### 1998 AMC 8 Exam Solutions

Problems used with permission of the Mathematical Association of America. Scroll down to view solutions, print PDF solutions, view answer key, or:

1.

For \(x=7,\) which of the following is the smallest?

\(\dfrac{6}{x} \)

\(\dfrac{6}{x+1} \)

\(\dfrac{6}{x-1} \)

\(\dfrac{x}{6} \)

\(\dfrac{x+1}{6} \)

###### Solution(s):

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

**A:** \(\dfrac 6x = \dfrac 67\)

**B:** \(\dfrac 6x = \dfrac 68\)

**C:** \(\dfrac 6x = 1\)

**D:** \(\dfrac x6 = \dfrac 76\)

**E:** \(\dfrac {x+1}6 = \dfrac{4}{3}\)

Thus, the correct answer is **E**.

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 \(~\begin{array}{r|l}3&4 \\ \hline 1&2\end{array}~?\)

\(-2 \)

\(-1 \)

\(0 \)

\(1 \)

\(2 \)

###### Solution(s):

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: \[\dfrac{\frac{3}{8} + \frac{7}{8}}{\frac{4}{5}}?\]

\(1 \)

\( \dfrac{25}{16} \)

\(2 \)

\(\dfrac{43}{20} \)

\(\dfrac{47}{16} \)

###### Solution(s):

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.)

\(9 \)

\(8 \)

\(7 \)

\(6 \)

\(5 \)

###### Solution(s):

First, we can clearly see the \(3\) small triangles, and the triangle that encompasses the entire figue. Also, there is \(1\) more triangle when combining the two rightmost smaller triangles.

This leads to us having \(5\) triangles.

Thus, the correct answer is **E**.

5.

Which of the following numbers is largest?

\(9.12344 \)

\(9.123\overline{4} \)

\(9.12\overline{34} \)

\(9.1\overline{234} \)

\(9.\overline{1234} \)

###### Solution(s):

Each of them start with \(9.1234.\) Thus, we need to look at the next digits.

For choices **A** and **B**, the next digit is \(4\) while it is \(3\) for **C**, \(2\) for **D**, and \(1\) for **E**.

This means we only have to look at **A** and **B** as possible solutions. After the \(9.12344,\) choice **A** terminates while the next term for **B** is \(4,\) so choice **B** is larger.

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

\(5 \)

\(6 \)

\(7 \)

\(8 \)

\(9 \)

###### Solution(s):

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.

\(100\times 19.98\times 1.998\times 1000=\)

\((1.998)^2 \)

\((19.98)^2 \)

\((199.8)^2 \)

\((1998)^2 \)

\((19980)^2 \)

###### Solution(s):

We will group the first two terms and the last two terms.

This will make the expression equal to \[(100\cdot 19.98)\cdot (1.998\cdot 1000)=\] \[1998\cdot 1998=1998^2.\]

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?

\(140 \)

\(170 \)

\(185 \)

\(198.5 \)

\(199.85 \)

###### Solution(s):

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

\( $2.00 \)

\($3.75 \)

\($4.00 \)

\($4.90 \)

\($6.40 \)

###### Solution(s):

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:

\(3 \)

\(4 \)

\(5 \)

\(6 \)

\(7 \)

###### Solution(s):

The fractions \(\dfrac{W}{X}\) and \(\dfrac YZ\) must be integers as there is no other fractional part that can there twice.

Thus, they are both integers, making \(1\) a denominator. The only other possible denominator could be \(2\) with its numerator being \(4.\)

Thus, we could get \[\dfrac 31 - \dfrac 42=1,\] making the sum equal to \[4+3=7.\]

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 \(\text{B}?\)

\(8 \)

\(10 \)

\(12 \)

\(15 \)

\(18 \)

###### Solution(s):

Since Harry has 3 sisters and 5 brothers, the family has 3 girls and 6 boys. Then, Harriet would be one or the girls, so she had \(2\) sisters and \(6\) brothers. Thus, the product of the number of brothers and sisters is \(2\cdot 6=12.\)

Thus, the correct answer is **C**.

12.

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

\(45 \)

\(49 \)

\(50 \)

\(54 \)

\(55 \)

###### Solution(s):

Directly evaluating, we get that: \[\begin{align*}&2\left(1-\dfrac{1}{2}\right) + 3\left(1-\dfrac{1}{3}\right) +\\ &4\left(1-\dfrac{1}{4}\right) + \cdots + 10\left(1-\dfrac{1}{10}\right)\\ &=(2-1)+\cdots +(10-1)\\ &=1+2 \cdots 8+9 \\ &= 45\end{align*}\]

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)

\(\dfrac{1}{6} \)

\(\dfrac{1}{7} \)

\(\dfrac{1}{8} \)

\(\dfrac{1}{12} \)

\(\dfrac{1}{16} \)

###### Solution(s):

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.

An 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?

\(12 \)

\(15 \)

\(30 \)

\(36 \)

\(40 \)

###### Solution(s):

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.

