AMC-8 Math Olympiad USA Math Olympiad

Probability | AMC-10A, 2003 | Problem 8

Try this beautiful problem from Probability: positive factors AMC-10A, 2003. You may use sequential hints to solve the problem

Try this beautiful problem from Probability based on Positive factors

Probability – AMC-10A, 2003- Problem 8

What is the probability that a randomly drawn positive factor of \(60\) is less than \(7\)?

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

Key Concepts




Check the Answer

But try the problem first…

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

Suggested Reading

AMC-10A (2003) Problem 8

Pre College Mathematics

Try with Hints

First hint

Now at first we find out the positive factors of \(60\) are \(1,2,3,4,5,6,10,12,15,20,30,60\).but the positive factors which are less than \(7\) are \(1,2,3,4,5,6\)

Can you now finish the problem ……….

Second Hint

so we may say that any For a positive number \(n\) which is not a perfect square, exactly half of the positive factors will be less than \(\sqrt{n}\).here \(60\) is not a perfect square and \(\sqrt 60 \approx 7.746\).Therefore half of the positive factors will be less than \(7\)

can you finish the problem……..

Final Step

Therefore the required probability=\(\frac{1}{2}\)

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