Monday, April 23, 2018

Algebra 2 Problems of the Day (open-ended)

Continuing with daily Algebra 2 questions and answers.

More Algebra 2 problems.

January 2018, Part II

Questions in Part II are worth 2 credits. All work must be shown or explained for full credit. A correct numerical answer without work is only worth 1 credit.

27.A formula for work problems involving two people is shown below.

1/t1 + 1/t2 = 1/tb
t1 = the time taken by the first person to complete the job
t2 = the time taken by the second person to complete the job
tb = the time it takes for them working together to complete the job

Fred and Barney are carpenters who build the same model desk. It takes Fred eight hours to build the desk while it only takes Barney six hours. Write an equation that can be used to find the time it would take both carpenters working together to build a desk.
Determine, to the nearest tenth of an hour, how long it would take Fred and Barney working together to build a desk.

The equation needed to solve this would be

1/8 + 1/6 = 1/tb or 1/8 + 1/6 = 1/x

Once you have this, you can solve for x (or tb).

3/24 + 4/26 = 1/x
7/24 = 1/x
x = 24/7 = 3.428571... = 3.4 hours

Alternatively, multiply the entire equation by (8)(6)(x) to eliminate the fractions:

(8)(6)(x)(1/8) + (8)(6)(x)(1/6) = (8)(6)(x)(1/x)
6x + 8x = 48
14x = 48
x = 48 / 14 = 3.4 hours.

Also of note, whoever wrote this questions obviously likes The Flintstones.

28.Completely factor the following expression:

x2 + 3xy + 3x3 + y

First, rewrite the expression as 3x3 + x2 + 3xy + y
Factor by grouping: x2(3x + 1) + y(3x + 1)
Factor again: (x2 + y)(3x + 1)

Comments and questions welcome.

More Algebra 2 problems.

No comments: