Helen and Ivan have the same number of coins. Helen has 64 20-cent coins, and Ivan has 104 20-cent coins. Helen has 40 more 50-cent coins than Ivan. Who has more money in coins, and by how much?

a) Who has more money in coins, and by how much? Step 1: Define the number of coins for Helen and Ivan. Let N_H50 be the number of 50-cent coins Helen has. Helen has N_H50 50-cent coins and 64 20-cent coins. Total number of coins for Helen: N_H = N_H50 + 64. Let N_I50 be the number of 50-cent coins Ivan has. Ivan has N_I50 50-cent coins and 104 20-cent coins. Total number of coins for Ivan: N_I = N_I50 + 104. Step 2: Use the condition that Helen and Ivan have the same number of coins. N_H = N_I N_H50 + 64 = N_I50 + 104 N_H50 = N_I50 + 104 - 64 N_H50 = N_I50 + 40 This means Helen has 40 more 50-cent coins than Ivan. Conversely, Ivan has 40 fewer 50-cent coins than Helen. Also, Helen has 104 - 64 = 40 fewer 20-cent coins than Ivan. Step 3: Calculate the difference in total money. The difference in their coin composition is that Helen has 40 more 50-cent coins and 40 fewer 20-cent coins compared to Ivan. This means Helen effectively exchanged 40 20-cent coins for 40 50-cent coins. The value difference per coin exchange is 50 cents - 20 cents = 30 cents. Since Helen has 40 such "exchanges" in her favor, the total difference in money is: Difference = 40 × (50 cents - 20 cents) Difference = 40 × 30 cents Difference = 1200 cents Converting to dollars: Difference = (1200)/(100) dollars = \12.00 Since Helen has the higher value coins in this exchange, Helen has more money. Helen has more money in coins, by \12.00. b) Given that each 50-cent coin is 2.7 g heavier than a 20-cent coin, what is the total mass of Ivan's coins in kilograms? Step 1: Define the masses of the coins. Let $m_50