1993 AMC 8 — Problems & Online Timed Exam

Timed

40:00

1.

Which pair of numbers does not have a product equal to $36?$

$\{-4, -9\}$

$\{-3, -12\}$

$\left\{\dfrac12, -72\right\}$

$\{1, 36\}$

$\left\{\dfrac32, 24\right\}$

Answer: C

Concepts:fraction

Difficulty rating: 560

Small Hint:

Multiply the two numbers in each pair and check which product is not $36$

Big Hint:

A negative times a negative is positive, so watch the signs of $\dfrac12 \times (-72)$

Solution:

Checking each pair: $(-4)(-9) = 36,$ $(-3)(-12) = 36,$ $\dfrac12 \times (-72) = -36,$ $(1)(36) = 36,$ and $\dfrac32 \times 24 = 36.$

Only $\dfrac12 \times (-72) = -36$ fails to equal $36.$

Thus, the correct answer is C .

2.

When the fraction $\dfrac{49}{84}$ is expressed in simplest form, then the sum of the numerator and the denominator will be

$11$

$17$

$19$

$33$

$133$

Answer: C

Concepts:fraction greatest common divisor

Difficulty rating: 450

Small Hint:

Find the greatest common factor of $49$ and $84$

Big Hint:

$49 = 7^2$ and $84 = 7 \times 12,$ so divide both by $7$

Solution:

Since $49 = 7 \times 7$ and $84 = 7 \times 12,$ the fraction reduces to $\dfrac{7}{12}.$

The sum of numerator and denominator is $7 + 12 = 19.$

Thus, the correct answer is C .

3.

Which of the following numbers has the largest prime factor?

$39$

$51$

$77$

$91$

$121$

Answer: B

Concepts:prime factorization prime

Difficulty rating: 660

Small Hint:

Factor each number completely into primes

Big Hint:

Compare the largest prime in each factorization; note $51 = 3 \times 17$

Solution:

Factoring: $39 = 3 \times 13,$ $51 = 3 \times 17,$ $77 = 7 \times 11,$ $91 = 7 \times 13,$ and $121 = 11 \times 11.$

The largest prime factor among these is $17,$ which is a factor of $51.$

Thus, the correct answer is B .

4.

$1000 \times 1993 \times 0.1993 \times 10 =$

$1.993 \times 10^3$

$1993.1993$

$(199.3)^2$

$1{,}993{,}001.993$

$(1993)^2$

Answer: E

Concepts:place value exponent

Difficulty rating: 730

Small Hint:

Group the powers of ten: $1000 \times 10 = 10000$

Big Hint:

$10000 \times 0.1993 = 1993,$ so the product is $1993 \times 1993$

Solution:

Regroup as $(1000 \times 10)$ $\times 0.1993$ $\times 1993$ $= 10000$ $\times 0.1993$ $\times 1993.$

Since $10000 \times 0.1993 = 1993,$ the product is $1993 \times 1993 = (1993)^2.$

Thus, the correct answer is E .

5.

Which one of the following bar graphs could represent the data from the circle graph shown?

Answer: C

Concepts:data and graph interpretation ratio and proportion

Difficulty rating: 660

Small Hint:

Decide what fraction of the whole circle each of the three regions takes up

Big Hint:

Each shaded region is a quarter and the unshaded region is a half, so the three bars should have heights in the ratio $1 : 1 : 2$

Solution:

The two shaded regions are each one quarter of the circle, and the unshaded region is one half. So the three quantities are in the ratio $\tfrac14 : \tfrac14 : \tfrac12,$ or $1 : 1 : 2.$

A matching bar graph must have the two shaded bars equal in height and the unshaded bar exactly twice as tall. Only one bar graph has two equal shaded bars with the white bar double their height.

Thus, the correct answer is C .

6.

A can of soup can feed $3$ adults or $5$ children. If there are $5$ cans of soup and $15$ children are fed, then how many adults would the remaining soup feed?

$5$

$6$

$7$

$8$

$10$

Answer: B

Concepts:rate

Difficulty rating: 730

Small Hint:

$15$ children need $15 \div 5 = 3$ cans, so find how many cans remain

Big Hint:

Each remaining can feeds $3$ adults

Solution:

Feeding $15$ children uses $15 \div 5 = 3$ cans, leaving $5 - 3 = 2$ cans.

