1998 AMC 8 Solutions | LIVE by Po-Shen Loh

1.

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

$\dfrac{6}{x}$

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

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

$\dfrac{x}{6}$

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

Concepts:substitution fraction

Difficulty rating: 450

Small Hint:

Substitute $x = 7$ first.

Big Hint:

With equal numerators, compare denominators.

Solution:

Substituting $x = 7$ gives the five values $\dfrac{6}{7},$ $\dfrac{6}{8},$ $1,$ $\dfrac{7}{6},$ and $\dfrac{4}{3}.$

The first two are the only values less than $1.$ Because they have the same numerator, the fraction with the larger denominator is smaller. Thus, the correct answer is B .

2.

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

$-2$

$-1$

$0$

$1$

$2$

Concepts:custom operation

Difficulty rating: 450

Small Hint:

Match $a,$ $b,$ $c,$ and $d$ to the four entries.

Big Hint:

Compute $3\cdot 2-4\cdot 1.$

Solution:

Substituting into the definition gives $3\cdot 2-4\cdot 1=2.$

Thus, the correct answer is E .

3.

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

$1$

$\dfrac{25}{16}$

$2$

$\dfrac{43}{20}$

$\dfrac{47}{16}$

Concepts:fraction

Difficulty rating: 560

Small Hint:

Add the numerator fractions first.

Big Hint:

Dividing by $\frac{4}{5}$ means multiplying by $\frac{5}{4}.$

Solution:

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

Thus, the correct answer is B .

4.

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

$9$

$8$

$7$

$6$

$5$

Concepts:counting shapes in figures casework

Difficulty rating: 720

Small Hint:

Count small and larger triangles separately.

Big Hint:

Do not forget triangles formed by combining smaller ones.

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?

$9.12344$

$9.123\overline{4}$

$9.12\overline{34}$

$9.1\overline{234}$

$9.\overline{1234}$

Concepts:repeating decimal place value

Difficulty rating: 730

Small Hint:

Write the repeating decimals out for a few places.

Big Hint:

Compare at the first digit where the decimals differ.

Solution:

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

Choice $A$ then terminates and is followed by zeros, while choice $B$ continues with more $4$ s. Thus, 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

$5$

$6$

$7$

$8$

$9$

Concepts:area decomposition lattice point

Difficulty rating: 870

Small Hint:

Move the slanted top triangle into the missing bottom space.

Big Hint:

The polygon has the area of a $2\times 3$ rectangle.

Solution:

Consider the $2 \times 3$ rectangle on the bottom. The triangular piece above that rectangle has the same area as the missing triangular piece below it. Rearranging one into the other gives a full $2 \times 3$ rectangle, so the polygon’s area is $6.$

Thus, the correct answer is B .

7.

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

$(1.998)^2$

$(19.98)^2$

$(199.8)^2$

$(1998)^2$

$(19980)^2$

Concepts:decimal exponent

Difficulty rating: 860

Small Hint:

Pair the factors with powers of $10.$

Big Hint:

Group the product as $(100\cdot 19.98)(1000\cdot 1.998).$

Solution:

Group the first two factors and the last two factors:

$\begin{aligned} &(100\cdot 19.98)(1.998\cdot 1000) \\ &= 1998\cdot1998 \\ &= (1998)^2. \end{aligned}$

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?

$140$

$170$

$185$

$198.5$

$199.85$

Concepts:rate

Difficulty rating: 730

Small Hint:

Find how much water evaporates in $30$ days.

Big Hint:

Subtract $0.5\cdot 30$ from the starting amount.

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

$\$2.00$

$\$3.75$

$\$4.00$

$\$4.90$

$\$6.40$

Concepts:percentage

Difficulty rating: 860

Small Hint:

A $20\%$ reduction leaves $80\%.$

Big Hint:

The second reduction halves the reduced price.

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. They satisfy $\dfrac{W}{X}-\dfrac{Y}{Z}=1$ What is the sum of $W$ and $Y?$

$3$

$4$

$5$

$6$

$7$

Concepts:fraction casework

Difficulty rating: 1070

Small Hint:

List possible fractions using $1,$ $2,$ $3,$ and $4.$

Big Hint:

The larger fraction must exceed the smaller one by exactly $1.$

Solution:

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

$\frac{3}{1}-\frac{4}{2}=1.$

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 $S$ sisters and $B$ brothers. What is the product of $S$ and $B?$

$8$

$10$

$12$

$15$

$18$

Concepts:basic counting logical deduction

Difficulty rating: 1020

Small Hint:

First count the total boys and girls in the family.

Big Hint:

Harriet is one of Harry’s sisters.

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, $SB=2\cdot 6=12.$

Thus, the correct answer is C .

12.

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

$45$

$49$

$50$

$54$

$55$

Concepts:summation algebraic manipulation

Difficulty rating: 1090

Small Hint:

Simplify each term $n(1-\frac{1}{n}).$

Big Hint:

The terms become consecutive integers.

Solution:

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

$n\left(1-\frac{1}{n}\right)=n-1.$

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.)

$\dfrac{1}{6}$

$\dfrac{1}{7}$

$\dfrac{1}{8}$

$\dfrac{1}{12}$

$\dfrac{1}{16}$

Concepts:area ratio area decomposition

Difficulty rating: 1150

Small Hint:

Extend the grid lines in the large square.

