# 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:

68 comments Page 1 of 7.
Harish udupa s said:
8 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.

SHEKHAR said:
1 decade ago

@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:)

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:)

Harish udupa s said:
8 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.

Rohit Joshi said:
6 months ago

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.

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.

(5)

Mohit Sharma said:
3 years ago

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

Now for the straight line, we know Î¸ = 180.

We can put all these values in the formula

Hour = 7

Î¸ = 180

min = ?

Î¸ = 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.

Now for the straight line, we know Î¸ = 180.

We can put all these values in the formula

Hour = 7

Î¸ = 180

min = ?

Î¸ = 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.

(3)

Himanshu dewaangan said:
1 decade ago

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

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

Prasanna said:
8 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.

Bangya said:
4 years ago

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.

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

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

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..!!

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