2019 AMC 8 Exam Solutions

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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.

Ike and Mike go into a sandwich shop with a total of $30.00 to spend. Sandwiches cost $4.50 each and soft drinks cost $1.00 each. Ike and Mike plan to buy as many sandwiches as they can, and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how many items will they buy?

66

77

88

99

1010

Solution:

If they buy 66 sandwiches, they would spend 6×4.5=276 \times 4.5 = 27 dollars. This means that they would not be able to buy any more sandwiches.

They would then have 3027=330 - 27 = 3 dollars left. With this, they could buy 33 sodas. They would therefore buy a total of 6+3=96 + 3 = 9 items.

Thus, the correct answer is D.

2.

Three identical rectangles are put together to form rectangle ABCD,ABCD, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 55 feet, what is the area in square feet of rectangle ABCD?ABCD?

4545

7575

100100

125125

150150

Solution:

From the figure, we can see that the longer side has the same length as two of the shorter sides. This makes it 25=102 \cdot 5 = 10 feet long.

This tells us that BC=10BC = 10 feet and DC=10+5=15DC = 10 + 5 = 15 feet. Therefore, the area would be 1015=150 ft.210 \cdot 15 = 150 \text{ ft.}^2

Thus, the correct answer is E.

3.

Which of the following is the correct order of the fractions 1511,1915,\dfrac{15}{11},\dfrac{19}{15}, and 1713,\dfrac{17}{13}, from least to greatest?

1511<1713<1915\dfrac{15}{11} \lt \dfrac{17}{13} \lt \dfrac{19}{15}

1511<1915<1713\dfrac{15}{11} \lt \dfrac{19}{15} \lt \dfrac{17}{13}

1713<1915<1511\dfrac{17}{13} \lt \dfrac{19}{15} \lt \dfrac{15}{11}

1915<1511<1713\dfrac{19}{15} \lt \dfrac{15}{11} \lt \dfrac{17}{13}

1915<1713<1511\dfrac{19}{15} \lt \dfrac{17}{13} \lt \dfrac{15}{11}

Solution:

We can rewrite the fraction as follows: 1511=1+4111713=1+4131915=1+415.\begin{align*} \dfrac{15}{11} &= 1 + \dfrac{4}{11} \\ \dfrac{17}{13} &= 1 + \dfrac{4}{13} \\ \dfrac{19}{15} &= 1 + \dfrac{4}{15}. \end{align*}

Recall that if two fractions have the same numerator, then the fraction with the larger denominator is smaller.

Using this fact, we can see that the correct ordering is 1915<1713<1511. \dfrac{19}{15} \lt \dfrac{17}{13} \lt \dfrac{15}{11}.

Thus, the correct answer is E.

4.

Quadrilateral ABCDABCD is a rhombus with perimeter 5252 meters. The length of diagonal AC\overline{AC} is 2424 meters. What is the area in square meters of rhombus ABCD?ABCD?

6060

9090

105105

120120

144144

Solution:

We can split the rhombus up into 44 triangles, each of which looks like this.

We know that this is a right triangle, since the diagonals of a rhombus are perpendicular, and we know the hypotenuse is 52÷4=13.52 \div 4 = 13.

Using the Pythagorean theorem, we get the other leg to be 132122=25=5. \sqrt{13^2 - 12^2} = \sqrt{25} = 5.

This means that the area of the rhombus is 45122=430=120. 4 \cdot \dfrac{5 \cdot 12}{2} = 4 \cdot 30 = 120. Thus, the correct answer is D.

5.

A tortoise challenges a hare to a race. The hare eagerly agrees and quickly runs ahead, leaving the slow-moving tortoise behind. Confident that he will win, the hare stops to take a nap. Meanwhile, the tortoise walks at a slow steady pace for the entire race. The hare awakes and runs to the finish line, only to find the tortoise already there. Which of the following graphs matches the description of the race, showing the distance dd traveled by the two animals over time tt from start to finish?

Solution:

The hare's path is the one with the horizontal line in the middle, which represent the period where the hare took a nap.

We also know that the time it takes the hare to reach the finish line is longer than the time it took the tortoise.

This means that the end of the hare's path should have a larger xx-value. This is exactly graph B.B.

Thus, the correct answer is B.

6.

