2018 AMC 8 Exam Problems

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All of the real AMC 8 and AMC 10 problems in our complete solution collection are used with official permission of the Mathematical Association of America (MAA).

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1.

An amusement park has a collection of scale models, with a ratio of 1:20, 1: 20, of buildings and other sights from around the country. The height of the United States Capitol is 289289 feet. What is the height in feet of its replica at this park, rounded to the nearest whole number?

14 14

15 15

16 16

18 18

20 20

Answer: A
Video solution:
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Written solution:

Let the height of the replica be h.h. Since the ratio of the scale model to the real world is 1:20,1:20, we know that h:289=1:20h:289 = 1:20 Therefore: h289=120h=28920h=14.4514\begin{align*} \dfrac{h}{289}&= \dfrac{1}{20} \\ h&=\dfrac{289}{20} \\ h&=14.45 \approx 14\end{align*}Thus, the correct answer is A.

2.

What is the value of the product (1+11)(1+12)(1+13)(1+14)(1+15)(1+16)?\begin{align*} &\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\\ \cdot&\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)? \end{align*}

76 \dfrac{7}{6}

43 \dfrac{4}{3}

72 \dfrac{7}{2}

7 7

8 8

Answer: D
Video solution:
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Written solution:

Let's first note that if we are given an expression of the form 1+1n,1 + \frac{1}{n}, we can rewrite this as nn+1n=n+1n.\frac{n}{n} + \frac{1}{n} = \frac{n+1}{n}. With that in mind, we can rewrite the expression given to us in the problem, as shown below: (1+11)(1+12)(1+13)(1+14)(1+15)(1+16)=213243546576=7\begin{align*} &\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\\ \cdot&\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)\\ &=\dfrac{2}{1} \cdot \dfrac{3}{2} \cdot \dfrac{4}{3} \cdot\dfrac{5}{4} \cdot\dfrac{6}{5} \cdot\dfrac{7}{6}\\ &=7 \end{align*} Thus, the correct answer is D.

3.

Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?

Arn \text{Arn}

Bob \text{Bob}

Cyd \text{Cyd}

Dan \text{Dan}

Eve \text{Eve}

Answer: D
Video solution:
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Written solution:

Notice that the first 5 numbers that contains a 77 as its digit or are a multiple of 77 are 7,14,17,21,27.7,14,17,21,27. Any player who lands on one of these numbers must leave the circle.

With this in mind, let's start counting. Initially, we have all 66 people, starting with AA(Arn). After everyone says a number, AA must say 7,7, so he leaves the circle.

The circle now has 55 members: BB (Bob), CC (Cyd), DD (Dan), EE (Eve), and FF (Fon) -- with BB restarting his counting at 8.8. Everyone in the circle counts without incident, and it loops around such that Bob says 13.13. However, this leaves Cyd to say 14,14, and he leaves the circle.

The circle now has 44 members: B,D,E,FB,D,E,F -- with DD continuing the counting at 15.15. EE says 16,16, and FF says 17,17, and therefore leaves the circle.

The circle now has 33 members: B,D,EB,D,E -- with BB continuing the counting at 18.18. The counting loops around, and BB says 21,21, and therefore leaves the circle.

The circle now has 22 members: D,ED,E -- with DD starting at 22.22. They go back and forth, with DD saying even numbers and EE saying odd numbers. As such, eventually, EE must say 27,27, and as such, leaves the circle. This makes DD -- Dan -- the last one left in the circle.

Thus, D is the correct answer.

4.

The twelve-sided figure shown has been drawn on 1 cm×1 cm1 \text{ cm}\times 1 \text{ cm} graph paper. What is the area of the figure in cm2?\text{cm}^2?

12 12

12.5 12.5

13 13

13.5 13.5

14 14

Answer: C
Video solution:
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Written solution:

To solve for the area of the figure, we seperate the compound shape into parts that are easier to work with, as such:

As is now clear, there is the center 3×33 \times 3 square (in pink), with 44 smaller triangles surrounding it (in blue).

