Here are the 4 test covered in this post:
Best of the Past
Best of the Past - Geometry
2011 AMC 10 A (problems 15 - 25)
2011 AMC 10 B (problems 15 - 25)
To make everything easier, BOTP questions will be labeled 1 - 10, BOTP - Geo questions will be labeled 11 - 20, AMC 10 A questions will be labeled 21 - 31, and AMC 10 B questions will be labeled 32 - 42.
Amongst the 14 members, I want each of you to write solutions to 14 problems. I know that this number is a lot and you cannot write symbols on this blog, but some students find the blog posts as helpful and even if that number is 1, the blog posts will continue. Here are the assigned problems:
Group G (notice how I don't use A) -> problems 1, 4, 7, 10, ... , 40
Maya Varma
Adiyan Kaul
Alice Yao
Stephanie Shi
Ming Li
Group E -> 2, 5, 8, 11, ... , 41
Christine Park
Vivian Qian
Anindit Gopalakrishnan
Allan Wu
Sanjana Kothuri
Group O -> 3, 6, 9, 12, ..., 42
Julianna Liu
Aishwarya Laddha
Jacky Lee
Christopher Chang
For those students not part of the 14, you can choose the problems assigned to a group and post.
I claim first comment!!!
ReplyDelete~Christine :D
2) Find the sum of this: 1/3 + 4/9 + 9/27 + 16/81 + 25/243 + ….
Answer: 3/2
Lets say that S=original sequence…
S=1/3+4/9+9/27+16/81+25/243+…
Then divide S by 3.
S/3= 1/9+4/27+9/81+16/243+…
Notice how I didn't simplify.
Now subtract S/3 from S.
S=1/3+4/9+9/27+16/81+25/243+…
- S/3= 1/9+4/27+9/81+16/243+…
2S/3=1/3+3/9+5/27+7/81+9/243+…
Now divide 2/S3 by 3 again..
2S/9=1/9+3/27+5/81+7/243+…
Now subtract 2S/9 from 2S/3
2S/3-2S/9=
4S/9=1/3+2/9+2/27+2/81+2/243+…
Using the Infinite Geometric Series Formula, a/(1-r), you can calculate 2/9+2/27+2/81+2/243+…
a=2/9, r=1/3
(2/9)/1-(1/3)
=(2/9)/(2/3)
=1/3
But we have to add 1/3
1/3+1/3=2/3
So…
4S/9=2/3
(4/9)S=2/3
S=3/2
Christine Again…
ReplyDeleteI claim 2nd comment!!!
5) Somebody else can do this..
8) 1. Anindit is running a marathon with a distance of 26 miles. He initially starts off with a speed of 12 miles per hour. Gradually, He begins to get tired and decelerates at a rate of 2 miles per hour squared. What is the instantaneous speed of Anindit at the finish line? Leave it in simplest radical form.
(Hint 1: Distance = Initial Velocity * time + (½)(acceleration)(time)2)
(Hint 2: Distance = average velocity * time)
(Hint 3: Initial, Final, and Average velocity are all unique)
(Hint 3: There is a better equation for this problem but I am not telling it!)
(Hint 4: There is only one answer to this)
Answer: 2√10
There is a formula for this…
v^2=u^2+2as
v=final velocity
u=initial velocity
a=acceleration (express as negative if deceleration)
s=distance
So..
v=?
u=12
a=-2 (because it's deceleration)
s=26
Now plug in the numbers…
v^2=12^2+2(-2)(26)
=144-104
=40
v=√40
v=2√10
I claim 3rd comment!!!
ReplyDelete11) Triangle ABC has area 80. Given that AB=20, what is the length of the altitude from C to AB?
Answer:8
The area of the triangle formula is.. (you should already know this)
R=(b x h)/2
Altitude means height.
AB can be the base
AB=20
R=80
Plug in the numbers...
80=(20h)/2
20h=160
h=8
The height/altitude is 8. :)
4) In parallelogram ABCD, AB = 9 and BC = 6. If m<BCD = 45 degrees, then
ReplyDeletewhat is the area of the parallelogram? Express your answer in simplest radical
form?
Answer: 27√2
This requires a paper and a pencil…
If angle BCD is 45º, then angle ABC and ADC is 135º.
We can extend line segment AB and create a new right triangle on the side of BC.
