Speed and distance Math

Speed and distance Online education teaching for math   speed and distance                          

Hello viewers I hope that all fine and m fine so today discussion speed of math . BCS and PSC exam shortcut math discussion I will do it. So let's go today's speed shortcut math solution.

At first a example : A boy goes to school with a speed of 200 km / hr it he takes 3 hours then , the distance?

         

            Before we will know

                                  Structure

SL       description.    Symbol.    Question

01.      Speed.               S.                200 km/h

02.       Time.                T.               3 hours

03.       Distance           D.                     ?

Formula D = ST

Solution D= 200×3 = 600 Answers

                       Practice now

1/A man goes to Dhaka a with a speed of 50 km/ hr. If he takes 4 hours then the distance is?

Solution D = 50×4 = 200 Answer.

2/Nazrul crossing street 600 long in 5 minute. His speed in km per hours is?

                             Structure

Formula : D = ST

Solution : 600×5/18 = s×5×60 ⇒s=600/5×60×18/5 km/ hr= 7.2 km/ hr

3/ Dhaka to Chittagong distance 300 km. From Dhaka to start this morning 7 am then arrive to Chittagong 3 pm .how do speed per hour the train?

300= s×8⇒s = 300/8 = 37.5 km/hr

4/ Dhaka to Chittagong distance 600 km.from Dhaka to start the train this morning 7 am . Arrive to Chittagong 1.00 pm. How do speed the train  per hours?

Solution : 600= s×6 ⇒s = 100 km/ hr

4/ Dhaka to Chittagong distance 300 km. From Dhaka to leave this morning 9.00 am and arrive to Chittagong 12.00 

Pm. How do speed the train per hours?

Solution : 300= s×3 ⇒s = 100 km/hr

5/Bimal sprinted 100 metres in 10.61 second. Find his velocity in kilometres / hours. Round to nearest kilometres?

Solution : 100 = s×10.61 ⇒s 9.43 m/ sec = 

9.43×18/5 km/ = 34 km.

6/ GM sharif covers 10.2 km with a speed of 3.4 km/ hr . The time takes cover this distance is?

                         

Structure

Stay fixes distance find out time

Formula : D = ST

Solution : 3.4× T ⇒ T = 3

                    Instant practice

6/ Mr. Habibi covers 30 km with a speed of 10 km/hr . Find the time taken to cover this distance?

Solution : 30 = 10 ×T ⇒ T = 3 Answer
7/ example : Gimu covers a certain distance by car driving at 70 km/ hr and he returns back at the starting point riding on scooter at 55 km/ hr find out his average speed the whole journey?

                        Structure

SL   Description.      Symbol.      Question

01.  Go speed.              X.               70 km/ hr

02.  Back speed.          Y.               55 minute

03. Find out average  s.                   ?

          Speed

Formula  S = 2xy/x+y

Solution : s = 2×70×55/70+55= 61.6

                                         km Answer

8/ A man travels a distance at a rate of 20 mile/ hours  & return at the rate of 30 mile / hours. What is he Avareg speed?
Solution : s = 2×20×30/20+30 = 14 answer.
9/ A man travels from A to B at speed of 20 km/ hr he makes the return journey at speed of 30 km / hr . what was his average speed?
Solution : s = 2×20×30/20+30 = 24 Answer
10/ the distance between Dhaka and bogra is 300 km. A car travels from Dhaka to bogra at 100 km per hours.and return from bogra to Dhaka along the same route at 60 kmper hours. What is the average speed of the round trip?
Solution : 2× 100×60/200+60 = 75 Answer

Speed and distance time Math

Time distance and speedHello viewers I that hope that all fine. Discussion for today how to math solution easy to succeed. Our how to knowledge can't math solution. As a result just fail any exam I think that math very simple and easy.we are think that math is heart As a result any exam fail.