The area of the square is 33=9.3 \cdot 3 = 9. The other triangles each have a base of 22 and a height of 1,1, so their area is equal to bh2=212=1.\dfrac{bh}{2} = \dfrac{2\cdot 1}{2} =1 . There are 44 of these triangles, so their total area is 14=4.1\cdot 4 = 4.

Therefore, the total area is 9+4=13.9+4 = 13.

Thus, the correct answer is C.

5.

What is the value of 1+3+5++2017+201924620162018?\begin{align*} &1+3+5+\cdots+2017+2019 \\ -&2-4-6-\cdots-2016-2018? \end{align*}

1010 -1010

1009 -1009

1008 1008

1009 1009

1010 1010

Answer: E
Video solution:
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Written solution:

Rearranging the terms, notice that the expression in the question is equal to: 1+(32)+(54)++(20172016)+(20192018).\begin{align*}&1 + (3-2) + (5-4) + \cdots +\\ &(2017-2016) + (2019-2018). \end{align*} Each term is equal to 1,1, and there are 201912+1=1010\frac{2019-1}2+1 = 1010 terms, so the total sum is 10101=1010.1010\cdot1 = 1010.

Thus, E is the correct answer.

6.

On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?

50 50

70 70

80 80

90 90

100 100

Answer: C
Video solution:
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Written solution:

Anh drove 1010 miles on the coastal road in 3030 minutes. Therefore, his speed on the coastal road (notated as vcv_c) is: vc=10miles30minutes=1mile3minutes=13milesminute.\begin{align*}v_c&=\dfrac{10\text{miles}}{30 \text{minutes}}\\ &= \dfrac{1\text{mile}}{3 \text{minutes}}\\ &=\dfrac13 \dfrac{\text{miles}}{\text{minute}}.\end{align*} Since he drives 33 times as fast on highway (i.e. vh=3vcv_h=3v_c), he drives at a speed of 313milesminute=1milesminute3\cdot \dfrac13 \dfrac{\text{miles}}{\text{minute}} = 1 \dfrac{\text{miles}}{\text{minute}} on the highway. Armed with these two facts, we know that Anh drove for 3030 minutes on the coastal road, and he drove 5050 miles at 11 mile per minute. This means it takes 5050 minutes to drive the 5050 miles on the highway.

As such, the total travel time is 50+30=8050+30=80 minutes.

Thus, the correct answer is C.

7.

The 55-digit number 2\underline{2} 0\underline{0} 1\underline{1} 8\underline{8} U\underline{U} is divisible by 9.9. What is the remainder when this number is divided by 8?8?

1 1

3 3

5 5

6 6

7 7

Answer: B
Video solution:
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Written solution:

Notice that a number is divisible by 99 if and only if the sum of its digits is also divisible by 9.9.

The sum of the digits of the 5-digit number in the problem is: 2+0+1+8+U=11+U.2+0+1+8+U= 11+U. As 2\underline{2} 0\underline{0} 1\underline{1} 8\underline{8} U\underline{U} is divisble by 9,9, 11+U11+U must also be divisible by 9.9. Also, as UU is a digit, we know that inuitively: 0U9.0\le U\le 9. This means that UU can only be 7.7.

Now we know that the 5-digit number in question is 20187, and we want to find the remainder when we divide 20187 by 8. To solve this, simply use long division to see that 20187=25238+3.20187=2523\cdot 8 + 3. Therefore, the remainder is 3.3.

Thus, the correct answer is B.

8.

Mr. Garcia asked the members of his health class how many days last week they exercised for at least 30 minutes. The results are summarized in the following bar graph, where the heights of the bars represent the number of students.

What was the mean number of days of exercise last week, rounded to the nearest hundredth, reported by the students in Mr. Garcia's class?

3.50 3.50

3.57 3.57

4.36 4.36

4.50 4.50

5.00 5.00

Answer: C
Video solution:
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Written solution:

Counting the number of each occurence, we can see that there are 1 11s, 3 22s, 2 33s, 6 44s, 8 55s, 3 66s, and 2 77s.

Therefore, there are 1+3+2+6+8+3+2=251+3+2+6+8+3+2 = 25 students in total.

