Question 846871
To solve, we need to know how to compute an average so we can correctly set up the equation.   An average is computed by added each number in a group together and then dividing that result by the total of numbers that were added together.  Let's look at what we know:


Average on 3 Tests:  78
Mark on First Test:  86
Mark on Second Test: x
Mark on Third Test:  y


We are told that the mark on the second test was 3 more than the mark on the 3rd test.  So, we can change the Mark on Second Test to read as:  y + 3:


Average on 3 Tests:  78
Mark on First Test:  86
Mark on Second Test: y + 3
Mark on Third Test:  y


Now, we can set this up as an equation, based on how we compute an average of a group of numbers:


{{{(86+(y+3)+y)/3=78}}}


Next, add everything in the numerator of the fraction on the left side of the equation, which gives us:


{{{(89+2y)/3=78}}}


Next, we need to get the y all by itself on the left side of the equal sign.  To do this, we can first multiply both sides of the equation by 3, which gives us:


{{{89+2y=78*3}}}----->


{{{89+2y=234}}}


Next, subtract 89 from both sides of the equation:


{{{89-89+2y=234-89}}}----->


{{{2y=145}}}


Then, divide both sides by 2:


{{{y=145/2}}}----->


{{{y=72.5}}}


We have just figured out what the 3rd test score is.  To find out the second test score, which we are told is 3 more than the 3rd test score, we will add 3 to 72.5, giving us 75.5.


Therefore Brian got a 75.5 and 72.5 on his second and third tests, respectively.