Question

There are some milk in three vessels.
1/3
part of the
first mixture is poured into second vessel.
1/4
part of
the second mixture is poured into third vessel.
1/10
part of the third mixture is poured into first ves-
sel. Now all three vessels contains 9 - 9 lt milk, then
find the initial quantity of milk in all three vessels.

Answer 1

"Given: There are some milk in three vessels. 1/3 part of the first mixture is poured into second vessel. 1/4

part of the second mixture is poured into third vessel.

1/10 part of the third mixture is poured into first vessel.

To find: The initial quantity of milk in all three vessels

Answer: 27 Litres"

Answer 2

Answer:

A=12lt, B=8lt, C=7lt

Step-by-step explanation:

let the initial quantity of mixture in 3 vessels A, B and C be x, y and z.

After pouring 1/3 part of A into B:

mixture remaining in A= 3x/4  

mixture in B= y+(x/3)....equ.i

After pouring 1/4 part of B in C:

mixture remaining in B= 3/4(y+(x/3))....equ.1

mixture in C=z+1/4(y+(x/3))

After pouring 1/10 part of C in A:

mixture remaining in C= 9/10*[z+1/4(y+(x/3))].....equ.2

mixture in A= (3x/4)+1/10*[z+1/4(y+(x/3))].....equ.3

equ.1=equ.2=equ.3=9 (given)

from equ1: 3/4(y+(x/3))=9 => (x/3)+y=12 => x+3y=36

from equ2: 9/10*[z+1/4(y+(x/3))]=9 => z+1/4((x/3)+y)=10 (put  (x/3)+y=12)

                  => z=7

from equ3: (3x/4)+1/10*[z+1/4(y+(x/3))]=9 (put (x/3)+y=12 and z=7)

                  => x=12

now put x=12 in equ.i => y=8