The triangle is ∆BCE where E is a new point not on the parallelogram.
Then angle EBC is 45º and angle BCE is 45º.
A 45-45-90 angle triangle has sides 1-1-√2
We know the measurement for BC, 6.
√2x=6
x=6/√2
x=6√2/2
x=3√2
Now we know the height of the parallelogram..
Now the area can be..
9 x 3√2
=27√2
LOL CHRISTINE XD
ReplyDeleteI will be making my posts ASAP. (cleaning my room)
~Stephanie
I did #5!!! T^T
ReplyDelete5) Wayne flips a fair coin 3 times and Maria flips a fair coin 5 times. What is the probability that they flip the same number of heads? Express your answer as a common fraction.
Answer: 7/128
For this, you can use the 3rd and the 5th row of the Pascal’s triangle.
3rd row: 1, 3, 3, 1
5th row: 1, 5, 10, 10, 5, 1
For one flip of a coin there is 1/2 the chance of getting a head.
Wayne: 1/(2 x 2 x 2)=1/8
Maria: 1/(2 x 2 x 2 x 2 x 2)=1/32
Chance of getting 0 heads..
1 x 1/8 x 32=1/1024
Chance of getting 1 head
3 x 5/8 x 32=15/1024
Chance of getting 2 heads
3 x 10/8 x 32=30/1024
Chance of getting 3 heads
1 x 10/8 x 32=10/1024
You can’t go on to 4 heads because Wayne only flipped his coin 3 times.
Then you add them all up..
1/1024+15/1024+30/1024+10/1024
=56/1024
=7/128
7) Will come later.. I'm almost there... something smells fishy with my solution...
ReplyDeleteAhhhh.. i see.. Ashish kindly reminded me that 8 x 32 is not 1024.
ReplyDeleteCorrection to #5...
Answer: 7/32
Chance of getting 0 heads..
1 x 1/8 x 32=1/256
Chance of getting 1 head
3 x 5/8 x 32=15/256
Chance of getting 2 heads
3 x 10/8 x 32=30/256
Chance of getting 3 heads
1 x 10/8 x 32=10/256
You can’t go on to 4 heads because Wayne only flipped his coin 3 times.
Then you add them all up..
1/256+15/256+30/256+10/256
=56/256
=7/32
10) You can refer to Julianna's solution...
ReplyDeletehttp://mathcountsproblems.blogspot.com/2011/11/mathleague-competition-group-1.html#comment-form
(second to last comment). Ashish said that there was a faster solution. I guess it would be faster, but it's difficult to understand.... Ashish said that he would explain it at the next meeting
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteAdiyan Kaul
ReplyDeleteBest of the Past Problems 1,4,7,10
1.What is the largest n so that n^15000 < 6^20000.
When you divide both of the powers by 5,000 you get n^3<6^4 so when you solve 6^4 it is n^3<1296
So the greatest n can be is 10
Ans- 10
4. In a bag, there are 2 blue marbles, 3 green marbles, 4 red marbles, and 6 black marbles. What is the probability to pick 1 green marbles, 2 red marble,and 2 black marbles without replacement?
3C1*4C2*6C2/15C5= 3*6*15/3003=270/3003=90/1001
Ans- 90/1001
7. The solutions to the equations x2 – 5x + 3 = 0 are a and b. Find a4 + b4.
x^2-5x+3=0
a+b=-b/a=5 This is the sum of the roots
a*b=c/a=3 This is the product of the roots
a= 3/b
(a^2+b^2)^2- 2a^2b^2
[(a+b)^2-2ab]^2-2(ab)^2
When you plug in the values it is this
[5^2-2*3]^2- 2(3)^2
[25-6]^2-18
19^2-18
361-18
Ans- 343
10.An ant is located on the origin of the Cartesian plane. During each second, the
ant moves from its current location (x, y) to one of the points (x - 1, y), (x, y - 1),
(x, y + 1), or (x + 1, y). Let p be the probability that after 6 seconds, the ant is
located at the point (1,1). Express as a common fraction.
