For the first blog post, every member of Group 1 will be required to post a solution to certain problems on the homework that said "Ten Essentials" to this blog post. Try to put all the assigned problem solutions in one blog post. Make these solutions easy to understand so everyone in group 1 can fully understand every one of these 50 problems. These problems will be due on Wednesday, November 16th by 10:00 p.m. I will be checking these thoroughly and will give NO snacks for the members of Group 1 who doesn't do this blog assignment and this time I mean it. For those who do all the problems they are assigned and the effort seems apparent, this will be an easy way to earn double snacks.
Here are the problems assigned to each person. To differentiate problems of different pages with the same number, I am using the position of the problem (e.g. Problem # 3 on page 5 will be labeled as 23):
Julianna : 1, 13, 25, 37, 49
Christopher : 2, 14, 26, 38, 50
Vivian : 3, 15, 27, 39
Maya : 4, 16, 28, 40
Anindit : 5, 17, 29, 41
Stephanie : 6, 18, 30, 42
Jacky : 7, 19, 31, 43
Sagaar : 8, 20, 32, 44
Ming : 9, 21, 33, 45
Alice : 10, 22, 34, 46
Allan : 11, 23, 35, 47
Christine : 12, 24, 36, 48
During Thursday and Friday, look over the solutions that everyone puts on the blog, not just yours. Remember that all the problems are due on paper during this week's meeting on the 18th.
First comment :D ~Stephanie
ReplyDeleteMathCounts Solutions (Week of 11/11/11)
Problem 6: {2, 4, 10, 12, 15, 20, 50}
Desired pairs:
2*50
4*50
10*20
10*50
12*50
15*20
20*50
Total number of pairs: 7C2 (21)
P = 7/21
P = 1/3
Thus, the probability of choosing 2 numbers and multiplying it together and getting a multiple of 100 is 1/3.
Problem 18: One (length) solution is to blunt force it and list out all the smallest seven distinct positive integers of 9 and add them together.
Another way is to realize that we are simply listing the multiples as:
9*1 + 9*2 + 9*3 + 9*4 + 9*5 + 9*6 + 9*7
Distribute:
9(1 + 2 + 3 + … + 6 + 7)
9(28) = 252
Thus, the sum of the seven distinct positive integers of 9 is 252.
Problem 30: Since equilateral triangles have the most area with the same amount of perimeter, a triangle with each side being 20/7 would make an equilateral triangle. However, the lengths need to be integers, so an isosceles triangle would be the next closest thing. The triangle is found to be leg length 7 and base length 6.
If we draw an altitude from the vertex to the midpoint of the base (this line is also known as the perpendicular bisector), we find that one of the legs of one of the right triangle formed is 3 and the hypotenuse is 7.
Pythagorean Theorem:
a^2 + b^2 = c^2
3^2 + b^2 = 7^2
9 + b^2 = 49
b^2 = 40
b = √40
b = 2 √10
Thus, the height of the triangle is 2 √10.
Area of the triangle:
A = ½ b*h
A = ½ 6 * 2 √10
A = 6 √10
Thus, the largest area of a triangle with integer side lengths and perimeter 20 is 6√10.
Problem 42:
Realize that 4x = 3y and that 4x(x+y) = 336 in.
If we multiply 4x(x+y) = 336 by 3:
4x(3x+3y) = 1008
4x(3x+4x) = 1008
4x(7x) = 1008
28x^2 = 1008
x^2 = 36
x = 6
4x = 3y
4(6) = 3y
24 = 3y
y = 8
The perimeter of the rectangle:
P = 2(l + w)
P = 2(4x + y + x)
P = 2(24 + 8 + 4)
P = 2(36)
P = 72
Thus, the perimeter is 72.
Sorry if these solutions are confusing.
~Stephanie
MathCounts
ReplyDelete~Christine T^T
Problems 12, 24, 36, an 48
*Sidenote: I didn't do #48. Will come later
12) A collection of nickels, dimes, and pennies has an average value of $0.07 per coin. If a nickel were replaced by 5 pennies, the average would drop to $0.06 per coin. What is the number of dimes in the collection?
Solution: 12 dimes
If A=Total Amount
N=Number of Coins
To make calculations easier multiply 0.07 by 100 to get 7.
Since we're expressing average, it's going to be..
A/N=7
A=7N
Now that a nickel is replace by 5 pennies, the total amount remains the same, but the number of coins increase by 4. (N-1+5) Subtract 1 because you take away 1 nickel, but add 5 because you replace with 5 pennies.