Therefore, the total number of days of exercise is 11+32+23+64+85+36+27=109\begin{align*}&1\cdot1+3\cdot2 + 2\cdot 3 + 6\cdot 4 \\ &+ 8\cdot 5 + 3\cdot 6 + 2\cdot 7 \\ &= 109\end{align*}

The mean number of days of exercise is defined as being equal to the total number of days of exercise divides by the total number of students, where: \dfrac{\text{total days}}{\text{# students}} = \dfrac{109}{25} \approx 4.36

Thus, C is the correct answer.

9.

Tyler is tiling the floor of his 12 foot by 16 foot living room. He plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will he use?

48 48

87 87

91 91

96 96

120 120

Answer: B
Video solution:
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Written solution:

Note that each square foot of the border would require one tile, meaning that the border will take 16+12+16+12=5616+12+16+12=56 tiles. However, notice that this will cause overlapping tiles in each of the four corners, so to fix this, we subtract 4.4. Therefore, the border will take 564=5256-4=52 1×11\times 1 square tiles to completely tile.

Since we have removed one foot from each side due to the border, the remaining rectangle is 10 feet×14 feet.10 \text{ feet} \times 14\text{ feet}. This must be tiled completely by 2×22 \times 2 tiles, so it will take 101422=35 \dfrac{10\cdot14}{2\cdot2} = 35 tiles in total to tile this area.

As it takes 5252 1×11\times 1 square tiles to tile the border, and 3535 2×22\times 2 square tiles to tile the remaining area, it will take 52+35=8752+35=87 tiles in total to fill in Tyler's entire living room floor.

Thus, the correct answer is B.

10.

The harmonic mean of a set of non-zero numbers is the reciprocal of the average of the reciprocals of the numbers. What is the harmonic mean of 1, 2, and 4?

37 \dfrac{3}{7}

712 \dfrac{7}{12}

127 \dfrac{12}{7}

74 \dfrac{7}{4}

73 \dfrac{7}{3}

Answer: C
Video solution:
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Written solution:

The reciprocals of 1,2, and 41,2, \text{ and }4 are 11,12, and 14\dfrac11, \dfrac12, \text{ and } \dfrac14 respectively. The average of these reciprocals is (1+12+14)3=(74)3=712.\begin{align*}\dfrac{\left(1 + \dfrac12 + \dfrac14\right)}{3} &= \dfrac{\left(\dfrac{7}{4}\right)}{3} \\&= \dfrac{7}{12}. \end{align*}

As the harmonic mean is the reciprocal of the average of the reciprocals of the numbers (which we just calculated to be 712\dfrac{7}{12}), we conclude that the harmonic mean is 127.\dfrac{12}{7}.

Thus, the correct answer is C.

11.

Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown. \begin{array}[ccc] \text{\text{X}}&\text{X}&\text{X} \\ \text{X}&\text{X}&\text{X} \end{array} If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?

13 \dfrac{1}{3}

25 \dfrac{2}{5}

715 \dfrac{7}{15}

12 \dfrac{1}{2}

23 \dfrac{2}{3}

Answer: C
Video solution:
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Written solution:

We can split the problem into two cases. In case 1, Abby is in one of the middle two seats, and in case 2, she is in one of the outer 4 seats.

Firstly notice that there is a 26=13 \dfrac26 = \dfrac13 probability of case 1 being true (i.e. Abby is in the middle two seats). For Bridget to be adjacent to Abby in this case, she must be in either of the two seats on the left or the two seats on the right of Abby, or she is in the same column as her. There are 33 ways to make this happen out of a possible 55 open seats, so there is a 35 \frac35 chance of this happening. Therefore, the total probability of this case is 1335=315. \dfrac13 \cdot \dfrac35 = \dfrac{3}{15} .

Next, notice that there is a 46=23 \dfrac46 = \dfrac23 probability of case 2 being true (i.e. Abby is in the outer four sets). For Bridget to be adjacent to Abby in this case, she must either be in the single seat next (to the left or right, depending on Abby's position), or she is in the same column as Abby. There are 22 ways to make this happen out of a possible 55 open seats, so there is a 25 \frac25 chance of this happening. Therefore, the total probability of this case is 2325=415. \frac23 \cdot \frac25 = \frac{4}{15} .

