Peter & Diane have the same average in mathematics. On the next test Diane scored 97% and her aver age improved to 90%, but Peter scored 73% and his average decreased to 87%. How many tests had proceeded this last test?

If there were already n tests, then if Diane scored d and Peter scored p, then with the current average of a,

n*a + 97 = (n+1)(90)
n*a + 73 = (n+1)(87)
so, we can eliminate the a and have

(n+1)(90)-97 = (n+1)(87)-73
(n+1)(3) = 24
n+1 = 8
n = 7

To find out how many tests had proceeded this last test, we need to compare the averages before and after the last test.

Let's name the number of tests taken before the last test as 'x'.

Before the last test:
- Diane's average was the same as Peter's average. Let's call it 'A'.
- Peter's average was 87%.

After the last test:
- Diane's average improved to 90%.
- Peter's average decreased to 87%.

To find the average, we can use the formula:
Average = (sum of scores) / (number of tests)

Based on the information given, we can set up two equations:

Diane's average:
(97% + sum of scores before) / (x + 1) = 90%

Peter's average:
(73% + sum of scores before) / (x + 1) = 87%

Now, let's solve these equations to find the value of 'x', which will give us the number of tests taken before the last test:

(97% + sum of scores before) / (x + 1) = 90%
97% + sum of scores before = 90% * (x + 1)
97% + sum of scores before = 90% * x + 90%

Similarly, for Peter's average:

73% + sum of scores before = 87% * x + 87%

Now, we can solve these two equations to find the value of 'x'.

97% + sum of scores before = 90% * x + 90%
73% + sum of scores before = 87% * x + 87%

Subtracting the second equation from the first:

97% - 73% = 90% * x + 90% - (87% * x + 87%)
24% = 3% * x + 3%

24% - 3% = 3% * x
21% = 3% * x

Dividing both sides by 3%:

(21% / 3%) = x
7 = x

Therefore, the number of tests that proceeded the last test is 7.