Those $2$ cans feed $2 \times 3 = 6$ adults.

Thus, the correct answer is B .

7.

$3^3 + 3^3 + 3^3 =$

$3^4$

$9^3$

$3^9$

$27^3$

$3^{27}$

Answer: A

Concepts:exponent

Difficulty rating: 660

Small Hint:

Adding three copies of $3^3$ is the same as $3 \times 3^3$

Big Hint:

$3 \times 3^3 = 3^{1+3}$

Solution:

Adding three equal terms, $3^3 + 3^3 + 3^3 = 3 \times 3^3 = 3^4 = 81.$

Thus, the correct answer is A .

8.

To control her blood pressure, Jill’s grandmother takes one half of a pill every other day. If one supply of medicine contains $60$ pills, then the supply of medicine will last approximately

$1$ month

$4$ months

$6$ months

$8$ months

$1$ year

Answer: D

Concepts:rate unit conversion

Difficulty rating: 860

Small Hint:

Half a pill every other day means one whole pill covers $4$ days

Big Hint:

Find the total number of days, then convert to months using about $30$ days per month

Solution:

She takes half a pill every two days, so one pill lasts $4$ days. Then $60$ pills last $60 \times 4 = 240$ days.

At about $30$ days per month, that is roughly $240 \div 30 = 8$ months.

Thus, the correct answer is D .

9.

Consider the operation $*$ defined by the following table:

$\begin{array}{c|cccc} * & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{array}$

For example, $3 * 2 = 1.$ Then $(2 * 4) * (1 * 3) =$

$1$

$2$

$3$

$4$

$5$

Answer: D

Concepts:custom operation

Difficulty rating: 730

Small Hint:

Read $2 * 4$ from row $2,$ column $4,$ and $1 * 3$ from row $1,$ column $3$

Big Hint:

Both inner values equal $3,$ so compute $3 * 3$ from the table

Solution:

From the table, $2 * 4 = 3$ and $1 * 3 = 3.$

Then $(2 * 4) * (1 * 3) = 3 * 3 = 4.$

Thus, the correct answer is D .

10.

This line graph represents the price of a trading card during the first $6$ months of $1993.$ The greatest monthly drop in price occurred during which month?

January

March

April

May

June

Answer: B

Concepts:data and graph interpretation

Difficulty rating: 660

Small Hint:

A drop happens where the line goes down from one month to the next; measure how far it falls each time

Big Hint:

Compare only the downward segments and find the steepest one

Solution:

The price changes month to month are: January $\$2.50 \to \$2.00$ (drop $\$0.50$ ), February $\$2.00 \to \$4.00$ (rise), March $\$4.00 \to \$1.50$ (drop $\$2.50$ ), April $\$1.50 \to \$4.50$ (rise), May $\$4.50 \to \$3.00$ (drop $\$1.50$ ), and June $\$3.00 \to \$1.00$ (drop $\$2.00$ ).

The largest drop is $\$2.50,$ which occurred during March.

Thus, the correct answer is B .

11.

Consider this histogram of the scores for $81$ students taking a test. The median is in the interval labeled which value?

$60$

$65$

$70$

$75$

$80$

Answer: C

Concepts:median (data)data and graph interpretation

Difficulty rating: 800

Small Hint:

With $81$ scores in order, the median is the middle one: the $41$ st score counting from the lowest

Big Hint:

Add the bar heights from left to right until the running total first reaches $41$

Solution:

Since $81$ students took the test, the median is the $41$ st score counting up from the lowest.

Adding the bar heights from the left gives running totals $1,$ $3,$ $7,$ $12,$ $18,$ $28,$ $42,$ $\ldots$ The total first passes $41$ at the interval labeled $70,$ which contains the $29$ th through $42$ nd scores. So the $41$ st score lies in the interval labeled $70.$

Thus, the correct answer is C .

12.

If each of the three operation signs, $+,$ $-,$ $\times,$ is used exactly once in one of the blanks in the expression

$5 \, \underline{\phantom{x}} \, 4 \, \underline{\phantom{x}} \, 6 \, \underline{\phantom{x}} \, 3$

then the value of the result could equal

$9$

$10$

$15$

$16$

$19$

Answer: E

Concepts:order of operations casework

Difficulty rating: 890

Small Hint:

Multiplication happens before addition and subtraction, so where you place $\times$ matters most

Big Hint:

Work through each way of assigning $+,$ $-,$ $\times$ to the three blanks and compute the result

Solution:

The six arrangements give $5 \times 4 + 6 - 3 = 23,$ $5 \times 4 - 6 + 3 = 17,$ $5 + 4 \times 6 - 3 = 26,$ $5 - 4 \times 6 + 3 = -16,$ $5 + 4 - 6 \times 3 = -9,$ and $5 - 4 + 6 \times 3 = 19.$

The only value among the choices is $19.$

Thus, the correct answer is E .