I don't know how to explain this
Adiyan Kaul
ReplyDeleteBest of the Past Geometry Problems 3
3. What is the radius of the incircle of a triangle with side lengths 28, 45, 53?
The formula for finding the radius of an incircle is always
sq rt of a+b+c/2(x-a)(x-b)(x-c)/x where a,b,c are the side lengths and x is the value of a+b+c/2
So now you plug in the values
sq rt 28+45+53/2(x-28)(x-45)(x-53)/x
sq rt 63(35)(18)(10)/63
sq rt 63*6300/63
sq rt 396900/63
630/63
10
Ans-10
My answers will be posted here:
ReplyDeletehttp://willyou.typewith.me/p/mathcounts
Please do not delete it ^^'
AMC 10A:
ReplyDelete17) In the eight-term sequence A, B, C, D, E, F, G, H, the value of C is 5 and the sum of any three consecutive terms is 30. What is A+H?
Answer: 25
A+B+C+D+E+F+G+H
=A+(B+C+D)+(E+F+G)+H
=A+30+30+H
=A+H+60
A+B+C+D+E+F+G+H
(A+B)+(C+D+E)+(F+G+H)
=25+30+30
=85
A+H+60=85
A+H=25
20) Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?
Answer: 1/3
http://www.artofproblemsolving.com/Wiki/index.php/2011_AMC_10A_Problems/Problem_20
23) Seven students count from 1 to 1000 as follows:
•Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
•Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.
•Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.
•Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.
•Finally, George says the only number that no one else says.
What number does George say?
Answer: 365
Numbers in parenthesis( ): Numbers not said
Numbers in brackets [ ]: patterns for unsaid numbers
Alice: 1, 3, 4, 6, 7, 9... (2, 5, 8, 11...) [+3]
Barbara: 2, 8, 11, 17, 20... (5, 14, 23, 32...) [+9]
Candice: 5, 23, 32, 50, 59... (14, 41, 68, 95, 122...) [+27]
Debbie: 14, 68, 95, 122... (41, 122, 203, 284, 365..) [+81]
Eliza: 41, 122, 203, 284...(122, 365, 608, 851) [+243]
Fatima: 122, 608, 851 (365)
At Fatima, the only inside the parenthesis is, 365, the only unsaid number.
Christopher-
ReplyDeleteBest of the Past Tests (normal and geometry):
#3: In 5 rolls of a standard die, what is the probability that the same number will be rolled exactly 4 times?
Solution: The number of possible numbers that the common number could be is 6, the number of possible numbers that the uncommon number could be is 5, and there are 5C4, or 5 ways for there to be 4 same rolls and one uncommon one. There are 6^5 or 7776 total ways to roll 5 dice. So, the probability is 6*5*5/7776 or 25/1296.
Answer: 25/1296
#6: A pair of whole numbers (a,b) is called complementary if a+b is a power of 2. John partitions the integers from 0 to 1999, inclusive, into two sets X and Y such that no two elements in X or Y are complementary. If 0 is in X, how many elements are in set X?
Solution: If 0 is in X, then the 11 powers of 2 (2^0 to 2^10) must be in Y, because 0 and 2^x is complementary. After that, we can divide the rest of the integers equally between the two sets. So, Y has 11-1 or 10 more elements than X, and Y+X = 2000. Solving this system, we find that X has 995 elements.
Answer: 995 elements
#9: Ming is bouncing a rubber ball. After the ball hits the ground, it bounces 80% of the initial height before the bounce. If he drops the ball from a height of 7 meters, what will the total distance the ball travels before coming to a stop?
Solution: The total distance can be written as 7 + 2*7(4/5*4/5^2*...), because the ball goes through every distance besides the initial bounce twice, once up and once down. This infinite sequence can be simplified to 7 + 2*7((4/5-1)), or 63 meters.
Answer: 63 meters
#12: What is the area enclosed by the graph |x| + |y| = 10?
Solution: This graph produces a rhombus. The dimensions of the kite can be found by finding the maximum and minimum values for x and y, which are 10 and -10 for both. So, the dimensions are 20 and 20, and the area is 1/2(20*20), or 200.
Answer: 200
#15: In isosceles right triangle ABC, AB=BC=3. Points D and E lie on AC such that AD=DE=EC=sqrt(2). Find BD^2.
Solution: We can find the height of the isosceles triangle as sqrt(3^2+(3*sqrt(2)/2)^2), or 3*sqrt(2)/2. So, the value of BD^2 is (sqrt(2)/2)^2 + (3*sqrt(2)/2)^2 = 5.
