Clock maths tricks Part: 2
Clock maths questions
Clock maths problems
How many times in a day hands of a clock coincides?

Clock Problems with solutions and answer

Clock coincides problems with solutions and tricks.

Q1.How many times in a day hands of a clock coincides?
a. 20 b. 30 c.22 d.24
Ans: c. 22
Trick In every hour both the minute hand and hour hand coincide ( i.e. they become straight in the same direction ) once (except 12hrs to 1hrs). So during the 12 hours, they coincide 11 times because from 12hrs to 1hrs they never coincide. In a complete day, there are 24 hours. Hence they will coincide total of 22 times.
Q2.How many times in a day do the hands of a clock become straight?
a. 22 b. 24 c. 44 d. 48
Ans: 44
Trick In every hour both the minute hand and hour hand become straight 2 times ( in the same direction one time and in opposite direction one time except 12hrs to 1hr and 6hrs to 7hrs ). Since 2 times every hour. So in a day, they become straight only 44times in 22hrs.
Q3.How many times in a day are the hands of a clock become straight but in opposite direction?
a. 21 c. 22 c. 44 d. 24
Ans: c 22
Trick In every hour both the minute hand and hour hand become straight in opposite directions 1 time (except 6hrs to 7hrs). So during the 12 hours, they become straight in opposite directions 11 times since during 6hrs to 7hrs they never become straight. In a complete day, there are 24 hours. Hence they will become straight in the opposite direction a total of 22 times.
Q4.At what time, in minutes between 3 O'Clock and 4 O'Clock both the needles will coincide with each other-
$a\cdot 5{\frac{1}{11}}'' b\cdot 12{\frac{4}{11}}'' c\cdot 13{\frac{4}{11}}'' d\cdot 16{\frac{4}{11}}''$

  Ans: d.
Trick: For this type of question we must count total minutes from 12 hrs to 3hrs i.e. from 12 to 3 there is 15 minutes space gap. So here M=15, Then we put M=15 in the formula
$\frac{12}{11}\times M$ then calculate the value as $\frac{12}{11} \times 15= \frac{180}{11}= 16{\frac{4}{11}}''$
That is after 3hrs and $16{\frac{4}{11}}''$ both the minute hand and hour hand will coincide.
Q4. At what time between 9 and 10 O'Clock will the minute hand and hour hand of a clock be together?

a. $45 min past 9$ .
b. $50 min past 9$
c. $49\frac{1}{11}min past 9$
d. $48\frac{2}{11}min past 9$
Ans :c $49\frac{1}{11}min past 9$
Trick:  Because here the total space gap from 12 hrs to 9 hrs in a minute is 45 min. So M=45, Then we put M=45 in the formula   $\frac{12}{11}\times M$ and get the result.

Q5. At what time between 7 and 8 O'Clock will the hands of a clock be in the same straight line but not together?
$a\cdot 5 min past 7$
$b\cdot 5\frac{2}{11} min past 7$
$c\cdot 5\frac{3}{11} min past 7$
$d\cdot 5 \frac{5}{11} min past 7$
Ans: $d\cdot 5 \frac{5}{11} min past 7$
Trick:
Here total space gap will be from 12 to 1 not from 12 to 7 bcz both hands are in a straight line but not together i.e. they will be straight in opposite direction, not in the same direction. So the opposite of 7 is 1. Hence we must count 12 to 1 space gap in a minute. So here M=5 min. Now put the value of M=5 in the formula. $\frac{12}{11}\times M= \frac{12}{11}\times 5= \frac{60}{11}= 5\frac{5}{11}min$ That is at $5\frac{5}{11}minpast 7$ both the hands of the clock become straight but not together.

Q6. At what time between 4 and 5 O'Clock hands ( Hands mean both minute hand and hour hand ) of a clock become straight in opposite direction?
$a. 4:45min. b. 4:40min. c. 4:50\frac{4}{11}min d. 4:54\frac{6}{11}min$
Ans: $d. 4:54\frac{6}{11}min{\color{Red} }{\color{Red} {\color{Red} }}$

Trick:  Here total space gap will be from 12 to 10 not from 12 to 4 bcz both the hands of the clock become straight in opposite direction, not in the same direction. So the opposite of 4 is 10. Hence we must count 12 to 10 space gaps in a minute. So here M=50 min. Now put the value of M=50 in the formula. $\frac{12}{11}\times M=\frac{12}{11}\times 50=\frac{600}{11}=54\frac{4}{11}\: \: \: \: min$ That is at $54\frac{4}{11}\: \: min\: \: \: past \: \: 4$ both the hands of the clock become straight in opposite directions.

