A bucket contains a mixture of two liquids A and B in the proportion 7 : 5. If 9 liters of mixture is replaced by 9 liters of liquid B, then the ratio of the two liquids becomes 7 : 9. How much of the liquid A was there in the bucket?

dsuc.created: 2 years ago | dsuc.updated: 1 year ago

dsuc.updated: 1 year ago

$T h e b u c k e t c o n t a i n e d x l i t e r s o f A a n d y l i t e r s o f B G i v e n t h a t, \frac{x}{y} = \frac{7}{3} ∴ x = \frac{7 y}{5} . . . . . . . . . . . (i) a n d y = \frac{5 x}{7} . . . . . . . . (i i) s u m o f r a t i o = 5 + 7 = 12 N o w, i n 9 l i t e r s o f r e p l a c e d m i x t u r e, t h e a m o u n t o f A w a s = \frac{7}{12} \times 9 = \frac{21}{4} p a r t a n d t h e a m o u n t o f B w a s = 9 - \frac{21}{4} = \frac{36 - 21}{4} = \frac{15}{4} p a r t A c c o r d i n g t o q u e s t i o n, \frac{x - \frac{21}{4}}{y - \frac{15}{4} + 9} = \frac{7}{9} = \frac{x - \frac{21}{4}}{\frac{5 x}{7} + \frac{21}{4}} = \frac{7}{9} = 9 x - \frac{189}{4} = 5 x + \frac{147}{4} = 4 x = \frac{189}{4} + \frac{147}{4} = \frac{189 + 147}{4} = \frac{336}{4} = 84 ∴ x = \frac{84}{4} = 21$