Math question involving mass, density, volume, and displacement?
Canadian 10¢ and 25¢ coins are made of pure nickel which has a density of 8.902 g/cm3. Suppose each dime weighs 2.00 g and each quarter weighs 5.00 g. A small collection of dimes and quarters are submerged in 10.00 mL of water in a graduated cylinder causing the water level to rise to the 15.95 mL mark. What is the combined monetary value (in Can$) of those coins?
Note that there is a special relationship between the masses of the two types of coins and their monetary values.
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density = mass / volume
Therefore volume = mass / density
volume 10c piece = 2.00 g / 8.902 g/ml = 0.2247 ml
volume 15c piece = 5.00 g / 8.902 g/ml = 0.5617 ml
combine volume of coins = volume of water displaced
= 15.95 ml - 10.00 ml
= 5.95 ml
Now let number 10c pieces = a
let number 25 c pieces = b
a x 0.2247 ml + b x 0.5617 ml = 5.95 ml
At this point, to do this question properly with similtaneous equations you actually need more information. You need information which would allow you to write an expression for a using only b, that you can sub back into the equation above. But you don’t have this. you should have been given the total mass of the coins as well, this would have helped, particulay seeing the hint you were given pertains to the masses.
But you can finish by trial and error. You know that your coins must be in whole numbers.
So first see if all coins are 25c
5.95 / 0.5617 ml = 10.59. (not a whole number so coins are not all 25c pieces
Now subtract the volume of 1 10 c piece and see if the rest are 25c pieces
(5.95 - 0.2247) / 05617 = 10.19 (no, this is not right)
Now subtract 2 10c pieces
(5.95 - (2 x 0.2247)) / 0.5617 = 9.79 (nope, not this either)
we continue like this untill we get to 9 10c pieces)
(5.95 - (9 x 0.2247)) / 05617 = 6.99 (which is pretty much the whole number 7.. so this should be correct.
We have 7 25 c pieces and 9 10 c pieces
check the volumes
( 7 x 0.5617) + (9 x .2247) = 5.95 yay
I’ll let you add up the coin values.