A/(N+4)=6
A=6(N+4)
A=6N+24
Since A=7N and A=6N+24,
7N=6N+24
N=24 coins
Now we can substitute N=24 for A=7N or A=6N+24.
A=7(24)
A=168
A=6(24)=24
A=144+24
A=168
But in the beginning we multiplied 100 to the money. Now we have to undo that action by dividing 100 from 168. The total amount is $1.86.
Plug in numbers and you will get..
12 Dimes, 9 Nickels, and 3 Pennies.
24) Rectangle ABDE is inscribed in a circle. The lengths of
segments AB and AE are 48cm and 20cm respectively. Point
C is on the circle, and BC=CD. What is the number of centimeters
in the perimeter of pentagon ABCDE? Express your
answer in simplified radical form.
Solution: 8^(1/2)26+116
(8 square root 26 plus 116)
First, find the diameter of the circle. You can do that using sides ED and BD. ED=48cm and BD=20cm.
Using the Pythagorean Theorem (A^2= B^2+C^2) you can do..
48^2+20^2=A^2
2304+400=A^2
2704=A^2
A=52
So the diameter is 52 cm. Save that information while we work on something else. Look at triangle BCD. Locate the midpoint of BD, and name it Point F. Look at triangle BFC. The measurement of BF is 10 cm (half of BD). If we can figure out FC then using the Pythagorean Theorem we can find out the measurement of BC.
Now draw an straight imaginary line from Point C, passing Point F, then a new point, Point G (midway between AE) all the way to the other side of the circle then name that Point K. From Point C to Point K is the diameter, 52 cm. Now subtract 48 from 52. Then you get the measurement from Point K to Point G and Point C to Point F. You only want the measurement from Point C to Point F so you do..
(52-48)/2
=4/2
=2 cm
For triangle BFC, to find out the hypotenuse, you can again use the Pythagorean Theorem.
10^2+2^2=A^2
100+4=A^2
104=A^2
A= 4^(1/2)26
(4 square root 26)
AE=20, AB=48, DE=48, BC=4^(1/2)26, CD=4^(1/2)26
If you add all those together, it becomes..
8^(1/2)26+116
(8 square root 26 plus 116)
36) If x/y=3/4, y/z=2/3, and z/w=5/8, what is the value of x+y+w/w?
Solution: 1_35/48
First, you have to simplify.
x/y=3/4
4x=3y
y/z=2/3
3y=2z
z/w=5/8
8z=5w
If we make z equal to w...
8z=5w
z=5w/8
Then we move on to making y equal to w.
3y=2z
3y(4)=2z(4)
12y=8z
(Combine with 8z=5w)
12y=5w
y=5w/12
Then we move on to making x equal to w.
4x=3y
4x(4)=3y(4)
16x=12y
(Combine with 12y=5w)
16x=5w
x=5w/16
Now substitute x=5w/16, y=5w/12, and
Then we move on to making y equal to w.
3y=2z
3y(4)=2z(4)
12y=8z
(Combine with 8z=5w)
12y=5w
y=5w/12
Then we move on to making x equal to w.
4x=3y
4x(4)=3y(4)
16x=12y
(Combine with 12y=5w)
16x=5w
x=5w/16
Now substitute x=5w/16, y=5w/12 for x+y+w/w?
(5w/16)+(5w/12)+w/w
=(15w/48+20w/48+48/48w)/w
=(83w/48)/w
=83w/48w
=1_35/48
So the answer is 1_35/48
MathCounts
ReplyDelete~Christine T^T
Problem 48 (I solved it!)
48) For positive integer n such that n<10,000, the number n+2005 has exactly 21 positive factors. What is the sum of all the possible values of n?
Solution: 16, 592
We know do get the number of factors, we can first prime factorize that number and add 1 to each of the exponents and multiply it.
Therefore to get 21 factors, we can have the prime factorizations..
n^20 or (a^6)(b^2)
First, let's look at n^20. For n, we can plug in the smallest prime number, which is 2.
2^20
=1, 048, 576
But n+2005 has 21 factors not n. To find out the value of n, we have to subtract 2005 from 1,048,576. We already know that it will exceed 10,000, so there is no point doing the calculation.