Q7. At what time between 6 and 7 o'clock both the hands of a clock will be 90° or right angle the first time?
$a\cdot 16\frac{4}{11}\, \, min\, \, past\, \, 6.$
b. $At 16\frac{1}{11}\, \, \, min\, \, \, past \, \, 6\, \, \, \,$
$c. At \, 15\frac{1}{11}\, \, \, min\, \, \, past \, \, 6\, \, \, \,$
$d. \, \, None$

Ans: $a. 16\frac{4}{11}\, \, \, min\, \, past \, \, 6$

Trick: There is two (2) formula to find out the time.
1. Required time $= \left ( Total\, \, \, gap\, \, \, in\, \, \, Minute\, \, \, from\, 12\right )\times \frac{12}{11}$ .
2. Required Time $= \left ( Total\, \, \, gap\, \, \, in\, \, \, degree\, \, \, from\, 12\right )\times \frac{2}{11}$
If we apply the 1 no formula,
Required TIME $= \left ( Total\, \, \, gap\, \, \, in\, \, \, Minute\, \, \, from\, 12\right )\times \frac{12}{11}$ $= M\times\frac{12}{11}$

We have to find the total gap in minutes, (in the case of the first time means the minute hand before crossing 6 both hands become one time 90 degrees) ) we have to count up to 12 hrs to 3 hrs. because at 6 O' Clock minute hand is at 12 and when the minute hand reached from 12 hrs to 3 hrs, approximately there will be 90 degrees gap between the hour hand and minute hand, So we must count gap from 12 to 3. Here from 12 hrs to 3 hrs total gap in the minute will be 15 minutes. Hence M= 15. And we will put the value of M in 1 no formula as-

Required Time = $15\times \frac{12}{11}= \frac{180}{11}=16\frac{4}{11}\, \, minute \, \, \, past\, \, 6.$

If we apply the 2 no formula,
Required Time $= \left ( Total\, \, \, gap\, \, \, in\, \, \, degree\, \, \, from\, 12\right )\times \frac{2}{11}$ $= D\times\frac{2}{11}$

We have to find the total gap in Degree, (in the case of the first time means the minute hand before crossing 6 both hands become one time 90 degrees) we have to count up to 12 hrs to 3 hrs. because at 6 O' Clock minute hand is at 12 and when the minute hand reached from 12 hrs to 3 hrs, approximately there will be 90 degrees gap between the hour hand and minute hand, So we must count gap from 12 to 3. Here from 12 hrs to 3 hrs total gap in degree will be 90°. Hence D= 90°. And we will put the value of D in 2 no formula as-

Required Time = $= D\times \frac{2}{11}= 90^{\circ}\times\frac{2}{11}=\frac{180}{11}= 16\frac{4}{11}\, \, min\, \, past\, \, 6.$

Q8.At what time between 6 and 7 o'clock both the hands of a clock will be 90° in the second time?

$a.\, \,At\6: 16\frac{4}{11}\, \, hrs$
$b.\, \,At\6: 49\frac{1}{11}\, \, hrs$
$c.\, \,At\6: 49\frac{3}{11}\, \, hrs$
$d.\, \,At\6: 49\frac{1}{12}\, \, hrs$
Ans:: $b.\, \,At\6: 49\frac{1}{11}\, \, hrs$ .
Trick: Using 1st formula-

Here 2nd time means, that the minute hand after crossing 6 whenever it reached 9 at that time approximately both the minute and hour hand become 90 degrees. So, we must count gap space in minutes from 12 to 9. That is, M= 45 Minutes and put the value of M in the 1st formula-

