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 7 of 7.
Prasanna said:
1 decade ago
I have a problem with this answer.
Both hh and mh will meet at 7 i.e. time is 7:35 but exact answer is 7:35+35/11 = 7:35+32/11 = 7:38 2/11.
Then how it will 5 5/11 past to 7?
Both hh and mh will meet at 7 i.e. time is 7:35 but exact answer is 7:35+35/11 = 7:35+32/11 = 7:38 2/11.
Then how it will 5 5/11 past to 7?
Sushil said:
1 decade ago
Given mh&hh are straight, so angle is 180.
As per formula:
(hh*30) + (x/2) - (x*6) = 180.
Here x is min and hh is 7 by solving x we will get x as 5+5/11. So D is the answer.
As per formula:
(hh*30) + (x/2) - (x*6) = 180.
Here x is min and hh is 7 by solving x we will get x as 5+5/11. So D is the answer.
Nithya said:
2 decades ago
How is that at 7 o'clock they are 25 min space apart ?
Vimal panchal said:
1 decade ago
Simple formula :
Given start time i.e. 7 o clock.
x=difference between both hand in minutes.
Formula =(60*x)/(60-x) past the given time.
Given start time i.e. 7 o clock.
x=difference between both hand in minutes.
Formula =(60*x)/(60-x) past the given time.
Skt said:
1 decade ago
Please clear that how you got 5 5/11 min from {60/55*5}min.
Shaik shareef said:
1 decade ago
Can anyone give brief explanation of 5*5/11min how this get?
I want simplification.
I want simplification.
Sangram said:
1 decade ago
Use simple technique.
Between 7-8 they're going straight at some 7 hrs x minute.
then,
Angle by hrs needle - angle by min needle = 180.
( 7 * 30 + x/2 ) - x * 6 = 180.
30 degree cover by hrs needle in one hrs.
6 degree cover by min needle in one min.
1/2 degree cover by hrs needle in one min.
Between 7-8 they're going straight at some 7 hrs x minute.
then,
Angle by hrs needle - angle by min needle = 180.
( 7 * 30 + x/2 ) - x * 6 = 180.
30 degree cover by hrs needle in one hrs.
6 degree cover by min needle in one min.
1/2 degree cover by hrs needle in one min.
Vidya said:
1 decade ago
Hello.
I have a doubt when comparing problem 6 and problem 14. i.e, both problems say clocks of a hand are in straight line but not together or it says in opposite direction. But in problem 6 - we subtract minutes apart (30-25 =5) but in problem 14 - we add the minutes (30+20 =50). Can anyone explain.
I have a doubt when comparing problem 6 and problem 14. i.e, both problems say clocks of a hand are in straight line but not together or it says in opposite direction. But in problem 6 - we subtract minutes apart (30-25 =5) but in problem 14 - we add the minutes (30+20 =50). Can anyone explain.
Pyare said:
1 decade ago
Rajesh.
x>6 or x<6 its fine. Then how about when x=6? can you please clarify it?
For these formula,
(5x+30)12/11 where x<6.
(5x-30)12/11 where x>6.
x>6 or x<6 its fine. Then how about when x=6? can you please clarify it?
For these formula,
(5x+30)12/11 where x<6.
(5x-30)12/11 where x>6.
Sharmistha said:
1 decade ago
How 55 min spaces are gained in 60 mins?
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