Fraction of the class that works alone

Question

In a class that contains both 5th and 6th graders, each student must do a class presentation. Each student may do the presentation alone or with a class partner. A 5th grader’s partner must be an 6th grader and an 5th grader’s partner must be a 6th grader. If two-thirds of the 5th graders and three-
fifths of the 6th graders work with partners, what fraction of the class works alone?

 

I am not really sure how to get started with this problem. 

 

Answer 1

Well this is not a simple fraction problem. You need to set up equations.

 

First let

x = fraction of 5th graders in the WHOLE class

y = fraction of 6th graders in the WHOLE class

 

Eq. 1 is:

 

x + y = 1 (5th graders plus 6th graders is the WHOLE class)

 

Eq. 2 is:

 

(2/3)x = (3/5)y (Since they are partners, the fraction OF THE FRACTIONS should be equal. Get it? Lol.)

 

Solving, x = (9/10)y, substituting that to the eq 1. you'll get:

 

(9/10)y + y = 1

 

y = 10/19, and by using any equations, you'll get x = 9/19.

 

From these you could ASSUME a population of 19 in the class: 9 5th graders and 10 6th graders.

 

Using the forever alone fractions 1/3 for 5th and 2/5 for 6th, you will get 3 and 4 respectively.

 

Adding them will be 7 over the total population of 19 = 7/19