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.



Observe the following,








By expressing them to product of their differences we have,


Cancel out terms from 2 to 2012 we only left with


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.


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,


But one of the angles is 40º



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




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.


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)


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

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