Now that n^20 is out of the game, (because if 2^20-2005 exceeds 10,000 then so will larger prime numbers), we can focus on a^6 x b^2
Again, let's look at the smallest prime number, 2, and plug it in for a. For b, we can put in the next smallest prime number, 3. (b can't be same as a because otherwise, the powers will add up to 8, giving us 9 factors.)
2^6 x 3^2
=64x9
=576
If we subtract 2005 from 576, the difference will be a negative, so again, there is no point even doing the calculation. So (2^6 x 3^2)-2005 is not an eligible value of n.
Lets keep a=2, and b=5 (the third smallest prime number).
2^6 x 5^2
=64x25
=1600
We have to subtract 2005 from 1600, but again we know that the difference is going to be a negative so again there is no point doing the calculation.
We can keep on plugging in numbers for b.
2^6 x 7^2=3136-2005=1131 (o)
2^6 x 11^2=7744-2005=5739 (o)
2^6 x 13^2=10,816-2005=8811 (o)
2^6 x 17^2=18496-2005=16491 (x) too big
Now we can plug in the next smallest prime number for a, 3. For b, we can plug in 2.
3^6 x 2^2=2916-2005=911 (o)
Now we can change b to 5.
3^6 x 5^2=18225-2005=16220 (x) too big.
Now we move on to a=5, b=2.
5^6 x 2^2=62500-2005=60495 (x) too big.
Then we know that the rest of the numbers are going to be too big.
So the values of n=925, 5739, 8811, 911
The sum of those numbers:
1131 + 5739 + 8811 + 911
=6870+9722
=16,592
Therefore, the sum of the values of n is 16,592.
Hi people :D ~julianna
ReplyDeleteMathcounts Homework Problems #1,13,25,37,49
#1- A math conference is presenting a lecture series with six different lecturers. If Dr.Smith's lecture depends on Dr.Jones' lecture, so that Dr.Smith must be at some time after Dr.Jones, in how many orders can the six lecturers be scheduled?
There are six different "time slots" for the six lecturers. Dr.Jones must have one of the first five, while Dr.Smith must be at one of the slots after Dr.Jones. Here are the ways Dr.Jones and Dr.Smith can be arranged (the numbers indicate their time slot positions ex. "1" represents going first, "2" represents going second, etc.):
1,2 2,3 3,4 4,5 5,6
1,3 2,4 3,5 4,6
1,4 2,5 3,6
1,5 2,6
1,6
There are a total of 15 ways to arrange Dr.Smith and Dr.Jones. Once we have those, we can find the number of ways to arrange the four other lecturers. These four lecturers have four time slots available, so there are 4P4=24 different ways to arrange them. Total, there are 15*24=360 ways to arrange the six lecturers.
ANSWER- 360 ways
#13- Reid took seven tests. On the first five tests he took, he averaged 86 points. On the last three tests, he averaged 95 points. If he averaged 88 points on all seven tests, how many points did he average on the last two tests?
Given that the average of the first five tests was 86, we can determine that the sum of the scores of his first five tests was 86*5=430. Also, the average of the last three tests was 95, so the sum of the scores of these three tests was 95*3=285. The last given condition is that the average of all seven tests was 88. The sum of the scores on all seven tests was 88*7=616. We can see that the first and second sums we found overlap because the score of his fifth tests was used twice. Therefore, we can find the score of his fifth test by subtracting the total sum from the sum of the sums of the first 5 and the last 3 tests. (430+285)-616=99. Thus, he got a score of 99 on his fifth test. From here, we can determine the sum of the scores of his last two tests by subtracting 99 from the sum of the scores of his last three test. 285-99=186. Therefore, the average of the last two tests was 186/2=93.
ANSWER- 93
#25- Steve has an isosceles triangle with base 8 inches and height 10 inches. He wants to cut it into eight pieces having equal areas. To the nearest hundredth of an inch, what is the number of inches in the greatest perimeter among the eight pieces?
