How it Works › Forums › Introductory Problems (AMC 8 Problems 1020) › AMC 8 Review Session #1! (Geometry and Counting/Probability)
This topic contains 2 replies, has 2 voices, and was last updated by Pratham S. 15 hours, 4 minutes ago.

AuthorPosts

MathJax TeX Test Page
Hey Math Circle Students!Here are the introductory problems for the next $3$ days! These are more difficult than AMC 8 Problems 1015 but they fit the range of $1621$. Each of these problems are worth $5$ points to your leaderboard score! If you don’t understand one of the questions, feel free to post in the questions forum here
1. (Problem 5) In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is $X$, in centimeters?
2. (Problem 10) How many integers between $1000$ and $9999$ have four distinct digits?
3. (Problem 19) A triangle with vertices as $A=(1,3)$, $B=(5,1)$, and $C=(4,4)$ is plotted on a $6\times5$ grid. What fraction of the grid is covered by the triangle?
4. (Problem 11) In the small country of Mathland, all automobile license plates have four symbols. The first must be a vowel ($A, E, I, O,$ or $U$), the second and third must be two different letters among the $21$ nonvowels, and the fourth must be a digit ($0$ through $9$). If the symbols are chosen at random subject to these conditions, what is the probability that the plate will read “$AMC8$”?
Remember that you can post your solutions below to these problems and I will keep a running leaderboards for the top 10 students who participate the most! I especially appreciate the same problem being solved in different ways!
Prem Dhoot1) 7
2) 50404) 21000
Pratham S.Question number 1: By counting the right sides vertical lengths you get a total of 10. On the left side you get 5 + x. So, 5+x=10, x = 5.
Question number 2: The first number has 9 possibilities, 19. The second number has 9 possibilities as it is 09, but not the first number. The third number has 8 possibilities, 09, but not the first two numbers. The last digit has 7 possibilities, 09, but not the other three numbers. So, 9*9*8*7 = 4536.
Question number 3: Let M be the midpoint of AB. The area of the triangle can be found by base AB and height MC, because the triangle is isosceles. Using Pythagorean theorem AB = 2sqrt5, and MC =
sqrt5. Then find the area by 1/2bh. (2sqrt*sqrt5)/2 = 5. The grid’s area is 30, so 5/30 = 1/6.
Question number 4: The first place is one of 5 letters. The second place is one of 21 letters, which means the third place is one of 20 letters. The last place is one of 10 numbers. This is very straightforward, so 5*21*20*10 = 21000 are the total number of solutions. AMC8 being one of the possibilities has a 1/21000 chance of being made. 
AuthorPosts