# Solving Problems in MTAP Saturday Class Session 6 for Grade 8

One of the Mathletes named Patrick John Uy, a grade 8 student  from F Torres High School sent problems of their session 6 Saturday MTAP program. As a medium of help to those students like him who are eager to earn. We will demonstrate the solutions to problems that bothers them.

Problem 1:

Express the following products as reduced fractions.

$(1-\displaystyle\frac{1}{2})(1-\displaystyle\frac{1}{3})(1-\displaystyle\frac{1}{4})\ldots(1-\displaystyle\frac{1}{2013})$

Solution:

Observe the following,

$1-\displaystyle\frac{1}{2}=\displaystyle\frac{1}{2}$

$1-\displaystyle\frac{1}{3}=\displaystyle\frac{2}{3}$

$1-\displaystyle\frac{1}{1}=\displaystyle\frac{3}{4}$

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$1-\displaystyle\frac{1}{2013}=\displaystyle\frac{2012}{2013}$

By expressing them to product of their differences we have,

$(\displaystyle\frac{1}{2})(\displaystyle\frac{2}{3})(\displaystyle\frac{3}{4})\ldots(\displaystyle\frac{2012}{2013})$

Cancel out terms from 2 to 2012 we only left with

$\displaystyle\frac{1}{2013}$

Problem 2:

One angle of the triangle has a measure of 40º. Find the measures of the other two angles if the difference between their measures is 10º.

Solution:

Use the fact that the sum of the interior of a triangle is 180º.

Let A, B, and C are the measures of the three angles,

$A+B+C=180$

But one of the angles is 40º

$40+B+C=180$

$B+C=140$

It is given in the remaining angles that the difference of their measures is 10º, hence

$C-10+C=140$

$2C=150$

$C=75^\circ$

The other angle (B) is 10 degrees less than the angle C. Thus B=75-10=65º.

Worked Problem 3:

The height of a cylindrical container of radius r  is 30 cm.  what will be the height of the water if it is poured in another cylindrical container which is four times the radius of the other.

Solution:

Let $r_1$ be the radius of the first container

Let $h_1$ be the height of the first container

Let $r_2$ be the radius of the second container

Let $h_2$ be the height of the water in the second container

Given:  $h_1=30 cm$,  $r_2=4r_1$

Using the formula for the volume of the 1st Cylinder we have,

$V_1=\pi r_1^2h_1$

$V_1=\pi r_1^2(30)$

$V_1=30\pi r_1^2$

Using the formula for the volume of the second Cylinder we have,

$V_2=\pi r_2^2h_2$

But,  $r_2=4r_1$

$V_2=\pi (4r_1)^2h_2$

$V_2=16\pi r_1^2h_2$

Using a little common sense, the volume of the water in both container must be the same. ( Math approach)

$V_1=V_2$

$30\pi r_1^2=16\pi r_1^2h_2$

Cancel out like terms in both sides of equation and solving for $h_2$ we have,

$h_2=\displaystyle\frac{30}{16}=\displaystyle\frac{15}{8} cm$