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:
71 comments Page 5 of 8.
Rajendra Sahu said:
1 decade ago
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
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
Marati Raju said:
1 decade ago
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 . . . . .
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 . . . . .
Puja ahirwal said:
1 decade ago
Please tell me about when minute space are add or subtract.
Rakesh said:
1 decade ago
Can some one give me a simple way to solve these watch question ?
Pritam said:
1 decade ago
How 55 min. spaces are gained in 60 min?
Can anybody explain?
Can anybody explain?
Rajesh said:
1 decade ago
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.
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.
Scientist said:
2 decades ago
Dear aparna how did you got 11/2 ?
Vivek jain said:
2 decades ago
How 55 min. spaces are gained in 60 min. ?
Please somebody tell me!
Please somebody tell me!
Harshad said:
2 decades ago
How do you know that at 7 the difference is 25min?
@ Nitesh: Assuming I don't have a watch to look at and count the minutes.
@ Nitesh: Assuming I don't have a watch to look at and count the minutes.
Fatima said:
2 decades ago
Please someone clear that what angle in 8o'clock9 between.
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