Question

4) A safe contains 120 coins, the value of which is $10. If the coins consist of nickels and dimes, how many of each kind are in it?

5) A sum of $14 is made of 92 coins, consisting of dimes and quarters.How many are there of each kind of coin?
6) A boy's coins, consisting of nickels and dimes, amount to 42.15. If the number of dimes exceeds 3 times the number of nickels by 4, find the number of each kind of coin?

Answer 1

We solved three separate systems of linear equations representing coin problems. By setting up equations using the known values of each coin and their total amount, we found the individual quantities of nickels and dimes for each scenario.

Step-by-step explanation:

Step-by-Step Solutions to Coin Problems

Let's solve each of the three questions separately using algebra.

Question 4 Solution

Let x be the number of nickels and y be the number of dimes. The total value in cents is 5x + 10y = 1000 (since $10 equals 1000 cents) and the total number of coins is x + y = 120. Solving these two equations will give us:

• x = 80 (nickels)
• y = 40 (dimes)

Question 5 Solution

Let d represent dimes and q represent quarters. We have the system: 10d + 25q = 1400 (for the total value in cents of $14), and d + q = 92. Solving these equations gives us:

• d = 32 (dimes)
• q = 60 (quarters)

Question 6 Solution

Let n be the number of nickels and d be the number of dimes. We have that 5n + 10d = 4215 (for the total amount in cents of $42.15) and d = 3n + 4. Solving this system of equations provides:

• n = 63 (nickels)
• d = 193 (dimes)

Answer 2

120=x+y and 10=.05x+.10y
120-x=y then subsitute 10=.05x+.10(120-x) distruibte and solve you can do it feel free to go off of my math.
x(nickels) y(dimes)