MHB Therefore, Kenny scores 31 points in the last test.

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Kenny's score in the last test is determined to be 31 points. After the fourth test, his average increased by 5 points, but it dropped by 9 points after the fifth test. The total score for the last two tests is 122 points, leading to the conclusion that his fourth test score is 91. Using algebraic equations derived from the problem's conditions, it is confirmed that the calculations align with the average changes specified. Therefore, the final score for Kenny in the last test is 31 points.
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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?
 
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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.
 
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The statement:

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

allows us to write:

$$\frac{a+b+c+d}{4}=\frac{a+b+c}{3}+5\tag{1}$$

Then the statement:

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

allows us to write:

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

And finally, the statement:

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

allows us to write:

$$d+e=122\tag{3}$$

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

$$e=122-d$$

And from (2), we find:

$$e=\frac{a+b+c+d}{4}-45$$

And from (1), we have:

$$a+b+c=3d-60$$

Hence:

$$e=d-60$$

Thus:

$$122-d=d-60\implies d=91$$

And so:

$$e=91-60=31$$
 
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