts ($18$).

So there are $18 - 12 = 6$ more girls than boys.

Thus, the correct answer is D .

Small Hint:

The total for seven tests is $7 \times 85,$ and the total for six tests is $6 \times 84$

Big Hint:

The seventh score is the difference between those two totals

Solution:

Seven tests averaging $85$ total $7 \times 85 = 595$ points; six tests averaging $84$ total $6 \times 84 = 504$ points.

The seventh score is $595 - 504 = 91.$

Thus, the correct answer is D .

Small Hint:

Let the price be $p;$ then $9p \lt 10$ and $10p \gt 11$

Big Hint:

Combine into $1.10 \lt p \lt 1.111\ldots$ and find the price in whole cents

Solution:

From $9p \lt 10$ we get $p \lt 1.111\ldots,$ and from $10p \gt 11$ we get $p \gt 1.10.$

The only price in whole cents between $\$1.10$ and $\$1.111\ldots$ is $\$1.11.$

Thus, the correct answer is E .

Small Hint:

The perimeter is $2(\ell + w)$

Big Hint:

Multiplying both $\ell$ and $w$ by $1.1$ multiplies the whole perimeter by $1.1$

Solution:

The new perimeter is $2(1.1\ell + 1.1w) = 1.1 \cdot 2(\ell + w),$ which is $1.1$ times the old perimeter.

That is a $10\%$ increase.

Thus, the correct answer is B .

Small Hint:

January has $31$ days, which is $4$ full weeks plus $3$ extra days

Big Hint:

The three weekdays that occur five times are the days of January $1,$ $2,$ and $3;$ Tuesday and Saturday must not be among them

Solution:

Since $31 = 4 \cdot 7 + 3,$ the weekdays falling on January $1,$ $2,$ and $3$ each occur five times that month, and every other weekday occurs four times.

For Tuesday and Saturday to occur only four times, neither may be January $1,$ $2,$ or $3.$ The only starting day that works is Wednesday: then Wednesday, Thursday, and Friday occur five times, while Tuesday and Saturday each occur four times.

Thus, the correct answer is C .

Small Hint:

Applying a $10\%$ raise multiplies the salary by $1.1$

Big Hint:

After four raises the salary is multiplied by $1.1^4;$ compare that to $1.45$

Solution:

After four raises the salary is multiplied by $1.1^4 = 1.4641.$

That is an increase of $46.41\%,$ which is more than $45\%.$

Thus, the correct answer is E .

Small Hint:

The first digit can be any of $2$ through $9,$ which is $8$ choices

Big Hint:

Multiply the fraction of telephone numbers that begin with $9$ by the fraction that end in $0$

Solution:

The first digit is one of $8$ allowed digits ($2$ through $9$), so $\dfrac18$ of the numbers begin with $9.$ The last digit is any of $10$ digits, so $\dfrac{1}{10}$ end in $0.$

These conditions are independent, so the fraction is $\dfrac18 \cdot \dfrac{1}{10} = \dfrac{1}{80}.$

Thus, the correct answer is B .

Small Hint:

Count the total number of student-class attendances: $1200 \times 5$

Big Hint:

Divide by $30$ to get the number of classes, then by $4$ to get the number of teachers

Solution:

Each day there are $1200 \times 5 = 6000$ student-class attendances. Since each class holds $30$ students, there are $\dfrac{6000}{30} = 200$ classes.

Each teacher teaches $4$ classes, so there are $\dfrac{200}{4} = 50$ teachers.

Thus, the correct answer is E .