Pre-Algebra Averages problem

Question

Justin threw a football as far as he could 3 times and had an average of 49 yards. He threw it again and his new average was 37 yards. How far was his fourth throw?

Answer 1

(49+49+49+x)/4=37

(147+x)/4=37

147+x=4*37

147+x=148

x=148-147

x=1 yard

his fourth throw was only 1 yard

look at it this way...

an average of 49 in 3 throws is the same as 3 throws of 49 each

49-12=37, 49-12=37, 49-12=37

each throw of 49 went down 12 yards

12+12+12=36 and 36+1=37, his new average

 

Answer 2

Let's call his throws a, b, c, and d.

First, let's write this equation:

 

(a + b + c) / 3 = 49

 

This is just the equation for the average of the first three throws.

 

Now let's go ahead and write the equation for the average of all four throws:

 

(a + b + c + d) / 4 = 37

 

Okay. Now let's simplify the first equation a little:

 

a + b + c = 49 * 3 = 147

 

And then the second:

 

a + b + c + d = 37 * 4 = 148

 

Now, let's subtract the two equations:

 

  a+b+c+d=148

-(a+b+c    =147)

               d=1

 

The fourth throw was just 1 yard.

 

We could've figured that out by seeing that the total distance thrown after three throws was 147 and after the fourth was 148, so his fourth throw had to be 1 yard.

Answer 3

The sum of his first four throws was 147 (3 x 49)

To this was added his fourth throw, f.

(147 + f) / 4 = 37

 

Can you solve for f?