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 \(2050.\)

\(600 \)

\(800 \)

\(1000 \)

\(2000 \)

\(3000 \)

###### Solution(s):

The population in \(2048,\) which is \(50\) years after \(1998,\) is \(3^2\cdot 200=1800.\)

Therefore, as \(2048\approx 2050,\) the population in \(2050\) is approximately \(1800,\) which is approximately equal to \(2000\)

Thus, the correct answer is **D**.

16.

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 year in which the population of Nisos will be approximately \(6000.\)

\(2050 \)

\(2075 \)

\(2100 \)

\(2125 \)

\(2150 \)

###### Solution(s):

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 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.

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

###### Solution(s):

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?

###### Solution(s):

When we undo the fold, the rectangle with the punched hole in the upper left is in the bottom right. The only such answer choice is **B**.

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?

\(\dfrac{4}{9} \)

\(\dfrac{5}{9} \)

\(\dfrac{1}{2} \)

\(\dfrac{1}{3} \)

\(\dfrac{2}{3} \)

###### Solution(s):

Tamika, with equal probability, can get one of \(17,18,19.\)

Carlos, with equal probability, can get one of \(15,18,30.\)

There is a \(\frac 13\) probability that Carlos gets \(15,\) which would always have Tamika having a higher number.

There is a \(\frac 13\) probability that Carlos gets \(18,\) which would have a \(\dfrac 13\) probability that Tamika gets a higer number of \(19.\)

There is a \(\frac 13\) probability that Carlos gets \(30,\) which would always have Tamika never a higher number.

Therefore, the total probability is: \[\dfrac 13 + \dfrac 13\cdot \dfrac 13 = \dfrac 49.\]

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 \(PQRS.\)

\(9 \)

\(16 \)

\(18 \)

\(24 \)

\(36 \)

###### Solution(s):

Let the side length be \(s.\) Then, folding \(P\) to \(R\) would give an isoceles right triangle with area \(\frac 12{s^2}.\) Here, \(Q\) and \(S\) are across from each other, so when it is folded, the area is \(\frac14{s^2}=9,\) making \(s=6.\)

As such, the area 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?

\(48 \)

\(52 \)

\(60 \)

\(64 \)

\(80 \)

###### Solution(s):

The entire \(4 \times 4 \times 1\) bottom with a volume of \(16\) is counted. Then, there are \(48\) cubes left. There are \(3\) layers, with each of being a \(4 \times 4 \) square. Only the \(2 \times 2\) interior doesn't touch the outside, so each layer has \[4^2-2^2=12\] cubes. This makes the total number of cubes equal to \[12\cdot 3+16=52.\]

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: \[98, 49, \ldots \]

\(6 \)

\(11 \)

\(22 \)

\(27 \)

\(54 \)

###### Solution(s):

The sequence starts with the following: \[98,49,44,22,11,\]\[6,54,27,22, \cdots\] Thus, after the first four, it has a cycle of length \(5.\) This makes the \(98^{th}\) term equal to the \(8^{th}\) term of the sequence, which is \(27.\)

Thus, the correct answer is **D**.

23.

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

\(\dfrac{3}{8} \)

\(\dfrac{5}{27} \)

\(\dfrac{7}{16} \)

\(\dfrac{9}{16} \)

\(\dfrac{11}{45} \)

###### Solution(s):

The total number of triangles in the \(n^{th}\) triangle is \(n^2.\)

The total number of shaded triangles in the \(n^{th}\) triangle is the \(n-1^{th}\) triangular number, which is: \[\dfrac{n^2-n}2.\]

This makes the ratio equal to: \[\dfrac{n-1}{2n} = \dfrac{7}{16}.\]

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?

\(36 \)

\(64 \)

\(78 \)

\(91 \)

\(120 \)

###### Solution(s):

The only shaded squares are the triangular numbers (the \(n^{th}\) triangular number is the sum of the first \(n\) numbers). As such, we have to find the first triangular number such that every possible value \(\mod 8\) is accounted for.

Note that the formula for triangular numbers is \[\dfrac{n^2+n}2.\]

Thus, we must find a triangular number for each value modulo \(8.\) For \(0\mod 8,\) the first one is \(120,\) meaning the answer is at least \(120.\) \(120\) is the greatest answer choice available, so we know the answer must be \(120.\) By inspection, we can also see that every other value modulo \(8\) comes before anyways, but that is omitted here.

Thus, the correct answer is **E**.

25.

Three generous friends, each with some money, redistribute the money as followed: Amy gives enough money to Jan and Toy to double each amount has. Jan then gives enough to Amy and Toy to double their amounts. Finally, Toy gives enough to Amy and Jan to double their amounts. If Toy had \($36\) at the beginning and \($36\) at the end, what is the total amount that all three friends have?

\($108 \)

\($180 \)

\($216 \)

\($252 \)

\($288 \)

###### Solution(s):

Since Toy doubles his money after the first two turns, he has \($144\) in the end. Then, he gives away \(144-36=$108.\) Since this doubles Any and Jan's money, they had \($108\) before Toy gives them money. Thus, the total amount of money is \[$144+$108=$252.\]

Thus, the correct answer is **D**.