Therefore, the final probability of either of these cases happening is 315+415=715. \dfrac{3}{15} + \dfrac{4}{15} = \dfrac{7}{15} .

Thus, C is the correct answer.

12.

The clock in Sri's car, which is not accurate, gains time at a constant rate. One day as he begins shopping he notes that his car clock and his watch (which is accurate) both say 12:00 noon. When he is done shopping, his watch says 12:30 and his car clock says 12:35. Later that day, Sri loses his watch. He looks at his car clock and it says 7:00. What is the actual time?

5 5:5050

6 6:0000

6 6:3030

6 6:5555

8 8:1010

Answer: B
Video solution:
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Written solution:

Starting from 1212:0000 noon, after 3030 minutes of time elapsed, the car clock went 3535 minutes ahead.

Therefore, for every minute the car clock goes ahead, 3035=67 \frac {30}{35} = \frac 67 minutes of actual time pass by. From the time 1212:0000 to 77:00,00, the car goes ahead 760=4207\cdot 60 = 420 minutes, and therefore, 42067=360420 \cdot \frac67 = 360 minutes, or 66 hours, of actual time have passed by. If we start at 1212:0000 and 66 hours pass by, the time is 66:00.00 .

Thus, B is the correct answer.

13.

Laila took five math tests, each worth a maximum of 100100 points. Laila's score on each test was an integer between 00 and 100,100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82.82. How many values are possible for Laila's score on the last test?

4 4

5 5

9 9

10 10

18 18

Answer: A
Video solution:
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Written solution:

Since the average score on the five tests is 82,82, the total score of those five tests must be 582=410.5\cdot82 = 410 .

Now, let ff be the score on the first 4 tests and let ll be the score for the last test.

We know that f<l<100f < l < 100 and 4f+l=410.4f + l = 410. And as 410=4f+l<5l,410 = 4f + l < 5l , we know 4105=82<l.\frac{410}{5} = 82 < l .

Also, since 4f+l=410,4f + l = 410 , and dividing 410410 by 44 gives us a remainder of 2, we know that dividing ll by 44 must leave a remainder of 22 as 4f4f will leave no remainder when divided by 4.4. Equivalently: l2mod4.l \equiv 2 \mod 4 . Since 82<l<10082 < l < 100 and l2mod4,l \equiv 2 \mod 4, the only options for ll are 86,90,94,98.86,90,94,98. This yields four distinct solutions as follows: (f,l)=(81,86);(80,90);(79,94);(78,98)\begin{align*} (f,l) =& (81,86);\\ &(80,90);\\ &(79,94);\\ &(78,98) \end{align*} Therefore, there are 44 solutions, and A is the correct answer.

14.

Let NN be the greatest five-digit number whose digits have a product of 120.120. What is the sum of the digits of N?N?

15 15

16 16

17 17

18 18

20 20

Answer: D
Video solution:
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Written solution:

To make the largest possible 55 digit number, we must maximize the first digit (the digit in the ten-thousands place).

The largest number that is strictly less than 1010 and divides 120120 is 8,8, so the first digit must be 8.8. Therefore, the product of the remaining number is 15.15.

Similarly, we must now maximize the second digit.

The largest number that is less than 1010 and divides 1515 is 5,5, so the second digit is 5.5. Therefore, the product of the remaining number is 3.3.

We must then maximize the third digit.

The largest number that is less than 1010 and divides 33 is 3,3, so the third digit is 3.3. Therefore, the product of the remaining number is 1.1. This means the 4th and 5th digits are 1.1.

This makes N=85311,N = 85311, so the sum of the digits is 8+5+3+1+1=188+5+3+1+1=18

Thus, D is the correct answer.

15.

In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of 11 square unit, then what is the area of the shaded region, in square units?

14 \dfrac{1}{4}

13 \dfrac{1}{3}

12 \dfrac{1}{2}

1 1

π2 \dfrac{\pi}{2}

Answer: D
Video solution:
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Written solution:

Let AA be the area of the large circle.

