Then 3n + 1 = number of dimes

and n + 10 = number of pennies

Multiply by how much each coin is worth and add to get the $4.52:

0.01(n + 10) + 0.05n + 0.10(3n + 1) = 4.52

0.01n + 0.10 + 0.05n + 0.30n + 0.10 = 4.52

0.36n + 0.20 = 4.52

0.36n = 4.32

Divide both sides by 0.36 to get:

n = 12

So 3n + 1 = 3*12 + 1 = 36+1 = 37 = number of dimes

and n + 10 = 12+10 = 22 = number of pennies

12 nickles, 37 dimes and 22 pennies

To check, Karen has:

12 nickles worth $0.60,

37 dimes worth $3.70, and

22 pennies worth $0.22.

Altogether, this is 0.60 + 3.70 + 0.22 = 4.52

But I can’t remember how to set up the problem algebraically. Since it came out with a 2 in the ones place and there were ten more pennies than nickels, the 2 had to be in the ones place for nickels. And then it only made sense. $3.40 + .60 + .22 = $4.52

In other words,

d(.10)+n(.05)+p(.01)=$4.52 {Main Problem}

“If there is one more dime than three times the number of nickels . . .”

3n+1=d {Equation #1}

“. . . and 10 more pennies than nickels”

n+10=p {Equation #2}

Okay, now the real work starts. You’ll want to find out the value of n by using Eqution #2.

n+10=p . . . n=p-10

Now you’ll want to find out the value of d in terms of p by plugging the value of n into Equation #1

3n+1=d . . . 3(p-10) +1=d

Now we know the value of n in terms of p is (p-10) and the value of d in terms of p is 3(p-10)+1

Plug those values in terms of p into the main problem so that d(.10) + n(.05) + p(.01) =$4.52

becomes

[3(p-10)+1](.10) + (p-10)(.o5) + p(.01) =$4.52

Now solve for p

(.30p-$2.90) + (.05p-.5) + (.01p) =$4.52

.36p-$3.40=$4.52

p=22

There are 22 pennies.

Use E#2 to find the number of nickels and E#1 to find the number of dimes.

22 pennies

12 nickels

37 dimes

37 dimes

22 pennies