49.6k views
4 votes
3) How many liters of a 20% acid solution must be mixed with a 60% solution to obtain 40 liters of a 35% solution? (25,15)THATS THE ANSWER I NEED THE STEPS

2 Answers

3 votes
the answer is 14


Have a great night and best of luck my friend :)

User Chandan Kumar
by
7.9k points
4 votes
let x = the number of liters of the 20% solution.

let y = the number of liters of the 60% solution.

you want x + y to be equal to 40 liters.

x is the number of liters total in the first solution.
y is the number of liters total in the second solution.

you want .2 * x + .6 * y to be equal to .35 * 40

.2 * x is the number of liters of acid in the first solution.
.6 * y is the number of liters of acid in the second solution.
.35 * 40 is the number of liters of acid in the final solution.

you have two equations that need to be solved simultaneously.

they are:

x + y = 40
.2x + .6y = .35*40

simplify these equations to get:


x + y = 40
.2x + .6y = 14

you can solve by substitution or by elimination or by graphing.
i will solve this one by graphing.
this means to graph both equations and find the intersection.

the graph looks like this:


the graph says the intersection is at the coordinate point of (25,15).

this means that x = 25 and y = 15.

x is the number of liters of the 20% solution.
y is the number of liters of the 60% solution.

the formula of .2x + .6y = 14 becomes .2 * 25 + .6 * 15 = 14.
simplify this equation to get 14 = 14.
this confirms the solution is good.
User Edudjr
by
8.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.