2007 AMC 8 Solutions

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https://live.poshenloh.com/past-contests/amc8/2007/solutions

Problems © Mathematical Association of America. Reproduced with permission.

1.

Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of \(10\) hours per week helping around the house for \(6\) weeks. For the first \(5\) weeks she helps around the house for \(8,\) \(11,\) \(7,\) \(12\) and \(10\) hours. How many hours must she work for the final week to earn the tickets?

\(9\)

\(10\)

\(11\)

\(12\)

\(13\)

Solution(s):

During the first \(5\) weeks, Theresa works for a total of \[ 8 + 11 + 7 + 12 + 10 = 48 \] hours.

She, however, promised to work for \[ 10 \cdot 6 = 60 \] hours.

This means that she has to work \[60 - 48 = 12\] hours during the final work to earn the tickets.

Thus, D is the correct answer.

2.

\(650\) students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?

\(\dfrac{2}{5}\)

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

\(\dfrac{5}{4}\)

\(\dfrac{5}{3}\)

\(\dfrac{5}{2}\)

Solution(s):

There are \(250\) students who preferred spaghetti and \(100\) that preferred manicotti.

The ratio is therefore \[ \dfrac{250}{100} = \dfrac{5}{2}. \]

Thus, E is the correct answer.

3.

What is the sum of the two smallest prime factors of \(250?\)

\(2\)

\(5\)

\(7\)

\(10\)

\(12\)

Solution(s):

We can prime factorize \(250\) to get \[ 250 = 2 \cdot 5^3. \] From this, we can see that \(250\) only has two prime factors: \(2\) and \(5.\) The sum of these is \(7.\)

Thus, C is the correct answer.

4.

A haunted house has six windows. In how many ways can Georgie the Ghost enter the house by one window and leave by a different window?

\(12\)

\(15\)

\(18\)

\(30\)

\(36\)

Solution(s):

Georgie has \(6\) options for which window he enters through. He, however, only has \(5\) options for the exit since it must be different from the entrance.

The total number of paths is therefore \(6 \cdot 5 = 30.\)

Thus, D is the correct answer.

5.

Chandler wants to buy a \(500\) dollar mountain bike. For his birthday, his grandparents send him \(50\) dollars, his aunt sends him \(35\) dollars and his cousin gives him \(15\) dollars. He earns \(16\) dollars per week for his paper route. He will use all of his birthday money and all of the money he earns from his paper route. In how many weeks will he be able to buy the mountain bike?

\(24\)

\(25\)

\(26\)

\(27\)

\(28\)

Solution(s):

The total amount of money that Chandler got from his birthday is \[ 50 + 35 + 15 = 100 \] dollars. Therefore, he only needs to raise \(500 - 100 = 400\) more dollars.

This can be achieved in \(400 \div 16 = 25\) weeks through his paper route.

Thus, B is the correct answer

6.

The average cost of a long-distance call in the USA in \(1985\) was \(41\) cents per minute, and the average cost of a long-distance call in the USA in \(2005\) was \(7\) cents per minute. Find the approximate percent decrease in the cost per minute of a long-distance call.

\(7\)

\(17\)

\(34\)

\(41\)

\(80\)

Solution(s):

The difference in the costs is \(41 - 7 = 34\) cents. The approximate percent decrease is therefore \[ 100 \cdot \dfrac{35}{41} \approx 100 \cdot \dfrac{32}{40}\]\[=100 \cdot \dfrac{4}{5} \]\[= 80 \%. \]

Thus, E is the correct answer.

7.

The average age of \(5\) people in a room is \(30\) years. An \(18\)-year-old person leaves the room. What is the average age of the four remaining people?

\(25\)

\(26\)

\(29\)

\(33\)

\(36\)

Solution(s):

Initially, the total age of everyone in the room is \(5 \cdot 30 = 150\) years.

After the person leaves, the total age is \(150 - 18 = 132.\) With \(4\) people remaining, the average age becomes \(\dfrac{132}{4} = 33.\)

Thus, D is the correct answer.

8.

