|
||||||||||||
|
||||||||||||
| ||||||||||||
Ken, use algebra to solve this one. Let a = the amount of Brand A in pounds that Stephen needs and let b = the amount of Brand B in pounds. Then a + b = 3 pounds, right? The total mixture is 3 lbs. Stephen wants an 8% mixture, so 8% of the 3 lbs is plant food. Let's write that as (3 x 8%). That's the number of pounds of plant food in the final mixture. The amount of plant food in the Brand A is 10%, so the amount of pounds of plant food he has in "a" pounds of it is (a x 10%). Similarly, the amount from Brand B will be (b x 7%). But if you add these two together, you get the mixture, so: Turn this into decimals and simplify: 0.10a + 0.07b = 0.24 So we have two equations in two unknowns. This is what we call a "system of equations" or "simultaneous equations". An easy way to solve it is by using "substitution". Since a + b = 3, then we know that a = 3 - b. But that means that wherever we see "a", we can replace it with "b-3", because they equal each other. So 0.10a + 0.07b = 0.24 becomes 0.10(b-3) + 0.07b = 0.24. Now you can solve this for b and once you know that, you can solve a = 3 - b and you have your answer. Cheers, | ||||||||||||
|
||||||||||||
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. |