Aptitude - Clock - Discussion

At 7o clock hour, the hand will be 210 degrees, at 5min minute hand will be 30 degrees, (same line).

So consider 5min 5/11 for hour hand and just 5/11min for minute hand.

For every 1min hour hand rotates by 0.5 degrees so for 60/11 min it rotates by 30/11 degree.

For every min minute hand rotates by 6 degrees for 5/11 min it rotates by 30/11 degree hence equal.

So at 7 past 5min 5/11 it will be 210+30/11 degree in hour hand and 30+30/11 for minute hand difference is 180 degree thus they are in same line.

@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:)

Angle = |11/2 * min-30 * hr| as said by someone.

Where || indicates take positive value.

How is this derived?

=> For every hour, hour hand rotates by 30 degrees (360/12 = 30).

=> For every minute, minute hand rotates by 6 degrees (360/60 = 6).

=> But for every minute, hour hand rotates by 1/2 degree (1 hr = 30 degrees 1 min = half degree).

=> So effective angle = (6 * min) - ((30 * hr) + (0.5 * min)).

=> Solving angle = 5.5 * min - 30 * hour.

As the clocks are in a straight line but not together, the angle formed by the hands of the clock will be 180.

Now, We know that Angle = |30*Hours - 11/2*Minutes|
Therefore, 180 = 30*7 * 11/2 * Minutes.
210 - 180 = 11/2 * Minutes.
30 = 11/2 * Minutes.
60/11 = Minutes.
55 * 5/11 = Minutes.
So, finally, the answer is D.

θ = { hour * 30 } diff { min * 11/2 } take difference between greater value and smaller value.

Now for the straight line, we know θ = 180.

We can put all these values in the formula

Hour = 7
θ = 180
min = ?

θ = hour * 30 diff min * 11/2.
180 = { 7 * 30 } diff { min* 11/2}.
180 = { 210 } diff { min* 11/2},
min = 2/11 * 210-180,
min = 60/11.

which is equal to 5 5/11.

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

@All.

As we know that;

1hr i.e. 60 min the the minute hand gains 55 minute spaces over the hour hand.
So in the minute formula just replace 60 by 55 to get the answer
Given:-
The min and hour hand are in a straight line so the angle between them is 180 degrees
So angle = 180 degree.
Formula :

= |55x7-360/11|,
= 385 - 360/11 = 25/11.
Hence, the answer = 5 5/11.

Here, is some Equation:

Angle between hands = [(Hour Hand * 30) + (Minute hand * 1/2)] - Minute hand * 6.

(Hour Hand * 30) -> Because At 1 o'clock it will be 30 deg so at 7 it will be 150 deg.

Minute hand * 1/2 -> Hour hand moves 1/2 deg per min.

Minute hand * 6 -> Minute hand moves 6 deg per min.

180 = [7 * 30 + x/2] - 6x.

So, the answer is 5 5/1.

Assume that the time is 1 O'clock. If we look in a clockwise direction, it's 55 min space but if we look in the anticlockwise direction, it's 5-minute space. Can somebody explain which direction should we look to know the correct minute space?

If both directions are correct, then 5-minute space will be equal to 55-minute space.

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