Alex's test scores

We know that Alex's average after the 5 tests (after rounding) was an $85$ (therefore, $\frac{x}{5} = 85$ ), but we would like to know the lowest possible exact average that when multiplied by 5 gives an integer (the sum of his test scores). Since $84.5 \times 5 = 422.5$ is not an integer, $84.6$ is the lowest possible average, which gives us a test score sum of $423$ .

In order to get the lowest possible grades on two tests (the 3rd, x, and 4th, y), we must maximize the sum score from the other three tests. Because each score must be distinct and under $95$ , those three grades are $92$ , $93$ , and $94$ . So now we have the equation $x + y + 92 + 93 + 94 = 423$ . We can reduce that to $x + y = 144$ .

In order to minimize the fourth grade, y, we must maximize the third grade, x. This gives us $x = 72$ and $y = 72$ but the grades must be distinct so the third grade becomes a $71$ and the fourth grade becomes a $73$ ...our answer. =)

He never scored 95 or more:

Let us take he scored 94 in first 93 in second and 92 in fifth. Let the scores in third and fourth be x and y. Let us take the average is exactly 85 so the sum of scores of all test must be $85 \times 5 = 425$

$94 + 93 + x + y + 92 = 425 \Rightarrow x + y = 146$

the lowest possible value of y is 74 as x less than y but then the average would be exactly 85.

So lowest possible fourth test score is 73

Since the scores are integers their sum is also an integer and therefore when we divide the sum by 5 ,we have to get an integer or a real number with with 2,4,6,8 in its first decimal place.since it is given in the problem that the average score was rounded to 85 the average score must be a real number having either 2 or 4 or 6 or 8 in its decimal position.

For minimizing the score of the fourth test we have to maximize the scores of the first,second and fifth test.

Since the scores are distinct and < 95 we the scores must be 94 93 and 92 .Their sum turns out to be 279.

Let his scores in test 3 and 4 be a and b we know that a<b and the average score must be 84.6 , 84.8 , 85.2, 85.4.(in order to round off to 85)

Since a+b should be minimum (since a is least and b should be minimized) a+b = 84.6*5 - 279

a+b = 144

since a<b

implies a=71 and b=73

Average scores that can be rounded to the nearest integer of 85 included 84.5, 84.6, 84.7, 84.8, 84.9, 85, 85.1, 85.2, 85.3 and 85.4. The total score will be 422.5, 423, 423.5, 424, 424.5, 425, 425.5, 426, 426.5 and 427 respectively.

Let the lowest score of third test be x. To obtain the lowest possible score for the fourth test, assume the score of fourth test is only slightly higher than x, that is x+1. Since all his scores are less than 95 and to get lowest possible score for the fourth test, assume the score for the remaining three tests be 92, 93 and 94.

Total score = 92+93+94+x+x+1=2x+280

Total scores of non integer are eliminated since all scores are integer, so do the total score. The possible total scores are now 423,424,425,426 and 427.Total scores of odd number are eliminated since they will result in non integer x. The possible total scores are now 424 and 426. To get lowest possible score for the fourth test, use the lowest total score, 424.

Total score = 2x+280 =424

x= 72

The fourth test has a score of x+1=72+1=73

Name the 5 scores $A > B > C > D > E$ . Combining the distinctness ( $A \neq B \neq C \neq D \neq E$ ) and upper bound ( $A < 95$ ) constraints with the goal of finding the lowest value for $D$ , we determine $A=94, \, B=93$ and $C=92$ because for $D$ to be as low as possible, $A, \, B$ and $C$ must be as high as possible.

We can write these values into the average equation: $\begin{aligned} \frac{94 + 93 + 92 + D + E}{5} &= 85 \\ 279 + D + E &= 425 \\ D + E &= 146 \end{aligned}$

The minimum $D>E$ is $E+1$ , which gives us $2E + 1 = 146$ . Solving shows that $D=73.5$ and $E=72.5$ . $D$ and $E$ must be integers, however. A quick check shows that $D=73$ is the minimum such integer satisfying all the constraints. $\square$

This is like my own exam results. Half of my best subjects plus another worst half. But my best subjects put 95+ papers on the table so an A average would still be achievable then. Even better when the school changed it into weighted CGPA system. C in History and Biology, but who cares?
If we want to minimize the second lowest marks, we have to push the three good ones as high as can be, and the average down to the lowest possible.
(Highest three) + (Lowest two) ≥ 5x(Lowest possible average).
(94 + 93 + 92) + (n + (n+1)) ≥ 5 * (85 - 0.5).
279 + 2n+1 ≥ 5 * 84.5.
2n + 280 ≥ 422.5.
2n ≥ 142.5, and n is integer.
n ≥ 71.25, so n = 72.
The question asked for the second lowest score, so answer is n+1 = 73.