Jo has 37 coins (all nickels, dimes and quarters) worth $5.50. She has 4 more quarters than nickels. How many of each type of coins does she have?
d=#dimes
n=#nickels
q=#quarters.
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d+n+q=37
n+4=q
0.1d + 0.05n + 0.25q = 5.50
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three equations.three unknowns.
Solve for d,n,q.
Post your work if you get stuck and need further assistance.
21
8 dime to 3 quarters
To solve this system of equations, let's use the substitution method.
Let's start with the equation n + 4 = q, which tells us that she has 4 more quarters than nickels. We can rewrite this equation as q = n + 4.
Now, let's substitute this expression for q in the other two equations:
Equation 1: d + n + q = 37
Substituting q = n + 4, we get:
d + n + (n + 4) = 37
Combining like terms, we have:
d + 2n + 4 = 37
d + 2n = 33 --> Equation 3
Equation 2: 0.1d + 0.05n + 0.25q = 5.50
Substituting q = n + 4, we get:
0.1d + 0.05n + 0.25(n + 4) = 5.50
Simplifying, we have:
0.1d + 0.05n + 0.25n + 1 = 5.50
Combining like terms, we get:
0.1d + 0.3n + 1 = 5.50
0.1d + 0.3n = 4.50 --> Equation 4
Now, we have a system of two equations:
d + 2n = 33 --> Equation 3
0.1d + 0.3n = 4.50 --> Equation 4
We can solve this system using substitution or elimination method. Let's use the substitution method.
From Equation 3, we can express d in terms of n:
d = 33 - 2n.
Now, substitute this expression for d in Equation 4:
0.1(33 - 2n) + 0.3n = 4.50
Simplifying, we get:
3.3 - 0.2n + 0.3n = 4.50
Combining like terms:
0.1n = 1.2
Dividing both sides by 0.1, we get:
n = 12
Now, substitute this value for n in Equation 3 to find d:
d + 2(12) = 33
d + 24 = 33
Subtracting 24 from both sides, we have:
d = 9
Now that we have the values of d and n, we can find q using the equation q = n + 4:
q = 12 + 4
q = 16
Therefore, Jo has 9 dimes, 12 nickels, and 16 quarters.
To solve this problem, we can use the system of equations you have already written:
Equation 1: d + n + q = 37
Equation 2: n + 4 = q
Equation 3: 0.1d + 0.05n + 0.25q = 5.50
Let's solve it step by step:
Step 1: Rearrange Equation 2 to express q in terms of n:
n + 4 = q ⇒ q = n + 4
Step 2: Substitute q in Equation 1 and Equation 3:
d + n + (n + 4) = 37 ⇒ d + 2n + 4 = 37 (substituting q in Equation 1)
0.1d + 0.05n + 0.25(n + 4) = 5.50 ⇒ 0.1d + 0.05n + 0.25n + 1 = 5.50 (substituting q in Equation 3)
Simplify the equations:
d + 2n = 33 (Equation 4)
0.1d + 0.3n = 4.50 (Equation 5)
Step 3: Multiply Equation 4 by 0.1 to eliminate the decimal coefficient:
0.1(d + 2n) = 0.1(33)
0.1d + 0.2n = 3.3 (Equation 6)
Step 4: Subtract Equation 5 from Equation 6 to eliminate d:
(0.1d + 0.2n) - (0.1d + 0.3n) = 3.3 - 4.50
0.1d - 0.1d + 0.2n - 0.3n = -1.2
-0.1n = -1.2
Simplify the equation:
n = -1.2 / -0.1
n = 12
Step 5: Substitute n = 12 into Equation 4 to find d:
d + 2(12) = 33
d + 24 = 33
d = 33 - 24
d = 9
Step 6: Substitute n = 12 into Equation 2 to find q:
q = 12 + 4
q = 16
So, Jo has 9 dimes, 12 nickels, and 16 quarters.