Answer: 5
#18: In tetrahedron ABCD, points E, F, and G are the midpoints of AB, AC, and AD, respectively. If tetrahedron has volume 72, then what is the volume of tetrahedron AEFG?
Solution: Since all of the dimensions are halved, the volume is 72/2^3, or 9.
Answer: 9
Christopher-
ReplyDeleteAMC 10A Part 1:
#21: A cell phone plan costs $20 each month, plus 5 cents per text message sent and 10 cents for each minute used over 30 hours. In January Michelle sent 100 text messages and talked for 30.5 hours. How much did she have to pay?
Solution: 5 cents*100 is $5, and 10 cents(60(30.5-30)) is $3. This adds up to 20+5+3=$28.
Answer: $28
#24: Let X and Y be the following sums of arithmetic sequences X=10+12+...100 and Y=12+14+...102. What is the value of Y-X?
Solution: Both sequences have a total of 46 numbers. Since for each number, the value of Y is increased by two, Y-X=2(46) or 92.
Answer: 92
#27: Which of the following equations does NOT have a solution?
A: (x+7)^2=0
B: |-3x|+5=0
C: sqrt(-x)-2=0
D: sqrt(x)-8=0
E: |-3x|-4=0
Solution: For A, the solution is -7. For B, there is no solution, since subtracting 5 from both sides leaves |-3x|=-5. For C, the solution is -4. For D, the solution is 64. For E, the solution is +or- 4/3.
Answer: |-3x|+5=0
#30: A majority of the 30 students in Ms. Deameanor's class bought pencils at the student bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $17.71. What was the cost of a pencil in cents?
Solution: The number of students who bought pencils, the number of pencils bought by each student, and the cost per pencil in cents must all be positive integers. So, we can start by finding the prime factorization of 1771, which is 7*11*23. The number of students who bought pencils must be 23, since no other number is more than half of 30. Since the cost of a pencil is more than the number of pencils each student bought, the cost must be 11 cents.
Answer: 11 cents
#33: How many even integers are there between 200 and 700 whose digits are all different and come from the set {1,2,5,7,8,9}?
Solution: If the numbers must be even, the units digits must be either 2 or 8. The hundreds digit must be either 2 or 5. So, if the hundreds digit is 2, then the tens digit could be any of 4 digits and the units digit must be 8. If the hundreds digit is 5, then the tens digit could be any of 4 digits, and the units digit could be any of the two even digits. This adds up to 1*4*1+1*4*2=12 integers.
Answer: 12 integers
#36: What is the value of sqrt(9-6*sqrt(2))+sqrt(9+6*sqrt(2))?
Solution: If we square this and then square root it, we get sqrt(9-6*sqrt(2)+2(sqrt(9)+9+6*sqrt(2)), or sqrt(24), or 2*sqrt(6).
Answer: 2*sqrt(6)
Christopher-
ReplyDeleteAMC 10A Part 2:
#39: In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?
Solution: If the population in 1991 is x^2, the population in 2001 is y^2+9, and the population in 2011 is z^2, then x^2+150=y^2+9, and x^2+141=y^2. You can rearrange this to make 141=y^2-x^2=(y-x)(y+x). The pairs of factors of 141 are (1,141), and (3,47). Trying both pairs, we find that if y-x=3 and y+x=47, y=25, x=22, and 22^2+300=784=28^2. Then, the percent increase from 1991 to 2011 is 300/484, or about 62%.
Answer: 62%
#42: Each vertex of convex pentagon ABCDE is to be assigned a color. There are 6 colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
Solution: If A is any point, B is a diagonal vertex to A, C is to B, and so on, then when C is the same color as A, there are 6 choices for A, 5 for B, 1 for C, 5 for D, and 4 for E, making 600 choices. If C and D are different from A, A has 6 choices, B has 5, C has 4, D has 4, and E has 4, making 1920 choices. When C isn't the same color as A but D is, A has 6 choices, B has 5, C has 4, D has 1, and E has 5, resulting in 600 choices. This adds up to 3120 combinations.
Answer: 3120 combinations
#45: Let R be a square region and n >or= 4 an integer. A point X in the interior of R is caled n-ray partitional if there are n rays emanating from X that divide R into n triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?
