2018 AMC 10A 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.
What is the value of
Solution:
We can simplify this as follows.
Thus, B is the correct answer.
2.
Liliane has more soda than Jacqueline, and Alice has more soda than Jacqueline. What is the relationship between the amounts of soda that Liliane and Alice have?
Liliane has more soda than Alice.
Liliane has more soda than Alice.
Liliane has more soda than Alice.
Liliane has more soda than Alice.
Liliane has more soda than Alice.
Solution:
Let be the number of gallons of soda that Jacqueline has. Then Alice has gallons, and Liliane has gallons.
Therefore, the relationship can be found by dividing the amount of soda that each has to yield which mean Liliane has more soda.
Thus, A is the correct answer.
3.
A unit of blood expires after seconds. Yasin donates a unit of blood at noon of January On what day does his unit of blood expire?
January
January
January
February
February
Solution:
We can divide by and to get the number of days that it takes for a unit of blood to expire.
The first division cancels a and The second division cancels and The final division cancels and turns the into a
This leaves and a which multiply to There are days in January, so by February the blood only has days left.
days from February would make the blood expire on February
Thus, E is the correct answer.
4.
How many ways can a student schedule mathematics courses — algebra, geometry, and number theory — in a -period day if no two mathematics courses can be taken in consecutive periods?
(What courses the student takes during the other periods is of no concern here.)
Solution:
The classes can occupy the following periods:
This means that there are ways to choose which periods the mathematics courses occur.
For each configuration, there are ways to determine the order of the courses, for a total of schedules.
Thus, E is the correct answer.
5.
Alice, Bob, and Charlie were on a hike and were wondering how far away the nearest town was. When Alice said, "We are at least miles away," Bob replied, "We are at most miles away." Charlie then remarked, "Actually the nearest town is at most miles away."
It turned out that none of the three statements were true. Let be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of
Solution:
Alice's statement tells us that Bob's statement tells us that Charlie's statement tells us that
Combining all of these tells us that and which means is in the interval
Thus, D is the correct answer.
6.
Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of and the score increases by for each like vote and decreases by for each dislike vote.
At one point Sangho saw that his video had a score of and that of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?
Solution:
If of votes were like votes, then of votes are dislike votes. Then Sangho's score is the total number of votes.
We know that Sangho's score is so the total number of votes is
Thus, B is the correct answer.
7.
For how many (not necessarily positive) integer values of is the following value an integer?
Solution:
We can rewrite the expression as
For this to be an integer, the exponents must be positive. This means that
This gives us values for
Thus, E is the correct answer.
8.
Joe has a collection of coins, consisting of -cent coins, -cent coins, and -cent coins. He has more -cent coins than -cent coins, and the total value of his collection is cents. How many more -cent coins does Joe have than -cent coins?
Solution:
Let be the number of -cent coins that Joe has. Then the number of -cent coins he has is
Therefore, Joe has -cent coins.
The total value of all these coins is
We know that
This means that Joe has -cent coins. Therefore, he has more -cent coins than -cent coins.
Thus, C is the correct answer.
9.
All of the triangles in the diagram below are similar to isosceles triangle in which Each of the smallest triangles has area and has area What is the area of trapezoid
Solution:
We know that the side length of the smaller triangles is times the length of the larger triangle from similar triangles.
Then the side length of is times the length of the side length of the larger triangle.
This makes the ratio of the areas
Therefore, the area of is The area of the trapezoid is then
Thus, E is the correct answer.
10.
Suppose that real number satisfies What is the value of
Solution:
Note that the left hand side of the equation and the desired expression are conjugates. Multiplying them would remove the square roots.
Multiplying them yields
This means that the product of the values of the expressions is equal to The desired value is therefore
Thus, A is the correct answer.
11.
When fair standard -sided dice are thrown, the probability that the sum of the numbers on the top faces is can be written as where is a positive integer. What is
Solution:
We can use stars and bars to find It is the same as finding the number of ways to put balls into boxes, where each box has at least one ball.
The formula for such a scenario is where is the number of balls and is the number of boxes.
The desired answer is therefore
Thus, E is the correct answer.
12.
How many ordered pairs of real numbers satisfy the following system of equations?
Solution:
We can graph these equations to easily find out where the intersection points are.
From the graph, we see that the line intersects the other graph three times.
Thus, C is the correct answer.
13.
A paper triangle with sides of lengths and inches, as shown, is folded so that point falls on point What is the length in inches of the crease?
Solution:
Note that the crease will the perpendicular bisector of Let be the crease.
By similarity, we know that Therefore, Plugging in values:
Simplifying gets us that
Thus, D is the correct answer.
14.
What is the greatest integer less than or equal to
Solution:
Let and Then the expression can be rewritten as
Note that
Therefore, the answer is less than is the only choice that satisfies this condition.
Thus, A is the correct answer.
15.
Two circles of radius are externally tangent to each other and are internally tangent to a circle of radius at points and as shown in the diagram. The distance can be written in the form where and are relatively prime positive integers. What is
Solution:
Let and be the centers of the small circles and be the center of the large circle.
