This is a question on my son’s math homework today:
The density of iron is 7.874 grams per cubic centimeter. Estimate the space that would be occupied by 50 grams of iron.
a) 16 cm3 b) 35 cm3 c) 350 cm3 d) 400 cm3
I realize that it has been a long time since I did any algebra, but this seemed like a very simple problem to me. Taking it step-by-step…
1) You start with this equation: 7.874g/1cm3 = 50g/Xcm3
2) Isolate the variable by multiplying both sides of the equation by Xcm3…
Xcm3(7.874g/1cm3) = Xcm3(50g/Xcm3)
X7.874g = 50g
3) …and then dividing both sides by 7.874.
X7.874g/7.874g = 50g/7.874g
X = ~6.35
His teacher says the correct answer is c) 350 cm3. Am I missing something very simple here, or is the teacher wrong and the correct answer not listed?
Triton? Mark? Will one of you guys (or someone else) double check my junior high algebra for me?
I agree with you, Jay. The problem is set up incorrectly. If I had to guess, I would say the question should read thusly:
The density of iron is 7.874 grams per cubic centimeter. Estimate the mass of iron that would occupy 50 cubic centimeters of space.
a) 16 g b) 35 g c) 350 g d) 400 g
The correct answer would then be ‘d’, 400 g.
The question as written doesn’t make much sense with the answer choices provided.
Unless, of course, your son is in a modern physics class and is dealing with relativistic mass.
Relativistic something…
Sorry, guys – you’re making a simple problem far too hard.
50 grams divided by:
7.8 grams / CC
yields 6.4 CC. (cm cubed)
So – you’re right, Jay – just not confident enough.
PS> That “350 cm3″ answer is really starting to bother me. This teacher should think about it a bit.
Multiply 350 CC x 7.9 gms/CC to check just how much such a cube would weigh.
How about 2.75 kg. That’s just shy of 7 lbs – quite a substantial block – and FAR more than a mere 50 gms – no matter how you mangle it.
On a more positive note (I hope!), I received his TAKS reading test results in the mail yesterday. Perfect score of 48 out of 48. I’ve heard bad things about that test and the way it’s used, but a perfect score is a perfect score!
Sorry, guys – you’re making a simple problem far too hard.
We got that part, Mark. The division is simple. The question is how or why did the teacher screw up. Usually in cases like this, all of the number values will still work if you just change the units or the wording or something. Hence my theory of what the question should have been.
You’re very kind, Triton (at least to the teacher!
My theory was simpler, too…
Oh…and congratulations, Jay!