Joseph always throws change into a can and takes it out every two weeks. This time it is all nickels and dimes. There are six times as many dimes as nickels, and the value of the dimes is $4.95 more than the value of the nickels. How many nickels and dimes does Joseph have. (please help me and please show the solution and equation of the problem. Thank you)
Solve mentally:
Start with 6 dimes and 1 nickel.
The difference is $0.55.
To get $4.95 as the difference, we have to multiply by the factor $4.95/0.55=9
So there are 54 dimes and 9 nickels.
Solve by algebra:
N=number of nickels
D=6N=number of dimes
0.10D-0.05N=4.95
0.10(6N)-0.05N=4.95
0.55N=4.95
N=4.95/0.55=9 (as before)
D=6*9=54
John sister has $4.95 in dimes and quarters in her piggy bank. There are 3 fewer dimes than quarters. How many quarters and dimes are in the piggy bank?
To solve this problem, we can use a system of equations. Let's define the following variables:
Let's say the number of nickels is "x".
The number of dimes is then "6x" (since there are six times as many dimes as nickels).
The value of each nickel is $0.05, and the value of each dime is $0.10. The total value of the nickels is 0.05x, and the total value of the dimes is 0.10 * 6x = 0.60x.
According to the problem statement, the value of the dimes is $4.95 more than the value of the nickels. So we can form the equation:
0.60x = 0.05x + 4.95
To solve this equation, we can simplify it:
0.60x - 0.05x = 4.95
0.55x = 4.95
Now we can solve for x by dividing both sides of the equation by 0.55:
x = 4.95 / 0.55
x ≈ 9
So, there are approximately 9 nickels. And since there are six times as many dimes, there would be:
6x = 6 * 9 = 54
Therefore, Joseph has approximately 9 nickels and 54 dimes.