Since the diameter of each of the two smaller circles is itself the radius of the larger circle, the radius of the smaller circles half that of the larger circle.

Symbolically, if we allow rlr_l to be the radius of the large circle and rsr_s to be the radius of each of the smaller circles: rs=12rlr_s = \dfrac12 r_l As the area of the larger circle is equal to A=πrl2,A=\pi r_l^2, the area of the smaller circles are equal to πrs2=π(12rl)2=14(πrl2)=14A.\begin{align*}\pi r_s^2 &= \pi \left(\dfrac12 r_l\right)^2 \\&= \dfrac14 (\pi r_l^2)\\&=\dfrac14 A.\end{align*} As the area of two of these smaller circles combined is equal to 1 square unit, then it follows that 214A=12\cdot \dfrac14 A=1 square unit, implying that A=2A=2 square units.

As the area of the shaded region is equal to the area of the larger circle (A)(A) minus the combined area of the two smaller circles (1),(1), the area of the shaded region is A1=21=1 A - 1=2-1=1 square unit.

Thus, the correct answer is D

16.

Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?

1440 1440

2880 2880

5760 5760

182,440 182,440

362,880 362,880

Answer: C
Video solution:
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Written solution:

Since we are keeping the Arabic books together and the Spanish books together, we can look at each of them as the same object (collections of books).

As such, there are 5 objects on the bookshelf: three German books, one collection of Arabic books, and one collection of Spanish books. There are 5!5! ways to order the 55 objects. As we already have the books together, there are 2!2! ways of ordering the Arabic books and 4!4! ways of ordering the Spanish books. Therefore, the total ways to order the books is 5!4!2!=120242=5760\begin{align*}5! \cdot 4! \cdot 2! &= 120 \cdot 24 \cdot 2 \\&= 5760 \end{align*}

Thus, the correct answer is C.

17.

Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 55 times as fast as Bella walks. The distance between their houses is 22 miles, which is 10,56010,560 feet, and Bella covers 2122 \tfrac{1}{2} feet with each step. How many steps will Bella take by the time she meets Ella?

704 704

845 845

1056 1056

1760 1760

3520 3520

Answer: A
Video solution:
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Written solution:

Since for every foot Bella walks, Ella walks 5 feet, we know that Bella will walk 16\frac16 of the cumulative distance between the two of them, and so she walks 1610560=1760\dfrac16 \cdot 10560=1760 feet. Since she walks 2.52.5 feet per step, she takes 17602.5=704\dfrac{1760}{2.5} = 704 steps by the time she meets Ella.

Thus, A is the correct answer.

18.

How many positive factors does 23,232 have?

9 9

12 12

28 28

36 36

42 42

Answer: E
Video solution:
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Written solution:

Begin by finding the prime factorization of 23232.23232. To do this, we repeatedly factor out the smallest prime factor from the number, a process that terminates when the number is a prime number. This process is outlined below: 23232=211616=225808=232904=241452=25726=26363=263121=263112\begin{align*}23232 &= 2\cdot 11616\\ &= 2^2\cdot 5808\\ &= 2^3\cdot 2904\\ &= 2^4\cdot 1452\\ &= 2^5\cdot 726\\ &= 2^6\cdot 363\\ &= 2^6\cdot 3 \cdot 121\\ &=2^6\cdot 3\cdot 11^2 \end{align*} An arbitrary factor of 2323223232 can be created by taking the product of any number of prime factors. More explicitly, as 2323223232 can be represented p1e1p2e2pmemp_1^{e_1} p_2^{e_2} \cdots p_m^{e_m} where pp is a prime number, each factor has (e1+1)(em+1)(e_1+1) \cdots (e_m+1) options of prime factorizations to choose from, and thus, there are (e1+1)(em+1)(e_1+1) \cdots (e_m+1) factors. Plugging in values, we can see that there are (6+1)(1+1)(2+1)=723=42\begin{align*}(6+1)(1+1)(2+1) &= 7\cdot2\cdot3\\ &= 42 \end{align*} factors of 23232.23232.

Thus, E is the correct answer.

19.

In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid?

