At what time between 4 and 5 will the hands of a watch coincide?
Minute hand must gain 20 minutes before two hands can be coincident.
But the minute-hand gains 55 minutes in 60 minutes.
Let minute hand will gain x minutes in 20 minutes.
\[\frac{55}{20}=\frac{60}{x}\Rightarrow x=\frac{20\times 60}{55}\]
\[=\frac{240}{11}\]
\[=21\frac{9}{11}\] min.
\[\therefore\] The hands will be together at
\[21\frac{9}{11}\] min past 4.
At what time between 4 and 5 will the hands of a watch point in opposite directions.
Hands will be opposite to each other when there is a space of 30 minutes between them.
This will happen when the minute hand gains (20 + 30) = 50 minutes.
Now, the minute hand gains 50 min in \[\frac{50\times60}{55}\] or \[54\frac{6}{11}\] min.
\[\therefore\] The hands are opposite to each other at \[54\frac{6}{11}\] min past 4.
How many times does the 29th day of the month occur in 400 consecutive years?
In 400 consecutive years, there are 97 leap years. Hence, in 400 consecutive years, February has the 29th day 97 times and the remaining eleven months have the 29th day 400 × 1100 = 4400 times
\[\therefore\] The 29th day of the month occurs (4400 + 97) or 4497 times.
Today is 5th February. The day of the week is Tuesday. This is a leap year. What will be the day of the week on this date after 5 years?
Today is 5th February. The day of the week is Tuesday. This is a leap year. What will be the day of the week on this date after 5 years?
This is a leap year. So, the next 3 years will give one odd day each.
Then leap year gives 2 odd days and then again next year give 1 odd day.
Therefore (3 + 2 + 1) = 6 odd days will be there.
Hence the day of the week will be 6 odd days beyond Tuesday,
i.e., it will be Monday.
At what time between 5 and 6 O' clock are the hands of a clock together?
The two hands of the clock will be together bbetween H and (H + 1) O' clock at \[\left(\frac{60H}{11}\right)\] mins past H O'clock
Here H = 5
\[\therefore \frac{60H}{11}=\frac{60}{11}\times5=\frac{300}{11}=27\frac{3}{11}\]
\[\therefore\] Hands of a clock are together at \[27\frac{3}{11}\] past 5 O' clock.
Find at what time between 2 and 3 O' clock will the hands of a clock be in the same straight line but not together.
The two hands of the clock will be in the same straight line but not together between H and (H + 1) O' clock at
\[\left(5H+30\right)\frac{12}{11}\] mins past H, when H > 6
Here H = 2 < 6
\[\therefore \left(5H+30\right)\frac{12}{11}\]
\[=\left(5\times 2+30\right)\frac{12}{11}\]
\[=\frac{480}{11}, \ i.e. \ ,43\frac{7}{11}\]
So, the hands will be in the same straight line but not together at \[43\frac{7}{11}\] mins past 2 O' clock.
Find at what time between 7 and 8 O' clock will the hands of a clock be in the same straight line but not together.
The two hands of the clock will be in the same straight line but not together between H and (H + 1) O' clock at
\[\left(5H-30\right)\frac{12}{11}\] mins past H, when H > 6
Here H = 7 > 6
\[\therefore \left(5H-30\right)\frac{12}{11}\]
\[=\left(5\times 7-30\right)\frac{12}{11}\]
\[=\frac{60}{11}, \ i.e. \ ,5\frac{5}{11}\]
So, the hands will be in the same straight line but not together at \[5\frac{5}{11}\] mins past 7 O' clock.
Find the time between 4 and 5 O' clock when the two hands of a clock are 4 mins apart.
Between H and (H + 1) O' clock, the two hands of a clock are M mins apart at \[\left(5H\pm M\right)\frac{12}{11}\] mins past H O' clock.
Here
\[\therefore \frac{12}{11}\left(5H\pm M\right)\]
\[=\frac{12}{11}\left(5\times 4\pm 4\right)\]
\[=26\frac{2}{11}\ and\ 17\frac{5}{11}\]
\[\therefore\] The hands will be 4 mins apart at \[26\frac{2}{11}\] mins past 4 O' clock and \[17\frac{5}{11}\] mins past 4 O' clock.
Find the angle between the two hands of a clock at 15 mins past 4 O' clock.
When the minute hand is behind the hour hand, the angle between the two hands at
M mins past H O' clock
\[=30\left(H-\frac{M}{5}\right)+\frac{M}{2}\] degrees.
Here
\[\therefore\] The required angle
\[=30\left(H-\frac{M}{5}\right)+\frac{M}{2}\]
\[=30\left(4-\frac{15}{5}\right)+\frac{15}{2}\]
\[=\frac{75}{2}\]
\[=37.5^°\]
Find the angle between the two hands of a clock at 30 mins past 4 O' clock.
When the minute hand is past the hour hand, the angle between the two hands at
M mins past H O' clock
\[=30\left(\frac{M}{5}-H\right)-\frac{M}{2}\] degrees.
Here
\[\therefore\] The required angle
\[30\left(\frac{30}{5}-4\right)-\frac{30}{2}\]
\[=30\left(\frac{30-20}{5}\right)-15\]
\[=\frac{300}{5}-15\]
\[=60-15\]
\[=45^∘\]
The minute hand of a clock overtakes the hour hand at intervals of 65 times. How much a day does the clock gain or lose?
The minute hand of a clock overtakes the hour hand at intervals of M mins of correct time. The clock gains or loses in a day by
\[=\left(\frac{720}{11}-M\right)\left(\frac{60\times 24}{M}\right)\]
Here M = 65
\[\therefore\] The clock gains or loses in a day by
\[=\left(\frac{720}{11}-65\right)\left(\frac{60\times 24}{65}\right)\]
\[=\frac{5}{11}\times \frac{12\times 24}{13}\]
\[=\frac{1440}{143}\]
\[=10\frac{10}{143}\] mins.
Since the sign is +ve, the clock gains by \[10\frac{10}{143}\] mins.
The length of a minute hand of a clock is 7cm. The area swept by the minute hand in 30 minutes is :
Area swept by the minute hand in an hour \[= π r^2\]
\[\therefore \text{Required area} = \frac{π r^2}{2}\]
\[=\frac{22 \times 7 \times 7}{7 \times 2} = 77 \ cm^2\]
If a clock started at noon, then the angle turned by hour hand at 3.45 PM is
The hour hand traces 30° in an hour.
\[\therefore \text{Angle traced in} \ 3\frac{3}{4} \] hours
\[=\frac{15}{4}\times 30°\]
\[=\frac{225}{2}^°\]
\[=112\frac{1}{2}^°\]