1998 AMC 8 Exam Problems

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1.

For $x=7,$ which of the following is the smallest?

$\dfrac{6}{x}$

$\dfrac{6}{x+1}$

$\dfrac{6}{x-1}$

$\dfrac{x}{6}$

$\dfrac{x+1}{6}$

Answer: B

Solution:

If we plug in the values into each answer choice, we get the following:

A: $\dfrac 6x = \dfrac 67$

B: $\dfrac 6{x+1} = \dfrac 68$

C: $\dfrac 6{x-1} = 1$

D: $\dfrac x6 = \dfrac 76$

E: $\dfrac {x+1}6 = \dfrac{4}{3}$

Thus, the correct answer is B .

2.

If $~\begin{array}{r|l}a&b \\ \hline c&d\end{array} = \text{a}\cdot \text{d} - \text{b}\cdot \text{c},$ what is the value of $~\begin{array}{r|l}3&4 \\ \hline 1&2\end{array}~?$

$-2$

$-1$

$0$

$1$

$2$

Answer: E

Solution:

Plugging into the formula above, we get: $3\cdot 2-1\cdot 4=2.$

Thus, the correct answer is E .

3.

What is the value of: $\dfrac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}?$

$1$

$\dfrac{25}{16}$

$2$

$\dfrac{43}{20}$

$\dfrac{47}{16}$

Answer: B

Solution:

This evaluates to: $\begin{align*} \dfrac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}} &= \dfrac{\frac{5}{4}}{\frac{4}{5}} \\ &=\left(\dfrac 54\right)^2 \\ &= \dfrac{25}{16}.\end{align*}$

Thus, the correct answer is B .

4.

How many triangles are in this figure? (Some triangles may overlap other triangles.)

$9$

$8$

$7$

$6$

$5$

Answer: E

Solution:

The figure contains three small triangles, the triangle made from the two rightmost small triangles, and the large outside triangle.

This gives $5$ triangles.

Thus, the correct answer is E .

5.

Which of the following numbers is largest?

$9.12344$

$9.123\overline{4}$

$9.12\overline{34}$

$9.1\overline{234}$

$9.\overline{1234}$

Answer: B

Solution:

Each number starts with $9.1234$ . The next digits are $4$ for choices A and B , $3$ for C , $2$ for D , and $1$ for E .

Choice A then terminates as zeros, while choice B continues with more $4$ s, so choice B is largest.

Thus, the correct answer is B .

6.

Dots are spaced one unit apart, horizontally and vertically. The number of square units enclosed by the polygon is

$5$

$6$

$7$

$8$

$9$

Answer: B

Solution:

Consider the $2 \times 3$ rectangle on the bottom. The figure is the same as when we take some area out on the bottom and add the same area on the top. Thus, the area is the same as the the $2 \times 3$ rectangle, which is $6$

Thus, the correct answer is B .

7.

$100\times 19.98\times 1.998\times 1000=$

$(1.998)^2$

$(19.98)^2$

$(199.8)^2$

$(1998)^2$

$(19980)^2$

Answer: D

Solution:

Group the first two factors and the last two factors:

$(100\cdot 19.98)(1.998\cdot 1000)=1998\cdot1998=(1998)^2.$

Thus, the correct answer is D .

8.

A child's wading pool contains $200$ gallons of water. If water evaporates at the rate of $0.5$ gallons per day and no other water is added or removed, how many gallons of water will be in the pool after $30$ days?

$140$

$170$

$185$

$198.5$

$199.85$

Answer: C

Solution:

The amount lost is $0.5\cdot 30=15$ gallons. Therefore, the amount left is $200-15=185.$

Thus, the correct answer is C .

9.

For a sale, a store owner reduces the price of a $\$10$ scarf by $20\% .$ Later the price is lowered again, this time by one-half the reduced price. The price is now

$\$2.00$

$\$3.75$

$\$4.00$

$\$4.90$

$\$6.40$

Answer: C

Solution:

After the $20\%$ reduction, the price is $\$10\cdot 0.8=\$8.$

Then, after halving the price, the price is $\dfrac {\$8}2 = \$4.$

Thus, the correct answer is C .

10.

Each of the letters $W,$ $X,$ $Y,$ and $Z$ represents a different integer in the set $\{ 1,2,3,4\},$ but not necessarily in that order. If $\dfrac{W}{X} - \dfrac{Y}{Z}=1,$ then the sum of $W$ and $Y$ is:

$3$

$4$

$5$

$6$

$7$

Answer: E

Solution:

The only way to get a difference of $1$ is

$\frac31-\frac42=1.$

Thus $W=3$ and $Y=4$ , so $W+Y=7$ .

Thus, the correct answer is E .

11.