There are 8181 grid points (uniformly spaced) in the square shown in the diagram below, including the points on the edges. Point PP is in the center of the square. Given that point QQ is randomly chosen among the other 8080 points, what is the probability that the line PQPQ is a line of symmetry for the square?

15\displaystyle \dfrac{1}{5}

14\displaystyle \dfrac{1}{4}

25\displaystyle \dfrac{2}{5}

920\displaystyle \dfrac{9}{20}

12\displaystyle \dfrac{1}{2}

Solution:

Note the only lines of symmetry for a square are the two diagonals and the two lines connecting opposite midpoints.

Therefore, QQ must be a point on any one of these four lines. Each line consists of 99 points, for a total of 49=364 \cdot 9 = 36 points.

Note that QQ cannot be the same as P,P, so we have to subtract out the 44 points which coincide with P,P, for a total of 364=3236 - 4 = 32 points.

The probability is therefore 3280=25.\dfrac{32}{80} = \dfrac{2}{5}.

Thus, the correct answer is C.

7.

Shauna takes five tests, each worth a maximum of 100100 points. Her scores on the first three tests are 76,76, 94,94, and 87.87. In order to average 8181 for all five tests, what is the lowest score she could earn on one of the other two tests?

4848

5252

6666

7070

7474

Solution:

To minimize one of the scores, we have to maximize the other score. Assume that Shauna gets a 100100 on her fourth test.

The sum of the 44 tests is then 76+94+87+100=357. 76 + 94 + 87 + 100 = 357. For an average of 81,81, Shauna's test scores must add to 581=405.5 \cdot 81 = 405. This means that she needs to get a 405357=48405 - 357 = 48 on her last test.

Thus, the correct answer is A.

8.

Gilda has a bag of marbles. She gives 20%20\% of them to her friend Pedro. Then Gilda gives 10%10\% of what is left to another friend, Ebony. Finally, Gilda gives 25%25\% of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself?

2020

331333\dfrac{1}{3}

3838

4545

5454

Solution:

Assume that Gilda starts off with 100100 marbles. After giving 2020 marbles to Pedro, she has 10020=80100 - 20 = 80 marbles left.

She then gives 80×.1=880 \times .1 = 8 marbles to Ebony, with her having 808=7280 - 8 = 72 marbles left.

Finally, Gilda gives 72×.25=1872 \times .25 = 18 marbles to Jimmy. She has a total of 7218=5472 - 18 = 54 marbles left, which is 54%54\% of her original total.

Thus, the correct answer is E.

9.

Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are 66 cm in diameter and 1212 cm high. Felicia buys cat food in cylindrical cans that are 1212 cm in diameter and 66 cm high. What is the ratio of the volume of one of Alex's cans to the volume of one of Felicia's cans?

1:41:4

1:21:2

1:11:1

2:12:1

4:14:1

Solution:

Felicia's diameter is twice as much as Alex's which means that her radius is also twice as much.

Recall that the formula for the area of a circle is πr2.\pi r^2. If the radius is doubled, then the area the is quadrupled.

This means that the base of Felicia's can has 44 times the area as the base of Alex's can.

We also see that Alex's can is twice as tall as Felicia's can. The formula for the volume of a can is base times height.

Using this formula, we see that Felicia's can will have twice the volume of Alex's can since 4÷2=2.4 \div 2 = 2.

Thus, the correct answer is B.

10.

The diagram shows the number of students at soccer practice each weekday during last week. After computing the mean and median values, Coach discovers that there were actually 2121 participants on Wednesday. Which of the following statements describes the change in the mean and median after the correction is made?

The mean increases by 11 and the median does not change.

The mean increases by 11 and the median increases by 1.1.

The mean increases by 11 and the median increases by 5.5.

The mean increases by 55 and the median increases by 1.1.

The mean increases by 55 and the median increases by 5.5.

Solution:

There are 55 more participants, which means the mean is increased by 5÷5=1.5 \div 5 = 1.

Right now, we can see that median 20.20. If 1616 gets replaced with 21,21, then the median becomes 21.21.

This shows that the median is also increased by 1.1.

Thus, the correct answer is B.

11.

The eighth grade class at Lincoln Middle School has 9393 students. Each student takes a math class or a foreign language class or both. There are 7070 eighth graders taking a math class, and there are 5454 eight graders taking a foreign language class. How many eight graders take only a math class and not a foreign language class?