The picture shows an isosceles triangle vertically split into 8 smaller triangles. The height of each of these triangles is still 10 inches, but the 8-inch base of the large isosceles triangle has been split into 8 1-inch segments. Each small triangle has a base of 1 inch and a height of 10 inches. The triangle with the largest perimeter is the one farthest left (or right, they're the same). The cut in the middle splits the isosceles triangle into 2 right triangles. The four small triangles to the very left form the right triangle on the left, and vice versa for the one on the right. Each right triangle has a height of 10 inches and a base of 4 inches. We can use the Pythagorean Theorem to find the length of the hypotenuse, which is basically one leg of the original isosceles triangle (THIS IS GETTING RATHER HARD TO EXPLAIN BECAUSE I DON'T HAVE THE DIAGRAM).
a^2+b^2=c^2
10^2+4^2=c^2
100+16=c^2
116=c^2
√116=c
The hypotenuse of that right triangle is 2√29. This is also the length of one side of the triangle with the largest perimeter. Another side of this triangle is 1 because it is on the base of the original isosceles. To find the length of the last side, we can find the hypotenuse of the right triangle formed by the second, third, and fourth small triangles from the left. The base of this triangle is 3, and its height is 10.
Unfortunately, if I write any more in this comment, it will be over the 4,096 character mark. So I will post the rest of this problem and the other problems in another comment.
PROBLEM #25 CONTINUED...
ReplyDeletea^2+b^2=c^2
3^2+10^2=c^2
9+100=c^2
109=c^2
c=√109
Therefore, the last side of the triangle whose perimeter we are looking for has a length of √109. Now we can find the actual perimeter of this triangle.
1+2√29+√109 is approximately 22.21 inches.
ANSWER- 22.21 inches
#37- The price of a stock decreased by 20% in 2001 and then decreased by 20% in 2002. What percent increase in 2003 will the price to its value at the beginning of 2001? Express your answer to the nearest hundredth.
I find it easier to do these problems by assuming a beginning price. So assume that the price at the beginning of 2001 was $100. It decreased by 20%, so the price went down $20. At the beginning of 2002, it was $80. Then it went down another 20%. 20% of $80 is $16. At the beginning of 2003, the price was $64. Now we are trying to raise this price back up to $100, which was what it was at the beginning of 2000. To do this, we need another $36. We have to find what percent of $64 is $36.
36/64=x/100
x=56.25
Therefore,
ANSWER- 56.25%
#49- How many ordered pairs of positive integers (x,y) will satisfy y=18/x?
y=18/x can be rearranged to say xy=18. When dealing with ordered pairs, (1,2) is different from (2,1). In this case, we can find the pairs
(1,18), (18,1), (2,9), (9,2), (3,6), and (6,3). We can only used positive integers, so there are 6 total ways.
ANSWER- 6 ways
THE END :D
This comment has been removed by the author.
ReplyDeleteJulianna: for future reference, for the problem about the lecturers, what you can do is 6!/2 b/c every arrangement will have either Dr. Smith ahead or Dr. Jones or vice versa. 1/2 the ways have what you want, so thats why you divide by 2. Your way also works but I think this way is a bit faster :)
ReplyDeleteAnindit:
ReplyDelete#5:
there are 5 ways you can arrange the digits 1-5 as such:
1,2,3,4,5
2,3,4,5,6
3,4,5,6,7
4,5,6,7,8
5,6,7,8,9
Now, for each case, you have to put the date and moth dividers in. You can put the sticks in for each like this:
1/23/45 or 12/3/45
2/34/56 or 23/4/56
3/45/67 or 34/5/67..and so on
However, as you will notice, only the 1st case will work as there aren't 34 months in a year or 45 days in a month.
so, for 5 numbers, there are 2 possible cases
The other way it can be is:
12/34/56
23/45/67..and so on
However, realize again that only the first way works. so, there are 3 consecutive dates a century.
Anindit: #17
ReplyDeleteLets first list a few terms to check for a pattern:
12, 15, 18, 21, 51, 81, 102, 105,108,111,114,117,120,
oh wait, all the terms from the 100-200 range will have 1s in them. term 39 is 198, meaning that term 40 is 201. term Therm 43 is 210, while term 46 is 219. term 47 is 231, 48 is 261, 49 is 291, and therefore 50 is 303.
Allan: 11, 23, 35, 47
ReplyDeleteProblem 11:
Six classes with 25 students each, four classes with 20 students each, and two classes with 35 students each. Each student fills out a questionnaire asking for the total number of students in his/her class.
6(25)+4(20)+2(35)= 300 total students
6+4+2=12 total classes
Mean: 300 students/12 classes
Mean: 25
The average of all the numbers turned in on the questionnaires is 25.
ANSWER: 25
Problem 23:
Square with vertices (-a,a), (a,-a), (-a,a), (a,a) is cut by line y=x/2. What is the perimeter of each quadrilateral?
You can start off by drawing a diagram where a square is cut by the line y=x/2.
