78 students recently took an assessment and earned either a B or a C....

Question

78 students recently took an assessment and earned either a B or a C. The number of students who earned a B is twice as many as those who earned a C. How any students earned a B and how many earned a C?

Answer 1

You could also draw a bar model out on this one. Go visual. If you read the problem, it's like saying 78= B+B+C.  If you drew a rectangle and above it, wrote 78.  There are 3 parts to this total of 78.  If you drew lines, dividing bar into thirds... and if you didn't know the division, you could deal out your 'students' by putting a tally mark in each of the 'thirds' sections. Once all 'students' were placed in the model, you could then go back and count up the total for the B's and then the C's.  Good luck!

Answer 2

So . . . 2/3 of students got Bs & 1/3 got Cs

78 /3 =26

26 x 2 = 52

check: 52 (Bs) + 26 (Cs) = 78

Answer 3

Alright, Nick, this is a system of equations.

 

B + C = 78

B = 2C  

 

The second equation may be backward from what you would think, but if there were 10 B's, then there were 5 C's, and 10 = 2 x 5

 

We can take the "value" for B from the 2nd equation and substitute it for B in the first equation.

 

2C + C = 78

      3C = 78

        C = 26

 

Then we can take that and plug it into either equation.

 

B = 2(26) = 52

 

So, there were 52 B's and 26 C's.

 

Hope this helps!