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Algebra AMC 8 Math Olympiad

Mixture | Algebra | AMC 8, 2002 | Problem 24

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= \(\frac{\frac{8}{3}}{\frac{8}{3}+4} \times 100 =40\)

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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 Math Olympiad USA Math Olympiad

Pattern Problem| AMC 8, 2002| Problem 23

Try this beautiful problem from Algebra based on Pattern.

Pattern – AMC-8, 2002- Problem 23


A corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is made of darker tiles?

Pattern
  • \(\frac{5}{9}\)
  • \(\frac{4}{9}\)
  • \(\frac{4}{7}\)

Key Concepts


Algebra

Pattern

Fraction

Check the Answer


Answer:\(\frac{4}{9}\)

AMC-8 (2002) Problem 23

Pre College Mathematics

Try with Hints


The same pattern is repeated for every \(6 \times 6 \) tile

Can you now finish the problem ……….

Looking closer, there is also symmetry of the top \(3 \times 3\) square

can you finish the problem……..

Pattern

If we look very carefully we must notice that,
The same pattern is repeated for every \( 6 \times 6 \) tile

pattern 6/6


Looking closer, there is also symmetry of the top \( 3 \times 3\) square,

Pattern 3/3


Therefore the fraction of the entire floor in dark tiles is the same as the fraction in the square
Counting the tiles, there are dark tiles, and total tiles, giving a fraction of \(\frac{4}{9}\).

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of the Trapezoid | AMC 8, 2002 | Problem 20

Try this beautiful problem from Geometry based on Area of Trapezoid.

Area of the Trapezoid – AMC- 8, 2002 – Problem 20


The area of triangle XYZ is 8  square inches. Points  A and B  are midpoints of congruent segments  XY and XZ . Altitude XC bisects YZ.What is the area (in square inches) of the shaded region?

aera of trapezoid
  • \( 6\)
  • \( 4\)
  • \( 3\)

Key Concepts


Geometry

Triangle

Trapezoid

Check the Answer


Answer:\(3\)

AMC-8 (2002) Problem 20

Pre College Mathematics

Try with Hints


Given that Points  A and B  are midpoints of congruent segments  XY and XZ and Altitude XC bisects YZ

Let us assume that the length of YZ=\(x\) and length of \(XC\)= \(y\)

Can you now finish the problem ……….

Therefore area of the trapezoid= \(\frac{1}{2} \times (YC+AO) \times OC\)

can you finish the problem……..

Triangle with centre O

Let us assume that the length of YZ=\(x\) and length of \(XC\)= \(y\)

Given that area of \(\triangle xyz\)=8

Therefore \(\frac{1}{2} \times YZ \times XC\)=8

\(\Rightarrow \frac{1}{2} \times x \times y\) =8

\(\Rightarrow xy=16\)

Given that Points  A and B  are midpoints of congruent segments  XY and XZ and Altitude XC bisects YZ

Then by the mid point theorm we can say that \(AO=\frac{1}{2} YC =\frac{1}{4} YZ =\frac{x}{4}\) and \(OC=\frac{1}{2} XC=\frac{y}{2}\)

Therefore area of the trapezoid shaded area = \(\frac{1}{2} \times (YC+AO) \times OC\)= \(\frac{1}{2} \times ( \frac{x}{2} + \frac{x}{4} ) \times \frac{y}{2}\) =\(\frac{3xy}{16}=3\) (as \(xy\)=16)

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Algebra AMC 8 Math Olympiad USA Math Olympiad

Problem related to Money | AMC 8, 2002 | Problem 25

Try this beautiful problem from AMC-8, 2002 related to money (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 =\(\frac{3y}{12y} = \frac{1}{4}\)

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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 Math Olympiad

Intersection of two Squares | AMC 8, 2004 | Problem 25

Try this beautiful problem from Geometry based on Intersection of two Squares.

When 2 Squares intersect | AMC-8, 2004 | Problem 25


Two \(4\times 4\) squares intersect at right angles, bisecting their intersecting sides, as shown. The circle’s diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?

Intersection of two Squares
  • \(28-2\pi\)
  • \(25-2\pi\)
  • \(30-2\pi\)

Key Concepts


Geometry

square

Circle

Check the Answer


Answer: \(28-2\pi\)

AMC-8, 2004 problem 25

Pre College Mathematics

Try with Hints


Area of the square is \(\pi (r)^2\),where \(r\)=radius of the circle

Can you now finish the problem ……….

Clearly, if 2 squares intersect, it would be a smaller square with half the side length, 2.

can you finish the problem……..

