# HELP CHEM!!!!!!!!!!10 PTS?

Canadian 10¢ and 25¢ coins are made of pure nickel which has a density of 8.902 g/cm3. Suppose each dime weighs 2.08 g and each quarter weighs 5.20 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 19.23 mL mark. What is the combined monetary value (in Can$) of those coins?

The volume of displaced water equals the total volume of the coins:

19.23 mL - 10.00 mL = 9.23 mL

of water was displaced. Since, by definition 1 mL water = 1 cm³ water, the volume of the coins was 9.23 cm³. The mass of this volume of coins is:

(9.23 cm³) (8.902 g/cm³) = 82.17 g.

Letting d = number of dimes, and q = number of quarters, we can write:

2.08d + 5.20q = 82.17 (equation 1).

The volume of one dime is:

(2.08 g) / (8.902 g/cm³) = 0.234 cm³

The volume of one quarter is;

(5.20 g) / (8.902 g/cm³) = 0.584 cm³

So we can write:

0.234d + 0.584q = 9.23 cm³ (equation 2).

Equations 1 and 2 are linear with two unknowns. You can solve this system of equations anyway you wish; I get d = 4.19 ≈ 4 dimes and q = 14.13 ≈ 14 quarters. To check,

(2.08) (4) + (5.20) (14) = 81.12;

(0.234) (4) + (0.584) (14) = 9.11;

Not too far from the original observations. The total value of the coins is:

(0.10) (4) + (0.25) (14) = $3.90.