Is p value sufficient to compare the significance What we have is the following:
Three groups of students are taking the same pre-test and post-test on a Math subject. In between they get to play with tutoring system A (for first group), tutoring system B (for second group), and tutoring system C (for last group). Here are the results:
Group A:      Pre-test  Post-tet
Student 1a       20%        34%
Student 2a       32%        45%
Student 3a       12%        40%
Student 4a       52%        55%
Student 5a       37%        39%
Group B:      Pre-test  Post-tet
Student 1b       16%        31%
Student 2b       22%        35%
Student 3b       32%        40%
Student 4b       25%        31%
Student 5b       27%        31%
Group B:      Pre-test  Post-tet
Student 1b       33%        51%
Student 2b       12%        25%
Student 3b       42%        45%
Student 4b       39%        45%
Student 5b       17%        33%
Question (1): Would it be sufficient to run a paired t-test for each group and use the p values to compare the effectiveness of three systems.
Question (2): If not, what additional test that I can run to make the comparison more convincing
 A: Doing three paired t tests would not tell you what you want
to know. Each paired test would tell you whether students
in one of the three groups (tutorial systems) shows
improvement. The answer to that is Yes for all groups,
because each group shows all five students with positive
improvement scores ($1/2^5 < .05$). But this says nothing at all about the relative
effectiveness of the three groups.
One appropriate procedure would be to find an improvement score for each
person 14 for student 1a. Then do a one-factor (one-way) ANOVA
on the three sets of improvement scores. If the F-test for
the ANOVA shows significance (i.e., that not all three
tutorial systems are equally effective), then you would need
to do some kind of multiple comparison procedure to see
what pattern of difference is plausible. 
A complication is that these are percentages and so not
really normally distributed. There are several possibilities
if you suspect (or find from a test on residuals) that 
data are far from normal: Kruskal-Wallis and permutation
tests are popular alternatives. 
The title of the Question asks whether the P-value is sufficiently small to declare
significant differences. But you don't show a P-value, so
that it not a good title for this Question.
I put these data into Minitab statistical software (quickly
and without proofreading, better check). The resulting
ANOVA is as follows:
 Source  DF     SS    MS     F      P
 Group    2   26.1  13.1  0.42  0.668
 Error   12  375.6  31.3
 Total   14  401.7

The P-value shown is not sufficient to reject the null
hypothesis that improvement scores are the same for all
three tutorial programs. I don't know whether you have
studied ANOVA or not. If not, I wonder about the purpose
of the question. If so, I suppose you will know how to
interpret the ANOVA table.
The largest difference among the three sample means is
a little over 3 (not shown). This ANOVA procedure assumes that the
variances are the same in the three groups. It estimates
the common group variance to be about 31 (Ms for Error). 
Note:  When significance is not found, it is good statistical
practice to do power computations to see whether the
study had any chance of success in detecting differences
among group population means.
(a) A one-way ANOVA with 3 groups, 5 replications within
each group, and a common variance of 31 has only about
one chance in ten of detecting a difference of 3 in
improvement scores, even if it is real. 
(b) It is difficult
to say how large a difference investigators would find
interesting. In order reliably to detect a difference
of 5 points between the best and worst tutorial system
one would need over 30 students in each group.