1616

2323

3131

3939

7070

Solution:

The number of kids that are taking both classes is 70+5493=31. 70 + 54 - 93 = 31. Subtracting this from the number of kids taking a math class will give us the number of kids taking only a math class.

The desired number is 7031=39.70 - 31 = 39.

Thus, the correct answer is D.

12.

The faces of a cube are painted in six different colors: red, white, green, purple, yellow, and black. Three views of the cube are shown below. What is the color of the face opposite the yellow face?

red

white

green

purple

black

Solution:

Using the first and third cubes, we can see that black is opposite purple (both have red and green connected).

Similarly, with the first and second cubes, we see that green and white are opposites.

This means that yellow must be opposite red.

Thus, the correct answer is A.

13.

A palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let NN be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of N?N?

22

33

44

55

66

Solution:

Note that 22-digit palindromes have the tens and units digits the same. This means that they are all multiples of 11.11.

If a 33-digit number is the sum of 33 22-digit palindromes, then it itself must also be a multiple of 11.11.

The smallest 33-digit multiple of 1111 that is not a palindrome is 110.110. This can be achieved by adding 11+22+77.11 + 22 + 77.

Therefore, N=110.N = 110. The sum of its digits is 1+1+0=2.1 + 1 + 0 = 2.

Thus, the correct answer is A.

14.

Isabella has 66 coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every 1010 days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the 66 dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?

Monday

Tuesday

Wednesday

Thursday

Friday

Solution:

Let us go through the answer choices.

If we she starts on Monday, her next day is Thursday, and the day after is Sunday.

If we start with Tuesday, the next day is Friday. The day after is Monday, which as we now from above will hit Sunday in 22 more days.

If we start with Wednesday, the next day is Saturday. The day after is Tuesday, which we know hits Sunday only after 44 more iterations.

Therefore, if we start with Wednesday, we loop through 66 iterations with hitting Sunday.

Thus, the correct answer is C.

15.

On a beach 5050 people are wearing sunglasses and 3535 people are wearing caps. Some people are wearing both sunglasses and caps. If one of the people wearing a cap is selected at random, the probability that this person is also wearing sunglasses is 25.\dfrac{2}{5}. If instead, someone wearing sunglasses is selected at random, what is the probability that this person is also wearing a cap?

1485\displaystyle \dfrac{14}{85}

725\displaystyle \dfrac{7}{25}

25\displaystyle \dfrac{2}{5}

47\displaystyle \dfrac{4}{7}

710\displaystyle \dfrac{7}{10}

Solution:

If the probability is 25,\dfrac{2}{5}, that means 25\dfrac{2}{5} of the people wearing caps are also wearing sunglasses.

This means that 3525=1435 \cdot \dfrac{2}{5} = 14 people are wearing both caps and sunglasses.

The probability of a person wearing sunglasses also wearing a cap is 1450=725.\dfrac{14}{50} = \dfrac{7}{25}.

Thus, the correct answer is B.

16.

Qiang drives 1515 miles at an average speed of 3030 miles per hour. How many additional miles will he have to drive at 5555 miles per hour to average 5050 miles per hour for the entire trip?

4545

6262

9090

110110

135135

Solution:

For the first 1515 miles, Qiang drove for 1530=12\dfrac{15}{30} = \dfrac{1}{2} an hour.

Let xx be the distance that Qiang must drive for to satisfy the condition. It will take him x55\dfrac{x}{55} hours to drive this distance.

The total trip is now 15+x15 + x miles. We know the average speed is 5050 miles per hour, so this will take him x+1550\dfrac{x + 15}{50} hours.

Setting the two times equal to each other, we get 12+x55=x+1550. \dfrac{1}{2} + \dfrac{x}{55} = \dfrac{x + 15}{50}.

Cross-multiplying and simplifying yields 15+x=25+10x11x=110. 15 + x = 25 + \dfrac{10x}{11} \Rightarrow x = 110.

Thus, the correct answer is D.

17.

What is the value of the product below?

