Question

A tanker used to transport oil is filled up by two pipes in 4 hours.If the larger pipe alone takes 6 hours less than the smaller pipe to fill up the tanker, find the time in which the tanker will be filled by each pipe operating individually.

Answer 1

6 hrs

Answer 2

can you explain how you did it?thanks a lot man.

Answer 3

(1/x)+(1/x+6)=1/4

Answer 4

(1/x)+[1/(x+6)]=1/4

Answer 5

I don't get it.Can you put it in a simpler way??

Answer 6

Let , larger pipe takes x hrs , individually.
so ,smaller pipe takes x+6 hrs,

Formula , if both pipes are operating at same time,
total time takes =
(1/x)+[1/(x+6)]=1/4

Answer 7

why do we need 1/x and not just x? sorry i am quite weak at math

Answer 8

1/x is taken because volume of the oil is considered 1. you can take any variable too

Answer 9

got it yet legend ?

Answer 10

let the larger pipe takes x hr and smaller takes x+6 hrs

the volume be v
in 4 hrs the larger pipe fills 4v/x and smaller 4v/(x+6)
so 4v/x+4v/(x+6)=v
1/x+1/(x+6)=1/4
4(2x+6)=x^2+6x
x^2-2x-24=0
(x-6)(x+4)=0
so x=6
so the larger one fills in 6 hrs and the smaller in 6+6=12 hrs

Answer 11

yeah got it..Thanks dispankarstudy.

Answer 12

6h and 12 is correct answer

Answer 13

yupp