              why don't math solution us.?
     A special point every person thinking is
     Math very heart it's very tension we are
     No tension just think very simple and
     Easy.
               How do think math solution?
(1) Yousef think math is very simple
(2) No tension math is very simple and easy.
(3) math will be try everyday day to night
    two hours.
(4) tension is very sad because fail the exam
(5) not fear of math I confidential can it .
(6) it need point of math regular try so
      successful exam
So hello viewers move on any examples
Today do it Relative speed job example
                             speed
Speed = distance/ time     
Avareg speed= distance/ time
⇛   A train lift from kh  station45 after kha stopped of station. Two station distance 85 km . But how to Avareg speed of two train?
Example :  1 hours = 60 minute
 1.   ,,.    ,,       ,,. 84/45. ,,
,, 60  ,,.   ,,.     ,, 84×60/45 = 112 km answer
⇛   A car speed per second 60 ft. How to speed per hours?
Example : 1 second gone = 60 ft.
1 hours or ( 60×60) second gone ( 60×60×60)
                       = ( 60×60×60)/3×1760
                       = 40.9 mile answer
⇛A train 46 minute A town from B town arrived . Two town distance 59.8 mile how to speed the train per hours?
Example : 46 minute gone = 59.8 mile
1.         ,,.      ,,.  59.8/46 mile
60       ,,.       ,,.  59.8×60/46
                      = 78 mile
⇛Distance of Chittagong to Dhaka 265 km a bus has arrived 7 hours  Rasthayto Khulna
A bus stopped 1 hours in a Road . how to Avareg speed s bus ?
Example : A bus speed =  distance/time
                      = 275/5 km
                      = 53 km.
⇛ distance of Rajshahy to Khulna 282 km. A bus has arrived 7 hours . But a bus stopped 1 hours how to Avareg speed a bus per hour?
Example :  bus move on = (7-1) = 6 hours
Avareg speed of bus = 282/6 km
                                      = 47 km Answer
⇛ A speeder speed of fevareble 1 hours 3 mile . And starting point of back . How to speed per hours?
Example : distance = (3+3) mile
                                   =  6 mile
Cost Tim.                  = ( 1+3)
                                   = 4 hours
Avareg speed.          = 6/4 mile. = 1 -1/2  mile

⇛ A speeder speed of fevareble 2 hours he gone 5 mile. And he back to 4 hours. But how to traveling Avareg speed per hour?
Example : distance = (5+5) mile
                                    = 10 mile
Cost time                   = (2+4)
                                    = 6 hours
Avareg speed           = distance/ time
                                   = 10/6 mile
                                   = 5/3 mile
                                   = 1-2/3 Answer
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Relative speed Math solve

Relative speedOnline teach for job preparation of math Relative speed ( part 2)

 I hope that all fine . I'm

taking about Relative speed how to solution train or car math example solution. So viewers hope that on my site very successful in our students life and job examination. Don't worry math solution Relative speed no heart very simple very  easy . Just we have to bad tension Very simple .so Don't more talking now  move on details Relative speed example solution . Okay viewers .
1/ Distance Dhaka to Chittagong 185 mile. By a bus from Chittagong per hours 2 hours and 85 mile going to 100 mile . how to speed by bus 50 km.?
⏩Answer : going to Dhaka fromChittagong  needed time t = distance/time = 185/50 = 3.7
85  mile 2 hours going to = (3.7-2) hours
        = 1.7 per hours = ( 1.7×60) = 102 minute
2/ A car speed per hour 25 mile.runing Chittagong from Dhaka.  Another A car speed per hour running Dhaka from Chittagong . Dhaka to Chittagong distance 200 mile. How to running between car After time face to face between car?
⏩Answer : when two car running opposition side
Therefore there Relative speed per hour
 ( 25+15)=  40 mile.
1  hoursspeed go on 40 mile
1   ,,    ,,    1/40
200    ,,     200/40 = 5 hours.
3/ traveling to fast 6 hours Avareg speed 40 km. And second position Avareg speed 60 km. If complete  traveling to car Avareg 55 km. how to traveling time is?
⏩Answer : the car traveling time = t .
       The cat crossing t hour 55 km.
       The car crossing 6 hour =
                       (40×60)=240km
The car (t-6) crossing per hours
                           =  60( t-6) km
Condition apply 240+60( t- 6) = 56 t
Or 240 +60 t - 360 = 55 t
Or 5 t = 120