13.

The word “HELP” in block letters is painted as a shaded region with strokes $1$ unit wide on a $5$ by $15$ rectangular sign. Each letter is $3$ units wide with a $1$ -unit gap between letters, as shown. The area of the unshaded portion of the sign, in square units, is

$30$

$32$

$34$

$36$

$38$

Answer: D

Concepts:area decomposition

Difficulty rating: 960

Small Hint:

The whole sign is $5 \times 15 = 75$ square units

Big Hint:

Count the shaded unit squares making up each letter, then subtract the total from $75$

Solution:

The full sign has area $5 \times 15 = 75$ square units. Counting the shaded unit squares in each letter gives $H = 11,$ $E = 11,$ $L = 7,$ and $P = 10,$ for a shaded total of $11 + 11 + 7 + 10 = 39.$

The unshaded area is $75 - 39 = 36.$

Thus, the correct answer is D .

14.

The nine squares in the table shown are to be filled so that every row and every column contains each of the numbers $1,$ $2,$ $3.$ Then $A + B =$

$\begin{array}{|c|c|c|} \hline 1 & \phantom{X} & \phantom{X} \\ \hline \phantom{X} & 2 & A \\ \hline \phantom{X} & \phantom{X} & B \\ \hline \end{array}$

$2$

$3$

$4$

$5$

$6$

Answer: C

Concepts:logical deduction

Difficulty rating: 930

Small Hint:

Each row and each column must contain $1,$ $2,$ $3$ exactly once

Big Hint:

Once two entries of a row or column are known, the third is forced; fill in step by step

Solution:

Filling the grid so each row and column has $1,$ $2,$ $3,$ the top row becomes $1,$ $3,$ $2,$ the middle row $3,$ $2,$ $A,$ and the bottom row $2,$ $1,$ $B.$ The middle row forces $A = 1,$ and the last column $2,$ $1,$ $B$ forces $B = 3.$

So $A + B = 1 + 3 = 4.$

Thus, the correct answer is C .

15.

The arithmetic mean (average) of four numbers is $85.$ If the largest of these numbers is $97,$ then the mean of the remaining three numbers is

$81.0$

$82.7$

$83.0$

$84.0$

$84.3$

Answer: A

Concepts:mean

Difficulty rating: 660

Small Hint:

The four numbers sum to $4 \times 85$

Big Hint:

Subtract the largest number, then divide the remaining sum by $3$

Solution:

The four numbers sum to $4 \times 85 = 340,$ so the remaining three sum to $340 - 97 = 243.$

Their mean is $243 \div 3 = 81.$

Thus, the correct answer is A .

16.

What is the value of the following expression?

$\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{3}}}$

$\dfrac16$

$\dfrac{3}{10}$

$\dfrac{7}{10}$

$\dfrac56$

$\dfrac{10}{3}$

Answer: C

Concepts:continued fraction fraction

Difficulty rating: 860

Small Hint:

Work from the bottom up: first simplify $2 + \dfrac13$

Big Hint:

Then $1 + \dfrac{1}{\frac{7}{3}} = 1 + \dfrac37,$ and take the reciprocal

Solution:

Starting inside, $2 + \dfrac13 = \dfrac73,$ so $\dfrac{1}{\frac{7}{3}} = \dfrac37.$

Then $1 + \dfrac37 = \dfrac{10}{7},$ and the whole expression is $\dfrac{1}{\frac{10}{7}} = \dfrac{7}{10}.$

Thus, the correct answer is C .

17.