Required Time $= \left ( Total\, \, \, gap\, \, \, in\, \, \, Minute\, \, \, from\, 12\right )\times \frac{12}{11}$ $=45\times \frac{12}{11}= \frac{540}{11}= 49\frac{1}{11}$
That is $At \, \, \, \, 6: 49\frac{1}{11} hrs$
OR
Using 2nd formula-
Here, the gap space from 12 hrs to 9 hrs in degree is 270°. So, D=270 and we put the value of D in the 2nd formula and get-
Required time = $D\times\frac{2}{11}= 270\times \frac{2}{11}= \frac{540}{11}= 49\frac{1}{11}$ .
That is $At \, \, \, \, 6: 49\frac{1}{11} hrs$ .
Q9. After how many minutes past 7, the hands of a clock form 90 degree angle for the 1st time?

a. $7:20\frac{10}{11} \, \, \, \, hrs$
b. $7:30\frac{10}{11} \, \, \, \, hrs$
c. $7:21\frac{9}{11} \, \, \, \, hrs$
d. $7:21\frac{7}{11} \, \, \, \, hrs$
Ans:c $7:21\frac{9}{11} \, \, \, \, hrs$ .
Trick: Here you must count gap space in minutes from 12 hrs to 4 hrs because whenever the minute hand reaches 4, the angle between the minute and hour hand will form approximately 90 degrees. So, M=20 (gap in a minute from 12hrs to 4hrs ). Now put the value of M in 1st formula and get the result.

Q10.After how many minutes past 7, the hands of a clock form 90 degree angle for the 2nd time?
*Practice yourself
Ans: $At\, \, 7:54\frac{6}{11}\, hrs$ .

Q11. How many times in a day do the hands of a clock become 90° / right angle?
a. 22 b. 24 c. 44 d. 48
Ans: c. 44 .

Clock Error or faulty Clock problems and solution

Q12.A clock gains 5 sec in every 3 minutes and was set right at 8 AM. What time will show at actual 10PM on the same day?
a. 10:23:01PM
b. 10:27:20PM
c. 10:23:20PM
d. 10:27:01PM
Ans:c. 10:23:20PM.
Trick:
Here clock ran exactly from 8AM to 10PM, that is total of 14 hrs.

$\because\, \, In\, \, 3\, min \, clock \, \, gain\, \, = 5\, \, sec.$

$\therefore \, \, In\, \, 1\, min \, clock \, \, gain\, \, = \frac{5}{3}\, \, sec.$
$\therefore \, \, In\, \, 60\, min \, clock \, \, gain\, \, = \frac{5}{3}\, \times 60\, \, sec.$
$\therefore \, \, In\, \, 1\, hr \, clock \, \, gain\, \, = 100\, \, sec.$
$\therefore \, \, In\, \, 14\, hr \, clock \, \, gain\, \, = 100\, \times 14\, \, sec.$
$= 1400\, \, sec.$
$= \frac{1400}{60}\, \, min.$
  = 23 min 20 sec.
Since in 14 hrs faulty clock gain 23 min 20sec. So, at original 10 PM, the faulty clock shows 23 min 20 sec more time than the original time. That is, the faulty clock will show 10: 23:20 PM.

Q13. A clock loses 5 sec in every 3 minutes and was set right at 8 AM. What time will show at actual 10 PM on the same day?

a. 09:23:01 PM
b. 09:27:20 PM
c. 10:36:20PM
d. 09:36:40 PM

Ans: d. 09:36:40 PM
Trick: As the above example from 8AM to 10 PM Clock gain 23min 20 sec, here also the clock will lose 23 min 20 sec. So, we must be back 23min 20 sec from 10PM. The result will be -
10PM=09 pm:59 min : 60 sec
    -23 min : 20 sec
09pm: 36 min: 40 sec.
Q14.A watch gains 10 min every hour and was set right at 8 AM. What was the actual time when it showed 10 PM on the same day?
a. 8:00 PM
b.8:40 PM
c. 8:43 PM
d. 8:20 PM
Ans: a. 8:00 PM.
Trick:  Here, the total time from 8AM to 10 PM is 14hrs. The wrong watch starts exactly at 8 PM and lastly, it shows 10 PM. That is the wrong moved total of 14 hrs.
Since every hour the wrong watch gain 10 min, we can reversely say that when the wrong watch moves 1hr + 10min=(1+10/60) hrs=(1+1/6)hrs=7/6 hrs, the right watch moves 1hrs.