(1322)(2433)(3544)(97999898)(981009999)\begin{align*}\displaystyle & \left(\dfrac{1\cdot 3}{2\cdot 2}\right)\left(\dfrac{2\cdot 4}{3\cdot 3}\right)\left(\dfrac{3\cdot 5}{4\cdot 4}\right) \\ & \dots\left(\dfrac{97\cdot 99}{98\cdot 98}\right)\left(\dfrac{98\cdot 100}{99\cdot 99}\right) \end{align*}

12\displaystyle \dfrac{1}{2}

5099\displaystyle \dfrac{50}{99}

98009801\displaystyle \dfrac{9800}{9801}

10099\displaystyle \dfrac{100}{99}

5050

Solution:

We can regroup all the number as follows. 12(3223)(4334)(99989899)10099\begin{align*} & \dfrac{1}{2} \cdot \left(\dfrac{3 \cdot 2}{2\cdot 3}\right)\left(\dfrac{4\cdot 3}{3\cdot 4}\right) \\ & \dots\left(\dfrac{99\cdot 98}{98\cdot 99}\right) \cdot \dfrac{100}{99} \end{align*}

From this representation, we can see that all the middle terms cancel leaving only 1210099=5099. \dfrac{1}{2} \cdot \dfrac{100}{99} = \dfrac{50}{99}.

Thus, the correct answer is B.

18.

The faces of each of two fair dice are numbered 1,1, 2,2, 3,3, 5,5, 7,7, and 8.8. When the two dice are tossed, what is the probability that their sum will be an even number?

49\displaystyle \dfrac{4}{9}

12\displaystyle \dfrac{1}{2}

59\displaystyle \dfrac{5}{9}

35\displaystyle \dfrac{3}{5}

23\displaystyle \dfrac{2}{3}

Solution:

The only two ways that the sum can be even is if both rolls are even or both are odd.

The probability that a roll is even is 26=13,\dfrac{2}{6} = \dfrac{1}{3}, and the probability that it is odd is 46=23.\dfrac{4}{6} = \dfrac{2}{3}.

Therefore, the desired probability is 1313+2323=59. \dfrac{1}{3} \cdot \dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{2}{3} = \dfrac{5}{9}.

Thus, the correct answer is C.

19.

In a tournament there are six teams that play each other twice. A team earns 33 points for a win, 11 point for a draw, and 00 points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?

2222

2323

2424

2626

3030

Solution:

We can assume that the top 33 teams won every game against every team not amongst themselves.

They play 32=63 \cdot 2 = 6 games in total, getting a total of 63=186 \cdot 3 = 18 points.

Now, amongst the top 3,3, each team plays each other time twice. To even out the scores, we can let one team win one game and let the other team win the other game.

This means that for every pair of the top 33 teams, each team in the pair gets 33 points. There are 33 such pairs, with each team appearing in 22 pairs.

This means that each team will get an extra 32=63 \cdot 2 = 6 points. Therefore, their maximum score is 18+6=24.18 + 6 = 24.

Thus, the correct answer is C.

20.

How many different real numbers xx satisfy the equation

(x25)2=16? (x^2-5)^2=16?

00

11

22

44

88

Solution:

Recall that a2b2=(a+b)(ab).a^2 - b^2 = (a + b)(a - b). We can do the following rearrangement. (x25)242=0(x25+4)(x254)=0(x+1)(x1)(x+3)(x3)=0 \begin{gather*} (x^2 - 5)^2 - 4^2 = 0 \\ (x^2 - 5 + 4)(x^2 - 5 - 4) = 0 \\ (x + 1)(x - 1)(x + 3)(x - 3) = 0 \end{gather*}

From this we can see that there are 44 possible values for x.x.

Thus, the correct answer is D.

21.

What is the area of the triangle formed by the lines y=5,y=5, y=1+x,y=1+x, and y=1x?y=1-x?

44

88

1010

1212

1616

Solution:

First, let us find the vertices of the triangles. The intersections between y=5y = 5 and the other two lines are (4,5) and (4,5). (4, 5) \text{ and } (-4, 5).

To find the third intersection, we can equate the two equations to get 1+x=1xx=0. 1 + x = 1 - x \Rightarrow x = 0.

This gives us the point (0,1).(0, 1).

Note that the first two intersection points are mirrored across the yy-axis. This tells us that the triangle is isosceles.

The base is therefore 24=8,2 \cdot 4 = 8, and the height is 51=4.5 - 1 = 4. The area of the triangle is therefore 1284=16. \dfrac{1}{2} \cdot 8 \cdot 4 = 16.

