Therefore, Kenny scores 31 points in the last test.

[Marcelo Arevalo]

After the fourth test, Kenny's average mark rises by 5 points, but after the fifth test, it drops by 9 points.
If his total score in the last two tests is 122 points,
How many points does he score in the last test?

 

[Marcelo Arevalo]

I used Trial & error method;
can somebody teach me how to do this in algebra method?

here's my solution; (After 17 trials, using calculator.. I am dis-appointed in myself.. shame on me! )

Since we know the fact that the last two tests is 122 points in total. I focused my attention to that given fact in Trial & error method.
122-31 = 91
91 is Kenny's fourth test score
the other three test I make them 71, 71, 71 = 71 is the average

solving for the average of four test
71+71+71+91 = 304 / 4 = 76

from the average of three test = 71 then after the fourth test Kenny's mark rises to 76 which satisfy the condition in the problem. it rises by 5 points.

then :
71+71+71+91+31 = 335/5 = 67

then ; from the average of 76 it drop by 9 points into 67

therefore the answer is 31 his score in the last test.

 

[MarkFL]

The statement:

"After the fourth test, Kenny's average mark rises by 5 points"

allows us to write:

\(\displaystyle \frac{a+b+c+d}{4}=\frac{a+b+c}{3}+5\tag{1}\)

Then the statement:

"after the fifth test, it drops by 9 points"

allows us to write:

\(\displaystyle \frac{a+b+c+d+e}{5}=\frac{a+b+c+d}{4}-9\tag{2}\)

And finally, the statement:

"his total score in the last two tests is 122 points"

allows us to write:

\(\displaystyle d+e=122\tag{3}\)

Now, we are being asked to find the value of $e$, so from (3), we obtain:

\(\displaystyle e=122-d\)

And from (2), we find:

\(\displaystyle e=\frac{a+b+c+d}{4}-45\)

And from (1), we have:

\(\displaystyle a+b+c=3d-60\)

Hence:

\(\displaystyle e=d-60\)

Thus:

\(\displaystyle 122-d=d-60\implies d=91\)

And so:

\(\displaystyle e=91-60=31\)