Solution: The first four rays must intersect the four vertices of the square. Starting with the bottom-leftmost point, we can divide the farthest two triangles into 49 triangles each using 48 rays each from the remaining 96 rays. If the distance from this point to the nearest side is a and the side length of the square is s, then a*s/2 = 1/49 * (s-a)*s/2, and a is s/50. So, the 100-ray partitional points make a grid 49*49 with points s/50 apart, making 2401 points. 60-ray partitional points have each point s/30 away from each other, making a 29*29 grid. The greatest common factor of 30 and 50 is 10, so the overlapping set of points is a 9*9 grid, or 81 points, and the total number of 100-ray partitional but not 60-ray partitional points is 2401-81 = 2320 points.
Answer: 2320 points
Adiyan Kaul
ReplyDeleteAmc A Problems 16,19,22,25
16.What is the value of sqrt
sqrt(9-6*sqrt(2))+sqrt(9+6*sqrt(2))?
When you square the expression than square root it you get sqrt(9-6 sqrt 2+ 2 sqrt(9)+9+6*sqrt(2)= sqrt 24= 2*sqrt 6
19.In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?
Solution- Let's say that the population in 1991 is x^2,the population in 2001 is y^2+9,and let's say that the population in 2011 is z^2. So x^2+150=y^2+9 When you simplify this equation you get 141=(y-x)(y+x). I did some guess and check and got x=22 and y=25. Now 22^2+300=784=28^2. So z=28. Now we find the percent of increase from 484 to 784=about 61%=62%
22.Each vertex of convex pentagon is to be assigned a color. There are colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
Solution- When you list all the possibilities that C is the same color as a,when c and d are different from a and when c is the same color but d is you get this possibilities.
6*5*1*5*4=600
6*5*4*4*4=1920
6*5*4*1*5=600
Adding all these up you get 3120
25.Let R be a square region and n >or= 4 an integer. A point X in the interior of R is caled n-ray partitional if there are n rays emanating from X that divide R into n triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?
I don't know how to do this problem look at Christopher's post.
Adiyan Kaul
ReplyDeleteAmc B problem 17
17.In the circle the diameter EB is parallel to DC and AB is parallel to ED. The angles AEB and ABE are are in the ratio 4:5. What is the degree measure of angle BCD?
Solution- We know that degree measure of A is 90 and the degree measures AEB and ABE are 4x and 5x. So 4x+5x+90=180 9x+90=180 9x=90 x=10 So AEB is 40. Since AED is 90 you can tell that BED is 50. Since BED and C is supplementary BED is 130.
I don't understand 20 and 23 in AMC B.
Christopher-
ReplyDeleteAMC 10B Part 1:
#48: At a store, when a length is reported as x inches that means the length is at least x - 0.5 in. and at most x + 0.5 in. If a rectangular tile is reported as 2 in. by 3 in., what is the minimum area for the rectangle?
Solution: The minimum lengths of the sides is 1.5 by 2.5 in., and the minimum area is 3.75 sq. in.
Answer: 3.75 sq. in.
#51: On Halloween Casper ate 1/3 of his candies and then gave 2 candies to his brother. The next day he ate 1/3 of his remaining candies and then gave 4 candies to his sister. On the third day he ate his final 8 candies. How many candies did Casper have at the beginning?
Solution: Before he gave 4 candies to his sister, he had 12 candies. Before he ate 1/3 of his candies on the second day, he had 18 candies. Before he gave 2 candies to his brother, he had 20 candies. Before he ate 1/3 of his original collection of candies, he had a total of 30 candies.
Answer: 30 candies
#54: The area of triangle EBD is one third of the area of 3-4-5 triangle ABC. Segment DE is perpendicular to segment AB. What is BD?
Solution: Since DE is perpendicular to AB, EBD is a right triangle, which is similar to ABC. Because they are similar, corresponding sides must be proportional, so if EBD is 1/3 of ABC, then each side is sqrt(1/3) of each corresponding side on ABC. Therefore sqrt(3)/3 * 4 = 4*sqrt(3)/3.
Answer: 4*sqrt(3)/3
#57: Keiko walks once around a track at exactly the same constant speed every day. The sides of the track are straight, and the ends are semicircles. The track has a width of 6 meters, and it takes her 36 seconds longer to walk around the outside edge than around the inside edge. What is Keiko's speed in meters per second?
