A child's bank account contained twice as many nickels as pennies and two-thirds as many dimesas nickels, the total value being at least $3.65.

Question

Find the smallest possible number of coins in the bank.   (Ans: 65 coins ) Solve. 

Show each step. 

Answer 1

2p=n and (2/3)n=d would be your starting two equations.

Notice that "twice as many nickels as pennies" is NOT 2n=p, as I've often seen written.  Pennies was the basis for the comparison.

The third "equation" here is where you link the coin's value to the number of coins.  I prefer to use the integer form of money (cents) rather than the dollar form so that if I get any solutions that have decimal values it's easier for me to see and figure out what may have gone wrong.

1p + 5n + 10d ≥ 365. I didn't need to put the "1" with the p but did it to be clear that the pennies are each worth 1 cent; the nickels, 5 cents ...  just don't forget to turn $3.65 into cents too!

Now substitute the correct math statements we wrote above into the inequality.  Any of the three coins could be chosen by you to keep.  I think p looks pretty good to go with, so...

p + 5(2p) + 10((2/3)(2p)) ≥ 365 Simplify using the rules for multiplication, then add using like denominators.

p + 10p + (40/3)p ≥ 365

(33/3)p + (40/3)p ≥ 365 and so on.

From this you should get p ≥ 15.

You might not get something this nice in a future problem and would have to use a value of p that would be the next greater integer when finding the possible solution to use. The rest falls out nicely from going with p = 15 for this one. Now the fun begins since you’ll just need to plug this lowest integer value in for the pennies to get the nickels, and then see if it will give you an integer number of dimes. In no time you should be able see how 65 total coins are involved and are the LEAST number of coins.