Aptitude - Clock - Discussion

How 55 min spaces are gained in 60 mins?

This type of problem
remember these formula:
(5x+30)12/11 where x<6
(5x-30)12/11 where x>6

In this problem 7 and 8 clock
So here 7 is grater than 6(x>6)
So we use this formula :(5x-30)12/11 where x>6
Here x=7 so :(5x7-30)12/11= 60/11=5+5/11
So the answer is: (5+5/11) minute and 7 past.

ANOTHER METHOD:
In 7 o'clock there are angle between hands= 5*30=150 degree
we have to make that angle = 180 degree
let after X min
min hand will increase angle by = X*6 =6X,on 150 degree

1 degree of min = 1/12 degree of hr
6X degree of min = 6X/12 = X/2 degree of hr (decrease angle on 150)
so angle have to be between hands=180
150+6X-X/2=180
x=5+5/11
ANS=(5+5/11) minute and 7 past

@amit:11/2min-30h this formula applicable only when h<6
if u put h=6 or above then the min>60 which is not acceptable(for the 180 degree),
in this problem if we put that formula then the ans is 77(10/11)
but in that case the h=7 so we apply the another formula which is 30h-11/2
in this case if we put that then formula the ans is 5(5/11)
if u apply that formula then u will find that min<60 which is acceptable

BASICALLY WE NEED THIS 2 FORMULAS BUT IN DIFF CASE...
hope u will understand:)

What is the formula? whether its 11/2*min-30hr or 30hr-11/2*min?

For straight line angle =180
The formula is for finding angle =30H-(11/2)*M
where H =>Hours
M => Minute.
Now

30H-(11/2)*M=180
30*7-(11/2)*M=180
210 - (11/2)*M=180
-(11/2)*M=180-210
-(11/2)*M=-30
11*M=60 (Cross Multiply)
M=60/11 minutes
means 5+5/11 minute

So Time will be

(5+5/11) minute and 7 past

This problem is very simple . . . .
For this problem
in clock 7 ki opposite 1 i.e 5min
hence at 5:[5/11]min past 7 O'clock at that time hands of a clock be in straight line
or
at 60/11 min past 7
the hands of a clock on straight line . . . . .

Please tell me about when minute space are add or subtract.

Can some one give me a simple way to solve these watch question ?

How 55 min. spaces are gained in 60 min?

Can anybody explain?