Harry has 3 sisters and 5 brothers. His sister Harriet has $\text{S}$ sisters and $\text{B}$ brothers. What is the product of $\text{S}$ and $\text{B}?$

$8$

$10$

$12$

$15$

$18$

Answer: C

Solution:

Since Harry has $3$ sisters and $5$ brothers, the family has $3$ girls and $6$ boys. Harriet is one of the girls, so she has $2$ sisters and $6$ brothers.

Therefore $SB=2\cdot6=12$ .

Thus, the correct answer is C .

12.

What is the value of the following expression? $2\left(1-\dfrac{1}{2}\right) + 3\left(1-\dfrac{1}{3}\right) +$ $4\left(1-\dfrac{1}{4}\right) + \cdots + 10\left(1-\dfrac{1}{10}\right)$

$45$

$49$

$50$

$54$

$55$

Answer: A

Solution:

For each integer $n$ from $2$ through $10$ ,

$n\left(1-\frac1n\right)=n-1.$

The expression is therefore $1+2+\cdots+9=45$ .

Thus, the correct answer is A .

13.

What is the ratio of the area of the shaded square to the area of the large square? (The figure is drawn to scale)

$\dfrac{1}{6}$

$\dfrac{1}{7}$

$\dfrac{1}{8}$

$\dfrac{1}{12}$

$\dfrac{1}{16}$

Answer: C

Solution:

We could extend the figure to the following:

Then, we could look at the bottom triangle that makes up a quarter of the figure.

Half of that area is the shaded area, so the entire shaded area is $\dfrac 18.$

Thus, the correct answer is C .

14.

At Annville Junior High School, $30\%$ of the students in the Math Club are in the Science Club, and $80\%$ of the students in the Science Club are in the Math Club. There are $15$ students in the Science Club. How many students are in the Math Club?

$12$

$15$

$30$

$36$

$40$

Answer: E

Solution:

Since $80\%$ of people are in both clubs, the number of people in both clubs is $0.8\cdot 15=12.$ This is $30\%$ of the math club, so the number of people in the math club is $\dfrac{12}{0.3}=40.$

Thus, the correct answer is E .

15.

Problems $15, 16,$ and $17$ all refer to the following:

Don’t Crowd The Isles

In the very center of the Irenic Sea lie the beautiful Nisos Isles. In $1998$ the number of people on these islands is only $200,$ but the population triples every $25$ years. Queen Irene has decreed that there must be at least $1.5$ square miles for every person living in the Isles. The total area of the Nisos Isles is $24900$ square miles.

Estimate the population of Nisos in the year $2050.$

$600$

$800$

$1000$

$2000$

$3000$

Answer: D

Solution:

The population in $2048$ , which is $50$ years after $1998$ , is $3^2\cdot200=1800$ .

Since $2048$ is close to $2050$ , the population in $2050$ is approximately $1800$ , and the closest choice is $2000$ .

Thus, the correct answer is D .

16.

Estimate the year in which the population of Nisos will be approximately $6000.$

$2050$

$2075$

$2100$

$2125$

$2150$

Answer: B

Solution:

This would be the year the population is $30$ times as much as in $1998.$ This means the population triples approximately $3$ times, making the year approximately $3\cdot 25=75$ years after $1998.$ This would be $2073,$ so $2075$ is the best approximation.

Thus, the correct answer is B .

17.

In how many years, approximately, from $1998$ will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support?

$50$ years

$75$ years

$100$ years

$125$ years

$150$ years

Answer: C

Solution:

The maximal population is $\dfrac{24900}{1.5}=16600.$ This is $83$ times as much as the population in $1998,$ so it would be about $4$ triples from $1998.$ That would be $25\cdot 4=100$ years.

Thus, the correct answer is C .

18.

As indicated by the diagram below, a rectangular piece of paper is folded bottom to top, then left to right, and finally, a hole is punched at $X$. What does the paper look like when unfolded?

Answer: B

Solution:

The final folded rectangle is the upper-right quarter of the original sheet, and the hole is punched in the upper-left part of that folded rectangle.

Unfolding reflects the hole across the horizontal and vertical fold lines. Only choice B has the four corresponding holes.

Thus, the correct answer is B .

19.

Tamika selects two different numbers at random from the set $\{ 8,9,10 \}$ and adds them. Carlos takes two different numbers at random from the set $\{3, 5, 6\}$ and multiplies them. What is the probability that Tamika's result is greater than Carlos' result?

$\dfrac{4}{9}$

$\dfrac{5}{9}$

$\dfrac{1}{2}$

$\dfrac{1}{3}$

$\dfrac{2}{3}$

Answer: A

Solution:

Tamika can get $17,18,19$ , and Carlos can get $15,18,30$ .

