Question

A dog after travelling 50 km meets a swami who counsels him to go slower.He then proceeds at 3/4th of his formal speed and arrives 35 min late.Had the meeting occured 24 km further the dog would have reached the destination 25 mins late.The speed of the dog is?
a.)48 km/hr
b.)36 km/hr
c.)54 km/hr
d.)58 km/hr

Answer 1

what meeting the dog is attending ?

Answer 2

he is attending a secret meeting with a swami

Answer 3

Also is it an indian dog ?

Answer 4

yep

Answer 5

https://www.kalimandir.org/wp-content/uploads/2014/06/full.jpg

Answer 6

lmao

Answer 7

lets say the original speed of dog is \(s\)

Answer 8

say the distance is \(d\) : \[\dfrac{50}{s} + \dfrac{d-50}{\frac{3}{4}s} = \dfrac{d}{s}+35\tag{1}\]

Answer 9

is dog travelling back 50 km at \(3s/4\) speed

Answer 10

Okay looks i misinterpreted the problem
let me try again :)

Answer 11

@mathmath333 hey i got the answer.....
and plz expalin where its written tht he is returning back

Answer 12

*explain

Answer 13

this page has a solution , which i didnt get

http://www.piit.testfunda.com/Answers/ViewQuestion.aspx?QID=0fac7f0e-ea91-4869-a85f-d73caca87114

Answer 14

@mathstudent55 mine answer is correct

Answer 15

oh sry ,it is not written that he is returning back

Answer 16

hmm i don't like interpreting other's solutions much... let me try once again

Answer 17

see let the whole distance be x km.....and his speed be "a" km/hr
so if he travels whole distance at tht speed he is gonna take time \[T _{1} = x/a\]

Answer 18

i am assuming now that he is going forward and not returning back

Answer 19

@mathmath333 u need to frame few equations using the data u r provided.......:)

Answer 20

yep i know, i tried and failed.

Answer 21

because i cant understand the question correctly

Answer 22

see the time he would take if he travelled at "a" km/hr is T = x/a
now see he travels 50 km with tht speed...so time he takes to travel tht distance
T1 = 50/a hr
now the distance left i.e (x - 50) he travells at (3/4)a km/hr....si time taken
T2 - (x - 50)4/3a

Answer 23

question says T1 + T2 = T + 35/60
u need to change 35 mins into hour

Answer 24

try to frame another equation urself........@mathmath333 :)))

Answer 25

u mean \(\Large T_2 = \frac{(x - 50)4}{3a}\) ?

Answer 26

it says "Math Processing error" lol...;P

Answer 27

reload the page

Answer 28

nah brother its failing.......hold on i m uploading a pic ogf the solution....:))

Answer 29

and second equation is this ?
\(\large \color{black}{\begin{align} \dfrac{74}{a}=\dfrac{4(x-74)}{3a}+\dfrac{25}{60}\hspace{.33em}\\~\\
\end{align}}\)

Answer 30

the first equation is
\[(x - 50)4/3a + 50/a = x/a + 35/60\]

Answer 31

give another try for the second equation
see this he travels 74 km with speed "a" km/hr

Answer 32

its not tht....see T3 = 74/a
and the rest distance i.e (x - 74) he travels at (3/4)a km/hr
so T4 = (x - 74)4/3a

Answer 33

i have attached by second equation in pic

Answer 34

T3 + T4 = x/a + 25/60

Answer 35

@mathmath333 ??????

Answer 36

yes

Answer 37

solve the equations u will get the answer.....:))

Answer 38

what answer did u get from there

Answer 39

option a . 48 ?

Answer 40

48 km/hr.....:))

Answer 41

wbu????

Answer 42

can u write again ur two equations in one line .

Answer 43

first equation
\[\frac{ (x - 50)4 }{ 3a } + \frac{ 50 }{ a } = \frac{ x }{ a } + \frac{ 35 }{ 60 }\]

Answer 44

ok

Answer 45

second
\[\frac{ (x - 74)4 }{ 3a } + \frac{ 74 }{ a } = \frac{ x }{ a } + \frac{ 25 }{ 60 }\]

Answer 46

ok thanks i will ponder this equations now

Answer 47

all the best