Let distance = $ x $ km and usual rate = $ y $ kmph.
| Then,$ \dfrac{x}{y} $-$ \dfrac{x}{y + 3} $=$ \dfrac{40}{60} $ $\Rightarrow$ 2 $y \left(y + 3\right)$ = 9$ x $ ....(i) |
| And,$ \dfrac{x}{y -2} $-$ \dfrac{x}{y} $=$ \dfrac{40}{60} $ $\Rightarrow$ $y \left(y- 2\right)$ = 3$ x $ ....(ii) |
On dividing (i) by (ii), we get: $ x $ = 40.
Total distance travelled = 10 + 12 = 22 km
Time taken to travel 10 km at an average speed of 12 km/hr = $\dfrac{distance}{speed} = \dfrac{10}{12}$hr
Time taken to travel 12 km at an average speed of 10 km/hr = $\dfrac{distance}{speed} = \dfrac{12}{10}$hr
Total time taken =$\dfrac{10}{12} + \dfrac{12}{10}$hr
Average speed =$\dfrac{distance}{time}= \dfrac{22 }{\left(\dfrac{10}{12} + \dfrac{12}{10}\right)} = \dfrac{22 \times 120 }{\left(10 \times 10\right) + \left(12 \times 12\right)}$
$\dfrac{22 \times 120 }{244}=\dfrac{11 \times 120 }{122}=\dfrac{11 \times 60}{61}=\dfrac{660}{61}\approx $10.8kmph
Given that time taken for riding both ways will be 2 hours lesser than
the time needed for waking one way and riding back
From this, we can understand that
time needed for riding one way = time needed for waking one way - 2 hours
Given that time taken in walking one way and riding back = 5 hours 45 min
Hence The time he would take to walk both ways = 5 hours 45 min + 2 hours = 7 hours 45 min
In fact, you can do all these calculations mentally and save a lot of time
which will be a real benefit for you.
Speed = 5 km/hr
Time = 15 minutes = $\dfrac{1}{4}$hour
Length of the bridge = Distance Travelled by the man = $Speed \times Time = 5 \times \dfrac{1}{4}$km
= 5$\dfrac{1}{4} \times 1000$ metre= 1250metre
Let the duration of the flight be $ x $ hours.
| Then,$ \dfrac{600}{x} $-$ \dfrac{600}{x + (1/2)} $= 200 |
| $\Rightarrow \dfrac{600}{x} $-$ \dfrac{1200}{2x + 1} $= 200 |
$\Rightarrow x \left(2x + 1\right)$ = 3
$\Rightarrow$ 2$ x $2 + $ x $ - 3 = 0
$\Rightarrow$ $\left(2x + 3\right)\left(x- 1\right)$ = 0
$\Rightarrow x $ = 1 hr. [neglecting the -ve value of $ x $]
Let speed of the car be $ x $ kmph.
| Then, speed of the train =$ \dfrac{150}{100} x =\left(\dfrac{3}{2}\right)$ x kmph. |
| $\therefore \dfrac{75}{x} $-$ \dfrac{75}{(3/2)x} $=$ \dfrac{125}{10 \times 60} $ |
| $\Rightarrow \dfrac{75}{x} $-$ \dfrac{50}{x} $=$ \dfrac{5}{24} $ |
| $\Rightarrow x $ =$ \left(\dfrac{25 \times 24}{5} \right) $= 120 kmph. |
Let the time in which he travelled on foot = x hr
Then the time in which he travelled on bicycle = $\left(9 - x\right)$ hr
distance = speed x time
$\Rightarrow 4x$ + 9$\left(9 - x\right)$ = 61
$\Rightarrow 4x$ + 81 - 9$x$ = 61
$\Rightarrow 5x $= 20
$\Rightarrow x$ = 4
$\Rightarrow $ distance travelled on foot = 4$x = 4 \times 4 $= 16km
Let the distance travelled by $ x $ km.
| Then,$ \dfrac{x}{10} $-$ \dfrac{x}{15} $= 2 |
$\Rightarrow$ 3$ x $ - 2$ x $ = 60
$\Rightarrow x $ = 60 km.
| Time taken to travel 60 km at 10 km/hr =$ \left(\dfrac{60}{10} \right) $hrs= 6 hrs. |
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
| $\therefore$ Required speed =$ \left(\dfrac{60}{5} \right) $kmph.= 12 kmph. |
The man in the train notices that he can count 21 telephone posts in one minute.
Number of gaps between 21 posts are 20 and Two posts are 50 metres apart.
It means 20 x 50 meters are covered in 1 minute.
Distance = $20 \times 50$meter = $\dfrac{20 \times 50}{1000}$km = 1 km
Time = 1 minute = $\dfrac{1}{60}$hour
Speed = $\dfrac{Distance}{Time} = \dfrac{1}{(\dfrac{1}{60})}$ = 60km/hr
Assume that they meet $ x $ hours after 8 a.m.
Then, train1,starting from A , travelling towards B, travels x hours till the trains meet
$\Rightarrow$ Distance travelled by train1 in x hours = Speed $\times$ Time = 60$x$
Then, train2, starting from B , travelling towards A, travels $\left(x-1\right)$ hours till the trains meet
$\Rightarrow$ Distance travelled by train2 in $\left(x-1\right)$ hours = Speed $\times$ Time = 75$\left(x-1\right)$
Total distance travelled = Distance travelled by train1 + Distance travelled by train2
=> 330 = 60$x$ + 75$\left(x-1\right)$
=> 12$x$ + 15$\left(x-1\right)$ = 66
=> 12$x$ + 15$x$ - 15 = 66
=> 27$x$ = 66 + 15 = 81
=> 3$x$ = 9
=> $x$ = 3
Hence the trains meet 3 hours after 8 a.m., i.e. at 11 a.m.