Categories

## 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

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

Categories

## 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

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=$\frac{2}{5}$

Categories

## 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?

• $\frac{5}{9}$
• $\frac{4}{9}$
• $\frac{4}{7}$

### Key Concepts

Algebra

Pattern

Fraction

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……..

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

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

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 $\frac{4}{9}$.

Categories

## 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?

• $6$
• $4$
• $3$

### Key Concepts

Geometry

Triangle

Trapezoid

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……..

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)

Categories

## 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

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 =$\frac{3y}{12y} = \frac{1}{4}$

Categories

## 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?

• $\frac{2}{3}$
• $\frac{1}{3}$
• $\frac{1}{4}$

### Key Concepts

probability

Equilly likely

Number counting

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…….

$1 – \frac{2}{4} \times \frac{2}{3}$= $1 – \frac{1}{3} = \frac{2}{3}$

Categories

## 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?

• $28-2\pi$
• $25-2\pi$
• $30-2\pi$

### Key Concepts

Geometry

square

Circle

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……..

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$\pi$

Categories

## 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

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 $\frac{6}{16}=\frac{3}{8}$

Categories

## 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?

• $7.1$
• $7.6$
• $7.8$

### Key Concepts

Geometry

Rectangle

Parallelogram

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 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

$\Rightarrow d=7.6$

Categories

## 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?

• $\frac{5}{\sqrt \pi}$
• $\frac{2}{\sqrt \pi}$
• $\sqrt \pi$

### Key Concepts

Geometry

Cube

square

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 ……….

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……..

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, $\pi r^2 – x=4-x$

$\Rightarrow \pi r^2=4$

$\Rightarrow r^2 = \frac{4}{\pi}$

$\Rightarrow r=\frac{2}{\sqrt \pi}$