View Full Version : Is this the correct way to solve...?
WhiteShadow
09-30-2010, 06:39 AM
Joe has a ring weighing 10 grams made of an alloy of 14% silver and the rest gold. He decides to melt down the rings and add enough silver to reduce the gold content to 72%. How many grams of silver should he add?
I'm trying:
(1.4 + x)/8.6=28/72
x=1.944444?
And how would I put the answer in 2 significant figures? Do the numbers before the decimal count? And when do they not count?
Thanks.
P.S I never took Calculus or Physics in High School and I'm now I'm taking them both in college. :{
Timer
09-30-2010, 07:27 PM
Joe has a ring weighing 10 grams made of an alloy of 14% silver and the rest gold. He decides to melt down the rings and add enough silver to reduce the gold content to 72%. How many grams of silver should he add?
Grams_of_Gold / (Initial Weight[g] + Grams_of_Silver_Needed_to_Reduce_Gold_Percentage[x g]) = Desired_Percentage_of_Gold
8.6 / (10 + x) = 0.72 (72%)
x = 1.94^_g
or
x = 35/18g
or
x = 1 17/18g
That's how I solved it. o.o
anonymity
09-30-2010, 07:45 PM
What is the purity of the silver and gold? If you are assuming 100% purity of substances:
find how many grams per %...
1.4g Ag + 8.6g Au = 10g ring
find how much silver to add to make 72% gold. Keep the total at ratio at 100
(1.4g + x)Ag / (8.6g)Au = 28 Ag / 72 Au
Solve for x
x = 1.94444
**************
same as what you did... both ways get the same answer...
and two sig figs would be 1.9 - here is a good teach http://www.sciencebyjones.com/sig_figs.htm
TomTuff
10-04-2010, 02:32 PM
small recap on sig figs:
sig figs when adding/subtracting:
there must be the same amount of places after the decimal points as the lowest number of places. so, 8.92 + 9.1 + 8.666 = 26.7
sig figs when multiplying/dividing:
the number of sig figs must be the same as the number with the lowest number of sig figs. so, 6.766 * 3.4456 / 1.2 = 19
Nava2
10-05-2010, 11:52 AM
With sigfigs and logs/exponentials, you keep the same number of decimal places as you had sig figs in the original question. (Odd eh?)
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