Clock into pieces !!?

Question

If the clock(Conventional clock with numbers from 1 to 12 in order) is cut into 3 pieces such that the sum of numbers on each piece are in Arithemetic Progression(A.P) with a common difference of 1.
What is the sum of even numbers in the group where 5 is present?
A. 4 B. 10 C. 12 D. 14

Answer 1

The sum of the numbers 1 to 12 = (1 + 12)(2 + 11)(3 + 10) ... (6 + 7). That's 6 sets of 13 = 78.

Divide that by 3 and you get an average of 26.

So you need each section to add to 25, 26 and 27 respectively.

I assume you have to keep consecutive numbers together... one way to do that is:

3, 4, 5, 6, 7 = 25
8, 9, 10 = 27
11, 12, 1, 2 = 26

The sum of the even numbers in the group where 5 is present {3, 4, 5, 6, 7} is 4 + 6 = 10.

The answer is B) 10.

Answer 2

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. In this case, they mean 1, 2, 3, ...

So:
1+2+3+... = a
... = b
... +12 =c

1+2+3=6
5+4=9
Try again...

1+2+3+4=10
5+6=11
7+8=15
nope...

1+2+3+4+5=15
6+7=13
...

1+2+3+4+5+6=21
7+8+9=24
...

1+2+3+4+5+6+7=28
8+9+10=27
11+12=23
...

1+2+3+4+5+6+7+8=36
9+10+11=30
.. oh no, what to do...

Now you see the devious nature of your teacher
What if you cut the twelve in half and seperated the one and the two...
10+11+1=22
2+1+2+3+4+5+6=23
7+8+9=24

Tricky...

I'll let you do the second part on your own.

Answer 3

14

Answer 4

b. 10

in the pieces, the numbers will be distributed in the following way
{3,4,5,6,7} [sum. 25]
{8,9,10} [sum. 27]
{11,12,1,2} [sum. 26]