In trapezoid \(ABCD\), \(\overline{AD}\) is perpendicular to \(\overline{DC},\) \(AD = AB = 3,\) and \(DC = 6.\) In addition, \(E\) is on \(\overline{DC},\) and \(\overline{BE}\) is parallel to \(\overline{AD}.\) Find the area of \(\triangle BEC.\)

\(3\)

\(4.5\)

\(6\)

\(9\)

\(18\)

Solution(s):

We know that \[ EC = DC - DE\]\[ = 6 - 3 \]\[= 3. \] We also know that \[ BE = AD = 3. \] Therefore, the area of \(\triangle BEC\) is \[ \dfrac{1}{2} \cdot 3 \cdot 3 = \dfrac{9}{2}. \]

Thus, B is the correct answer.

9.

To complete the grid below, each of the digits \(1\) through \(4\) must occur once in each row and once in each column. What number will occupy the lower right-hand square?

\(1\)

\(2\)

\(3\)

\(4\)

\(\text{cannot be determined}\)

Solution(s):

Consider the last element in the second row. This has to be a \(1\) since \(4\) cannot be in that column.

Then, the number in the top right square must be \(3\) since it cannot be \(2.\)

This forces the \(2\) to be in the bottom right square.

Thus, B is the correct answer.

10.

For any positive integer \(n,\) define \(\boxed{n}\) to be the sum of the positive factors of \(n.\) For example, \[ \boxed{6} = 1 + 2 + 3 + 6 = 12. \] Find \(\boxed{\boxed{11}}.\)

\(13\)

\(20\)

\(24\)

\(28\)

\(30\)

Solution(s):

First, we find that \[ \boxed{11} = 1 + 11 = 12 \] since \(11\) is prime. Then we need to find \(\boxed{12}.\) The factors of \(12\) are \[ 1, 2, 3, 4, 6, 12. \] Adding these yields \(\boxed{12} = 28.\)

Thus, D is the correct answer.

11.

Tiles \(I, II, III\) and \(IV\) are translated so one tile coincides with each of the rectangles \(A, B, C\) and \(D.\) In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle \(C?\)

\(I\)

\(II\)

\(III\)

\(IV\)

\(\text{cannot be determined}\)

Solution(s):

Note that only Tile \(III\) has the number of \(0.\) This forces it to be either \(C\) or \(D.\) This tile also is the only one with \(5.\) This makes sure that Tile \(III\) is \(D.\)

The only tile that can match with \(1\) on Tile \(III\) is Tile \(IV.\) Therefore, Tile \(IV\) is \(C.\)

Similarly, we get that Tile \(II\) is \(A\) and Tile \(I\) is \(B.\)

Thus, the correct answer is D

12.

A unit hexagram is composed of a regular hexagon of side length \(1\) and its \(6\) equilateral triangular extensions, as shown in the diagram. What is the ratio of the area of the extensions to the area of the original hexagon?

\(1:1\)

\(6:5\)

\(3:2\)

\(2:1\)

\(3:1\)

Solution(s):

Note that we can split the hexagon into \(6\) congruent equilateral triangles as follows.

Since each of them share an edge with an exterior triangle, all the triangles are congruent. Therefore, the ratio of areas is \(1:1.\)

Thus, A is the correct answer.

13.

Sets \(A\) and \(B,\) shown in the Venn diagram, have the same number of elements. Their union has \(2007\) elements and their intersection has \(1001\) elements. Find the number of elements in \(A.\)

\(503\)

\(1006\)

\(1504\)

\(1507\)

\(1510 \)

Solution(s):

Let \(x\) be the number of elements in each \(A\) and \(B.\) Also let \(y\) be the number of elements in their intersection.

The conditions give us that \[ 2x - y = 2007 \] and \[ y = 1001. \] Plugging \(y\) into the first equation yields \[ \begin{align*} 2x - 1001 &= 2007 \\ 2x &= 3008 \\ x &= 1504. \end{align*} \]

Thus, C is the correct answer.

14.

The base of isosceles \(\triangle ABC\) is \(24\) and its area is \(60.\) What is the length of one of the congruent sides?

\(5\)

\(8\)

\(13\)

\(14\)

\(18\)

Solution(s):

Construct \(\overline{BD}\) as the altitude from \(B\) to \(\overline{AC}.\)

Then \[ 60 = \dfrac{1}{2} \cdot BD \cdot 24, \] which gives us that \(BD = 5.\)

From this, we apply the Pythagorean Theorem on \(\triangle ABD:\) \[ AB^2 = 5^2 + 12^2 = 169 = 13^2. \] This gives us that \(AB = 13.\)

Thus, C is the correct answer.

15.

Let \(a, b\) and \(c\) be numbers with \(0 < a < b < c.\) Which of the following is impossible?