2 2

4 4

8 8

12 12

16 16

Answer: C
Video solution:
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Written solution:

Suppose we have two cells and the cell above them. If we are given the bottom left cell and the top cell, we can always find the bottom right cell as follows:

If the top cell is +,+, then the bottom right cell must be the same as the bottom left cell, and if the top cell is ,-, the bottom right cell must be the opposite of the bottom left cell.

Now, suppose we are given a row. Then, suppose we choose a value for the cell below and to the left of the leftmost cell in our given row. We then can inductively determine the entire row below our given by first finding the bottom-right cell of the leftmost cell in our row, and using that newly found cell as the bottom-left reference for the second to the left cell in the given row to find its bottom-right counterpart. The process continues on until the row below the given row is fully solved.

Therefore, since we know that the top row has a cell labelled +,+, we have 22 choices for the row below -- depending on our choice of the bottom-left cell. Similarly, we have 22 choices for the third row, and thus 22 choices for the fourth row. This makes 222=82*2*2 = 8 total choices for the bottom row of the sign pyramid.

Thus, the correct answer is C.

20.

In ABC,\triangle ABC, a point EE is on AB\overline{AB} with AE=1AE=1 and EB=2.EB=2. Point DD is on AC\overline{AC} so that DEBC\overline{DE} \parallel \overline{BC} and point FF is on BC\overline{BC} so that EFAC.\overline{EF} \parallel \overline{AC}. What is the ratio of the area of CDEFCDEF to the area of ABC?\triangle ABC?

49 \dfrac{4}{9}

12 \dfrac{1}{2}

59 \dfrac{5}{9}

35 \dfrac{3}{5}

23 \dfrac{2}{3}

Answer: A
Video solution:
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Written solution:

Let the area of ABC\triangle ABC be equal to t.t. Since DEBCDE \parallel BC and FECA,FE \parallel CA , we can deduce that ADEABC and EFBABC.ADE \sim ABC\text{ and } EFB \sim ABC. Since AE=AB3,AE = \dfrac{AB}3, the area of ADEADE is equal to (13)2t=t9.\left(\dfrac13\right)^2 t = \dfrac{t}{9} . Since EB=AB3,EB = \dfrac{AB}3, the area of EFBEFB is equal to (23)2t=49t.\left(\dfrac23\right)^2 t = \dfrac{4}{9}t . Finally, to find the area of CDEF,CDEF, we take the area of ABC=tABC =t and subtract the areas of ADEADE and EFB.EFB. This is equivalent to the expression tt94t9=4t9.t- \frac{t}{9} - \frac{4t}{9} = \frac{4t}{9} . Therefore, the ratio of the area of CDEFCDEF and ABCABC is (4a9)a=49.\dfrac{\left(\dfrac{4a}{9}\right)}{a} = \dfrac{4}{9} .

Thus, A is the correct answer.

21.

How many positive three-digit integers have a remainder of 22 when divided by 6,6, a remainder of 55 when divided by 9, and a remainder of 77 when divided by 11?11?

1 1

2 2

3 3

4 4

5 5

Answer: E
Video solution:
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Written solution:

Suppose xx is a number that satisfies these conditions. We know that 100x999.100 \leq x \leq 999.

The first statement implies that x2mod64mod6\begin{align*}x &\equiv 2 \mod 6\\&\equiv -4 \mod 6\end{align*} This, in turn, implies that x+40mod6.x+4 \equiv 0 \mod 6 .

Similarly, the second statement implies that x5mod94mod9\begin{align*}x &\equiv 5 \mod 9 \\&\equiv -4 \mod 9 \end{align*} This, in turn, implies that x+40mod9.x+4 \equiv 0 \mod 9 .

Finally, the third statement implies that x7mod114mod11\begin{align*}x &\equiv 7 \mod 11 \\&\equiv -4 \mod 11\end{align*} This, in turn, implies that x+40mod11.x+4 \equiv 0 \mod 11.

