Ratio of LCM & GCF | Algebra | AMC 8, 2013 | Problem 10
Try this beautiful problem from Algebra based on the ratio of LCM & GCF from AMC-8, 2013.
Ratio of LCM & GCF | AMC-8, 2013 | Problem 10
What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
310
330
360
Key Concepts
Algebra
Ratio
LCM & GCF
Check the Answer
Answer:$330$
AMC-8, 2013 problem 10
Challenges and Thrills in Pre College Mathematics
Try with Hints
We have to find out the ratio of least common multiple and greatest common factor of 180 and 594. So at first, we have to find out prime factors of 180 & 594. Now.......
\(180=3^2\times 5 \times 2^2\)
\(594=3^3 \times 11 \times 2\)
Can you now finish the problem ..........
Now lcm of two numbers i.e multiplications of the greatest power of all the numbers
Therefore LCM of 180 & 594=\(3^3\times2^2 \times 11 \times 5\)=\(5940\)
For the GCF of 180 and 594, multiplications of the least power of all of the numbers i.e \(3^2\times 2\)=\(18\)
can you finish the problem........
Therefore the ratio of Lcm & gcf of 180 and 594 =\(\frac{5940}{18}\)=\(330\)
Try this beautiful problem from Algebra based on Mixture from AMC-8, 2002.
Mixture | AMC-8, 2002 | Problem 24
Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice?
34%
40%
26%
Key Concepts
Algebra
Mixture
Percentage
Check the Answer
Answer:40%
AMC-8, 2002 problem 24
Challenges and Thrills in Pre College Mathematics
Try with Hints
Find the amount of juice that a pear and a orange can gives...
Can you now finish the problem ..........
Find total mixture
can you finish the problem........
3 pear gives 8 ounces of juice .
A pear gives \(\frac {8}{3}\) ounces of juice per pear
2 orange gives 8 ounces of juice per orange
An orange gives \(\frac {8}{2}\)=4 ounces of juice per orange.
Therefore the total mixer =\({\frac{8}{3}+4}\)
If She makes a pear-orange juice blend from an equal number of pears and oranges then percent of the blend is pear juice= \(\frac{\frac{8}{3}}{\frac{8}{3}+4} \times 100 =40\)
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Arrangement.
Arrangement - AIME 2012
Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.
is 107
is 216
is 840
cannot be determined from the given information
Key Concepts
Arrangements
Algebra
Number Theory
Check the Answer
Answer: is 216.
AIME, 2012, Question 3
Combinatorics by Brualdi
Try with Hints
Here the number of ways to order the string BBBCCCFFF, such that one B is in first three positions
one C is in 4th to 6th positions, and one F is for last three positions. There are (3)(3)(3)=27 ways for first 3. Then for next two, 2 ways.
Problem related to Money | AMC 8, 2002 | Problem 25
Try this beautiful problem from AMC-8, 2002 related to money (problem 25).
Problem related to Money - AMC-8, 2002 - Problem 25
Loki, Moe, Nick and Ott are good friends. Ott had no money, but the others did. Moe gave Ott one-fifth of his money, Loki gave Ott one-fourth of his money and Nick gave Ott one-third of his money. Each gave Ott the same amount of money. What fractional part of the group's money does Ott now have?
\(\frac{1}{3}\)
\(\frac{1}{4}\)
\(\frac{3}{4}\)
Key Concepts
Algebra
Number theory
fraction
Check the Answer
Answer:\(\frac{1}{4}\)
AMC-8 (2002) Problem 25
Challenges and Thrills in Pre College Mathematics
Try with Hints
Each Friend gave Ott the equal amount of money
Can you now finish the problem ..........
Assume that ott gets y dollars from each friend
Can you finish the problem........
Given that Ott gets equal amounts of money from each friend, we can say that he gets y dollars from each friend. This means that Moe has 5y dollars, Loki has 4y dollars, and Nick has 3y dollars. The total amount is 12y dollars, Therefore Ott gets 3y dollars total, Required fraction =\(\frac{3y}{12y} = \frac{1}{4}\)
Problem from Probability | AMC 8, 2004 | Problem no. 21
Try this beautiful problem from Probability from AMC 8, 2004.
Problem from Probability | AMC-8, 2004 | Problem 21
Spinners A and B are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners' numbers is even?
\(\frac{2}{3}\)
\(\frac{1}{3}\)
\(\frac{1}{4}\)
Key Concepts
probability
Equilly likely
Number counting
Check the Answer
Answer: \(\frac{2}{3}\)
AMC-8, 2004 problem 21
Challenges and Thrills in Pre College Mathematics
Try with Hints
Even number comes from multiplying an even and even, even and odd, or odd and even
Can you now finish the problem ..........
A odd number only comes from multiplying an odd and odd..............
can you finish the problem........
We know that even number comes from multiplying an even and even, even and odd, or odd and even
and also a odd number only comes from multiplying an odd and odd,
There are few cases to find the probability of spinning two odd numbers from 1
Multiply the independent probabilities of each spinner getting an odd number together and subtract it from 1 we get.......
Try this beautiful problem from Probability .You may use sequential hints to solve the problem.
Probability | AMC-8, 2004 |Problem 22
At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is\(\frac{2}{5} \). What fraction of the people in the room are married men?
\(\frac{3}{8}\)
\(\frac{1}{2}\)
\(\frac{1}{4}\)
Key Concepts
probability
combination
Number counting
Check the Answer
Answer: \(\frac{3}{8}\)
AMC-8, 2007 problem 24
Challenges and Thrills in Pre College Mathematics
Try with Hints
Find the married men in the room ...
Can you now finish the problem ..........
Find the total people
can you finish the problem........
Assume that there are 10 women in the room, of which \(10 \times \frac{2}{5}\)=4 are single and 10-4=6 are married. Each married woman came with her husband,
so there are 6 married men in the room
Total man=10+6=16 people
Now The fraction of the people that are married men is \(\frac{6}{16}=\frac{3}{8}\)
Try this beautiful problem from Algebra based on Prime numbers.
Algebra based on Number theory - AMC-8, 2009 - Problem 23
Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers?
$14$
$12$
$16$
Key Concepts
Algebra
Number theory
card number
Check the Answer
Answer:$14$
AMC-8 (2006) Problem 25
Pre College Mathematics
Try with Hints
Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained.
Can you now finish the problem ..........
Obtain this even number would be to add another even number to 44
Can you finish the problem........
Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number would be to add another even number to 44 , and a different one to 38. Since there is only one even prime ( 2 ), the middle card's hidden number cannot be an odd prime, and so must be even. Therefore, the middle card's hidden number must be 2, so the constant sum is 59+2=61. Thus, the first card's hidden number is 61-44=17, and the last card's hidden number is 61-38=23
Since the sum of the hidden primes is 2+17+23=42, the average of the primes is \(\frac{42}{3}=14\)