\(a + c < b \)

\(a \cdot b < c\)

\(a + b < c\)

\(a \cdot c < b\)

\(\dfrac{b}{c} = a\)

Solution(s):

We know that \(b < c\) and \(0 < a.\) Adding these two inequalities together yields \[ b \lt c + a. \] This shows that A is impossible, and therefore the right answer.

To ensure that this is correct, we can show that the other options are possible.

B and C: \(a = 1,\) \(b = 2,\) and \(c = 4\)

D: \(a = \dfrac{1}{3},\) \(b = \dfrac{1}{2},\) and \(c = 1\)

E: \(a = \dfrac{1}{2},\) \(b = 1,\) and \(c = 2\)

Thus, A is the correct answer.

16.

Amanda Reckonwith draws five circles with radii \(1, 2, 3, 4\) and \(5.\) Then for each circle she plots the point \((C, A),\) where \(C\) is its circumference and \(A\) is its area. Which of the following could be her graph?

Solution(s):

The cirumferences of circles with radii \(1\) through \(5\) are \[ 2\pi, 4\pi, 6\pi, 8\pi, 10\pi. \] Their respective areas are \[ \pi, 4\pi, 9\pi, 16\pi, 25\pi. \] The only graph that shows these points is A, making it the correct answer.

This is the only graph showing a quadratic function.

Thus, A is the correct answer.

17.

A mixture of \(30\) liters of paint is \(25\%\) red tint, \(30\%\) yellow tint and \(45\%\) water. Five liters of yellow tint are added to the original mixture. What is the percent of yellow tint in the new mixture?

\(25\)

\(35\)

\(40\)

\(45\)

\(50\)

Solution(s):

The amount of yellow tint in the original mixture is \(.3 \times 30 = 9\) liters. Adding \(5\) liters results in a total of \(14\) liters of yellow tint.

The new mixture has a total of \(35\) liters, so the percent of yellow tint is \[ 100 \cdot \dfrac{14}{35} = 100 \cdot \dfrac{2}{5} = 40 \%. \]

Thus, C is the correct answer.

18.

The product of the two \(99\)-digit numbers \[303,030,303,...,030,303\] and \[505,050,505,...,050,505\] has thousands digit \(A\) and units digit \(B.\) What is the sum of \(A\) and \(B?\)

\(3\)

\(5\)

\(6\)

\(8\)

\(10 \)

Solution(s):

We only care about the last \(4\) digits, so we can calculate \(303 \cdot 505\) to find them (the thousands digit is \(0\) for both numbers).

\[ \begin{array}{r} &\cdots 303 \\ \times \hspace{-4mm} &\cdots 505 \\ \hline &\cdots 1515 \\ &\cdots 1500 \\ \hline &\cdots 3015 \end{array} \]

This gives us that \(A = 3\) and \(B = 5.\) Therefore, \(A + B = 8.\)

Thus, D is the correct answer.

19.

Pick two consecutive positive integers whose sum is less than \(100.\) Square both of those integers and then find the difference of the squares. Which of the following could be the difference?

\(2\)

\(64\)

\(79\)

\(96\)

\(131\)

Solution(s):

Let \(x\) and \(x + 1\) the two integers, where \(2x + 1 \lt 100.\) The difference of the squares is \[ (x + 1)^2 - x^2\]\[=(x + 1 + x)(x + 1 - x) \]\[= 2x + 1. \]

From this, we see that the desired difference is less than \(100\) and is odd.

Thus, the correct answer is C.

20.

Before district play, the Unicorns had won \(45 \%\) of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?

\(48\)

\(50\)

\(52\)

\(54\)

\(60\)

Solution(s):

Let \(x\) be the number of games the Unicorns had played before district play. Then after the entire season, they won a total of \(.45x + 6\) games.

In total, they played \(x + 8\) games. The problem statement also tell us that \[ (x + 8) \div 2 = .45x + 6. \] Solving, \[ \begin{align*} x + 8 &= .9x + 12 \\ .1x &= 4 \\ x &= 40. \end{align*} \] Therefore, the Unicorns played a total of \(40 + 8 = 48\) games.

Thus, A is the correct answer.

21.

Two cards are dealt from a deck of four red cards labeled \(A,\) \(B,\) \(C,\) \(D\) and four green cards labeled \(A,\) \(B,\) \(C,\) \(D.\) A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?