Big Hint:

The shaded square is made from four half-unit squares.

Solution:

Extend the figure to a $4$ by $4$ grid as shown:

The large square consists of $16$ unit squares. The shaded square is made from four half-unit squares, so its area is $4\cdot\dfrac{1}{2}=2.$ Therefore, the required ratio is $\dfrac{2}{16}=\dfrac{1}{8}.$

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?

$12$

$15$

$30$

$36$

$40$

Concepts:percentage

Difficulty rating: 1180

Small Hint:

Find the number of students in both clubs first.

Big Hint:

That overlap is $30\%$ of the Math Club.

Solution:

Since $80\%$ of the $15$ Science Club students are also in the Math Club, the two clubs overlap in $0.8\cdot 15=12$ students. This is $30\%$ of the Math Club, so the Math Club has $\dfrac{12}{0.3}=40$ students.

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 $24{,}900$ square miles.

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

$600$

$800$

$1000$

$2000$

$3000$

Concepts:geometric sequence estimation

Difficulty rating: 960

Small Hint:

Tripling twice gets from $1998$ to $2048.$

Big Hint:

Use $2048$ as the closest benchmark to $2050.$

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 $6000.$

$2050$

$2075$

$2100$

$2125$

$2150$

Concepts:geometric sequence estimation

Difficulty rating: 1030

Small Hint:

Compare $6000$ with the starting population $200.$

Big Hint:

Three triplings are close to a factor of $30.$

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

Concepts:geometric sequence estimation

Difficulty rating: 1120

Small Hint:

Compute how many people the islands can support.

Big Hint:

Compare the capacity with the population after each tripling.

Solution:

The maximal population is $\dfrac{24{,}900}{1.5}=16{,}600.$ 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?

Concepts:paper folding reflection (geometry)

Difficulty rating: 1280

Small Hint:

Unfold the last fold first.

Big Hint:

Each fold reflects the hole across the fold line.

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?

$\dfrac{4}{9}$

$\dfrac{5}{9}$

$\dfrac{1}{2}$

$\dfrac{1}{3}$

$\dfrac{2}{3}$

Concepts:basic probability casework

Difficulty rating: 1430

Small Hint:

List Tamika’s possible sums and Carlos’ possible products.

Big Hint:

Count the favorable ordered pairs among the nine equally likely pairs.

Solution:

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

The nine equally likely pairs are formed by choosing one result from each person. Tamika’s result is greater in $(17,15),$ $(18,15),$ $(19,15),$ and $(19,18),$ 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 $PQRS.$

$9$

$16$

$18$

$24$

$36$

Concepts:paper folding area

Difficulty rating: 1360

Small Hint:

After the two folds, four congruent pieces make the original square.

Big Hint:

Use the resulting area to find the original side length.

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 $4\cdot 9=36,$ giving side length $6.$ Its perimeter is $4\cdot 6=24.$

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?

$48$

$52$

$60$

$64$

$80$

Concepts:complementary counting 3D geometry

Difficulty rating: 1410

Small Hint:

Count the cubes that do not touch a side or the bottom.

Big Hint:

The untouched interior core is $2\times 2\times 3.$

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 $2\times 2\times 3,$ 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 $98$ th term of the sequence that begins $98,$ $49,$ $\ldots.$

$6$

$11$

$22$

$27$

$54$

Concepts:recursion modular arithmetic

Difficulty rating: 1480

Small Hint:

Generate terms until a value repeats.

Big Hint:

After the repeat begins, reduce the index modulo the cycle length.

Solution:

The sequence begins

$\begin{gathered} 98,49,44,22,11, \\ 6,54,27,22,\ldots. \end{gathered}$

After the first three terms, the cycle $(22,11,6,54,27)$ repeats. Since $98-3=95$ is a multiple of $5,$ the $98$ 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?

$\dfrac{3}{8}$

$\dfrac{5}{27}$

$\dfrac{7}{16}$

$\dfrac{9}{16}$

$\dfrac{11}{45}$

Concepts:triangular number pattern recognition

Difficulty rating: 1410

Small Hint:

At step $n,$ there are $n^2$ small triangles.

Big Hint:

The shaded count is $1+2+\cdots+(n-1).$

Solution:

The $n$ 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 $\dfrac{\frac{8\cdot 7}{2}}{8^2}=\dfrac{7}{16}.$

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?

$36$

$64$

$78$

$91$

$120$

Concepts:triangular number modular arithmetic

Difficulty rating: 1630

Small Hint:

Columns correspond to residues modulo $8.$

Big Hint:

The shaded squares are triangular numbers.

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 $1,$ $3,$ $6,$ $2,$ $7,$ $5,$ $4,$ $4,$ $5,$ $7,$ $2,$ $6,$ $3,$ and $1.$ Thus, the eighth-column residue has not appeared yet.

The next triangular number is $120,$ and $120\equiv 0\pmod{8}.$ 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?

$\$108$

$\$180$

$\$216$

$\$252$

$\$288$

Concepts:work backwards invariant

Difficulty rating: 1620

Small Hint:

Track Toy’s amount through the first two exchanges.

Big Hint:

Before Toy’s final turn, the amount he gives equals Ami and Jan’s combined amount.

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 .