Three solutions contain an acid 1 has 15acid 2 has 20 acid 3

Solution

let amount of acid volume in 20% acid be x, amount of acid volume in 30% acid be y

mixture is to contain 100L more of 15% acid than 20% =>amount of acid volume in 20% acid =100+x

A chemist wishes to use all 3 solutions to obtain 600L of acid with 25% acid.

=>total volume of acid =x+(100+x)+y =600

=>2x+y=500

=>y=500-2x---------->(1)

total quantity of acid=(15/100)*x+ (20/100)(100+x) +(30/100)y =600(25/100)

15x+2000+20x+30y=600*25

35x+30y=(600*25)-2000

35x+30y=13000

7x+6y=2600--------------------->(2)

substitute (1)in(2)

7x+6(500-2x)=2600

7x+3000-12x=2600

5x=400

x=80

=>y=500-2*80

=>y=340

finally amount of acid volume in 15% acid 100+80=180L, amount of acid volume in 20% acid be 80L, amount of acid volume in 30% acid be =340L,