Intersection of two Squares

Clearly the intersection of 2 squares would be a smaller square with half the side length, 2.

The area of this region =Total area of larger two squares – the area of the intersection, the smaller square i.e \(4^2+4^2 -2^2=28 \)

Now The diagonal of this smaller square created by connecting the two points of intersection of the squares is the diameter of the circle

Using the Pythagorean th. diameter of the circle be \(\sqrt{2^2 +2^2}=2\sqrt 2\)

Radius=\(\sqrt 2\)

area of the square=\(\pi (\sqrt2)^2\)=\(2\pi\)

Area of the shaded region= 28-2\(\pi\)

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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 Geometry Math Olympiad USA Math Olympiad

Area of Rectangle Problem | AMC 8, 2004 | Problem 24

Try this beautiful problem from Geometry from AMC-8, 2004 ,Problem-24, based on area of Rectangle.

Rectangle | AMC-8, 2004 | Problem 24


In the figure ABCD is a rectangle and EFGH  is a parallelogram. Using the measurements given in the figure, what is the length  d  of the segment that is perpendicular to  HE and FG?

Area of Rectangle Problem
  • $7.1$
  • $7.6$
  • $7.8$

Key Concepts


Geometry

Rectangle

Parallelogram

Check the Answer


Answer:$7.6$

AMC-8, 2004 problem 24

Pre College Mathematics

Try with Hints


Find Area of the Rectangle and area of the Triangles i.e \((\triangle AHE ,\triangle EBF , \triangle FCG , \triangle DHG) \)

Can you now finish the problem ……….

Area of the Parallelogram EFGH=Area Of Rectangle ABCD-Area of\((\triangle AHE +\triangle EBF + \triangle FCG + \triangle DHG) \)

can you finish the problem……..

Area of Rectangle Problem

Area of the Rectangle =\(CD \times AD \)=\(10 \times 8\)=80 sq.unit

Area of the \(\triangle AHE\) =\(\frac{1}{2} \times AH \times AE \)= \(\frac{1}{2} \times 4 \times 3\) =6 sq.unit

Area of the \(\triangle EBF\) =\(\frac{1}{2} \times EB \times BE \)= \(\frac{1}{2} \times 6 \times 5\) =15 sq.unit

Area of the \(\triangle FCG\) =\(\frac{1}{2} \times GC \times FC\)= \(\frac{1}{2} \times 4\times 3\) =6 sq.unit

Area of the \(\triangle DHG\) =\(\frac{1}{2} \times DG \times DH \)= \(\frac{1}{2} \times 6 \times 5\) =15 sq.unit

Area of the Parallelogram EFGH=Area Of Rectangle ABCD-Area of\((\triangle AHE +\triangle EBF + \triangle FCG + \triangle DHG) \)=\(80-(6+15+6+15)=80-42=38\)

As ABCD is a Rectangle ,\(\triangle GCF\) is a Right-angle triangle,

Therefore GF=\(\sqrt{4^2 + 3^2}\)=5 sq.unit

Now Area of the parallelogram EFGH=\( GF \times d\)=38

\(\Rightarrow 5 \times d\)=38

\(\Rightarrow d=7.6\)

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AMC 8 Geometry Math Olympiad USA Math Olympiad

Radius of the Circle | AMC-8, 2005 | Problem 25

Try this beautiful problem from Geometry: Radius of a circle

Radius of a circle – AMC-8, 2005- Problem 25


A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?

radius of the circle
  • $\frac{5}{\sqrt \pi}$
  • $ \frac{2}{\sqrt \pi} $
  • $\sqrt \pi$

Key Concepts


Geometry

Cube

square

Check the Answer


Answer: $ \frac{2}{\sqrt \pi} $

AMC-8 (2005) Problem 25

Pre College Mathematics

Try with Hints


The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square

Can you now finish the problem ……….

Shaded region of the figure

Region within the circle and square be \(x\) i.e In other words, it is the area inside the circle and the square

The area of the circle -x=Area of the square – x

can you finish the problem……..

Shaded region of the figure

Given that The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square

Let the region within the circle and square be \(x\) i.e In other words, it is the area inside the circle and the square .

Let r be the radius of the circle

Therefore, The area of the circle -x=Area of the square – x

so, \(\pi r^2 – x=4-x\)

\(\Rightarrow \pi r^2=4\)

\(\Rightarrow r^2 = \frac{4}{\pi}\)

\(\Rightarrow r=\frac{2}{\sqrt \pi}\)

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