Kinetic graphs - class-X

Suppose he move 4 Km downstream in x hours.Then 
Speed downstream=$\frac{4}{n}$Km\hr
Speed upstream=$\frac{3}{n}$Km\hr
$\therefore \frac{48}{$\frac{4}{n}$}+\frac{48}{\frac{3}{x}}=14$ 
$\frac{48x}{4}+\frac{48x}{3}=14$
Or x=$\frac{336}{168}=2$
So spead downstream=$\frac{4}{\frac{1}{2}}=8$Km\hr
Or speed upstream =$\frac{3}{\frac{1}{2}}=6$Km\hr
So rate of stream =$\frac{1}{2}(8-6)=\frac{2}{2}=1$Km\hr

According to Abdul Aligarh is more than 10km and less than 15 kms
According to Nagma, Aligarh is more than 12 km but less than 14 kms
If both are correct, then the only common number between them is 13 km
Answer is Option A

By definition,

If a car travels a distance d at rate $r _1$  and returns the same distance at rate $r _2$ , then 
Average speed = $\dfrac{total distance}{total time} = \dfrac{2d}{d/r _1+ d/r _2}= \dfrac{2r _1r _2}{r _1+r _2}$; 
$\therefore  x = \dfrac{2.30.40}{70} = \dfrac{240}{7} ~ 34 km/hr$.

Let the principal be $P$ and rate of interest be $r$%.
According to question,
$1012 = P + \dfrac {P\times r \times 5}{100\times 2} .... (1)$
Interest in $\dfrac {3}{2} years = 1067.20 - 1012 = Rs. 55.20$
$P = \dfrac {I\times 100}{R\times T} = \dfrac {55.20\times 100}{3} = \dfrac {3680}{r}$
Putting values in equation $(1)$,
$1012 = \dfrac {3680}{r} + \dfrac {3680\times r\times 5}{r\times 100\times 2}$
$1012 = \dfrac {3680}{r} + 92$
$\dfrac {3680}{r} = 1012 - 92 = 920$
$r = \dfrac {3680}{920} = 4$% per annum.
Hence, the rate of interest is $4$% per annum.

Using the formula, $Speed=\frac { Distance }{ time } $
Mohit speed=
${ S } _{ m }=\frac { 225 }{ 4.5 } =50km/h$
${ S } _{ r }=50+10=60km/h$
Distance=225km
$Time=\frac { Distance }{ Speed } =\frac { 225 }{ 60 } =3\frac { 3 }{ 4 } hr$
Answer (C) $3\frac { 3 }{ 4 } hr$

$\Rightarrow$  Distance covered by grasshopper in one jump = $91.4,cm$.

Let distance between two consecutive trees be x m.
So, distance between 1st and 25th tree = 24x m = $30$ m
Distance between 3rd and 15th = 12x m = $15$

Time taken in downstream $= \displaystyle \frac{d}{15 + 5} = \frac{d}{20}$
Time taken in upstream $= \displaystyle \frac{d}{15 - 5} = \frac{d}{10}$
$\therefore$ Average speed $ = \displaystyle \frac{2d}{\displaystyle \frac{d}{20} + \frac{d}{10}} = \frac{40}{3} km/hr$
So, $\displaystyle \frac{\text{Average speed}}{\text{Speed in still water}} = \frac{40/3}{15} = \frac{40}{45} = \frac{8}{9}$.

Relative speed = (45 + 30) km/hr
$ = \left ( 75 \times \dfrac{5}{18} \right ) m/sec$
$= \left( \dfrac{125}{6} \right) m/sec.$
We have to find the time taken by the slower train to pass the Driver of the faster train and not the complete train.
So, distance covered = Length of the slower train.
Therefore, Distance covered = 500 m.
$\therefore$ Required time $\left( 500 \times \dfrac{6}{125} \right ) = 24 sec.$

Distance travelled by the train at 80 km/hr in 30 min. = 40 km. This distance of 40 km is to be covered with a relative speed of
(90 - 80) = 10 km/hr.
Time taken to cover this distance = $\frac{40}{10}$ = 4 hr.
Hence, the trains will be together after 4 $\times $ 90=360 km from Delhi.

A boy takes $20\ min$ to reach the school at an average speed $12\ km/hr$.

Let the length of Journey = x km, and
Speed of the train = v km/hr

$\displaystyle\therefore30:min.+45:min.+\frac{\displaystyle\left(x-\frac{v}{2}\right)}{\displaystyle\frac{2}{3}v}$

$\displaystyle=\frac{x}{v}+1:hour:30:min.$

$\displaystyle\implies\frac{1}{2}+\frac{3}{4}+\frac{\displaystyle\frac{2x-v}{2}}{\displaystyle\frac{2v}{3}}=\frac{x}{v}+\frac{3}{2}$

$\displaystyle\implies\frac{3(2\times-v)}{4v}=\frac{x}{v}+\frac{1}{4}$

$\displaystyle=\frac{4x+v}{4v}$

$2x=4v$ ....(i)

Again, it accident occurred 60 kms later,

$\displaystyle\therefore30:min.+\frac{60}{v}+45:min.+\frac{\displaystyle\left(x-60-\frac{v}{2}\right)}{\displaystyle\frac{2}{3}v}$

$\displaystyle=\frac{x}{v}+1:hour:30:min.-30:min.$

$\displaystyle\implies\frac{1}{2}+\frac{60}{v}+\frac{3}{4}+\frac{3(2x-120-v)}{4v}=\frac{x}{v}+1$

$\displaystyle\implies\frac{x}{v}-\frac{60}{v}-\frac{3(2x-120-v)}{4v}=\frac{5}{4}-1$

$\displaystyle\implies\frac{4x-240-6x+360+3v}{4v}=\frac{1}{4}$

$\implies2x=2v+120$ ....(ii)
Solving eq. (i) and eq. (ii), we get
$\therefore x=120:km$