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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Ratio and Inequalities | AIME I, 1992 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Ratio and Inequalities.

Ratio and Inequalities – AIME I, 1992


A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly 0.500. During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than 0.503. Find the largest number of matches she could have won before the weekend began.

  • is 107
  • is 164
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Ratios

Inequalities

Check the Answer


Answer: is 164.

AIME I, 1992, Question 3

Elementary Algebra by Hall and Knight

Try with Hints


Let x be number of matches she has played and won then \(\frac{x}{2x}=\frac{1}{2}\)

and \(\frac{x+3}{2x+4}>\frac{503}{1000}\)

\(\Rightarrow 1000x+3000 > 1006x+2012\)

\(\Rightarrow x<\frac{988}{6}\)

\(\Rightarrow\) x=164.

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Categories
AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Percentage Problem | AIME I, 2008 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Percentage.

Percentage Problem – AIME I, 2008


Of the students attending a party, 60% of the students are girls, and 40% of the students like to dance. After these students are joined by 20 more boy students, all of whom like to dance, the party is now 58% girls, find number of students now at the party like to dance.

  • is 107
  • is 252
  • is 840
  • cannot be determined from the given information

Key Concepts


Ratios

Percentage

Numbers

Check the Answer


Answer: is 252.

AIME I, 2008, Question 1

Elementary Number Theory by David Burton

Try with Hints


Let number of girls and boys be 3k and 2k, out of 3k girls, 2k likes to dance and 2k+20(boys) like to dance

here given that \(\frac{3k}{5k+20}\)=\(\frac{58}{100}\) then k=116

then 2k+20=252.

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