Square corners, $5$ units on a side, are removed from a $20$ unit by $30$ unit rectangular sheet of cardboard. The sides are then folded to form an open box. The surface area, in square units, of the interior of the box is

$300$

$500$

$550$

$600$

$1000$

Answer: B

Concepts:surface area net (3D geometry)

Difficulty rating: 980

Small Hint:

Folding does not change area; the interior surface equals the sheet minus the four corner squares

Big Hint:

The sheet is $20 \times 30;$ subtract four $5 \times 5$ corner squares

Solution:

The interior surface is exactly one face of the cardboard after the corners are removed. The sheet has area $20 \times 30 = 600,$ and each removed corner has area $5 \times 5 = 25.$

So the interior surface area is $600 - 4 \times 25 = 500.$

Thus, the correct answer is B .

18.

The rectangle shown has length $AC = 32,$ width $AE = 20,$ and $B$ and $F$ are midpoints of $\overline{AC}$ and $\overline{AE},$ respectively. The area of the quadrilateral $ABDF$ is

$320$

$325$

$330$

$335$

$340$

Answer: A

Concepts:area decomposition triangle area

Difficulty rating: 1090

Small Hint:

The rectangle $ACDE$ has area $32 \times 20;$ remove the parts outside $ABDF$

Big Hint:

Subtract triangle $BCD$ and triangle $DEF$ from the full rectangle

Solution:

Rectangle $ACDE$ has area $32 \times 20 = 640.$ Triangle $BCD$ has area $\dfrac{16 \times 20}{2} = 160,$ and triangle $DEF$ has area $\dfrac{10 \times 32}{2} = 160.$

The remaining region $ABDF$ has area $640 - (160 + 160) = 320.$

Thus, the correct answer is A .

19.

What is the value of the following expression?

$\begin{gathered} \small (1901 + 1902 + 1903 + \cdots + 1993) \\ \small {}- (101 + 102 + 103 + \cdots + 193) \end{gathered}$

$167{,}400$

$172{,}050$

$181{,}071$

$199{,}300$

$362{,}142$

Answer: A

Concepts:pairing and grouping arithmetic sequence

Difficulty rating: 960

Small Hint:

Pair each term with the one below it: $1901 - 101,$ $1902 - 102,$ $\ldots$

Big Hint:

Each of the $93$ pairs has difference $1800$

Solution:

Each number in the first sum is exactly $1800$ more than the matching number in the second sum, and there are $93$ such pairs.

So the difference is $93 \times 1800 = 167{,}400.$

Thus, the correct answer is A .

20.

When $10^{93} - 93$ is expressed as a single whole number, the sum of the digits is

$10$

$93$

$819$

$826$

$833$

Answer: D

Concepts:place value digits

Difficulty rating: 1140

Small Hint:

$10^{93}$ is a $1$ followed by $93$ zeros; subtracting $93$ changes only the right end

Big Hint:

The result is a block of nines ending in $07;$ determine how many nines there are

Solution:

Subtracting $93$ from $10^{93}$ (a $1$ followed by $93$ zeros) gives a number that is $91$ nines followed by $07.$

The digit sum is $91 \times 9 + 0 + 7 = 819 + 7 = 826.$

Thus, the correct answer is D .

21.

If the length of a rectangle is increased by $20\%$ and its width is increased by $50\%,$ then the area is increased by

$10\%$

$30\%$

$70\%$

$80\%$

$100\%$

Answer: D

Concepts:percentage

Difficulty rating: 820

Small Hint:

New area $=$ (new length) $\times$ (new width) $= 1.2 \times 1.5$ times the old area

Big Hint:

Multiply the two growth factors $1.2$ and $1.5$ to get the factor by which the area grows

Solution:

The new length is $1.2$ times the old and the new width is $1.5$ times the old, so the new area is $1.2 \times 1.5 = 1.8$ times the old area.

That is an increase of $80\%.$

Thus, the correct answer is D .

22.

Pat Peano has plenty of $0$ ’s, $1$ ’s, $3$ ’s, $4$ ’s, $5$ ’s, $6$ ’s, $7$ ’s, $8$ ’s and $9$ ’s, but he has only twenty-two $2$ ’s. How far can he number the pages of his scrapbook with these digits?

$22$

$99$

$112$

$119$

$199$

Answer: D

Concepts:digits systematic listing

Difficulty rating: 1200

Small Hint:

Count how many $2$ ’s appear when numbering pages $1$ through $99$

Big Hint:

Ten $2$ ’s appear in the units place and ten in the tens place through $99;$ then track the remaining twos past $100$

Solution:

Numbering $1$ through $99$ uses ten $2$ ’s in the units place and ten in the tens place, a total of twenty $2$ ’s. Pages $100$ and $101$ use none.

