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AMC 8 Math Olympiad USA Math Olympiad

Problem based on LCM | AMC 8, 2016 | Problem 20

Try this beautiful problem from Algebra based on LCM from AMC-8, 2016.

Problem based on LCM – AMC 8, 2016


The least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$. What is the least possible value of the least common multiple of $a$ and $c$?

  • \(26\)
  • \(20\)
  • \(28\)

Key Concepts


Algebra

Divisor

multiplication

Check the Answer


Answer:20

AMC-8, 2016 problem 20

Challenges and Thrills of Pre College Mathematics

Try with Hints


We have to find out the least common multiple of $a$ and $c$.if you know the value of \(a\) and \(c\) then you can easily find out the required LCM. Can you find out the value of \(a\) and \(c\)?

Can you now finish the problem ……….

Given that the least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$ .then b must divide 12 and 15. There is only one possibility that b=3 which divide 12 and 15. therefore \(a\)=\(\frac{12}{3}=4\)

can you finish the problem……..

so\(b\)=3. Given that LCM of \(b\) and \(c\) is 15. Therefore c=5

Now lcm of \(a\) and \(c\) that is lcm of 4 and 5=20

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AMC 8 Math Olympiad

Linear Equations | AMC 8, 2007 | Problem 20

Try this beautiful problem from Algebra based on Linear equations from AMC-8, 2007.

Linear equations – AMC 8, 2007


Before the district play, the Unicorns had won $45$% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?

  • \(40\)
  • \(48\)
  • \(58\)

Key Concepts


Algebra

linear equation

multiplication

Check the Answer


Answer:48

AMC-8, 2007 problem 20

Challenges and Thrills of Pre College Mathematics

Try with Hints


At first, we have to Calculate the number of won games and lost games. Unicorns had won $45$% of their basketball game.so we may assume that out of 20 unicorns woned 9.

Can you now finish the problem ……….

Next unicorns won six more games and lost two.so find out the total numbers of won game and total numbers of games i.e won=\(9x+6\) and the total number of games become \(20x+8\)

can you finish the problem……..

Given that Unicorns had won \(45\)% of their basketball games i.e \(\frac{45}{100}=\frac{9}{20}\)

During district play, they won six more games and lost two,

Therefore they won\(9x+6\) and the total number of games becomes \(20x+8\)

According to the question, Unicorns finish the season having won half their games. …

Therefore,\(\frac{9x+6}{20x+8}=\frac{1}{2}\)

\(\Rightarrow 18x+12=20x+8\)

\(\Rightarrow 2x=4\)

\(\Rightarrow x=2\)

Total number of games becomes \(20x+8\) =\((20 \times 2) +8=48\)

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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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Algebra AMC 8 Math Olympiad USA Math Olympiad

Quadratic equation Problem | AMC 8, 2009 | Problem 23

Try this beautiful problem from Algebra based on Quadratic equation.

Algebra based on Quadratic equation – AMC-8, 2009 – Problem 23


On the last day of school, Mrs. Awesome gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought  400 jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class?

  • $34$
  • $28$
  • $25$

Key Concepts


Algebra

quadratic equation

Factorization

Check the Answer


Answer:$28$

AMC-8 (2009) Problem 23

Pre College Mathematics

Try with Hints


Let the number of girls be x

so the number of boys be x+2

Can you now finish the problem ……….

She gave away  400-6=394  jelly beans

\(x^2 + (x+2)^2=394\)

can you finish the problem……..

Let the number of girls be x

so the number of boys be x+2

she gave each girl x jellybeans and each boy x+2 jellybeans,

Therefore she gave total number of  jelly beans to girls be \(x^2\)

Therefore she gave total number of  jelly beans to boys be \((x+2)^2\)

 She gave away  400-6=394  jelly beans

\(x^2 + (x+2)^2=394\)

\(\Rightarrow x^2 + x^2 +4x+4=394\)

\(\Rightarrow 2x^2 +4x-390=0\)

\(\Rightarrow x^2 +2x -195=0\)

\((x+15)(x-13)=0\)

i.e x=-15 , 13 (we neglect negetive as number of students can not be negetive )

Therefore x=13 i.e number os girls be 13 and number of boys be 13+2=15

total number of students ne 13+15=28

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AMC 8 Math Olympiad Number Theory

Page number counting |AMC 8- 2010 -|Problem 21

Try this beautiful problem from Algebra about Page number counting

Page number counting | AMC-8, 2010 |Problem 21


Hui is an avid reader. She bought a copy of the best seller Math is Beautiful. On the first day, Hui read \(\frac{1}{5}\) of the pages plus more, and on the second day she read  \(\frac{1}{4}\) of the remaining pages plus 15 pages. On the third day she read \(\frac{1}{3}\) of the remaining pages plus 18 pages. She then realized that there were only  62 pages left to read, which she read the next day. How many pages are in this book?

