A child has a bottle full of pennies, nickels, dimes, and quarters. There are four as many quarters as pennies, two times as many as nickels as pennies, and seven times as many dimes as nickels. How many more dimes does the child have than nickels?
A.
7 times as many
B.
5 times as many
C.
2 times as many
D.
10 times as many
The correct answer and Explanation is :
Let’s break down the problem using algebra.
Step 1: Define the variables
Let:
- ( p ) = the number of pennies
- ( n ) = the number of nickels
- ( d ) = the number of dimes
- ( q ) = the number of quarters
Step 2: Translate the information into equations
From the problem, we are given three key relationships:
- There are four times as many quarters as pennies:
[
q = 4p
] - There are two times as many nickels as pennies:
[
n = 2p
] - There are seven times as many dimes as nickels:
[
d = 7n
]
Step 3: Solve for the number of dimes in terms of pennies
Since ( n = 2p ), we can substitute this into the equation for the number of dimes:
[
d = 7n = 7(2p) = 14p
]
So the number of dimes is ( d = 14p ).
Step 4: Compare the number of dimes and nickels
We already know that the number of nickels is ( n = 2p ) and the number of dimes is ( d = 14p ).
Now, let’s find how many more dimes there are than nickels:
[
\frac{d}{n} = \frac{14p}{2p} = 7
]
Thus, the child has 7 times as many dimes as nickels.
Conclusion
The correct answer is A. 7 times as many. This means the child has seven times more dimes than nickels.
Explanation Summary
We defined the number of each coin using the relationships provided. By expressing all the coins in terms of pennies, we solved for the number of dimes and nickels and found that the child has 7 times more dimes than nickels. This approach helps us understand how the quantities relate to each other algebraically and leads us to the correct answer.