Math help with liner applications ?

Question

how many liters of 40% and 80% acids must be mixed in order to make 20 liters of a 55%acid solution ?

Answer 1

Suppose x liters of 40% and 20-x liters of 80% are taken. Then

solve 40x + 80(20-x) = 55(20) you will get it.

Answer 2

Liters of 40% solution (x):

0.4x + 0.8(20 - x) = 0.55(20)

0.4x + 16 - 0.8x = 11

0.4x = 5

x = 12.5

Liters of 80% solution:

= 20 - 12.5

= 7.5

Answer: 40%, 12.5 liters; 80%, 7.5 liters

Proof (mixture is 55% solution):

= 100%([0.4{12.5 liters} + 0.8{7.5 liters}]/20 liters)

= 100%([5 liters + 6 liters]/20 liters)

= 100%(11/20)

= (100%/20)(11/[20/20])

= 5%(11)

= 55%

Answer 3

let quantity of acid1=x litres and quantity of acid2=y litres

given that 40x+80y=55*20 and x+y=20

i.e., x+2y=27.5 and x+y=20

on subtracting you get, x=7.5 and hence, y=12.5

so, quantity of acid1=7.5 litres and quantity of acid2=12.5 litres

Answer 4

let the number of liters be x

then the equation formed is-

40% x + 80% x = 55 % * 20 liters

solving the equation now this is also equal to

=>0.4x + 0.8x= 0.55 * 20

=>1.2x = 11

=> x= 11/1.2

=>x= 9.167

nw puting back the value of x in original equation

solution1 = .4 * 9.167

              = 3.668 lts of 40 % solution

solution2 = 0.8 * 9.167

             = 7.33 lts of 80% solution

now if you add the solution 1 & 2 u get 11 lts (55% of 20 litrs)