Question

In a collection of coins there are twice as many nickles as dimes and 7 less quarters than dimes find the number of each coin if the collection is worth 16.25

Answer 1

Using algebra, the collection contains 40 dimes, with twice as many nickels making it 80 nickels, and seven less quarters than dimes resulting in 33 quarters.

Step-by-step explanation:

To solve the question of determining the number of coins in a collection worth $16.25, with twice as many nickels as dimes and seven less quarters than dimes, we will use algebraic methods. Let's define d as the number of dimes. Therefore, we can express the number of nickels as 2d and the number of quarters as d - 7. Using the values of each coin (nickel: $0.05, dime: $0.10, and quarter: $0.25), we can create the following equation:

0.10d + (2d × 0.05) + (d - 7) × 0.25 = 16.25

Simplifying the equation gives:

0.10d + 0.10d + 0.25d - 1.75 = 16.25

We combine like terms:

0.45d - 1.75 = 16.25

Adding 1.75 to both sides:

0.45d = 18.00

Finally, dividing by 0.45 gives:

d = 40

So, we have 40 dimes. Now, calculate twice as many nickels:

2 × 40 = 80 nickels

And the number of quarters:

40 - 7 = 33 quarters

To summarize, the collection contains 40 dimes, 80 nickels, and 33 quarters.