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}

.

.

.

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

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Dan

Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.
Dan

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