Together, these three conditions mean that x+46,x+49,x+411x+4 |6, x+4 |9, x+4 |11 we know that x+4lcm(6,9,11).x+4 | \text{lcm}(6,9,11) . Therefore, x+4198.x+4 | 198. We also know 104x+41003,104 \leq x+4 \leq 1003 , so we can see that there are 55 possible values in this interval such that x+4198.x+4 |198 .

Thus, E is the correct answer.

22.

Point EE is the midpoint of side CD\overline{CD} in square ABCD,ABCD, and BE\overline{BE} meets diagonal AC\overline{AC} at F.F. The area of quadrilateral AFEDAFED is 45.45. What is the area of ABCD?ABCD?

100 100

108 108

120 120

135 135

144 144

Answer: B
Video solution:
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Written solution:

Let HH be the point on BC\overline{BC} where the altitude from FF to BC\overline{BC} meets BC.\overline{BC}. This altitude, FH\overline{FH} is illustrated above. Then, by angle-angle similarity, we can see that CABCFH\triangle CAB \sim \triangle CFH and BFHBEC. \triangle BFH \sim \triangle BEC .

Since the sides of similar triangles are proportional, we know that FHBH=ECBC\dfrac{FH}{BH} = \dfrac{EC}{BC}andFHHC=ABBC.\dfrac{FH}{HC} = \dfrac{AB}{BC} . Thus, FHEC=BHBC\dfrac{FH}{EC} = \dfrac{BH}{BC}andFHAB=HCBC.\dfrac{FH}{AB} = \dfrac{HC}{BC} . Adding these equations yields: FHEC+FHAB=BHBC+HCBC=BH+HCBC=BCBC=1.\begin{align*}\dfrac{FH}{EC} + \dfrac{FH}{AB} &= \dfrac{BH}{BC} + \dfrac{HC}{BC}\\ &= \frac{BH+HC}{BC} \\&= \frac{BC}{BC} \\&= 1.\end{align*} This, in turn, goes to show that 1EC+1AB=1FH.\frac{1}{EC} + \frac{1}{AB} = \frac{1}{FH} .

Now, let ss be the side length of the square. We know AB=2EC=s.AB = 2\cdot EC = s. This means 1FH=1EC+1AB=1a+2a=3a.\begin{align*}\dfrac{1}{FH} &= \dfrac{1}{EC} + \frac{1}{AB} \\&= \frac{1}{a} + \frac{2}{a} \\&= \frac{3}{a} .\end{align*} Therefore, FH=s3.FH = \frac{s}{3} .

Now, to compute the area of EFC,\triangle EFC, we take the area of BCE\triangle BCE and subtract the area of BFC.\triangle BFC. This is equal to

BCEC2BCFH2=BC(ECFH)2=s(s2s3)2=ss62=s212.\begin{align*} \dfrac{BC\cdot EC}{2} - \dfrac{BC\cdot FH}{2} &= \dfrac{BC\cdot(EC-FH)}{2} \\ &= \dfrac{s\cdot\left(\dfrac{s}{2} - \dfrac{s}{3}\right)}{2} \\ &= \dfrac{s\cdot\frac{s}{6}}{2} \\ &= \dfrac{s^2}{12}. \end{align*}

The area of AFEDAFED is the area of ACD\triangle ACD minus the area of EFC,\triangle EFC, which is equal tos22s212=5s212=45.\begin{align*} \dfrac{s^2}{2}-\dfrac{s^2}{12} &= \dfrac{5s^2}{12} \\&= 45.\end{align*} With 512s2=45,\frac{5}{12} s^2 = 45 , we get s2=108,s^2 = 108, which is the area of the full square.

23.

From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?

27 \dfrac{2}{7}

542 \dfrac{5}{42}

1114 \dfrac{11}{14}

57 \dfrac{5}{7}

67 \dfrac{6}{7}

Answer: D
Video solution:
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Written solution:

Without loss of generality, allow AA to be a vertex of the triangle. Suppose we also have points B,CB,C of the triangle with A,B,CA,B,C being in clockwise order. Let xx be the number of points (i.e. vertices of the octagon) in between the points of AA and B,B, yy be the number of points in between the points of BB and C,C, and zz be the number of points in between the points of BB and C.C. We know x+y+z=5x+y+z = 5 as it encompasses every vertex of the octagon except A,B,C.A,B,C.

