comparing the means of multiple distributions I wish to compare the means of three independent data sets.  These are exam scores that are from class A and class B and class C.  Class A and B are normal and tested ok for homogeneity.  A T-Test between class A and B revealed that the sets are significantly different and looking at the mean and median, and I determined class A performed better than B (examining the means).  When comparing the means between A and C and then B and C using the Mann Whitney U test, the both scenarios didn't show significance.  I thought that the comparison between A and C would test significant being b/c of the results of the T-Test.
I recently learned that when doing a multiple amount of comparisons, I have to apply the Bonferroni correction to my p value (p => .05/3 => .0167).  When comparing the means of A/B, B/C and A/C using p = .0167 all of the result comparison showed no significance.  My question is,..when I write my results, should I determine there was no significant relationship found between the settings.  That is to say, it wasn't found class A, B and C performed alike?
I don't think I should ignore the T Test result b/c a parametric test is always better.  What do you think?
Thanks for your help in advance.
 A: Moving yesterday's comments to Answer format to avoid abbreviations and make
room for some illustrations:
It seems you're saying that the sample means are arranged $\bar X_A>\bar X_C>\bar X_B.$ Using a t test, you have found a significant difference between groups A an B with a t test. But using nonparametric tests, you did not find significant differences not between A and C or between C and B.
It may be annoying not to know the exact pattern of differences, but it is not unusual that one is not able to resolve the exact pattern. 
I hope you can try a Kruskal-Wallis test on data for all three groups. It may show not all equal, but ad hoc rank-based two-sample tests may still not resolve whether C differs form A or from B. 
Also, the rank-based tests (Mann-Whitney-Wilcoxon and Kruskal-Wallis) may give different results from t tests and standard ANOVAs. They use different criteria. For normal data, t tests and standard ANOVA are more powerful (more likely to find true differences), but they are different tests using different criteria. Especially for non-normal data, the rank based tests may find significant differences where the normal-based t tests and ANOVA do not. 
Addendum (in response to Comment):
Thanks for the clarifications and additional detail. Roughly speaking, the Kruskal-Wallis test is a generalization of the two-sample Wilcoxon rank sum (equivalent to Mann-Whitney U) to handle $k$ groups--just as a one-way ANOVA is a generalization of the pooled 2-sample t test to handle $k$ groups. 
I am not sure what you see in group C to prevent your doing one-way ANOVA. In case you fear unequal variances, there is a Welch version of one one-way ANOVA that doesn't require variances to be equal.

Here are simulated data somewhat similar to yours, which I use to illustrate some
procedures in R statistical software.
set.seed(718)  # retain 'set.seed' for exactly same fake data; omit for fresh dataset
A = rnorm(30, 70, 15);  B = rnorm(30, 65, 15);  C = rnorm(30, 67, 15)
Scores = c(A, B, C);  Classes = as.factor(rep(1:3, each=30))  # 'as.factor' crucial
boxplot(Scores ~ Classes, col="skyblue2", pch=20)

mean(A); mean(B); mean(C)
[1] 69.70417
[1] 60.35609
[1] 64.89248


A one-way ANOVA finds differences:
anova.out = aov(Scores ~ Classes)
summary(anova.out)
            Df Sum Sq Mean Sq F value Pr(>F)  
Classes      2   1214   607.0     3.3 0.0415 *
Residuals   87  16002   183.9                 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

A comparison of A with B is almost significant, but C is not significantly
different from either A or B. (A rather conservative Bonferroni adjustment ensures that
the family of comparisons has an error rate below 5%.) 
[See this demo for a discussion on the procedure so far.]
A t test comparing A and B, not
shown, has P-value about 0.02.
pairwise.t.test(Scores, Classes, p.adj="bonf")

        Pairwise comparisons using t tests with pooled SD 

data:  Scores and Classes 

  1     2    
2 0.056 -    
3 0.663 0.745

A Kruskal-Wallis test shows differences in medians.
        Kruskal-Wallis rank sum test

data:  Scores by Classes
Kruskal-Wallis chi-squared = 6.3386, df = 2, p-value = 0.04203

A 2-sample Wilcoxon test finds a significant difference between A and B.
wilcox.test(A,B)

        Wilcoxon rank sum test

data:  A and B
W = 597, p-value = 0.02959
alternative hypothesis: true location shift is not equal to 0

Wilcoxon tests, not shown, find a significant difference between A and C, but not
between B and C. (The accumulated error rate looking at several Wilcoxon tests is not controlled.)