You can find the length of all sides except for the side lying on the line, y=x/2.
The length of the three sides are: 2a, 1.5a, and 0.5a.
The Pythagorean Theorem is then needed to figure out the length of the last side. We can find the length of the last side since we know that the other two lengths of the right triangle it forms is 2a and a.
a^2+b^2=c^2
(2a)^2+(a)^2=c^2
4a^2+a^2=5a^2
c=a√5
So the length of the last side is a√5.
P= w+L+w+L
P=2a+1.5a+(√5)a+.5a
P=4a+(√5)a
P=4√5a
ANSWER: 4√5a (a is not included within the square root symbol)
Problem 35:
An edgy spider walks only along the edges from A to B of the dodecahedron formed by the net shown. What is the number of edges in the shortest path that the spider could take?
The easiest way to solve this problem is probably to draw on the net. There are a lot of redundant points on the net, for example there are two point B's, the point B shown and also the one on the other section of the net towards the left. They are both the same point.
If we were to label the bottom most pentagon (the one with point B) with points (label them counterclockwise) C, D, E, F, they would match up the the four other outermost tips of the pentagons furthest from the middle. The shortest path would have 3 edges as it takes the spider one edge to move to point F (If you labeled as I did, point F will be the point horizontally right of point B. Point F will also be on the rightmost tip of the the whole net). There are two point A's shown as point A is also the first horizontally left point of F, so point A is two edges away from point F. We then add up the total edges (1+2=3) and we get 3 edges as our final answer.
Sorry if this was confusing as there needs to be a lot of drawing and not much calculation done for this problem.
ANSWER: 3 edges
Problem 47:
What is the product of the two smallest prime factors of (2^1024)-1?
You can factor out this problem to begin with so you'll get (2^512-1)(2^512+1), then with more factoring you'll get (2^128-1)(2^128+1)(2^512+1). One can realize that there's a pattern going on so the final two smallest prime factors are (2^2-1)(2^2+1).
(2^2)-1=3
(2^2)+1=5
Therefore 3 and 5 are the smallest prime factors of (2^1024)-1, and the product of 3 and 5 is 15.
ANSWER: 15
Anindit #29:
ReplyDeleteTo solve this problem, inscribe the triangle into a rectangle bounded by the vertices of the triangle. This rectangle is a 7*9 rectangle. Then, find the areas of the smaller triangles formed after inscribing the triangle into the rectangle. These areas are 10.5, 9, and 18. Subtract these from the total area of the rectangle and you get the area of the triangle to be 25.5
I thought I posted a correction to #42.
ReplyDeleteThe perimeter is actually 76, not 72 since x = 6.
In this problem, you do:
ReplyDeleteLet x = # of boys in school
Let y = # of girls in school
Therefore,
x + y =1200
1/2 x + 2/3 y = 730
Solve the system of equations to get y = 780.
Christopher-
ReplyDelete#2: On the refridgerator, MATHCOUNTS is spelled out with 10 magnets, one letter per magnet. Two vowels and three consonants fall off and are put away in a bag. If the Ts are indistinguishable, how many distinct possible collections of letters could be in the bag?
Solution: There are 3 vowels and 6 distinguishable consonants in MATHCOUNTS. The number of combinations of 2 vowels and 3 consonants would be 3C2*6C3 or 60.
Answer: 60 collections
#14: The arithmetic mean of three numbers x, y, and z is 24. The arithmetic mean of x, 2y, and z - 7 is 34. What is the arithmetic mean of x and z? Express your answer as a decimal to the nearest tenth.
Solution: If the mean of x, y, and z is 24, x + y + z is 72. Likewise x + 2y + (z - 7) is 102. The difference between (x + 2y + (z - 7)) and (x + y + z) is the same as 102 - 72. This means that y - 7 = 30, and y is 37. In this case, the sum of x, z, and 37 is 72, so x + z is 35 and their mean is 17.5.
Answer: 17.5
#26: Square ABCD has sides of length 1 cm. Triangle CFE is an isosceles right triangle tangent to arc BD at G. Arc BD is a quarter-circle with its center at A. What is the total area of the two shaded regions? Express your answer as a decimal to the nearest thousandth.
