Hi guys, I am not sure how to solve this math problem:
Students in a science lab are working in groups to build both a small and a large electrical circuit. A large circuit uses 4 resistors and 2 capacitors, and a small circuit uses 3 resistors and 1 capacitor. There are 100 resistors and 70 capacitors available, and each group must have enough resistors and capacitors to make one large and one small circuit. What is the maximum number of groups that could work on this lab project?
It would probably help a lot if i had any idea where my graphing calculator was. So don’t take this as anything close to certain.
To pull the info out of paragraph form:
large = 4r+2c
small = 3r + 1c
each group: 7r + 3c
r<=100, c<=70
max large + small
Without a calculator, I’m going to look at it logically: it’s probably either a lot of larges, a lot of smalls, or a somewhat equal number.
And, since it’s SAT, I want to do it quickly; too much time spent on it is a huge problem.
Each group has to make at least one of each… didn’t see that before.
7r<=100, r= 14== that would work with the c’s
3c <=70== not enough r’s
I’m gong with 14.
Too late to edit. Ignore this part:
“Without a calculator, I’m going to look at it logically: it’s probably either a lot of larges, a lot of smalls, or a somewhat equal number.”
Okay that makes sense thanks!
You don’t really need to write anything algebraic…
To make their circuits, each group needs 7 resistors and 3 capacitors. To see how many teams we can equip, we divide and round down.
You have 100 resistors. 100/7 = 14, drop the remainder…you have enough resistors for 14 teams.
You have 70 capacitors. 70/3 = 23, drop the remainder…you have enough for 23 teams.
So after you equip 14 teams, you run out of resistors before you run out of capacitors.
(If this was chemistry class, we would say that the resistors were the “limiting reagent”.)
|1/2w|=|w|+1
please help with this
|1/2w|=|w|+1
OK, for starters, you’re NOT Prepping for tomorrow’s SAT right now, right??? All it will do is make you a basket case-- I can pretty much promise that this particular problem won’t appear.
So let’s assume you’re prepping for October:
Plan A: Forget the algebra, let’s do this logically.
Take half a number, and make it positive. That should equal your original number, made positive, and increased by one.
How is that possible? The absolute value of half a number will be less than or equal to the absolute value of the original number, forget about adding one to the original.
No solution.
Plan B: If you insist on doing it algebraically, set up 2 equations:
.5w = w +1 and.5(-w) = (-w) +1
You’ll get 2 answers. But both end up as extraneous roots.
Plan C: Graph both halves on your graphing calculator, then see that they don’t intersect.