1988 AMC 8 Problems

The marks between $10$ and $11$ divide that unit into four equal parts, so each small division is $0.25$

The arrow sits just past the first mark after $10,$ so the reading is a little more than $10.25$

The scale is divided into fourths between $10$ and $11,$ so the interior tick marks fall at $10.25,$ $10.5,$ and $10.75.$

The arrow points just past the $10.25$ mark, so the reading is between $10.25$ and $10.5.$ Of the choices, only $10.3$ lies in that range.

Thus, the correct answer is D .

Rewrite the decimals as fractions: $0.25 = \dfrac14$ and $0.125 = \dfrac18$

Pair $8 \times \dfrac18$ and $\dfrac14 \times 2$ to simplify

Write the decimals as fractions: $8 \times \dfrac14 \times 2 \times \dfrac18.$

Grouping, $\left(8 \times \dfrac18\right)\left(\dfrac14 \times 2\right)$ $= 1 \times \dfrac12$ $= \dfrac12.$

Thus, the correct answer is C .

Each fraction reduces to the same value

$\dfrac{2}{20} = \dfrac{1}{10}$ and $\dfrac{3}{30} = \dfrac{1}{10}$

Each fraction equals $\dfrac{1}{10} = 0.1,$ since $\dfrac{2}{20} = \dfrac{3}{30} = \dfrac{1}{10}.$

So the sum is $0.1 + 0.1 + 0.1 = 0.3.$

Thus, the correct answer is D .

Look at the figure one horizontal row at a time

In each row the squares alternate but both ends are dark, so every row has exactly one more dark square than light square. Count how many rows there are.

In each horizontal row the squares alternate dark, light, dark, and so on, beginning and ending with a dark square. So every row has exactly one more dark square than light square.

The figure has $8$ rows, so the dark squares exceed the light squares by $8.$

Thus, the correct answer is B .

On the protractor, ray $BA$ reads about $20^\circ$ and ray $BD$ reads about $160^\circ$

Since $\angle CBD = 90^\circ$ and $BD$ reads $160^\circ,$ ray $BC$ reads $160^\circ - 90^\circ.$ Then $\angle ABC$ is the difference between the readings of $BC$ and $BA.$

Reading the protractor, ray $BA$ is at about $20^\circ$ and ray $BD$ is at about $160^\circ.$

Because $\angle CBD$ is a right angle, ray $BC$ is at $160^\circ - 90^\circ = 70^\circ.$

Therefore $\angle ABC = 70^\circ - 20^\circ = 50^\circ.$

Thus, the correct answer is C .

Compute the numerator and denominator separately: $(0.2)^3 = 0.008$

$(0.02)^2 = 0.0004,$ then divide

The numerator is $(0.2)^3 = 0.008$ and the denominator is $(0.02)^2 = 0.0004.$

So the value is $\dfrac{0.008}{0.0004} = \dfrac{80}{4} = 20.$

Thus, the correct answer is E .

Round each factor to a convenient number: $2.46 \approx 2.5,$ $8.163 \approx 8$

The sum $5.17 + 4.829 \approx 10,$ so the product is about $2.5 \times 8 \times 10$

Round each factor: $2.46 \approx 2.5,$ $8.163 \approx 8,$ and $5.17 + 4.829 \approx 10.$

The product is about $2.5 \times 8 \times 10 = 200.$

Thus, the correct answer is B .

Count the total number of decimal places in $0.075$ and $2.56$

There are $3 + 2 = 5$ decimal places, so place the decimal point that many digits from the right of $19200$

The factors $0.075$ and $2.56$ have $3$ and $2$ decimal places, for a total of $5.$

Placing the decimal point $5$ digits from the right of $19200$ gives $0.19200 = 0.192.$

Thus, the correct answer is B .

Find each triangle’s side lengths from the grid: a slanted side spanning $a$ across and $b$ up has length $\sqrt{a^2 + b^2}.$ A triangle is isosceles when two of its sides are equal.

Work out all three side lengths for each of the five triangles and look for a matching pair; only one triangle ends up with three different side lengths.

Read each triangle’s side lengths from the grid. A horizontal or vertical side is counted directly, and a slanted side spanning $a$ units across and $b$ units up has length $\sqrt{a^2 + b^2}.$

Top-left triangle: its two slanted sides each span $1$ across and $2$ up, so each has length $\sqrt{5},$ with a base of length $2.$ Two equal sides, so it is isosceles.

