Aptitude - Problems on Numbers - Discussion

Anyone can explain this problem? Please anyone help me to solve this problem.

Consecutive mean 1 after one. i.e 1, 3, 5, 7etc.

3 times of 1st consecutive means 3x.
Twice of 3rd consecutive 2(x+4).

3 more than twice of 3rd consecutive 2(x+4)+3, then.

3x = 2(x+4)+3.
3x-2x = 8+3.
x = 11.

3rd integer is x+4 = 11+4 = 15.

Three consecutive integer are 2x+1, 2x+3, 2x+5.

Now according to question.

3(2x+1) = 2(2x+5)+3.

6x+3 = 4x+10+3.

2x = 10.

x = 5.

And now third number is 2x+5 = 2*5+5 = 15.

15 is a answer.

Three consecutive odd numbers x+1, x+3, x+5.

(x+1)+(x+3)+(x+5) = 3+2(x+5).

x = 4. So the third integer x+5 = 4+5 = 9.

I think the answer will be 9 because in this problem three consecutive odd numbers is asked then, how can we start with 0 is my doubt.

If you take x+1 as first odd no. Then put 1 in the place of x, it becomes 1+1=2 its not odd integer you got it.

Can any one say how to solve this in 32 seconds?

But whats the reason to multiply 3 on both sides?

At first three number is x, x+2, x+4 these are three consecutive number.

Then according to question (three times the first of three consecutive odd number) here is our first number is x then,

Our first condition will be 3x----- (1).

And then 3 more than twice the third (there third is x+4).

3 more means 3+, twice means 2.

Then twice of third is 2(x+4).

Then again 3+2(2x+4) -------- (2).

Then from equation first and second we get:

3x = 3+2(x+4).
3x = 3+(2x+8).

3x = 3+2x+8.
3x-2x = 11.
x = 11.

Then our third integer is x+4 ==> 11+4 = 15.

Our third integer is (x+4) = 15.

I got idea about this solution. But if we assume the solution by question as,

3 times the x of three consecutive integers whether is my assumption also possible na.

3(x+X+2+x+4) = 3+2(x+4) // I consider it for first 3 con int respectively.

Anyone can explain my logic please.