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lidiya [134]
3 years ago
7

CAN SOMEONE HELP ME WITH QUESTION 3. PLEASE SHOW YOUR WORK

Mathematics
1 answer:
oee [108]3 years ago
8 0
Let 
x = number of liters of the 40% salt solution 
y = number of liters of the 20% salt solution 

The two amounts (x and y) must combine to 1500 liters, so 
x+y = 1500
we can solve for y to get
y = 1500-x
after subtracting x from both sides

---------------------------------------------------

If we have x liters of the 40% salt solution (composed of pure salt plus other stuff) then we have exactly 0.40*x liters of pure salt. Simply multiply the decimal form of the percentage with the amount of solution. 

Similarly, if we have y liters of the 20% solution, then we have 0.20*y liters of pure salt

Combined, we have 0.40*x + 0.20*y liters of pure salt all together. 

We want 1500 liters of a 28% solution, so we want 1500*0.28 = 420 liters of pure salt

Equate the two expressions (0.40*x + 0.20*y and 420) to get
0.40*x + 0.20*y = 420

---------------------------------------------------

We have the equation 0.40*x + 0.20*y = 420 and we also know that y = 1500-x

Let's use the substitution property now

0.40*x + 0.20*y = 420
0.40*x + 0.20*( y ) = 420
0.40*x + 0.20*( 1500 - x ) = 420 ... note how y is replaced with 1500-x

Now we can solve for x
0.40*x + 0.20*( 1500 - x ) = 420
0.40*x + 0.20*(1500) + 0.20*(-x) = 420
0.40*x + 300 - 0.20x = 420
0.40*x - 0.20x + 300 = 420
0.20x + 300 = 420
0.20x + 300 - 300 = 420 - 300
0.20x = 120
0.20x/0.20 = 120/0.20
x = 600

Now that we know x, use this to find y
y = 1500-x
y = 1500-600 ... plug in x = 600 (ie replace x with 600)
y = 900
---------------------------------------------------
---------------------------------------------------

Answers:

We need 600 liters of the 40% solution
We need 900 liters of the 20% solution

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