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Diane has quarters, dimes, and

pennies in her pocket. The value of her
money is $1.55. There are 4 more
dimes than quarters and the number of
pennies equals the sum of the number
of quarters and dimes.
How many pennies does Diane have?

User Eksapsy
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1 Answer

25 votes
25 votes

Answer: The solution here is found by establishing a system of linear equations.

A system of equations with x unknowns can be evaluated for an exact solution if the system has x equations.

Within the question, we are asked to discover the values of two different unknowns - dimes and nickels.

Therefore, we'll need two equations to obtain an exact solution, both of these will come directly from the problem itself.

Let d represent the number of dimes Diane has and let n represent the number of nickels.

Equation 1

We know from the given information that the total number of coins (both dimes and nickels) is equal to 19.

Subsequently, we can generate the equation d + n = 19.

Equation 2

We know that the individual value of a dime is $0.10 or 0.1 dollars, therefore the total value of the dimes is equal to the individual value of a dime multiplied by the total number of dimes.

The resultant expression for the total value of the dimes is (0.1)*d

We know that the individual value of a nickel is $0.05 or 0.05 dollars, therefore the total value of the nickels is equal to the individual value of a nickel multiplied by the total number of nickels.

The resultant expression for the total value of the nickels is (0.05)*n

We know that Diane only has dimes and nickels, and the total value of all her coins must be equal to the total value of the dimes plus the total value of the nickels. This value is given within the problem to be $1.55 or 1.55 dollars.

Subsequently, our second equation within the system is (0.1)*d + (0.05)*n = 1.55.

Now, we can solve the system of equations using whichever means is most convenient. For this example, I will solve the system by substitution

First, solve for one of the variables —> Equation 1 —> d + n = 19 —> d = 19 - n

Second, substitute the value of d into Equation 2 —> (0.1)*(19 - n) + (0.05)*n = 1.55

Recall that 0.1 = 1/10 and 0.05 = 1/20

Equation 2 —> (1/10)*19 - (1/10)*n + (1/20)*n = (31/20) —> (19/10) - (n/10) + (n/20) = (31/20)

Find the least common denominator of all fractions in the equation (20 in this case), convert the fractions to have like denominators, and solve.

(19/10)*(2/2) - (n/10)*(2/2) + (n/20) = (31/20) —> (38/20) - (2n/20) + (n/20) = (31/20) —> (38/20) - (n/20) = (31/20) —> (n/20) =(38/20) - (31/20) =(7/20) —>n = 7

Substitute solution into Equation 1 to solve for d

d + 7 = 19 —> d = 12

q+4+q + 10q+40 + 25q = 155

37q + 44 = 155

37q = 111

q = 3 (# of quarters)

d = q+4 = 7 (# of dimes)

p = d+q = 7+3 = 10 (# of pennies)

Explanation:

User Dynom
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