Question

water in a canal is 6m wide and 1.5m wide..it flowing with a speed of 10km/hr.how much area will it irrigate in 30 mins...if 8 cm if standing water is needed?

BIKASH travels 300 km to his home paetly by train and paetly by bus..he takes 4 hrs if he travels 60km by train and remaining by bus... if he travels 100 km by train and remaining by bus..he takes 10 mins longer..find the speed of train and the bus separately..

HLW GUYS THOSE WHO R IN 10 THEY WOULD BE ABLE TO ANSWER MORE NICELY AND PLZ GUYS WHO COULD BE ABLE TO ANSWER BOTH CORRWCTLY HE OR SHE COULD GET THE BRAINLIST ONE.....BEAT OF LUCK GUYS

Answer 1

(1)

Given that the length of canal = 6m.

Given that the breadth of canal = 1.5.

We know that Area = length * breadth

                                 = 6 * 1.5

                                 = 9m^2.


Given that Water is flowing with a speed = 10km/hr

                                                                     = 10000/60 m/minute.


Then the volume of water flowing in 1 minute = 

9 * 10000/60

= 1500m^3.


Then the volume of water in 30 minutes = 

30 * 1500

45000m^3.


Now,

The area irrigated by this water if 8cm(8/100 = 0.08m) standing water is desired = 


45000/0.08

= 562500m^2

= 56.25 hectares.


Therefore Area irrigated = 56.25 hectares.



(2)

Let the speed of the train = x km/hr.

Let the speed of the bus = y km/hr.

Given that Total distance traveled = 300km.

Now,

Condition (1):

He takes 4 hours if he travels 60 km by train and remaining by bus.

That means Distance traveled by Train = 60km.

                     Distance traveled by Bus = Total distance  - Train Distance

                             = 300 - 60

                             = 240km.


This can be written in equation as 

60/x + 240/y = 4

60y + 240x =  4xy  ---------------- (1)



Condition (2):

he travels 100km by train and remaining by bus, he takes 4 hours and 10 minutes longer = 4.1666 hours (or) 25/6.

That means Train Distance = 100km.

                     Bus Distance = 300 - 100

                                            = 200km.


This can be written in equation as,

100/x + 200/y = 25/6

100y + 200x = 25xy/6 

600y + 1200x = 25xy ------------------ (2)


On solving (1)*10 & (2), we get

600y + 2400x = 40xy

600y + 1200x = 25xy
------------------------------
             1200x = 15xy

               y = 1200/15

                  = 80.


Substitute y = 80 in (1), we get

60y + 240x = 4xy

60(80) + 240x = 4 * x * 80

480 + 240x = 320x

480 = 320x - 240x

480 = 80x

x = 480/80

x = 60.


Therefore the speed of the train = 60km/hr.

Therefore the speed of the bus = 80km/hr.



Hope this helps!  ------------ Good Luck

Answer 2

HELLO DEAR,

WIDTH OF THE CANAL=6 meter

DEPTH OF THE CANAL=1.5 meter

FLOW RATE OF WATER =10km/h

WE KNOW THAT DISTANCE=SPEED×TIME

DISTANCE TRAVELED BY WATER IN 30minute
=>
$1000 \times \frac{1}{2} = 500meter$
AMOUNT OF WATER IRRIGATED IN 20 MINUTES

=> 6 x 1.5 x 500 m3

= >4500 m3

AREA IRRIGATED BY THIS WATER IF 8 CM STANDING WATER IS DESIRED I.E 8CM = 0.08 M

= 4500 / 0. 08 M2

= 56250 M2

= 0.05625 KM2.

(2). TOTAL DISTANCE TRAVELLED BE BIKAS=300


LET SPEED OF TRAIN BE X


LET SPEED OF BUS BE Y

TRAVELLING 60KM BY

TRAIN AND REST
BY BUS S
HE TOOK 4 HOURS TO COVER 300km


TRAVELLING 100KM BY

TRAIN AND REST BY TRAIN HE TOOK 4 HOURS

AND 10 MINUTES.


PUTTING THIS

RELATIONS INTO EQN, WE GET,
=>60/x+240/y=300
=>100/x+200/y=300


REARRANGING THE

ABOVE EQNS 
=>60y+240x=300xy→(1)


=>100y+200x=300xy→(2)
=

DIVIDE FIRST EQN BY 60


=>y+4x=5xy→(3)


Divide second by 50


=>2y+4x=6xy→(4)


SUBTRACTING (3) FROM (4)
=>2y+4x=6xy
-y-4x=-5xy
Gives x=1km/min
=60km/hr
FROM EQN 

=>60/x+240/y=4
=>60/60+240/y=4
=>240/y=4-1
=>240/3=y
y=80km/hr


Speed Of Train Is 60KM/HR

SPEED OF BUS IS 80KM/HR


I HOPE ITS HELP YOU DEAR,
THANKS