So,    $when \, \, \frac{7}{6}hrs \, \, moves\, \, W/watch= 1hrs\, \, \, \, \, moves\, \, \, \, \, R/watch$

$\therefore\, \, \, when \, \, 1\, hrs \, \, moves\, \, W/watch= \frac{1}{\frac{7}{6}}hrs\, \, \, \, \, moves\, \, \, \, \, R/watch$
$\therefore\, \, \, when \, \, 14\, hrs \, \, moves\, \, W/watch= \frac{6}{7}\times 14\, \, hrs\, \, \, \, \, moves\, \, \, \, \, R/watch$
$= 6\times 2\, \, hrs\, \, \, \, \, moves\, \, \, \, \, R/watch$
$= 1 2\, \, hrs\, \, \, \, \, moves\, \, \, \, \, R/watch$ .
Now it is seen that whenever the wrong watch moves 10hrs, right watch moves 2 hrs less than the wrong watch. So, we must go back 2hrs from 10PM and get the actual time. So the actual time will be 10-2=8 PM=8:00PM.
Q15.A watch loses 5 min in every 1 hour and was set right at 5 AM. What was actual time when it was showing 6PM the next day?
a. $8\frac{3}{11}\, \, \, b.9\frac{6}{13}\, \, \, c.9\frac{4}{11}\, \, \, d.8\frac{4}{11}$
Ans: $c.\, \, \, \, 9\frac{4}{11}$
Trick: Here, the total time from 5AM to the next day 6PM is 37 hrs and the watch loses 5 min every hour, that means-
$When \, \, wrong \, \, watch\, \, moves\, \, 55\, \, min, right\, \, watch\, \, moves=60\, \, min$
$\therefore When \, \, wrong \, \, watch\, \, moves\, \, \frac{55}{60}\, \, hrs, right\, \, watch\, \, moves=1\, \, hrs$
$\therefore When \, \, wrong \, \, watch\, \, moves\, \, \frac{11}{12}\, \, hrs, right\, \, watch\, \, moves=1\, \, hrs$
$\therefore When \, \, wrong \, \, watch\, \, moves\, \, 1\, \, hrs, right\, \, watch\, \, moves=\frac{1}{\frac{11}{12}}\, \, hrs$
$\therefore When \, \, wrong \, \, watch\, \, moves\, \, 37\, \, hrs, right\, \, watch\, \, moves=\frac{12}{11}\times 37 hrs$
   $= \frac{444}{11} hrs$
$=40\frac{4}{11} hrs$
That is, the right watch moves $3\frac{4}{11} hrs$ more than the wrong watch. So, the actual time was 6PM+ $3\frac{4}{11} hrs$ = $9\frac{4}{11}PM$ .
Q16. A watch which gains uniformly is 2 minutes slow at noon Monday and is 4min 48sec fast at 2PM on the following Monday. When was it correct?
a. 2PM on Tuesday.
b.2PM on Wednesday.
c. 3PM on Thursday.
d 1PM on Friday.
Ans:b.2PM on Wednesday.
*Practice yourself

Clock Mirror image Problem and solution-

▭: 1. In a mirror image, the figure change from left to right but the top to bottom remains the same.

2. If the given time is between 1 O'clock to 11 O, clock, then we must subtract the given time from 12:00hrs =11:60hrs.

3. If the given time is between 11 O'clock to 1O, clock, then we must subtract the given time from 24:00hrs =23:60hrs.

4. Letter in mirror image remain same are- A,I,H,M,O,T,U,V,W,X,Y.

Q17. What time will show by mirror if the real-time shows by the clock are 9: 27?

a. 3:33

b.2:33

c.3:27

d.2:23

Ans:b.2:33.

Tk: Since the given time is9:27 which is between 1 to 11 so we subtract it from 11:60 and get the result.

11:60

- 9:27

2:33

Q18.If time appears in the mirror at 11:09 then what time will appear on the clock?

a. 1:33

b.12:33

c.12:51

d.2:51

Ans: c.12:51.

Tk: Since the given time is11:09 which is between 11 to 1 so we subtract it from 23:60 and get the result.

23:60

-11:09

12:51
Q19. The time shown in the mirror is 6:00, then what will be the real image on the clock?

a.5:58
b.6:60
c.6:00
d.6:01
Ans: c. 6:00

Tk: Since the given time is 6:00 which is between 1 to 11 so we subtract it from 11:60 and get the result.