Solution: The area of the entire square is 1 sq. cm, and the area of the quarter-circle is 1/4(pi*1^2), or 1/4 pi. Since the triangle is an isosceles right triangle, it is a 90-45-45 triangle, and the hypotenuse is twice as long as either leg. The diagonal of the square is 2^1/2, so the height of the triangle (if the base is the hypotenuse) is 2^1/2 - 1 (which is the radius of the quarter-circle). Therefore, the base is 2*2^1/2 - 2, and the area of the triangle is 3 - 2*2^1/2. The area of the shaded region is the area of the square minus the area of the quarter-circle and the area of the triangle, or 1 - 1/4 pi - (3 - 2*2^1/2), which is rounded to 1 - 0.7854 - 0.1716, which is about 0.043 sq. cm.
Answer: 0.043 sq. cm
#38: The line y = b - x with 0<b<4 intersects the y-axis at P and the line x = 4 at S. If the ratio of the area of triangle QRS to the area of triangle QOP is 9:25, what is the value of b? Express the answer as a decimal to the nearest tenth.
Solution: If the ratio of 1/2*(4-b)^2 to 1/2*b^2 (the areas of the small and large triangles respectively) is 9:25, then the ratios of (4-b) to b is 3:5. Since b + (4-b) = 4, b must equal 2.5.
Answer: 2.5
#50: N^2 is a divisor of 8!. What is the greatest possible integer value of N?
Solution: The prime factorization of 8! is 2^7*3*5*7. From here, the largest square factor is 2^6 (the square of 2^3). If N^2 is 2^6, N is 2^3, or 8.
Answer: 8
Christopher-
ReplyDeleteI made a mistake in problem #26. I said the hypotenuse of a right isosceles triangle is twice either leg, but I meant that it is twice the height (assuming the hypotenuse is the base). Sorry...
Ming
ReplyDelete9, 21, 33, 45
9) To know if a number is a multiple of 8, you look at the last 3 digits and see if they are a multiple of 8. So we only need to look at 3 rolls of a die. The last digit must be an even number for the number to be a multiple of 8. The probability of that is ½. If the last digit is 2, the other two must be 1 and 1, 1 and 5, 2 and 3, 3 and 1, 3 and 5, 4 and 3, 5 and 1, 5 and 5, and 6 and 3. There are 9 possible combinations if the last digit is 2. There are the same number for 4 and 6, so there is a total of 27 combinations. There are 216 possible combinations, so the probability is ⅛.
21) From the diagram, we know that the sum of FA and FD is 9. With the Pythagorean Theorem, FD^2 + D^2 = FA^2. I found FA to be 5 and FD to be 4. Then I drew a right triangle with hypotenuse FE and one leg on the rectangle. Using the Pythagorean Theorem again, I found FE to equal the square root of 10.
33) Using the formula V=Bh/3, I found the height of the cone to be 9. Then with the Pythagorean Theorem I found the lateral height to be 15. This is length of BC. The area of the shaded region is pi x rl, which equals 180 pi. The area of the whole circle is 225 pi. The fraction of circle unshaded is ⅕, so the angle measure of the sector is 72.
45) (a + b)^3 equals a^3 + 3a^2b + 3ab^2 + b^3, which is also 343. If you subtract the equation by a^3 + b^3 = 42, you get 3a^2b + 3ab^2 = 301. By simplifying, I get ab = 301/21. I know 1/a + 1/b = (a+b)/(ab), and by plugging those equations in I get 21/43.
#10
ReplyDeleteYou have to find what number Chris has to pick to so Ashely has a bigger chance to get prime
For this problem you could list all the possibilities if one number is picked.
If 1 was picked there would be 2 possibilities (2,4) (3,5) there would be 1/2 chance that Ashely would get a prime number
If 2 was picked there would be 2 possibilities (1,5) (3,5) there would be 1/2 chance that Ashely would get a prime number
If 3 was picked there would be 2 possibilities (1,5) (2,4) there would be 1/2 chance that Ashely would get a prime number
If 4 was picked there would be 2 possibilities (1,5) (3,5) there would be 1/2 chance that Ashely would get a prime number
If 5 was picked there would be 1 possibilitiy (2,4) there would be 3/4 chance that Ashely would get a prime number
Thus making 5 the best choice
#22
You first use the Pythagorean Theorem to write an equation so you come up with:
(6a+1)2 = (a+1)2 + (6a)2
Then you simplifiy it to a2-10a-1=0, you know a cannot be 0 or less.