\(\dfrac{2}{7}\)

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

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

\(\dfrac{4}{7}\)

\(\dfrac{5}{8}\)

Solution(s):

After drawing the first card, there are \(7\) left. \(3\) of them have the same color, and \(1\) of them has the same letter. Therefore, there are \(4\) out of \(7\) possibilities that result in a winning pair.

The probability of drawing a pair is then \(\dfrac{4}{7}.\)

Thus, D is the correct answer.

22.

A lemming sits at a corner of a square with side length \(10\) meters. The lemming runs \(6.2\) meters along a diagonal toward the opposite corner. It stops, makes a \(90^{\circ}\) right turn and runs \(2\) more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?

\(2\)

\(4.5\)

\(5\)

\(6.2\)

\(7\)

Solution(s):

Based on the lengths given in the problem, the lemming is still in the square after it stops.

Since the lemming is still in the square, the sum of the distances to the horizontal sides is \(10\) meters and the same for the vertical sides. Therefore, the \(4\) distances sum to \(20\) meters, making the average \(20 \div 4 = 5\) meters.

Thus, C is the correct answer.

23.

What is the area of the shaded pinwheel shown in the \(5 \times 5\) grid?

\(4\)

\(6\)

\(8\)

\(10\)

\(12\)

Solution(s):

We can find the area of the shaded part by subtracting the area of the unshaded part from the whole area.

There are \(4\) squares with side length \(1,\) contributing \(1\) each to the unshaded area, for a total of \(4.\)

There are also \(4\) triangles with base \(3\) and height \(\dfrac{5}{2}.\) This contributes a total area of \[ 4 \cdot \dfrac{1}{2} \cdot 3 \cdot \dfrac{5}{2} = 15. \]

The total unshaded area is therefore \(4 + 15 = 19.\)

The total area is \(5^2 = 25,\) and subtracting the unshaded area yields \(25 - 19 = 6\) as the area of the shaded region.

Thus, B is the correct answer.

24.

A bag contains four pieces of paper, each labeled with one of the digits \(1,\) \(2,\) \(3\) or \(4,\) with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of \(3?\)

\(\dfrac{1}{4}\)

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

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

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

\(\dfrac{3}{4}\)

Solution(s):

Recall that a number is divisible if the sum of its digits is divisible by \(3.\)

The only triples of distinct numbers that satisfy this condition are \((1, 2, 3)\) and \((2, 3, 4).\)

This means that the numbers form a three-digit number when either \(1\) or \(4\) is left in the bag.

Each of these events happens with \(\dfrac{1}{4}\) chance, for a total probability of \(2 \cdot \dfrac{1}{4} = \dfrac{1}{2}.\)

Thus, C is the correct answer.

25.

On the dart board shown in the figure below, the outer circle has radius \(6\) and the inner circle has radius \(3\). Three radii divide each circle into three congruent regions, with point values shown. The probability that a dart will hit a given region is proportional to the area of the region. When two darts hit this board, the score is the sum of the point values of the regions hit. What is the probability that the score is odd?

\(\dfrac{17}{36}\)

\(\dfrac{35}{72}\)

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

\(\dfrac{37}{72}\)

\(\dfrac{19}{36}\)

Solution(s):

The area of the outer circle is \(6^2\pi = 36\pi,\) and the area of the inner circle is \(3^2\pi = 9\pi.\) Therefore, the area of the outer ring is \(36\pi - 9\pi = 27\pi.\)

This means that the probability of hitting an outer segment is \(\dfrac{9\pi}{36\pi} = \dfrac{1}{4}.\) Similarly, the probability of hitting an inner segment is \(\dfrac{3\pi}{36\pi} = \dfrac{1}{12}.\)

Summing over all possibilities, the probability of hitting a \(1\) is \[ \dfrac{1}{12} + 2 \cdot \dfrac{1}{4} = \dfrac{7}{12}. \] Similarly, the probability of hitting a \(2\) is \[ 2 \cdot \dfrac{1}{12} + \dfrac{1}{4} = \dfrac{5}{12}. \]

The only way to get an odd score is to hit one \(1\) and one \(2\).

The probability of hitting these numbers in either order is \[ 2 \cdot \dfrac{5}{12} \cdot \dfrac{7}{12} = \dfrac{35}{72}. \]

Thus, B is the correct answer.

Problems: https://live.poshenloh.com/past-contests/amc8/2007