Top-middle triangle: it has a vertical side of length $2$ and a horizontal side of length $2$ (its third side is the slant $2\sqrt{2}$). Two equal sides, so it is isosceles.

Top-right triangle: its sides are $2,$ $\sqrt{5},$ and $\sqrt{13},$ which are all different, so it is not isosceles.

Bottom-left (wide, flat) triangle: its two slanted sides each span $3$ across and $1$ up, so each has length $\sqrt{10}.$ Two equal sides, so it is isosceles.

Bottom-right triangle: two of its sides each span $2$ and $1$ (giving length $\sqrt{5}$), while the third is $\sqrt{10}.$ Two equal sides, so it is isosceles.

Four of the five triangles are isosceles.

Thus, the correct answer is D .

The day of the week repeats every $7$ days, so only the remainder of $60$ upon division by $7$ matters

$60 = 8 \times 7 + 4,$ so the day is $4$ days after Thursday

Days of the week repeat every $7$ days. Since $60 = 8 \times 7 + 4,$ the day $60$ days later is $4$ days after Thursday.

Counting forward from Thursday: Friday, Saturday, Sunday, Monday. So it is a Monday.

Thus, the correct answer is A .

Find perfect squares just below and just above $164$

$12^2 = 144$ and $13^2 = 169$

Since $12^2 = 144$ and $13^2 = 169,$ and $144 \lt 164 \lt 169,$ the value $\sqrt{164}$ lies between $12$ and $13.$

Thus, the correct answer is E .

Write the numbers with powers of ten: $20$ billion $= 2 \times 10^{10}$ and $250$ million $= 2.5 \times 10^{8}$

Divide the total cost by the number of people

Each person’s share is $$\dfrac{20 \times 10^9}{250 \times 10^6} = \dfrac{2 \times 10^{10}}{2.5 \times 10^{8}}.$$

This equals $0.8 \times 10^{2} = 80.$

Thus, the correct answer is C .

The bushes go around the edge, so find the circumference $2\pi r$

With $\pi \approx 3.14,$ compute $2 \times 3.14 \times 12$

The bushes surround the circular edge, so their number is about the circumference $2\pi r = 2\pi(12).$

Using $\pi \approx 3.14,$ this is about $2 \times 3.14 \times 12 \approx 75$ feet, so roughly $75$ bushes are needed.

Thus, the correct answer is D .

List the factor pairs of $36$

The sum is largest when the two factors are as far apart as possible

The factor pairs of $36$ are $1 \times 36,$ $2 \times 18,$ $3 \times 12,$ $4 \times 9,$ and $6 \times 6.$

The sum is largest for the most spread-out pair, $1 + 36 = 37.$

Thus, the correct answer is E .

First add the fractions using a common denominator of $6$

The reciprocal flips the numerator and denominator

Adding, $\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{3}{6} + \dfrac{2}{6} = \dfrac{5}{6}.$

The reciprocal of $\dfrac{5}{6}$ is $\dfrac{6}{5}.$

Thus, the correct answer is C .

Try to leave one square empty in each row and column so no line gets filled

With $7$ X’s only two squares are empty, but two empty squares cannot break all three rows, so some row would be full

A placement of $6$ X’s works: leave the three squares along one main diagonal empty and fill the other six. Then each row and each column is missing one square, the used diagonal has an empty square, and the other diagonal passes through the empty center, so no line of three is complete.

Seven X’s is impossible: only two squares would be empty, and two empty squares can lie in at most two of the three rows, forcing the remaining row to be completely filled with three X’s.

So the greatest number is $6.$

Thus, the correct answer is E .

Add the areas of the two rectangles, but the overlapping region gets counted twice

The horizontal rectangle is $10 \times 2,$ the vertical one is $3 \times 8,$ and the overlap is $3 \times 2$

The horizontal rectangle has area $10 \times 2 = 20,$ and the vertical rectangle has area $3 \times 8 = 24.$ Their overlap, a $3 \times 2$ rectangle, has area $6.$

Adding the two rectangles counts the overlap twice, so the shaded area is $20 + 24 - 6 = 38.$

Thus, the correct answer is B .

Find the total weight of the boys and the total weight of the girls separately

Divide the combined total weight by $10$

The total weight is $6 \times 150$ $+ 4 \times 120$ $= 900 + 480$ $= 1380$ pounds.

The average of the $10$ children is $\dfrac{1380}{10} = 138$ pounds.

Thus, the correct answer is C .

