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 2 of 7.
Faiz said:
1 decade ago
@Harshad
Man at 7 o clock, the hour hand is at 7 and the minute hand is in 12 i.e 60th min. assume the display in your watch is like 5 10 15.. 60.. so at 7 the hour hand will be at 35 and the min hand will be in 60.. now jus go on adding the minute space clockwise or els simple (60-35) gives 25 mins space..!!
Man at 7 o clock, the hour hand is at 7 and the minute hand is in 12 i.e 60th min. assume the display in your watch is like 5 10 15.. 60.. so at 7 the hour hand will be at 35 and the min hand will be in 60.. now jus go on adding the minute space clockwise or els simple (60-35) gives 25 mins space..!!
Swaraj said:
7 months ago
The angle required is 180°.
At 7 the angle would be 25×6°=150°
So let's assume it moves x mins after 7.
So, keeping the movement of the hour hand and minute hand during these x mins, we can write the angle between both the hands as;
180°=150°+ (X. 6°- X. 1/2°).
=>30°=X(5.5) =>X=30°/5.5=5 5/11mins(Ans).
At 7 the angle would be 25×6°=150°
So let's assume it moves x mins after 7.
So, keeping the movement of the hour hand and minute hand during these x mins, we can write the angle between both the hands as;
180°=150°+ (X. 6°- X. 1/2°).
=>30°=X(5.5) =>X=30°/5.5=5 5/11mins(Ans).
(4)
Aparna said:
1 decade ago
Simple method:
In one hour,once minute hand and hour hand comes in straight line
so our answer is something 7 past,not 8 past
use formula,
titha(0)=30h-11/2m
titha is 180,hour hand is on 7
put values
we get titha=60/11
convert into quotitent*remainder/divisor format
we get
5*5/11
So answer is 5*5/11 past 7.
In one hour,once minute hand and hour hand comes in straight line
so our answer is something 7 past,not 8 past
use formula,
titha(0)=30h-11/2m
titha is 180,hour hand is on 7
put values
we get titha=60/11
convert into quotitent*remainder/divisor format
we get
5*5/11
So answer is 5*5/11 past 7.
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.
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.
Tushar said:
3 years ago
To calculate the angle between hr hand and min hand.
we have formula ->> 30*Hours-11/2*minutes.
In the above question, they said that minute and hour hands are in one line but not together means Angel between them should be 180°.
:: 30 * 7 - 11/2 * x = 180°.
Then, x = 5.4545 ~5 + 5/11.
we have formula ->> 30*Hours-11/2*minutes.
In the above question, they said that minute and hour hands are in one line but not together means Angel between them should be 180°.
:: 30 * 7 - 11/2 * x = 180°.
Then, x = 5.4545 ~5 + 5/11.
(34)
Jaanu said:
8 years ago
Use this formula to find angle :
Angle=((11/2)*minutes )-30* hour.
Between 7 and 8 means that the time would be something around 7. So consider 7 hours and x minutes.
180 = (11* x )/2 - 30 * 7.
210 - 180 = 11*x /2.
30 = 11*x /2.
30*2 = 11x,
60 = 11x,
X = 5.5/11.
That is 7 hours and 5 5/11 min.
Angle=((11/2)*minutes )-30* hour.
Between 7 and 8 means that the time would be something around 7. So consider 7 hours and x minutes.
180 = (11* x )/2 - 30 * 7.
210 - 180 = 11*x /2.
30 = 11*x /2.
30*2 = 11x,
60 = 11x,
X = 5.5/11.
That is 7 hours and 5 5/11 min.
Bombiyah said:
6 years ago
The angle formed from 12 o'clock to 7 o'clock is 210°.
Use this formula:
2/11 (angle formed (+) or (-) 180°)
Add 180 when the clock hands coincide, subtract when they are not.
Since, not coincide subtract it,
=2/11(210 * 180°)
= 60/11 or 5 5/11.
Use this formula:
2/11 (angle formed (+) or (-) 180°)
Add 180 when the clock hands coincide, subtract when they are not.
Since, not coincide subtract it,
=2/11(210 * 180°)
= 60/11 or 5 5/11.
(1)
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 . . . . .
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