Time and Clock Problems - Class VI

$5\dfrac {2}{3} hrs = \dfrac {17}{3}hrs$
$1\ hr = 60\ mins$
$\therefore \dfrac {17}{3} hrs = \dfrac {17}{3} \times 60\ mins = 340\ mins$.

$1$ hour $=60$ minutes

Given that one hour,

Now,

$1$ hour=$60\times 60$ second

$ 1s=\dfrac{1}{60\times 60}h $

$ =0.0002777777... $

$ =0.0002\overline{7} $

Hence, this is the answer.

$1$ year $=12$ months 

$i)$ $12$ hours $:$ $30$ hours$=8Km:20Km$

Gain in 60 min=5min
in 1 min =$\dfrac{5}{60}min=\dfrac{5}{60}\times 60 sec=5 sec$
Angle traversed by second hand=$360^0+\dfrac{5}{60}\times 360^0=390^0$

$3$ hr $12$ min $=3\times 60\times 60+12\times 60$

$\Rightarrow$ All the 12 months have 31 or 30 days.

As we know, $1$ hour $=60$ minutes
and $1$ minute $=60$ seconds
So, $1$ hour $=60\times 60=3600$ seconds.
Hence, an hour has $3600$ seconds.

The angle between $2$ successive numbers in the clock is $\dfrac{360}{12}=30^{\circ}$

In order to strike 12 there are 11 intervals of equal time = $\displaystyle \frac{33}{11}$ = 3 seconds each
Therefore to strike 6 it has 5 equal intervals, it requires 5 $\displaystyle \times $ 3 =15 sec.

As we know, $1$ hour $=60$ minutes
So, $5$ hours $=5\times 60=300$ minutes.
Hence, $5$ hours $=300$ minutes.

Let the distance between $A$ and $B$ be $x$ km.
Upstream speed $= 9 - 2 = 6\ kmph$
Downstream speed $= 9 + 3 = 12\ kmph$
$\dfrac {x}{6} + \dfrac {x}{12} = 3\Rightarrow \dfrac {2x + x}{12} = 3\Rightarrow 3x = 36 \Rightarrow x = 12\ km$.

$Speed\>=\>14\>\dfrac{m}{s}$

$=14\dfrac{\dfrac{1km}{1000}}{\dfrac{1hr}{3600}}$

$=\dfrac{14\cdot3600}{1000}\dfrac{km}{hr}=50.4\>\dfrac{km}{hr}$