The remaining two $2$ ’s are used on pages $102$ and $112.$ After that, pages $113$ through $119$ need no $2,$ but $120$ would require another $2,$ so he can number up to $119.$

Thus, the correct answer is D .

23.

Five runners, $P,$ $Q,$ $R,$ $S,$ $T,$ have a race, and $P$ beats $Q,$ $P$ beats $R,$ $Q$ beats $S,$ and $T$ finishes after $P$ and before $Q.$ Who could not have finished third in the race?

$P$ and $Q$

$P$ and $R$

$P$ and $S$

$P$ and $T$

$P,$ $S$ and $T$

Answer: C

Concepts:logical deduction

Difficulty rating: 1070

Small Hint:

$P$ finishes ahead of $Q,$ $R,$ and $T,$ while $Q$ finishes ahead of $S;$ determine $P$ ’s place

Big Hint:

If several runners must all finish ahead of a given runner, that runner cannot be as high as third

Solution:

Since $P$ beats $Q$ and $R,$ finishes ahead of $T,$ and $Q$ beats $S,$ runner $P$ finishes first and so cannot be third.

The clues give the chain $P$ before $T$ before $Q$ before $S.$ So $P,$ $T,$ and $Q$ all finish ahead of $S,$ meaning $S$ is no better than fourth and cannot be third either.

Each of $Q,$ $R,$ $T$ can finish third: for example $P,$ $T,$ $Q,$ $R,$ $S$ puts $Q$ third; $P,$ $R,$ $T,$ $Q,$ $S$ puts $T$ third; and $P,$ $T,$ $R,$ $Q,$ $S$ puts $R$ third. So only $P$ and $S$ cannot be third.

Thus, the correct answer is C .

24.

What number is directly above $142$ in this array of numbers?

$\begin{array}{ccccccccc} & & & & 1 & & & & \\ & & & 2 & 3 & 4 & & & \\ & & 5 & 6 & 7 & 8 & 9 & & \\ & 10 & 11 & 12 & \cdots & & & & \end{array}$

$99$

$119$

$120$

$121$

$122$

Answer: C

Concepts:perfect square pattern recognition

Difficulty rating: 1140

Small Hint:

Each row ends at a perfect square: row $n$ ends at $n^2$

Big Hint:

$142$ is in the row ending at $144 = 12^2;$ the row above ends at $121 = 11^2,$ and the rows line up at their right edges

Solution:

Each row ends at a perfect square, so the row containing $142$ ends at $144 = 12^2,$ and the row above it ends at $121 = 11^2.$

Since the rows are aligned at their right edges, $121$ sits directly above $143,$ and therefore $120$ sits directly above $142.$

Thus, the correct answer is C .

25.

A checkerboard consists of one-inch squares. A square card, $1.5$ inches on a side, is placed on the board so that it covers part or all of the area of each of $n$ squares. The maximum possible value of $n$ is

$4$ or $5$

$6$ or $7$

$8$ or $9$

$10$ or $11$

$12$ or more

Answer: E

Concepts:optimization Pythagorean Theorem

Difficulty rating: 1270

Small Hint:

A tilted card can cross more grid lines than a card lined up with the squares; the card’s diagonal has length $\sqrt{1.5^2 + 1.5^2} = \sqrt{4.5} \approx 2.1$

Big Hint:

Try tilting the card $45^\circ$ with its center on a grid corner, and count how many squares each corner of the card pokes into

Solution:

Tilt the card $45^\circ$ and center it on a corner where four grid squares meet, as shown. Because the card’s diagonal, $\sqrt{1.5^2 + 1.5^2} = \sqrt{4.5} \approx 2.1,$ is longer than $2,$ each of the four corners of the card reaches past a grid line into the next square.

The card covers the central $2 \times 2$ block of $4$ squares and pokes into $2$ more squares on each of its four sides, giving $4 + 4 \times 2 = 12$ squares.

This is also the most possible. The card is only $1.5$ inches wide, so its overall width and height are each at most $2.1$ inches; it therefore lies within a $4 \times 4$ block of $16$ squares. Its four pointed corners are the only parts that reach the edge of that block, so it can never reach the four corner squares of the block, leaving at most $12.$ Since $12$ is achievable, the maximum is $12,$ which falls in the range “ $12$ or more.”

Thus, the correct answer is E .