So t = 24 minute .
4/ house of Karim from Dhaka distance to 356 k.m. he has gone from Dhaka 319 km After bus is out of command so going to by Rickshaw. The speed of bus 22 km. Per hour. And Rickshaw speed 6 km. Per hour. So how did time arrive korim of house?
⏩Answer : Karim go to by bus 319 km
                    By Rickshaw ( 355-319)
                                             = 36 km.
                    Time = distance/ speed
By bus.       = 319/22 = 14 hour 30 minute
By Rickshaw= 36/22 = 6 hours.
Arrive to home time = 14 hour 30 minute+6 hour = 20 hours 30 minute.
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Relative speed train Math suggestions

Speed

Online math preparation Relative speed ( train)

we Done relative speed by train. Relative speed very simple and easy solution. So viewers we will try train speed how to knowing Relative speed solution. So

viewers let's go discuss at math solution.

                        Relative speed

Suppose between two train speed  u and v(>v)

1/ When  two car or between train move on opposite then Relative speed both equal .

         Relative speed = u+v

2/ when two car or train move on same side then Relative speed  = u-v

         Calculation the speed of a train

A signals post crossings x length and needed t time 

                A train t time crossings x mitres

So ,,      ,,       ,,   X/t

          Train speed = x/t

Y length a object ( when pull and hole, platform.) Others stand a car x length  needed t time .  

A train t time crossings (x+y). metre Dist.

Metres    ,,    ,,   ,,  x+y/t

                      Example solution

We Done example solution Relative speed A train. Okay . 

1 / A train speed per hour 60 metre how do time crossings A lamp post  100 metre train?

Answer:     1 hours = (60×60) second

                   60 km     = (60×1000) metre

(60×1000) crossings time (60×60) second.

1   ,,    ,,    ,,    ,,  60×60/60×100

100    ,,    ,,    ,,  60×60×100/60×1000

       

   Answer 6 second.

We saw more examples

2/ 50 metres length A train speed per hour 36 km. How do crossings time the train a lamp post ?

Answer: 36 km  =  (36× 1000) metre =3600 m

1 hour (60×60) = 3600 second.

The train crossings a post and self length 

          36000 metre needed time 3600 second

1            ,,     ,,    ,,   ,,   3600/36000

50         ,,     ,,    ,,    ,,   3600×50/36000

            Answer 5 second.

3/ it hase a plam tree a side on railline . Per hour continue a train speed 45km . The train length is 150  how do crossings time this plam tree?

Answer :  (45×1000) metre =45000 metre.

1hour = (60×60) = 3600 second

The train crossings and plam tree 45000 m

3600 second.

     1.  ,,     ,,    ,,   3600/45000

   150 ,,    ,,    ,,   3600×150/45000

              Answer = 12 second.

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Time and distance Math suggestions

Distance

Secondary school education teaching for students class time and distance (part 05)

We will be knowing time and distance relative math solutions so let's go come to the point. And details time and distance math policy. Okay friend.

                                  (Time)

⇨60  ( Second)  =  1 (minute)

⇨60  (minute).    =   1 ঘণ্টা (Hour)

⇨24  Hour                       =   1 .(Day)

⇨7   Day                           =   1 (week)

⇨4 week or  28 day       =. 1 (Lunar 

                                                               Month)

                    Name of shuchak

        15                                         15

⇨10.     Peta    1 peta metre = 10.     Metre

        12                                          12

⇨10.     Tera     1 tear gm     = 10.      gm

        9.                                            9

⇨10     geaga      1 geagajhu    =  10 jhul

        6.                                        6

⇨10     mega     1 mega wat  10 wat 

        3.                                           3

⇨10 k.l        1 k.l volt  10 volt

         2.                                               2

⇨ 10  Hector      1 Hector Pascale 10 Pascale

 

    1

⇨10  deka         1 deka Newton 10 Newton

       -1.                                        -1

⇨10  D.c         1 d.c Wham 10 Newton

       -2.                                          -2

⇨10  c.m         1 cm.10 metre

       -3.                                                 -3

⇨10  m.l.           1 m.l ampowar 10 metres

       -6.                                               -6

⇨10 mycro         1 mycro volt  10 volt

        -9.                                              -9

⇨10  nano .          1 nano second 10 second

        -12.                                             -13

⇨10  pk           1 pkf 10   

       -15.                                                -15

⇨10  famto         1 famtoletre 10  Lt.