Thus, the correct answer is E.

22.

A store increased the original price of a shirt by a certain percent and then decreased the new price by the same amount. Given that the resulting price was 84%84\% of the original price, by what percent was the price increased and decreased?

1616

2020

2828

3636

4040

Solution:

Let xx be the percent converted to a real number. Then the percent increase would multiply the original by 1+x.1 + x.

The percent decrease would multiply the new price by 1x.1 - x. The final price will the be the original price multiplied by (1+x)(1x)=1x2=.84. (1 + x)(1 - x) = 1 - x^2 = .84.

This tells us that x2=.16x=.4. x^2 = .16 \Rightarrow x = .4.

This tells us that the percent would be 40%.40\%.

Thus, the correct answer is E.

23.

After Euclid High School's last basketball game, it was determined that 14\dfrac{1}{4} of the team's points were scored by Alexa and 27\dfrac{2}{7} were scored by Brittany. Chelsea scored 1515 points. None of the other 77 team members scored more than 22 points. What was the total number of points scored by the other 77 team members?

1010

1111

1212

1313

1414

Solution:

Let xx be the total number of points and yy be the desired answer.

Then from the problem statement we get that x4+2x7+15+y=x. \dfrac{x}{4} + \dfrac{2x}{7} + 15 + y = x.

Simplifying yields y+15=13x28. y + 15 = \dfrac{13x}{28}.

We know that y14y \leq 14 since none of the 77 team members scored more than 22 points.

We also know that xx must be a multiple of 28.28. If x=28,x = 28, then we get that y+15=13y=2, y + 15 = 13 \Rightarrow y = -2, which is not allowed.

If x=228=56,x = 2 \cdot 28 = 56, then we have that y+15=26y=11, y + 15 = 26 \Rightarrow y = 11, which works.

Thus, the correct answer is B.

24.

In triangle ABC,ABC, point DD divides side AC\overline{AC} so that AD:DC=1:2.AD:DC=1:2. Let EE be the midpoint of BD\overline{BD} and let FF be the point of intersection of line BCBC and line AE.AE. Given that the area of ABC\triangle ABC is 360,360, what is the area of EBF?\triangle EBF?

2424

3030

3232

3636

4040

Solution:

We know that the area of BCD\triangle BCD is twice that of ABD\triangle ABD since its base is twice as long and they have the same heights.

This tells us that the area of ABD\triangle ABD is 360÷3=120.360 \div 3 = 120. Similarly, we get that the area of ABE\triangle ABE is 120÷2=60.120 \div 2 = 60.

Note that CD=23CA.CD = \dfrac{2}{3} CA. This tells us that the altitude of BCD\triangle BCD is 23\dfrac{2}{3} the altitude of ABC.\triangle ABC.

We also know that BD=12BD,BD = \dfrac{1}{2}BD, which tells us that the altitude of BEF\triangle BEF is 12\dfrac{1}{2} the altitude of BCD.\triangle BCD.

Finally, we get that the altitude of BEF\triangle BEF is 2312=13\dfrac{2}{3} \cdot \dfrac{1}{2} = \dfrac{1}{3} the altitude of ABC.\triangle ABC.

Note that the altitude of ABF\triangle ABF is the same as the altitude of ABC.\triangle ABC.

This tells us that the area of BEF\triangle BEF is 13\dfrac{1}{3} the area of ABF\triangle ABF since they have the same base but different altitudes.

This gives us the following equation, where xx equals the area of BEF.\triangle BEF.

xx+60=13. \dfrac{x}{x + 60} = \dfrac{1}{3}.

Cross-multiplying yields 3x=x+60x=30. 3x = x + 60 \Rightarrow x = 30.

Thus, the correct answer is B.

25.

Alice has 2424 apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?

105105

114114

190190

210210

380380

Solution:

Let us assign everybody 22 apples. This leaves us with 2432=1824 - 3 \cdot 2 = 18 apples.

Then, we want to solve a+b+c=18, a + b + c = 18, where aa is the number of apples that Alice gets and so on.

Note that these are all nonnegative, since we already satisfied the only condition we needed to.

We can use stars and bars to get the number of solutions as (18+3131)=(202)=190. \binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = 190.

Thus, the correct answer is C.