Question

If a 7% saline solution and a 4% saline solution are mixed to make 500 milliliters of a 5% saline solution, how much of each solution, to the nearest milliliter, is needed?
A. 167 milliliters of 7% solution and 333 milliliters of 4% solution
B. 110 milliliters of the 10% solution and 210 milliliters of 4% solution
C. 155 milliliters of the 10% solution and 300 milliliters of 4% solution
D. 170 milliliters of the 10% solution and 340 milliliters of 4% solution

Answer 1

Alg. 2?

Answer 2

yea

Answer 3

lol, same.

Answer 4

FLVS?

Answer 5

Or what?

Answer 6

what

Answer 7

NVM

Answer 8

I'm not sure.

Answer 9

I haven't gotten up to this.

Answer 10

Either that or I forgot it.

Answer 11

Let x be the number of milliliters of the 7% saline solution and y be the number of millilitres of the 4% saline solution.

They need to add up to 500 milliliters total, so x + y = 500.

The amount of salt in the x milliliter 7% solution is 0.07x milliliters. The amount of salt in the y millilitre 4% solution is 0.04y milliliters. Together, they need to add up to 5% of the total 500 milliliters, which is 0.05(500) = 25 milliliters, so 0.07x + 0.04y = 25.

Solve by substitution:

x + y = 500
x = 500 - y

0.07x + 0.04y = 25 (substitute 500 - y for x)
0.07(500 - y) + 0.04y = 25
35 - 0.07y + 0.04y = 25
-0.03y + 35 = 25
-0.03y = -10
y = 333.333...
y = about 333

x = 500 - y = 500 - 333 = 167
get it?

Answer 12

A.

Answer 13

lol

Answer 14

Okay them

Answer 15

Then*