KelseyWoods
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Michael has some coins in his pocket consisting of dimes, nickels, and pennies. He has two more nickels than dimes, and three times as many pennies as nickels. How many of each kind of coins does he have if the total value is 52 cents?

Respuesta :

mathstudent55
He has d dimes.
He has 2 more nickles than dimes, so he has d + 2 nickles.
He has 3 times as many pennies as nickles. He has 3(d + 2) pennies.

Now we figure out the value of each type of coin in cents.
d dimes are worth 10d cents.
d + 2 nickels are worth 5(d + 2) cents.
3(d + 2) pennies are worth 3(d + 2) cents.

Add all expressions of value and set equal to 52 cents.

10d + 5(d + 2) + 3(d + 2) = 52

10d + 5d + 10 + 3d + 6 = 52

18d + 16 = 52

18d = 36

d = 2

He has 2 dimes.
He has 2 more nickles than dimes, so he has 2 + 2 nickles, or 4 nickles.
He has three times as many pennies as nickels, or 3(4) = 12 pennies.

Answer: 2 dimes, 4 nickles, 12 pennies.

Using a system of equations, it is found that he has 2 dimes, 5 nickels and 12 pennies.

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In the system of equations, I am going to say that:

  • x is the number of dimes.
  • y is the number of nickels.
  • z is the number of pennies.

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Two more nickels than dimes, thus:

[tex]y = x + 2[/tex]

Three times as many pennies as nickels, thus:

[tex]z = 3y = 3(x + 2)[/tex]

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A dime is worth $0.1, a nickel is worth $0.05, and a penny is worth $0.01. Total of $0.52, thus:

[tex]0.1x + 0.05y + 0.01z = 0.52[/tex]

Solving for x:

[tex]0.1x + 0.05(x + 2) + 0.03(x + 2) = 0.52[/tex]

[tex]0.1x + 0.05x + 0.1 + 0.03x + 0.06 = 0.52[/tex]

[tex]0.18x = 0.36[/tex]

[tex]x = \frac{0.36}{0.18}[/tex]

[tex]x = 2[/tex]

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Solving for y and z:

[tex]y = x + 2 = 2 + 2 = 4[/tex]

[tex]z = 3(x + 2) = 3(2 + 2) = 12[/tex]

He has 2 dimes, 5 nickels and 12 pennies.

A similar problem is given at https://brainly.com/question/17096268

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