MATH SOLVE

3 months ago

Q:
# 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?

Accepted Solution

A:

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.

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.