11:60

-6:00

5:60= 6:00
Q20. If the reflecting time is 3:43 then the real-time of the clock is
a. 3:17
b. 7:17
c. 8:17
d. 8:43
Ans: c.8:17

Tk: Since the given time is 3:43 which is between 1 to 11 so we subtract it from 11:60 and get the result.

11:60

- 3:43

8:17
Q21 If the time on the clock is 12:37. What is the time in the mirror?
a. 12:33
b. 11:37
c.12:37
d. 1:37
Ans: b.11:37

Tk: Since the given time is 12:23 which is between 11 to 1 so we subtract it from 23:60 and get the result.

23:60

-12:23

11:37

Clock Water image Problem and solution-

▭: 1. In the water image, the figure change from top to bottom but the left to the right remains the same.
2. If in the given time hour H≤5 then the resultant hour value will be 5-H If H> 5 then the resultant hour value will be 17-H.
3. If in the given time Minute M≤30 then the resultant minute value will be 30-M, If M>30 then the resultant minute value will be 90-M.
4. Letter in the water image that remains the same are B, C, D, E, K, O, X, I, and H.
Q22. By looking in the water it appears that it is 3:10 in the clock. What was the real-time in the clock?
a.2:20
b.2:17
c.3:27
d.4:20
Ans: a.2:20

Tk: Since the given time is 3: 10, here H=3<5 and M=10<30 so we subtract it from 5:30 and get the result.

5:30

-3:10

2:20
Q23. By looking in the water it appears that it is 9:40 in the clock. What is the real-time in the clock?
a.8:20
b.8:50
c.3:50
d.8:20
Ans: b.8:50

Tk: Since the given time is 9: 40, here H=9>5 and M=40>30 so we subtract it from 17:90 and get the result.

17:90

-9:40

8:50

Q24. By looking in the water it appears that it is 3:13 in the clock. What is the real-time in the clock?
a.2:20
b.2:17
c.3:27
d.4:20
Ans: b.2:17

Tk: Since the given time is 3: 13, here H=3<5 and M=13<30 so we subtract it from 5:30 and get the result.

5:30

-3:13

2:17
Q25. By looking in the water it appears that it is 4:42 on the clock. What is the real time in the clock?
a.1:48
b.2:48
c.12:48
d.12:42
Ans: a.1:48

Tk: Since the given time is 4: 42, here H=4<5 and M=42>30 so we subtract it from 5:90 and get the result.

5:90

-4:42

1:48
Q26.The time that appears in the water is 5:47, what will be the correct time on the clock?
a. 12:47
b.1:47
c.12:43
d.1:43
Ans: c. 12:43

Tk: Since the given time is 5: 47, here H=5 and M=47>30 so we subtract it from 05:90 and get the result.

05:90

-5:47

00:43⩸12:43

Q27. After how many minutes past 7, the hands of a clock form 45 degree angle for the 1st time?

a. 7:30hrs

b. $7:30\frac{10}{11} \, \, \, \, hrs$

c. $7:21\frac{9}{11} \, \, \, \, hrs$

d. $7:21\frac{7}{11} \, \, \, \, hrs$

Ans:c .7:30

*Practice Paper

Q1. At what time between 7:00 and 8:00 are the hands of a clock at the angle of 45 degrees for the second time?

Ans: $7:46\frac{4}{11}$

Q2.What angle is made by minute hand in 29 sec?

Ans 2.9 degree.

Q3.What angle is made by minute hand in 59sec?

Ans 5.9 degree.
Q4.A clock which moves continuously fast, it lags 10 minutes on Sunday 7am, it is ahead 5min on Tuesday 7am. When was the clock correct?
Ans: Monday 11 PM.
Q5.What is the angle between the hands of a clock at 3:56?
Ans:142 degrees.

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How many times in a day hands of a clock coincides?

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Clock maths tricks I Clock maths questions I Clock maths problems I How many times in a day hands of a clock coincides?

Clock maths tricks Part: 2 Clock maths questions Clock maths problems How many times in a day hands of a clock coincides?

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Clock coincides problems with solutions and tricks.

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Clock maths tricks Part: 2
Clock maths questions
Clock maths problems
How many times in a day hands of a clock coincides?