A piggy bank contains nickels ,dimes and quarters and they total%.90.if there is 3 times as many quarters than nickels and 14 more dimes than nickels.how many of each coin is in the bank?

Let's solve the problem step by step by introducing variables:

Let:
- $ n $ represent the number of nickels.
- $ d $ represent the number of dimes.
- $ q $ represent the number of quarters.

### Step 1: Set up relationships from the problem

1. The value of the coins adds up to 90 cents:  
   $ 0.05n + 0.10d + 0.25q = 0.90 $  
   (where 0.05, 0.10, and 0.25 are the values of a nickel, dime, and quarter respectively, in dollars).

2. There are 3 times as many quarters as nickels:  
   $ q = 3n $.

3. There are 14 more dimes than nickels:  
   $ d = n + 14 $.

### Step 2: Substitute the relationships into the value equation

Substitute $ q = 3n $ and $ d = n + 14 $ into the value equation $ 0.05n + 0.10d + 0.25q = 0.90 $:

\[
0.05n + 0.10(n + 14) + 0.25(3n) = 0.90
\]

### Step 3: Simplify the equation

First, distribute the constants:

\[
0.05n + 0.10n + 1.40 + 0.75n = 0.90
\]

Combine like terms:

\[
0.90n + 1.40 = 0.90
\]

### Step 4: Solve for $ n $

Subtract 1.40 from both sides:

\[
0.90n = 0.90 - 1.40
\]

\[
0.90n = -0.50
\]

Divide both sides by 0.90:

\[
n = \frac{-0.50}{0.90} = -\frac{5}{9}
\]

It appears there was an error in calculations or misinterpretation of constraints. Let's retry

A piggy bank contains nickels, dimes, and quarters, and they total $0.90. If there are 3 times as many quarters as nickels and 14 more dimes than nickels, how many of each coin is in the bank?

Learn how to solve a coin word problem involving nickels, dimes, and quarters totaling $0.90. This problem includes key algebraic concepts like substitution and linear equations. Suitable for students in Grades 6-8.

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