Solution: The difference in the lengths of the tracks can be found by finding the difference between the circumference of a circle with diameter d+12 and a circle with diameter d, since the runs are congruent. The difference is 12 pi, and the speed is 12 pi/36 or pi/3 meters/second.
Answer: pi/3 meters/second
Christopher-
ReplyDeleteAMC 10B Part 2:
#60: If @ denotes the "averaged" operation, which distributive laws hold for all number x, y, and z?
1. x@(y+z)=(x@y)+(x@z)
2. x+(y@z)=(x+y)@(x+z)
3. x@(y@z)=(x@y)@(x@z)
Solution: If you simplify the first one, you get x+y+z/2=2x+y+z/2, which is incorrect. The second one yields x + y+z/2 = 2x+y+z/2, which is correct. The last one yields x/2 + y+z/4 = 2x+y+z/4, which is correct. So, number 2 and 3 are both correct.
Answer: #2, #3
#63: Rectangle ABCD has AB=6 and BC=3. Point M is chosen on side AB so that the measure of angle AMD = CMD. What is the degree measure of angle AMD?
Solution: Since AB || CD, angle AMD is congruent to angle CDM, and so angle CMD is congruent to CDM. So, triangle CDM is isosceles, and side MC is equal to side DC, which measures 6 units. So, triangle BCM has side 3 and hypotenuse 6, making it a 30-60-90 triangle, and making angle BMC 30 degrees. So, angle AMD is 180-30/2 or 75 degrees.
Answer: 75 degrees
#66: Brian writes down four integers w>x>y>z whose sum is 44. The pairwise positive differences of these numbers are 1,3,4,5,6,and 9. What is the sum of the possible values for w?
Solution: The difference of 9 has to be the difference between w and z. The difference of 1 must be between two adjacent integers, so the remaining difference of 8 must be divided into two differences between two other pairs of integers. The only two remaining possible differences are 3 and 5. Since there is no difference of 8, these differences must be separated by 1, so the differece between x and y is 1, and the differences of z and y or x and w can be either 5 or 3 each. If the difference between y and z is 3 and the difference between w and x is 5, then w+w-5+w-6+w-9=44, and w = 16. If the differences between y and z and w and x are reversed, then w+w-3+w-4+w-9=44, and w = 15. These possible values add up to 31.
Answer: 31
#69: A lattice point in an xy-coordinate system is any point (x,y) where both x and y are integers. The graph of y = mx+2 passes through no lattice point with 0<x<or=100 for all m such that 1/2<m<a. What is the maximum possible value of a?
Solution: For any integer value of the denominator d of fraction m <or= to 100, there will be a lattice point in y = mx+2 at the point (d,n+2), where n is the numerator. If there are no lattice points, d must be more than 100. So, the greatest value of a is the smallest fraction that is greater than 1/2 and has a maximum denominator that is less than or equal to 100, which is 50/99.
Answer: 50/99
Aishwarya Laddha
ReplyDelete3.In 5 rolls of a standard die, what is the probability that the same number will be rolled exactly 4 times?
answer:25/1296
why: probability: 6*5*5/7776 or 25/1296.
6.
9.
12. What is the area enclosed by the graph |x| + |y| = 10?
solution: when graphed, gives a rhombus, or a kite. you need to find the minimum and maximum values for x and y. The values will be 10 and −10, for both of them. Giving the dimensions are 20 and 20. and the area is 1/2(20*20), or 200.
answer: 200
Aishwarya Laddha
ReplyDeletenvrm i got #9.
9. Ming is bouncing a rubber ball. After the ball hits the ground, it bounces 80% of the initial height before the bounce. If he drops the ball from a height of 7 meters, what will the total distance the ball travels before coming to a stop?
its an infinite sequence: 7 + 2*7(4/5*4/5^2*…)
4/5 because .8 is nothing but 4/5
height will be 7 thats why its outside the parentheses.
simplified it will be 7 + 2*7((4/5-1))
solved will be 63 meters.
answer: 63 meters
Aishwarya Laddha
ReplyDelete6. A pair of whole numbers (a,b) is called complementary if a+b is a power of 2. John partitions the integers from 0 to 1999, inclusive, into two sets X and Y such that no two elements in X or Y are complementary. If 0 is in X, how many elements are in set X?
solution: 1999 not divisible by 3. closest is 1998. 1998/3 = 666
including 0 there will be 667
answer: 667 elements
Aishwarya Laddha
ReplyDelete12. What is the area enclosed by the graph |x| + |y| = 10
two ways can be used to find the area, area of 4 right triangles, or the area of a rhombus.
i did area of four right triangles.