       -18.                                           -18

⇨10  At           1 at.metre 10.  Metres
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Local and British weight system

Weight system British local and international.weight system British and international

Hello viewers how are ? I hope that all goods
Today I will be discussing international and local

weight system. Okay let's go come to the point.
                             

    weight

                 

local system

⇨5  Tolas            = 1 chatak
⇨16 Chataks      = 1  Seer
⇨40  Seer.           =  1  Mound

                    British system

⇨16  Ounce.          = 1 Pound
⇨28 pound.        = 1Quarter
⇨4 Quater.         = 1 Handar
⇨20 Handar         = 1 British ton

              metres system

⇨100  Kg.       = 1Quintal
⇨1000  Kg.     = 1 Metres
   

                                                             

⇨1 gm. = 0.0022pound = 0.086
                                                          (tola)
⇨1pound = 0.4536  K.g
⇨1 K.g = 2.2  pound = 1.07
                                                                     Seer
⇨1 Seer = 0.93  g.m
⇨1 Mound = 37.2 K.g
⇨1Handar = 1moun 14 seer 7 sotak  = 50.8
                                                        K.g
⇨1  British ton = 27 moun 9 seer = 1
                           Metres ton 16g.m

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Volume and balanced system secondary school Math suggestions

VolumeSecondary school education teaching balanced system volume (part 3)

We are knowing how do it balanced volume system local and international

         (Volume)
⇒1 .(Cubic metre) = 1. ( Stayor)
⇒1 . (Cubic metre) = 💯 (Litres)
⇒1 . (Litres).            = 💯(Cubic cen)
                  Related volume
⇒1 .(Stayor).              = 35.3 (Cubic
                                                                            ft)
⇒1 .(Cubic ft).            = 2867 (Litres)
                          (Liquidm.ca)
                                  British
⇒4 gills.                                      = 1pint (pt)
⇒2pints.                                     = 1quart(qt)
⇒4quart.                                    = 1gullons(gal)
⇒42U.S Gllon                            = 1 Barrel
                                  U.S
⇒1 pint (pt)                               = 1.201 pints
⇒1quart(qt)                              = 1.201quarts
⇒1gallon(gal)                           = 1.201 gallon
                               Metre
⇒1.202 pints.                            = 0.5679 pints
⇒1.201 quarts                          = 1.1359 litres
⇒1.201 gallon                           = 4.5437 litres
                          (D.M.C)
                     British & U.S

     
⇒2gallon.                                  = 1peck
⇒4peck.                                     = 1 bushel
⇒8bushel                                  = 1 quarts
                             U.S & British
⇒1gallon.                                  = 4.5434 Litres
⇒1 peck                                     = 9.0870 litres
⇒1bushel                                  = 36.3477litres
⇒1quarts.                                  = 290.7816 litre
                             British & U.S
⇒4.5434 Litres.                         = 4.404 litres
⇒9.0870 litres.                          = 8.810 litres
⇒36.3477 litres.                        = 35.238 litres
⇒290.7816 litres.                      = 281.904 litres
 So okay Now  stay good everybody

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Area and international balanced system secondary school Math suggestions

Math of Area Secondary school certificate teaching international and local balanced system Area( part 2)

  We will be knowing how do solve it local balanced system. So let's go come to the point. Okay 

                         (Area)

                     (Local Sytem)
⇛1  (Sq.inch).              1 (Ganda)
⇛20(Ganda).                1 (Satak)
⇛16 (Satak).                 1 (Katha)
⇛20 (Katha).                1(Bigha)
             

      (British system)
⇛144  (Sq.inch)      1 (Sq.ft)
⇛9 (Sq.ft)                     1(Sq.yrd)
⇛4840 (Sq.yrd).          1 (Acre)
⇛100 (Decimals).       1(Acre)
(Relationship between British local and metric System)
⇛4 (Gunda)                      1 (Sq.yrd)
⇛5 (Sq.yrd).                      1 (Satak)
⇛80 (Sq.yrd)                720  (Sq.ft)
⇛1 (Decimals).               435.6 (sq.ft)
⇛1600 (Sq.yrd)        1338(Sq.m)
⇛3Biga 4 sotak           4047 (Sq.m)
⇛640 (Acres).          (Sq.miles)
⇛247 (Acres).        1.(Sq.km)
⇛10000 (m✓) 100 (Ayor)=2.47

Measurements math secondary school education

Measurements Secondary school education teaching international system units measurements ( part 1)

Hello viewer's Assalamu alaykum I hope that all fine . Today discussing shall be international system unit .