The common difference is $4,$ so each term adds $4$ to the previous one

Starting from $1,$ reaching the $100$th term means adding $4$ a total of $99$ times

The sequence starts at $1$ with common difference $4.$ The $100$th term is reached by adding $4$ to the first term $99$ times.

So the $100$th term is $1 + 99 \times 4 = 1 + 396 = 397.$

Thus, the correct answer is A .

Set up a proportion: $45$ cups is to the full amount as $36$ is to $100$

If $36\%$ is $45$ cups, first find how many cups $4\%$ is

Let $n$ be the full capacity. Then $\dfrac{45}{n} = \dfrac{36}{100}.$

Solving, $n = \dfrac{45 \times 100}{36} = 125$ cups.

Thus, the correct answer is C .

The mean of the five numbers is $\dfrac{28 + n}{5};$ the median depends on where $n$ lands in the sorted order

Split into cases: the median is $6,$ is $n$ itself, or is $9,$ and solve the equation mean $=$ median in each

The five numbers have sum $28 + n$ and mean $\dfrac{28 + n}{5}.$ The median is the middle value when sorted, which is $6,$ $n,$ or $9$ depending on the size of $n.$

If the median is $6$: $\dfrac{28 + n}{5} = 6$ gives $n = 2,$ which is indeed less than $6.$ If the median is $n$: $\dfrac{28 + n}{5} = n$ gives $n = 7,$ which is between $6$ and $9.$ If the median is $9$: $\dfrac{28 + n}{5} = 9$ gives $n = 17,$ which is indeed greater than $9.$

Each case yields a valid value, so $n$ can be $2,$ $7,$ or $17$ — three possible values.

Thus, the correct answer is C .

Increasing by $25\%$ multiplies the price by $1.25;$ taking $20\%$ off multiplies by $0.80$

Multiply the two factors together and compare the result to $1$

Increasing by $25\%$ multiplies the price by $1.25,$ and then taking $20\%$ off multiplies by $0.80.$

The overall factor is $1.25 \times 0.80 = 1.00,$ so the sale price equals the original price.

Thus, the correct answer is E .

Work with a convenient batch of $12$ disks, since $12$ is a multiple of both $4$ and $3$

A dozen disks costs $\$15$ and sells for $\$20,$ a profit of $\$5$ per dozen

Consider disks in dozens, since $12$ is a multiple of both $4$ and $3.$ A dozen disks costs $3 \times \$5 = \$15$ to buy and sells for $4 \times \$5 = \$20,$ for a profit of $\$5$ per dozen.

To make $\$100$ profit she needs $\dfrac{100}{5} = 20$ dozen, which is $20 \times 12 = 240$ disks.

Thus, the correct answer is D .

Each time the square rolls over one edge of the hexagon it pivots about a shared corner and turns $150^\circ$ clockwise

Going from the top edge to the bottom edge is $3$ rolls, a total of $3 \times 150^\circ = 450^\circ,$ which is one full turn plus an extra quarter turn clockwise

Each roll of the square over one edge of the hexagon turns it $150^\circ$ clockwise. (At the pivot corner the square turns through $360^\circ$ minus the square’s own $90^\circ$ corner minus the hexagon’s $120^\circ$ interior angle, which leaves $150^\circ.$)

From the top of the hexagon to the bottom is $3$ rolls, for a total of $3 \times 150^\circ = 450^\circ.$ That is one complete revolution ($360^\circ$) plus an extra $90^\circ$ clockwise.

The triangle starts pointing straight up, so after the extra quarter turn clockwise it points to the right.

Thus, the correct answer is A .

Split into single-digit hours ($1$ through $9$) and two-digit hours ($10,$ $11,$ $12$), since the number of digits differs

For a single-digit hour $h,$ the time $h\!:\!m_1 m_2$ is a palindrome when $m_2 = h,$ leaving $m_1$ free among $0$ through $5$

For a single-digit hour $h$ (from $1$ to $9$), the display $h\!:\!m_1 m_2$ reads the same backwards when $m_2 = h.$ The tens digit $m_1$ can be $0,$ $1,$ $2,$ $3,$ $4,$ or $5,$ giving $6$ palindromes per hour, so $9 \times 6 = 54.$

For the two-digit hours $10,$ $11,$ $12,$ the display $h_1 h_2\!:\!m_1 m_2$ is a palindrome when $m_1 = h_2$ and $m_2 = h_1,$ giving exactly $10{:}01,$ $11{:}11,$ and $12{:}21$ — one each, for $3$ more.

The total is $54 + 3 = 57.$

Thus, the correct answer is A .