2013 AMC 10A Problems/Problem 1
A taxi ride costs $1.50 plus $0.25 per mile traveled. How much does a 5-mile taxi ride cost?
There are five miles wh
ich need to be traveled. The cost of these five miles is get . Adding this to , we
2013 AMC 10A Problems/Problem 2
Alice is making a batch of cookies and needs cups of sugar. Unfortunately, her measuring cup holds only cup of sugar. How many
times must she fill that cup to get the correct amount of sugar?
To get how many cups we need, we realize that we simply need to divide the number of cups needed by the number of cups collected in her measuring cup each time. Thus, we need to evaluate the fraction . Simplifying, this is equal to
2013 AMC 10A Problems/Problem 3
Square has side length . Point is on , and the area of is . What is ?
We know that the area of is equal to . Plugging in , we get . Dividing, we find
2013 AMC 10A Problems/Problem 4
A softball team played ten games, scoring 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?
We know that, for the games where they scored an odd number of runs, they cannot have scored twice as many runs as their opponents, as odd numbers are not divisible by 2. Thus, from this, we know that the five games where they lost by one run were when they scored 1, 3, 5, 7, and 9 runs, and the others are where they scored twice as many runs. We can make the following chart:
The sum of their opponent's scores is
2013 AMC 10A Problems/Problem 5
Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid $105, Dorothy paid $125, and Sammy paid $175. In order to share costs equally, Tom gave Sammy dollars, and Dorothy gave Sammy dollars. What is ?
The total amount paid is
. To get how much each should have paid, we do
Thus, we know that Tom needs to give Sammy 30 dollars, and Dorothy 10 dollars. This means that .
============================================== 2013 AMC 10A Problems/Problem 6
Joey and his five brothers are ages 3, 5, 7, 9, 11, and 13. One afternoon two of his brothers whose ages sum to 16 went to the movies, two brothers younger than 10 went to play baseball, and Joey and the 5-year-old stayed home. How old is Joey?
Because the 5-year-old stayed home, we know that the 11-year-old did not go to the movies, as the 5-year-old did not and . Also, the 11-year-old could not have gone to play baseball, as he is older than 10. Thus, the 11-year-old must have stayed home, so Joey is
2013 AMC 10A Problems/Problem 7
1 Prob 2 Solu tion 1 3 Solu tion 2 4 See Also
? ? ?
A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
Let us split this up into two cases. Case : The student chooses both algebra and geometry. courses have already been chosen. We have more options for the last course, so there are possibilities here.
This means that Case
: The student chooses one or the other. , and then add to the previous. of History, Art, and Latin, which is simply .
Here, we simply count how many ways we can do one, multiply by
WLOG assume the mathematics course is algebra. This means that we can choose
If it is geometry, we have another
options, so we have a total of
options if only one mathematics course is chosen.
Thus, overall, we can choose a program in
We can use complementary counting. Since there must be an English class, we will add that to our list of classes for remaining spots for the classes. We are also told that there needs to be at least one math class. This calls for complementary counting. The total number of ways of choosing classes out of the is . The total number of ways of choosing only non-mathematical classes is . Therefore the
amount of ways you can pick classes with at least one math class is
2013 AMC 10A Problems/Problem 8
What is the value of
Factoring out, we get:
Cancelling out the
from the numerator and denominator, we see that it simplifies to
2013 AMC 10A Problems/Problem 9
In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on of her three-point shots and of her two-point shots. Shenille attempted shots. How many points did she score?
Let the number of attempted three-point shots made be and the number of attempted two-point shots be . We know that , and we need to evaluate , as we know that the three-point shots are worth 3 points and that she made 20% of them and that the two-point shots are worth 2 and that she made 30% of them. Simplifying, we see that this is equal to we get . Plugging in ,
2013 AMC 10A Problems/Problem 10
A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?
Let the total amount of flowers be pink carnations is . Thus, the number of pink flowers is and the number of red carnations is , and the number of red flowers is . The number of . Summing these, the total number of
. Dividing, we see that
2013 AMC 10A Problems/Problem 11
A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 10 ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?
Let the number of students on the council be . We know that there are ways to choose a two person team. This gives
that If there are
, which has a positive integer solution of people on the welcoming committee, then there are
. ways to choose a three-person committee.
2013 AMC 10A Problems/Problem 12
In such that , and are parallel to and and . Points and are on sides , , and , respectively, ? , respectively. What is the perimeter of parallelogram
Note that because to It follows that and are parallel to the sides of , the internal triangles and are similar , and are therefore also isosceles triangles. . Thus, . .
Since opposite sides of parallelograms are equal, the perimeter is
2013 AMC 10A Problems/Problem 13
How many three-digit numbers are not divisible by , have digits that sum to less than , and have the first digit equal to the third digit?
These three digit numbers are of the form and yields a number divisible by 5. . When is , , , or , y can be any digit from to , as . This . We see that and , as does not yield a three-digit integer
The second condition is that the sum yields When When When numbers. , we see that , , so so so
. This yields
. This yields . This yields
more numbers. more numbers.
. This yields
Summing, we get