Question

In a liquid acid and water is inthe ratio 4:3 10 l acid adedthe new ratio is 3:1 find the water and acid

Answer 1

let initial quantity of Acid=4A,

initial quantity of water=3A,

after adding 10 ltr acid,

(4A+10)/3A = 3/1,

4A+10=9A,

then

10=9A-4A,

10=5A,

then

A=2,

then

initial acid=4×2=8 ltr,

initial water=3×2=6 ltr,

final acid=8 ltr+10 ltr=18 ltr,

and final water will be 6 ltr coz there is no change in the quantity of water.

short approach:- (Orally approach to solve ratios based questions),

Initial ratio of Acid and water = 4 : 3,

final ratio of Acid and water = 3 : 1,

but it is clearly mentioned in the question that the quantity of water has no changed then just equate the ratios of water in initial and final conditions so just multiply the final ratios by 3 so that the ratio of water will be same mean 3.

hence

new final ratio of Acid and water = (3 : 1) × 3,

=9 : 3,

so it's clear that the gap between the initial ratio and new final ratio of Acid is 5 unit but according to the question actuall difference is 10 ltr,

then

5 units = 10 ltr,

so

1 unit = 2 ltr,

therefore

final quantity of Acid = 9 unit,

= 18 ltr,

final quantity of water = 3 unit,

= 6 ltr