Then and
We also know that Therefore, Plugging in values:
Multiplying yields Thus, D is the correct answer.
16.
Right triangle has leg lengths and Including and how many line segments with integer length can be drawn from vertex to a point on hypotenuse
Solution:
Let be the foot of the altitude from to We also get that
This tells us that by calculating the area in two ways. This value is between and
Note that as we move the line segment from to the line segment's length ranges from to
The integer values it covers therefore goes from to Similarly, as the line segment moves from to its takes on the values from to
This gives us unique line segments that have an integer value length.
Thus, D is the correct answer.
17.
Let be a set of integers taken from with the property that if and are elements of with then is not a multiple of What is the least possible value of an element in
Solution:
We proceed by casing on possible values for
cannot be the smallest element since that would mean that no other number can be in the set.
cannot be the smallest element since we would have to include every odd number except This would make and violate the rule.
Let be the smallest element. Then we can include and We can finally include either or and or
Either way, the maximum number of elements that we can include is so cannot be the smallest element.
Starting with we can include and Finally, we can add either or creating a -element set.
Thus, C is the correct answer.
18.
How many nonnegative integers can be written in the form where for
Solution:
Note that every number formed by this sum is either positive, negative, or zero.
The number of positive numbers equals the number of negative numbers due to symmetry (flip the s to s and s to s).
The only way for the sum to be is if all the coefficients are
The total number of numbers is Because each power of is larger than the sum of all previous powers of three, each combination of coefficients yields different numbers.
Therefore, there are distinct nonnegative integers.
Thus, D is the correct answer.
19.
A number is randomly selected from the set and a number is randomly selected from What is the probability that has a units digit of
Solution:
Since we only care about the units digit, we can turn the set into Then we can case on the value of
Any value of works. This occurs with a probability.
Looking at powers of we see that this sequence of units digits repeats:
This means that must be a multiple of There are such values. This means that works of the time. The total probability is
Powers of always end in which means that this case will never work.
The units digits repeat in this pattern: This means that must be a multiple of to work. As when this case works with a probability of
The units digit alternates between and This means that has to be even. This happens with a chance. The total probability is then
The total probability is therefore
Thus, E is the correct answer.
20.
A scanning code consists of a grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of squares.
A scanning code is called symmetric if its look does not change when the entire square is rotated by a multiple of counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides.
What is the total number of possible symmetric scanning codes?
Solution:
Note that all the line of symmetry split the grid into congruent regions that like look the figure below.
If we analyze one of these pieces, we can see that what we color this region determines the coloring of the other regions.
There are sections in this figure, and we have options for each of them, for a total of colorings.
We have to subtract since all the colors cannot be the same ().
Thus, B is the correct answer.
21.
Which of the following describes the set of values of for which the curves and in the real -plane intersect at exactly points?
Solution:
We can substitute from the second curve into the first curve.
From this, we see that always yields a solution (which is in fact a double root).
For the other two roots to exist and be unique, we need to have a positive discriminant.
For this to happen, or equivalently Thus, E is the correct answer.
22.
Let and be positive integers such that and Which of the following must be a divisor of
Solution:
The problem statement tells us that divides and divides and and divides and
Note that these are the following prime factorizations:
This means that we can express and as follows:
Then we get that
Note that cannot have a factor of since that would mean has an extra factor of
Similarly, we can see that does not have a factor of
This means that only has prime factors that are greater than or equal to
The only number of the form where satisfies the above criteria, between and is
Thus, D is the correct answer.
23.
Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths and units. In the corner where those sides meet at a right angle, he leaves a small unplanted square so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from to the hypotenuse is units. What fraction of the field is planted?
Solution:
Let be the side length of Then we can split the field up into the following shapes.
We can express the area of the field in two ways:
Simplifying yields
The desired fraction is
Thus, D is the correct answer.
24.
Triangle with and has area Let be the midpoint of and let be the midpoint of The angle bisector of intersects and at and respectively. What is the area of quadrilateral
Solution:
Let and be the length of the altitude through
By the angle bisector theorem, we get that where Substituting yields We also know that due to similar triangles.
Note that the height of the trapezoid is and The area of the trapezoid is Thus, D is the correct answer.
25.
For a positive integer and nonzero digits and let be the -digit integer each of whose digits is equal to ; let be the -digit integer each of whose digits is equal to ; and let be the -digit (not -digit) integer each of whose digits is equal to What is the greatest possible value of for which there are at least two values of such that
Solution:
We can use the formula for the sum of a geometric sequence to rewrite and
Similarly, we get that and
We can substitute these expressions into our condition to get
Simplifying yields
From the last line, we see that and are constants.
For there to be at least unique values of that satisfy the equation, both sides must equal zero.
We can see this by realizing that this equation is linear with respect to If both sides are non-zero, then there cannot exist unique solutions to a linear equation.
This tell us that and
The first equation gives us Plugging this into the second equation gives us
This tells us that must be divisible by This gives us the following triples:
The last triple is not allowed, so the maximum sum is Thus, D is the correct answer.