1/2 *b*h*4=
1/2*10*10*4=
1/2*100*4=
1/2*400=
200
answer: 200
Aishwarya Laddha
ReplyDelete15. In isosceles right triangle ABC, AB=BC=3. Points D and E lie on AC such that AD=DE=EC=sqrt(2). Find BD^2.
solution: You can draw a diagram to help, if you want. (i thought it was easier to understand that way)
first you need to find Height of the whole triangle. sin of 45 = 1/sq rt of 2
(say the height of triangle is B to F) length of AB is 3. pythogeream theorem a^2 + b^2= c^2. AB^2= 9
AB^2=BF^2 + AF^2
9= 3*1/sq.rt.2 + AF^2
AF^2= 9- 9/2= 9/2
AF^2= 9/2
AF= 3/Sq. rt 2
EF= AF- sq rt 2
EF=1/sq rt 2
pythogeream theorem again..
BF^2+ EF^2 = BE^2
3/sqrt2 + 9/2= 5
answer: 5
Aishwarya Laddha
ReplyDelete18. In tetrahedron ABCD, points E, F, and G are the midpoints of AB, AC, and AD, respectively. If tetrahedron has volume 72, then what is the volume of tetrahedron AEFG?
Solution: because being halved three times
72*1/8= 9
Answer: 9
Aishwarya Laddha
ReplyDelete21.
Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged 55 miles per gallon. How long was the trip in miles?
solution: total miles
———— = 55
total gas
Let x be the distance the car traveled during the time it ran on gasoline, then the amount of gas used is .
40 + x = total distance travelled
40 + x
——— = 55
0.02x
when solved x= 400
400 + 40 = 440
answer: *C*
Aishwarya Laddha
ReplyDelete24. Circles and each have radius 1. Circles and share one point of tangency. Circle has a point of tangency with the midpoint of . What is the area inside Circle but outside circle and circle ?
solution: use the diagram to find the answer
use the top right hand quarter of circle A and Top left hand quarter of circle b. connect to form rectangle. do the same for circle c.
pi/2 +(2-pi/2)
pi/2 cancel
you should get 2
answer: area of the shaded region is 2.
Aishwarya Laddha
ReplyDelete27. Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?
id know how to solve this problem
Aishwarya Laddha
ReplyDelete30. Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?
solution: The intersection of the two tetrahedra is thus 1/2 * 1/3 =1/6
Answer: 1/6
This comment has been removed by the author.
ReplyDeleteAishwarya Laddha
ReplyDelete33. If @ denotes the "averaged" operation, which distributive laws hold for all number x, y, and z?
1. x@(y+z)=(x@y)+(x@z)
2. x+(y@z)=(x+y)@(x+z)
3. x@(y@z)=(x@y)@(x@z)
solution: plug in values, 1. not possible, 2. possible, 3. possible.
answer: 2 and 3 work
Aishwarya Laddha
ReplyDelete39. Brian writes down four integers w>x>y>z. whose sum is 44. The pairwise positive differences of these numbers are 1,3,4,5,6,9. What is the sum of the possible values for w ?
solution:
The largest difference, 9 must be between w and z
The smallest difference,1 must be directly between two integers.
The only remaining differences that would make this possible are 3 and 5 However, those two differences can't be right next to each other because they would make a difference of 8. This means 1 must be the difference between x and y. two possibilities for w then are 16 or 15
16+15=31
answer: 31
Aishwarya Laddha
ReplyDelete42. problem 25 on AMC B.
idk how to solve this
Aishwarya Laddha
ReplyDelete36. Rectangle ABCD has AB=6 and BC=3. Point M is chosen on side AB so that <AMD = <CMD . What is the degree measure of <AMD?
solution: Draw a figure.
because of alternate interior angles, and perpendicular lines
<BMC = 30 degrees
2x + 30= 180
2x= 150
x= 75
answer: 75 degrees
I'm into this type of designs cuz they can verily give contrast to rooms or to any designs imo.
ReplyDeleteming green marble Tile