After you figure that out you use the equation:
x=(-b±√b2 -4ac)/2a, you bring the equation into the formula making it= (10±√100-4)/2. You simplify that into (10±4√6)/2. Then you simplify it again to get 5±2√6. Then you finally get the answer: a=±2√6
#34
The problem gives the height of the cone, 8 inches tall, and the radius, 2 inches. To figure out the volume of the ice cream you use the formula (1/3) * pi * radius2 * height So it would be (8*pi*22)/3. Thus the answer would be 8 pi
#46
You already know that 12 and 15 are factors so the number has to be 60 or more. (3*4*5) But 60 doesn't work because there are only 10 positive factors. So then you can try 120. But 120 doesn't work either because there are only 14 factors. And then you try 180, then you find out that there are 16 factors:1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 45, 90
So the answer is 180
Problem #4:
ReplyDelete3/4 out of the 28 students in the class have brown hair, which is equal to 21 students. 6/7 out of the 28 students in the class are right-handed, which is equal to 24 students. Adding 21 and 24 gives a result of 45. However, there are only 28 students in the class. So, to find the least number of students that are both right-handed and have brown hair, you subtract 45-28, and get an answer of 17.
Problem #16:
To solve this problem, I added a couple more terms to the sequence.
a1=1, a2=13, a3=135, a4=1357, a5=13579, a6=1357911
For any multiple of nine, the digit sum must be equal to 9 or a multiple of 9. In this sequence, a3 and a6 are multiples of 9. This shows that every third term of the sequence is divisible by 9. Therefore the value of “m” for the 23rd term that is divisible by 9 is: 23*3=69. So the answer is 69.
Problem #28:
The total area of the square is 5*5, which equals 25. This will be the denominator of our fraction. The area of the smallest shaded square is 1*1=1. The area of the second largest shaded square is (3*3) - (2*2)=5. The area of the largest shaded square is (5*5)-(4*4)=9. Therefore, the total shaded area is 1+4+9=15, so the fraction of the total area that is shaded is 15/25 or 3/5. However, the problem asks for percentage, so the answer is 60%.
Problem #40:
Pipe A will fill a tank in 6 hours. So in one hour, it can fill 1/6 of the tank. Pipe B can fill a tank in 4 hours. In one hour, pipe B can fill 1/4 of the tank. Working together, the two pipes can fill (1/6)+(1/4) of the tank in an hour, which equals 5/12 of the tank. Therefore, Pipe C can fill 5/12 of the tank in one hour. If all three tanks work together for an hour, (5/12) + (1/4) + (1/6) of the tank will be filled. So the final answer is 10/12, which simplifies to 5/6.
Problem #3
ReplyDeleteThere are 3 cases:
1. All 4 walls are different. (6 ways)
2. Opposite walls are the same. (6 ways)
3. 1 pair of opposite walls is the same. (12)
Add it up to get 24 ways.
Problem #15
Since the distance from Brad's home and where he got the flat tire is always the same, we can call the time he takes from his home to where he stopped y, and the time from where he stopped to his home x. Now we have 2 equations:
3x = 9y
x + y = 6
We solve x to get 9/2, which is the time Brad takes to get from his home to the place where he got the flat tire. We multiply this by his speed to get 27/2, which is the distance. Therefore, the total distance is 27 miles, and his average speed is 27/6, or 4.5 miles per hour.
Problem #27
ABC is a 45-45-90 right triangle, so the hypotenuse is sqrt120, or 2sqrt30. Therefore, the hypotenuse of DEF is also 2sqrt30. In DEF, the two mini isosceles triangles are congruent, therefore the length of each of their legs is the same as the sidelength of the square. That means the length of the square is 1/3 of the hypotenuse, which is 2sqrt30. Thus, the sidelength of the square is (2sqrt30)/3. The area is then obviously 40/3.
Problem #39
If 1 billion heartbeats is the average lift span of an animal, then an elephant's heart beats 800-25 = 775 times less than a shrew per minute. Therefore, an elephant lives approximately 1290332 minutes longer, which is about 2 years more.
-Jacky-
ReplyDelete7)The probability is (3*2*1*2*1*4*3*2*1)/(9!) then you multiply it by 3! because the colors can be in any order so it is (3*2*1*2*1*4*3*2*1*6)/(1*2*3*4*5*6*7*8*9) and when simplified, it is 1/210 so the answer is 1/210.
19) So, the sequence is 1,1,2,4,8,16,32.....
