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Algebra Combinatorics Math Olympiad USA Math Olympiad

Algebra and Combination | AIME I, 2000 Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Algebra and Combination.

Algebra and combination – AIME 2000


In expansion \((ax+b)^{2000}\) where a and b are relatively prime positive integers the coefficient of \(x^{2}\) and \(x^{3}\) are equal, find a+b

  • is 107
  • is 667
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Equations

Combination

Check the Answer


Answer: is 667.

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

Try with Hints


 here coefficient of \(x^{2}\)= coefficient of \(x^{3}\) in the same expression

then \({2000 \choose 1998}a^{2}b^{1998}\)=\({2000 \choose 1997}a^{3}b^{1997}\)

then \(b=\frac{1998}{3}\)a=666a where a and b are relatively prime that is a=1,b=666 then a+b=666+1=667.

.

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AMC 8 Math Olympiad USA Math Olympiad

Time and Work | PRMO-2017 | Problem 3

Try this beautiful problem from PRMO, 2017 based on Time and work.

Time and work | PRMO | Problem-3


A contractor has two teams of workers : team A and team B. Team A can complete a job in 12 days and team B can do the same job in 36 days. Team A starts working on the job and team B joins team A after four days. The team A withdraws after two more days. For how many more days should team B work to complete the job ?

  • $20$
  • $16$
  • $13$

Key Concepts


Arithmetic

multiplication

unitary method

Check the Answer


Answer:$16$

PRMO-2017, Problem 3

Pre College Mathematics

Try with Hints


In the problem,we notice that first 4 days only A did the work.so we have to find out A’s first 4 days work done.next 2 days (A+B) did the work together,so we have to find out (A+B)’s 2 days work.

so we may take the total work =1

A’s 1 day’s work= \(\frac{1}{12}\) and B’s 1 day’s work=\(\frac{1}{36}\)

Can you now finish the problem ……….

Now B did complete the remaining work.so you have to find out the remaining work and find out how many more days taken….

so to find the remaining work subtract (A’s 4 day;s work + (A+B)’S 2 days work)) from the total work

Can you finish the problem……..

Let the total work be 1

A can complete the total work in 12 days,so A’S 1 day’s work=\(\frac{1}{12}\)

B can complete the total work in 36 days, so B’s 1 day’s work=\(\frac{1}{36}\)

First 4 days A’s workdone=\(\frac{4}{12}=\frac{1}{3}\)

After 4 days B joined and do the work with A 2 days

So \((A+B)\)’s 2 day’s workdone=\(2 \times( \frac{1}{12}+\frac{1}{36})\)=\(\frac{2}{9}\)

Remaining workdone=\((1-\frac{1}{3}-\frac{2}{9}\))=\(\frac{4}{9}\)

B will take the time to complete the Remaining work=\(36 \times \frac{4}{9}\)=16

Hence more time taken=16

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AMC 8 Combinatorics Math Olympiad Probability

Probability Problem | AMC 8, 2016 | Problem no. 21

Try this beautiful problem from Probability.

Problem based on Probability | AMC-8, 2016 | Problem 21


A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?

  • \(\frac{3}{5}\)
  • \(\frac{2}{5}\)
  • \(\frac{1}{4}\)

Key Concepts


probability

combination

fraction

Check the Answer


Answer: \(\frac{2}{5}\)

AMC-8, 2016 problem 21

Challenges and Thrills in Pre College Mathematics

Try with Hints


There are 5 Chips, 3 red and 2 green

Can you now finish the problem ……….

We draw the chips boxes in such a way that we do not stop when the last chip of color is drawn. one at a time without replacement

Can you finish the problem……..

There are 5 Chips, 3 red and 2 green

we draw the chips boxes in such a way that we do not stop when the last chip of color is drawn.

if we draw all the green chip boxes then the last box be red or if we draw all red boxes then the last box be green

but we draw randomly. there are 3 red boxes and 2 green boxes

Therefore the probability that the 3 reds are drawn=\(\frac{2}{5}\)

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AMC 8 Combinatorics Math Olympiad Probability

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?

Problem from Probability

  • \(\frac{2}{3}\)
  • \(\frac{1}{3}\)
  • \(\frac{1}{4}\)

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…….

\(1 – \frac{2}{4} \times \frac{2}{3}\)= \(1 – \frac{1}{3} = \frac{2}{3} \)  

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AMC 8 Combinatorics Math Olympiad Probability USA Math Olympiad

Probability | AMC 8, 2004 | Problem no. 22

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 \(\frac{6}{16}=\frac{3}{8}\)

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AMC 8 Combinatorics Math Olympiad Probability

Probability | AMC 8, 2010 | Problem no. 24

Try this beautiful problem from Probability .You may use sequential hints to solve the problem.

Probability | AMC-8, 2007 |Problem 24


A bag contains four pieces of paper, each labeled with one of the digits 1,2,3 or 4.  with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of 3?

  • \(\frac{3}{4}\)
  • \(\frac{1}{2}\)
  • \(\frac{1}{4}\)

Key Concepts


probability

combination

Number counting

Check the Answer


Answer: \(\frac{1}{2}\)

AMC-8, 2007 problem 24

Challenges and Thrills in Pre College Mathematics

Try with Hints


there are two ways that The combination of digits that give multiples of 3

Can you now finish the problem ……….

The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4)

can you finish the problem……..

The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4) . The number of ways to choose three digits out of four is 4. Therefore, the probability is \(\frac{1}{2}\)

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