Question

4 gallons are drawn from a vessel full of wine it is then filled with water 4 gallons of mixture and again run and vessel is again filled with water the quantity of 1 hour left in a vessel to that of water is 36: 13 how much does the vessel hold​

Answer 1

Answer:

Vessel capacity = 28 Gallon

Step-by-step explanation:

Let say vessel has capacity of C gallons

Wine initially  = C  gallons

Removed 4 gallons

Wine = C -4 Gallons

Water = 4 Gallons

Again 4 Gallons removed which have

4 *(C-4)/C  Wine

Water  16/C

Net wine remaining = (C - 4)   -  4(C-4)/C

=(C -4 )(C - 4)/C

Water = 4 - 16/C  + 4

= 8 - 16/C

= 8(C - 2)/C

(C -4 )(C - 4)/C : 8(C - 2)/C :: 36/13

= 13 (C -4)²  = 36*8(C-2)

=> 13 ( C² + 16 - 8C) = 288C - 576

=> 13C² - 392C + 784 = 0

=> 13C² - 28C - 364C + 784 = 0

=> C(13C - 28) - 28( 13C - 28) = 0

=> (C - 28) (13C - 28) = 0

C = 28   C = 28/13 not possible as can not be less than 4

Vessel capacity = 28 Gallon