The nine equally likely pairs are formed by choosing one of each. Tamika is greater in $(17,15),(18,15),(19,15),(19,18)$ , so $4$ of the $9$ pairs work.

Thus, the correct answer is A .

20.

Let $PQRS$ be a square piece of paper. $P$ is folded onto $R$ and then $Q$ is folded onto $S.$ The area of the resulting figure is 9 square inches. Find the perimeter of square $PQRS.$

$9$

$16$

$18$

$24$

$36$

Answer: D

Solution:

After the two folds, the resulting triangle has area $9$ . Four congruent copies of this triangle make the original square.

So the square has area $4\cdot9=36$ , giving side length $6$ . Its perimeter is $4\cdot6=24$ .

Thus, the correct answer is D .

21.

A $4\times 4\times 4$ cubical box contains 64 identical small cubes that exactly fill the box. How many of these small cubes touch a side or the bottom of the box?

$48$

$52$

$60$

$64$

$80$

Answer: B

Solution:

The only cubes that do not touch a side or the bottom form the interior core above the bottom layer. This core has dimensions $2\times2\times3$ , so it contains $12$ cubes.

Thus $64-12=52$ cubes touch a side or the bottom.

Thus, the correct answer is B .

22.

Terri produces a sequence of positive integers by following three rules. She starts with a positive integer, then applies the appropriate rule to the result, and continues in this fashion.

Rule 1: If the integer is less than 10, multiply it by 9.

Rule 2: If the integer is even and greater than 9, divide it by 2.

Rule 3: If the integer is odd and greater than 9, subtract 5 from it.

For example, consider the sample sequence: $23, 18, 9, 81, 76, \ldots .$

Find the $98^\text{th}$ term of the sequence that begins with: $98, 49, \ldots$

$6$

$11$

$22$

$27$

$54$

Answer: D

Solution:

The sequence begins

$98,49,44,22,11,6,54,27,22,\ldots .$

After the first three terms, the cycle $(22,11,6,54,27)$ repeats. Since $98-3=95$ is a multiple of $5$ , the $98^{\text{th}}$ term is the fifth term of the cycle, $27$ .

Thus, the correct answer is D .

23.

If the pattern in the diagram continues, what fraction of the interior would be shaded in the eighth triangle?

$\dfrac{3}{8}$

$\dfrac{5}{27}$

$\dfrac{7}{16}$

$\dfrac{9}{16}$

$\dfrac{11}{45}$

Answer: C

Solution:

The $n^{\text{th}}$ triangle has $n^2$ small triangles.

The number of shaded small triangles is $1+2+\cdots+(n-1)=\dfrac{n(n-1)}2$ .

For $n=8$ , the shaded fraction is $\dfrac{8\cdot7/2}{8^2}=\dfrac7{16}$ .

Thus, the correct answer is C .

24.

A rectangular board of 8 columns has squares numbered beginning in the upper left corner and moving left to right so row one is numbered 1 through 8, row two is 9 through 16, and so on. A student shades square 1, then skips one square and shades square 3, skips two squares and shades square 6, skips 3 squares and shades square 10, and continues in this way until there is at least one shaded square in each column.

What is the number of the shaded square that first achieves this result?

$36$

$64$

$78$

$91$

$120$

Answer: E

Solution:

The shaded squares are the triangular numbers $1,3,6,10,15,\ldots$ . Columns correspond to residues modulo $8$ , with residue $0$ representing the eighth column.

The triangular numbers through $105$ have residues $1,3,6,2,7,5,4,4,5,7,2,6,3,1$ , so the eighth-column residue has not appeared yet.

The next triangular number is $120$ , and $120\equiv0\pmod8$ . This is the first time every column has a shaded square.

Thus, the correct answer is E .

25.

Three generous friends, each with some cash, redistribute their money as follows: Ami gives enough money to Jan and Toy to double the amount that each has. Jan then gives enough to Ami and Toy to double their amounts. Finally, Toy gives Ami and Jan enough to double their amounts. If Toy has $\$36$ when they begin and $\$36$ when they end, what is the total amount that all three friends have?

$\$108$

$\$180$

$\$216$

$\$252$

$\$288$

Answer: D

Solution:

Toy begins with $\$36$ . After Ami doubles Toy's amount, Toy has $\$72$ . After Jan doubles Toy's amount, Toy has $\$144$ .

On Toy's final turn, Toy ends with $\$36$ , so Toy gives away $\$144-\$36=\$108$ . That gift doubles the combined amount of Ami and Jan, so Ami and Jan together had $\$108$ just before Toy's final turn.

The total amount of money is constant, so the total is $\$144+\$108=\$252$ .

Thus, the correct answer is D .