Created at 18 Dec 2015 11:46 Last update at 24 Jun 2016 06:06
The final term exam problem I created have raised questions among my students, especially the base counting problem. For their note, the whole exam was a trick in itself. Carry on reading.
Table of Contents

Said exam allowed all students to carry their own handwritten note (A4 paper), time limit of 120 minutes, and have the following questions:
 Points are given based on how well their assignment is (25 points, can be up to 30 points on extraordinary students)
 Identify 5 UML Diagrams from given images (10 points)
 Identify 10 video game controllers from given images (10 points)
 Identify 10 video game consoles from given images (10 points)
 Name all phases of Software Development Life Cycle (5 points)
 Name all phases of Unified Process (5 points)
 Convert all hexadecimal values into binary and decimal (20 points)
 Create an activity diagram based on the student's typical Sunday (10 pts)
From the first glance, #2 to #6 as well as #8 can be done in more less 20 minutes. Students gain points for #1 automatically just for doing the exam. Even if they skip #7, they already gain 80 points, more than enough to guarantee A for the majority of students (around 80% have the potential to reach A).
Dreaded #7
However, students' feedback was mostly negative to #7, the base counting problem. Below is the problem of #7 as well as the reason for the number itself.
Exam Problem
Convert the following hexadecimal values into binary and decimal, conversion method must be provided (20 points)!
#  Hexadecimal 

A  0000 000F h 
B  0000 00F0 h 
C  0000 0F00 h 
D  0000 F000 h 
E  0001 0000 h 
F  0000 0063 h 
G  0000 03E7 h 
H  0000 270F h 
I  0001 869F h 
J  05F5 E0FF h 
Trick 1: Time Soak
#7 was intended as a time soak, i.e. to make students waste all their precious time before they do the easy #8. Students who were keen to this, will do #7 last. As stated above, 80% of this exam can be done in 20 minutes. The other 20% requires student to muster their concentration. From efficiency point, #7 is plain trap.
Trick 2: The Easy Points
#7 point A to F, 6 points total, are relatively easy compared to the other 4. Keen students will focus on this first 6, gaining a hefty 12 points and skip the rest.
Trick 3: Imposing Length
From point A to E, students will notice an increasing number of writing required since counting method is required. Peeking their neighbor did not help since the answer length that their friend wrote was imposing enough to make them reluctant. In reality, only converting to decimals required lengthy calculations, converting to binary has its own trick.
Trick 4: Direct Binary Conversion
I only taught how to convert hexadecimal to decimal (and vice versa), and binary to decimal (also vice versa). By this available methods, students automatically convert hexadecimal to decimal, then decimal to binary, requiring a twostep method which is not only redundant, but also very long.
Of all 70 students (both night and day class), only 1 students succeeded in directly converting hexadecimal to binary. The exam problem itself even foreshadow this by stating "hexadecimal values into binary and decimal".
Hexadecimal to Binary: The Quick Conversion
Step 1
First, make a table to convert Decimal to Hexadecimal and Binary:
Decimal  Hexadecimal  Binary 

0  0  0000 
1  1  0001 
2  2  0010 
3  3  0011 
4  4  0100 
5  5  0101 
6  6  0110 
7  7  0111 
8  8  1000 
9  9  1001 
10  A  1010 
11  B  1011 
12  C  1100 
13  D  1101 
14  E  1110 
15  F  1111 
Step 2: Notice the pattern on A to D
If you convert hexadecimal to decimal, to binary, here's what you get from the first 4 numbers:
#  Hexadecimal  Decimal  Binary 

A  0000 000F  15  0000 0000 0000 1111 
B  0000 00F0  240  0000 0000 1111 0000 
C  0000 0F00  3840  0000 1111 0000 0000 
D  0000 F000  61440  1111 0000 0000 0000 
Noticed something? Here's a detailed table:
Problem A
0  0  0  0  0  0  0  F  

0000  0000  0000  0000  0000  0000  0000  1111 
Problem B
0  0  0  0  0  0  F  0  

0000  0000  0000  0000  0000  0000  1111  0000 
Problem C
0  0  0  0  0  F  0  0  

0000  0000  0000  0000  0000  1111  0000  0000 
Problem D
0  0  0  0  F  0  0  0  

0000  0000  0000  0000  1111  0000  0000  0000 
Each 4 binary digits exactly represents a single hexadecimal digits.
Step 3: Crosschecks with E and F
You can crosscheck your theory with E and F as follows:
#  Hexadecimal  Binary 

E  0001 0000  0000 0000 0000 0001 0000 0000 0000 0000 
F  0000 0063  0000 0000 0000 0000 0000 0000 0110 0011 
Detailed table:
Problem E
0  0  0  1  0  0  0  0  

0000  0000  0000  0001  0000  0000  0000  0000 
Problem F
0  0  0  0  0  0  6  3  

0000  0000  0000  0000  0000  0000  0110  0011 
Step 4: Solve the Remaining Problems
#  Hexadecimal  Binary 

G  0000 03E7  0000 0000 0000 0000 0000 0011 1110 0111 
H  0000 270F  0000 0000 0000 0000 0010 0111 0000 1111 
I  0001 869F  0000 0000 0000 0001 1000 0110 1001 1111 
J  05F5 E0FF  0000 0101 1111 0101 1110 0000 1111 1111 
This is an extremely quick way, since J's decimal value is 99 999 999. Imagine doing 99 999 999 MOD 2 all the way until you find the remaining dividend is 3 or 2 or 1.
Here's a detailed table on problem J:
Problem J
0  5  F  5  E  0  F  F  

0000  0101  1111  0101  1110  0000  1111  1111 
That was not so hard, was it?