If two sides form the sides of an octagon, the distance between them would be 0.0.

Therefore, if we use complementary counting to find how many have x,y,z>0,x,y,z > 0, we can deduce out how many triangles are formed with no sides of the triangle being a side of the octagon. This would make x,y,zx,y,z whole numbers whose sum is 5.5. Using the stars and bars method, we can see that there are (5131)=6 \binom {5-1}{3-1} = 6 ways to place B,CB,C such that x,y,z>0.x,y,z > 0. Now to find the total number of cases, since there are 77 points that aren't A,A, there are (72)=21\binom{7}{2} = 21 ways to place B,CB,C such that they run counter clockwise.

This means there is a 621=27\dfrac{6}{21} = \dfrac{2}{7} probability of the triangle not having sides on the octagon. Therefore, there is a 127=571- \dfrac{2}{7} = \dfrac{5}{7} probability of the triangle having at least one side on the octagon.

Thus, D is the correct answer.

24.

In the cube ABCDEFGHABCDEFGH with opposite vertices CC and E,E, JJ and II are the midpoints of segments FB\overline{FB} and HD,\overline{HD}, respectively. Let RR be the ratio of the area of the cross-section EJCIEJCI to the area of one of the faces of the cube. What is R2?R^2?

54 \dfrac{5}{4}

43 \dfrac{4}{3}

32 \dfrac{3}{2}

2516 \dfrac{25}{16}

94 \dfrac{9}{4}

Answer: C
Video solution:
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Written solution:

Allow ss to represent the length of an edge of the cube. Noting that each side of the cross section is equal in length, we conclude that EJCIEJCI is a rhombus. The area of this rhombus can be calculated as 12IJCE,\frac12 IJ\cdot CE, as the area of a rhombus is equal to half the product of its diagonals. Using the Pythagorean Theorem: IJ=FH=s2.IJ=FH=s\sqrt{2}. Similarly, using the Pythagorean Theorem again lets us see that: CE=AC2+AE2=(s2)2+s2=2s2+s2=s3\begin{align*}CE&=\sqrt{AC^2+AE^2}\\&=\sqrt{(s\sqrt{2})^2+s^2}\\&=\sqrt{2s^2+s^2}\\&=s\sqrt{3}\end{align*} Therefore, R=12IJCEs2=12s223s2=32\begin{align*}R&=\dfrac{\frac12 IJ\cdot CE}{s^2}\\&=\dfrac{\frac12 s^2\sqrt{2}\sqrt{3}}{s^2}\\&=\sqrt{\dfrac32}\end{align*} Thus, R2=32,R^2=\dfrac32, and the correct answer is C.

25.

How many perfect cubes lie between 28+12^8+1 and 218+1,2^{18}+1, inclusive?

4 4

9 9

10 10

57 57

58 58

Answer: E
Video solution:
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Written solution:

Suppose xx is any perfect cube in this range.

If x3218+1,x^3 \leq 2^{18}+1, then x3=218+1x^3 = 2^{18}+1 or x3218=263x^3 \leq 2^{18} = {2^6}^3    x26=64. \implies x \leq 2^6 = 64 . If x3=218+1=643+1,x^3 = 2^{18}+1 = 64^3 +1, then it follows that 643<x3<65364^3 < x^3 < 65^3     64<x<65\implies 64 < x < 65 This would mean that xx isn't an integer. This is a contradiction, so we know x64.x \leq 64 . We also know 28+1=257x3.2^8+1= 257 \leq x^3 . Now, suppose 257x3<343. 257\leq x^3 < 343. Then, we know 216<x3<343.216 < x^3 < 343 .

This means 6<x<7,6 < x < 7 , which also means that xx isn't an integer. This is a contradiction, so 7x.7\le x.

Therefore, all xx which satisfy 343x3(26)3343 \leq x^3 \leq (2^6)^3 must also satisfy 7x64.7 \leq x \leq 64 . Therefore, the number of possible xx's is 647+1=58.64-7+1 = 58.

Thus, E is the correct answer.