a. | 20 L |
b. | 34 L |
c. | 15 L |
d. | 32 L |
Here's another way to do this with only one variable.
Call the amount of the 63% solution = "x"...then the amount of the 43% solution is just "40-x"..we have
.63x + .43(40-x) = .53(40) multiply through by 100
63x + 43(40-x) = 53(40)
63x + 1720 - 43x = 2120 simplify
20x = 400 divide through by 20
x= 20 L
Let X be the amount of 63% soln used and let Y be the amount of 43% solution used.
$$\begin{array}{rlll}
X+Y&=&40\qquad \qquad & (1)\\
0.63X+0.43Y&=&0.53(X+Y)& (2)\\
0.20X+(0.43X+0.43Y)&=&0.53*40\qquad $Error fixed - thanks Alan$\\
0.20X+0.43(X+Y)&=&21.2\\
0.20X+0.43*40&=&21.2\\
0.20X+17.2&=&21.2\\
0.20X&=&4\\
X&=&4/0.20\\
X&=&20\\
\end{array}$$
so it seems that 20L of the 63% solution was needed.
Maybe you were supposed to ESTIMATE the answer
A solution of 63% fertilizer is to be mixed with a solution of 43% fertilizer to form 40 liters of a 53% solution. How much of the 63% solution must she use?
Choices 20L, 34L, 15L, 32L
The 63% solution is too strong by 10% and the 43% is too week by 10% so if i had to guess I would say 50% of each is needed. That is 20L each.
Again: This answer has been edited.
Melody, the first term in the third line of your derivation should be 0.2X not 0.23X.
.
Here's another way to do this with only one variable.
Call the amount of the 63% solution = "x"...then the amount of the 43% solution is just "40-x"..we have
.63x + .43(40-x) = .53(40) multiply through by 100
63x + 43(40-x) = 53(40)
63x + 1720 - 43x = 2120 simplify
20x = 400 divide through by 20
x= 20 L