Aptitude - Clock - Discussion
Discussion Forum : Clock - General Questions (Q.No. 6)
6.
At what time between 7 and 8 o'clock will the hands of a clock be in the same straight line but, not together?
Answer: Option
Explanation:
When the hands of the clock are in the same straight line but not together, they are 30 minute spaces apart.
At 7 o'clock, they are 25 min. spaces apart.
Minute hand will have to gain only 5 min. spaces.
55 min. spaces are gained in 60 min.
| 5 min. spaces are gained in |  |
60 | x 5 |  min |
= 5 | 5 | min. |
| 55 | 11 |
| Required time = 5 |
5 | min. past 7. |
| 11 |
Discussion:
70 comments Page 4 of 7.
Renz said:
1 decade ago
Hai, what is the basic steps or formula in solving problems?
Vash said:
10 years ago
Let x = to minutes and x/12 = to hour.
x = 35+x/12-30 = 12x = 420+x-360.
11x = 60. 35 minutes = 7 o'clock.
I put -30 because x = 60/11 directly opposite to each other = 5.45 or 5 5/11.
180 degrees.
x = 35+x/12-30 = 12x = 420+x-360.
11x = 60. 35 minutes = 7 o'clock.
I put -30 because x = 60/11 directly opposite to each other = 5.45 or 5 5/11.
180 degrees.
Paddu said:
10 years ago
Hello friends please explain in shortcut. I am still in confused.
Riya shah said:
9 years ago
Shortcut formula:
ANGLE = 30*H - 11/2M.
Together Means angle consider as zero.
Put into the formula.
0 = 30 * 7 - 11/2M.
11/2M = 210.
Imp together consider as zero.
ANGLE = 30*H - 11/2M.
Together Means angle consider as zero.
Put into the formula.
0 = 30 * 7 - 11/2M.
11/2M = 210.
Imp together consider as zero.
Riya said:
9 years ago
Not together so angle 180.
Angle = 30 * H + 11/2 * M.
Angle = 30 * H + 11/2 * M.
Harish udupa s said:
9 years ago
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.
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.
Harish udupa s said:
9 years ago
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.
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.
Vikram kumar said:
9 years ago
Use formula.
(5A + or - 30) * 12/11min past A.
Where A = + whan A < 6 and - when a > 6.
According to the question A = 5.
(5A + or - 30) * 12/11min past A.
Where A = + whan A < 6 and - when a > 6.
According to the question A = 5.
Sandun said:
9 years ago
Another way to do this.
The angle created when the time is 7 o' clock is 150*. So we have to add the extra gained by minute hand and to subtract the angle lost by hour hand.
If we put this in to an equation and we get x = 5 5\11.
The angle created when the time is 7 o' clock is 150*. So we have to add the extra gained by minute hand and to subtract the angle lost by hour hand.
If we put this in to an equation and we get x = 5 5\11.
Prasanna said:
9 years ago
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.
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.
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