১/ C.G.S System:⇒
  C.G.S meaning of method, c.m , gm, second it's system of metres

2/ M.KS system
  M.K.S it's meaning km, second. It's metre balanced
3/ F.P.S system
    F.P.S meaning of method fit , pound and second
  C.G.S           M.K.S         F.P.S
 Length       c.m.             fit
 Weight       gm              kg    pound
 Time    =    second
S.I  System:
Stat from  in 1960 in the word same processing all over the country system of (SI) Balanced . it's International System of Units . (S.I) Decided all over the country worldwide in 1982 use of  SI system length , weight,
⇒ meaning of metre
⇒ 1 metre = distance 1 core  divided 1.
Local system: (Local System)
⇒ 12   (Inch)  = 1  (foot)
⇒ 3     (Feet)   = 1 (yard)
⇒1760(Yard)  = 1 (mile)
( British system for measuring land)
⇒ 7.92  (inch)  = 1  ( link)
⇒ 25     (link)   =  1    (Rod)
⇒ 4       (Rod)   =  1 (Chain)
⇒ 10   (Chain) =  1(farling)

⇒ 8 (farlong)= 1 (mile)
(Relationship between British and metric System)
⇒ 1  (inch ) = 2.54 (c.m)
⇒ 39.37 (inch)  = 1 (metre)
⇒ 1        ( mile) = 1.61( k.m)
⇒ 0.62   ( mile) = 1 (k.m)
Balanced of sea  :
          British & US
⇒ 6 Feet = 1 fathom            = 1.8288 metres
⇒ 608 feet = 1 cable.           = 185.313 metres
⇒ 6080 feet = 1 sea(n.m)    = 1853.18 metres
⇒ 60 sea mile = 1 degree
⇒ 360 degree = 1circle
Note
1 Noticeable mile  = 2025 yeard
1 Noticeable mile = 1853.18 metre = 1.85318 km
1 Noticeable = 1.15 mile
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Compound and interest Math suggestions

Compound and circle

Secondary school education teaching for math ( compound interest and circle)

We will be learning compound interest so  let's go all dear my friend move on details

1/  population of the town 80 lakho . So this town growth of population rates 30 and after 3 years rate of population. How may do this people?
Example: population of a town , P= 8000000  Tim ,t= 3 year.
Circle of years, n= 2; growth of population rate, r= 30/1000×100%=33%
There applied growth of population rate formulation
Population C= (1+3/100×1)1×3
                             = 8000000×(1.03)3 tk
                      Answer  = 8741816 tk.
Example:
2/ 4% rate of interest how much do this compound and circle interest , will be difference 2 year's of 1 tk?.
P = 100 and  4%  rates year's Compoun interested  = (💯×2×4/💯) 8 tk.
Circle interest = {💯(1+4/100)2 -100} tk.
                   = {💯×(1.04)2 -100} tk.
                   =(১০৮.১৬-১০০) টাকা = ৮.১৬
Difference of interest (8.16-8) or 0.16 tk.
P 💯 tk.
1 "    "      "   100×100/16
                  = 625 tk.
Example:
3/ rate of annual interest 10 tk. Interest of 500 tk rate of 3 years. How will be interest and circle difference?
 There, P= 500 tk.
                            Time, t= 3 years.
  Circle of years,n= 1
Annual interest rate r, =10
Compound interest , = p+i
                                       = p+ p×r×t/100
                         

        =(5000+ 5000×10×3/100
                                    =65000 tk
Circle interest
         C= p(1+ r/n)n×t
           = 5000(1+ 10/100×1)1×3
           = 5000×(1.1)3
           = 6655 tk.
Difference of compound interest
           = (6655-6500)
           = 155 tk.

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