  • 320
  • 240
  • 200

Key Concepts


Algebra

Arithmetic

multiplication

Check the Answer


Answer:$240$

AMC-8, 2010 problem 21

Challenges and Thrills in Pre College Mathematics

Try with Hints


assume that the number of all pages be \(x\)

Can you now finish the problem ……….

count day by day

can you finish the problem……..

Let x be the number of pages in the book

First day ,Hui Read \(\frac{x}{5} + 12\) pages

After first day Remaining pages=\(\{x-(\frac{x}{5}+12)\}\)=\(\frac{4x}{5} -12\)

Second day ,Hui Read \(\frac{1}{4} (\frac{4x}{5} -12) +15=\frac{x}{5} +12\)

After Second day Remaining pages= \((\frac{4x}{5} -12) -(\frac{x}{5} +12)\)=\(\frac{4x}{5} -\frac{x}{5}-24\)=\(\frac{3x}{5} -24\)

Third day,Hui read \(\frac {1}{3} (\frac{3x}{5} -24) +18\) =\((\frac{x}{5} -8+18)\)=\(\frac{x}{5} +10\)

After Third day Remaining pages = \((\frac{3x}{5} -24) -(\frac{x}{5} +10)\) =\(\frac{2x}{5} – 34\)

Now by the condition, \(\frac{2x}{5} – 34 = 62\)

\(\Rightarrow 2x-170=310\)

\(\Rightarrow 2x=480\)

\(\Rightarrow x=240\)

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Algebra AMC 8 Math Olympiad

Divisibility | AMC 8, 2014 |Problem 21

Try this beautiful problem from Algebra based on multiplication and divisibility of two given numbers.

Multiplication and Divisibility- AMC 8, 2014


TheĀ  7-digit numbers 74A52B1 ana 326AB4C are each multiples of 3.which of the following could be the value of c ?

  • 1
  • 2
  • 3

Key Concepts


Algebra

Division algorithm

Integer

Check the Answer


Answer:1

AMC-8, 2014 problem 21

Challenges and Thrills of Pre College Mathematics

Try with Hints


Use the rules of Divisibility ……..

Can you now finish the problem ……….

If both numbers are divisible by 3 then the sum of their digits has to be divisible by 3……

can you finish the problem……..

Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. 7 + 4 + 5 + 2 + 1 = 19. In order to be a multiple of 3, A + B has to be either 2 or 5 or 8… and so on. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add two of the selected values, 5 to 15, to get 20. We then see that C = 1, 4 or 7, 10… and so on, otherwise the number will not be divisible by three. We then add 8 to 15, to get 23, which shows us that C = 1 or 4 or 7… and so on. In order to be a multiple of three, we select a few of the common numbers we got from both these equations, which could be 1, 4, and 7. However, in the answer choices, there is no 7 or 4 or anything greater than 7, but there is a 1. so the answer is 1

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Algebra AMC 8 Math Olympiad Number Theory

Least common multiple | AMC 8, 2016 – Problem 20

LCM – AMC 8, 2016 – Problem 20


The least common multiple of a and b is 12 .and the lest common multiple of b and c is 15.what is the least possible value of the least common multiple of a and c?

  • 30
  • 60
  • 20

Key Concepts


Algebra

Division algorithm

Integer

Check the Answer


Answer:20

AMC-8, 2016 problem 20

Challenges and Thrills of Pre College Mathematics

Try with Hints


Find greatest common factors

Can you now finish the problem ……….

Find Least common multiple….

can you finish the problem……..

we wish to find possible values of a,b and c .By finding the greatest common factor 12 and 15, algebrically ,it’s some multiple of b and from looking at the numbers ,we are sure that it is 3.Moving on to a and c ,in order to minimize them,we wish to find the least such that the LCM of a and 3 is 12,$\to 4$.similarly with 3 and c,we obtain 5.the LCM of 4 and 5 is 20 .

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