A bag contains twice as many pennies as nickels and four more dimes than quarters. Find all possibilities for the number of each coin if their total value is $2.01. I have pennies = 4n nickels = n dimes = 4 + q quarters = q but I don't know where to go from there?
"twice as many pennies as nickels"
number of nickels ----- x
number of pennies --- 2x
"four more dimes than quarters"
number of quarters --- y
number of dimes ---- y+4
5x + 2x + 25y + 10(y+4) = 201
7x + 35y = 161
divide by 7
x + 5y = 23
y = (23-x)/5
since y must be a whole number, 23-x must be a multiple of 5
x y, there are only 4 cases
3 4
8 3
13 10
18 1
check:
case 1: x=3, y = 4
nickels = 3, pennies= 6, quarters = 4, dimes = 8
value = 15+ 6 + 100 + 80 = 201 , yup!!
case 2: x = 8, y = 3
nickels = 8 , pennies = 16, quarters = 3, dimes = 7
value = 40 + 16 + 75 + 70 = 201 , yup!!
I will leave it up to you to verify the other two cases
and then to state a conclusion
the answers are so bad bro
you forgot to add up all the values.
p+5n+10d+25q = 201
Now, as you say, you also have
p = 2n (twice as many, not 4x as many)
d = 4+q
Now, there may not be a unique solution, since you have only three equations for four unknowns. But let's start substituting and see where we get:
p+5n+10d+25q = 201
2n+5n+10d+25q = 201
7n + 10(4+q)+25q = 201
7n + 40 + 10q + 25q = 201
7n+35q = 161
n+5q = 23
Clearly, q <= 4, so check the possibilities:
q=4, n=3
q=3, n=8
q=2, n=13
q=1, n=18
q=0, n=23
Check the combinations:
(p,n,d,q) = (6,3,8,4): 6+3*5+8*10+4*25 = 6+15+80+100 = 201
and you can check the others...
To solve this problem, you have correctly set up the variables. Let's continue solving it step by step:
1. Assign variables: Let's call the number of nickels "n," the number of pennies "2n," the number of dimes "4+q," and the number of quarters "q."
2. Determine the value of each coin: We know that the value of a penny is 1 cent, the value of a nickel is 5 cents, the value of a dime is 10 cents, and the value of a quarter is 25 cents.
3. Set up the equation: The total value of the coins is $2.01, which is equal to 201 cents. We can write the equation as follows:
(2n * 1) + (n * 5) + (4+q) * 10 + q * 25 = 201
4. Simplify the equation: Distribute and combine like terms:
2n + 5n + 40 + 10q + 25q = 201
7n + 35q + 40 = 201
5. Rearrange the equation: Move constant terms to the other side of the equation:
7n + 35q = 201 - 40
7n + 35q = 161
6. Reduce the equation: Divide both sides of the equation by 7 to simplify it:
n + 5q = 23
7. Find possible values for n and q: To find all the possibilities for the number of each coin, we need to find all the integer solutions for the equation by trial and error:
n + 5q = 23
If we let n = 23 and q = 0, the equation is satisfied.
If we let n = 18 and q = 1, the equation is satisfied.
And so on.
By trying different values for n and q, you can find all the possible combinations that satisfy the equation.
For example:
- If n = 13 and q = 2, the equation is satisfied.
- If n = 8 and q = 3, the equation is satisfied.
- If n = 3 and q = 4, the equation is satisfied.
- If n = -2 and q = 5, the equation is satisfied.
Note that in this case, when n becomes negative, it means there are too many quarters and the number of pennies becomes negative. Therefore, we should only consider the positive solutions.
So, the possible combinations for the number of each coin are:
n = 23, q = 0 (23 nickels, 46 pennies, 4 dimes, 0 quarters)
n = 18, q = 1 (18 nickels, 36 pennies, 5 dimes, 1 quarter)
n = 13, q = 2 (13 nickels, 26 pennies, 6 dimes, 2 quarters)
n = 8, q = 3 (8 nickels, 16 pennies, 7 dimes, 3 quarters)
n = 3, q = 4 (3 nickels, 6 pennies, 8 dimes, 4 quarters)
Remember to check your answers by substituting the values back into the equation to verify if they satisfy it.