As you can see, they are powers of 2. So, I tried to find the first power of 2 bigger than 5000 and found that it was 71. There is an extra 1 in the beginning of the sequence so the answer is 71+1 or 72.
31) The height and diameter of the cylinder is equal and I will use x to represent the value. The fraction with the area of the 2 bases over the total area is
(2πr^2)/(2πr^2)+(2πrh)
When simplified, it is
r/r+h
Since the radius is half of the diameter, r is x/2 and the height, h, is x. So plug in the values and you get
(x/2)/(x/2)+x
Simplify it and you get
x/x+2x
x/3x
and the answer is
1/3
43) 2x-9y=14 is the same as 6x=27y+42. Since 6x=42+y, 42+y=27y+42. When simplified, you get y=27y so y has to be 0. Anything times 0 is 0 so xy=0.
Vivian, as far as I'm concerned, your solution for number 3 is incorrect. I got 84. You see, the first wall could be chosen from 4 finishes. The wall across from it can be any finish it likes as well. If this wall is the same finish as the first wall, there are 3 possible finishes for each of the two remaining walls. If that wall is not (there are three ways this could happen), then there are 2 possible finishes for each of the two remaining walls. This adds up to 4*1*3*3 + 4*3*2*2 = 84 ways.
ReplyDeleteIn addition, I think you miscalculated on problem 39. I have no idea how you got 1290332 minutes. I found out that elephants live 40 million minutes and shrews live 1.25 million, so elephants live 38.75 million minutes more or about 74 years.
Jacky, although you had good intentions on problem 19, you said that 71 was larger than 5,000. The smallest power of 2 more than 5,000 is actually 8,192.
Christopher, i think YOUR number 3 is incorrect. Its just 4*3*3*2/4. your reasoning is incorrect because the wall across from the first one cannot be any that it would like to be because if the side between them is the same as the one picked to be across, it can't be (kinda confusing)
ReplyDeleteHey Alice,
ReplyDeleteI think that 34 is incorrect, because I got an answer of 16 pi. I think you miscalculated in finding the volume of the cone. Also, remember that there is a hemisphere of ice cream on top, meaning that there is another half of a sphere with radius 2 to add in. For the cone part, I got 1/3*pi*2^2*8, which came out as 32/2pi. For the hemisphere, it is half of a sphere with radius 2. The volume of the sphere would be 4/3*pi*2^3=32/3pi. But a hemisphere is half of that, so the volume of that is 32/3pi*1/2=16/3pi. Add 32/3pi+16/3pi, and you get 16pi.
Also...
ReplyDeleteChristopher,
For problem 50, I got an answer of 24. The prime factorization of 8! is 2^7*3^2*5*7. I think you missed 3^2. So the answer should actually be 2^3*3=24.
This comment has been removed by the author.
ReplyDeleteAlice:
ReplyDeleteFor #22:
I actually simplified
(a+1)^2 + (6a)^2 = (6a+1)^2
to be
a^2 - 10a = 0
a(a - 10) = 0
a = 0 or a - 10 = 0
Therefore a = 10.
Christopher:
ReplyDeleteFor #2 I got a different answer.
There are 3 distinguishable vowels and 6 distinguishable consonants (7 consonants total)
Therefore:
3!/2! + 7!/4!*3!/2!
= 3 + 7*5*2
= 73
(I don't know if this is correct or not... maybe you're correct :])
[Saagar]
ReplyDelete#32 (Ratio #2)
2000 $100
2001 $100 * 0.8 = $80
2002 $80 * 0.8 = $64
$36 / $64 = 56.25%
#8 (Probability #3)
ReplyDeleteP(exactly one of the 5 visible faces is painted)
= P(selecting a cube with one face painted and that face is not at the bottom) + P(selecting a cube with two faces painted and that one of the faces is at the bottom)
= 6/27 * 5/6 + 12/27 * 2/6
= 1/3 (ANSWER)
#20 (Pythagora #5)
ReplyDeleteLargest perimeter = sqrt(116) + sqrt(109) + 1
= 10.77 + 10.44 + 1 = 22.21
ANSWER = 22.21 in
#44 (Algebra #4)
ReplyDeletex = number of T-shirts sold at $10
y = number of T-shirts sold at $12
10x + 12y + 378 = 12x + 10y
x - y = 189
x + y = 661
2x = 850
x = 425
ANSWER: 425 SHIRTS
you can check a good solution here:
ReplyDeletehttp://www.getquiz.co/q/q_8039sO9tN/