Expert: Josh - 2/22/2011

Question
A student had 100 coins totaling $5.00. He had no nickels. What coins did he have if 60% of the coins were one of the kind?

Answer
Hi Chris,

I think this problem has no solution. There is something wrong with the numbers. As a sanity check, consider dividing five dollars (or 500 cents) by 100. This shows that each coin "on average" has a value of 5 cents. A penny, nickel, dime and quarter are worth 1, 5, 10 and 25 cents, respectively. Taking out the nickel (as instructed by the question), we are left with 1 cent, 10 cent and 25 cent coins. Recall at this point that to make up 5 dollars with 100 coins, each coin will have an average value of 5 cents. With the three types of coins carrying a weight of 1, 10 and 25, there is no way that we can bring the average down to 5 if we have a majority (60 out of 100 coins) having a value of 10 cent or 25 cents. This leaves only one viable option, i.e., to have 60 pennies, with 40 coins consisting of dimes and quarters. Subtracting 60 from 500, we have 440 cents remaining. It is not possible to have this coming from any combination of dimes and quarters totaling 40 coins. If we remove the constraint on the number of coins, then 39 dimes and 2 quarters would produce a viable solution [60 x 1 + 39 x 10 + 2 x 25 = 500].

-----The following is an attempt at solving the problem-----

Excluding nickel, the common types of coins we have are penny, dime and quarter. These are worth 1, 10 and 25 cents, respectively.

Suppose we have X pennies, Y dimes and Z quarters. The total is given by multiplying the value of each type of coin by its number, and adding them up.

Expressing everything in cents (with 100 cents = 1 dollar), we obtain the first equation:
500 = X + 10Y + 25Z  ...[1]

We also know that the total number of coins must add up to 100. So, we get the second equation
100 = X + Y + Z ...[2]

Since 60% of the coins were one of a kind, we know that for one variety, its number (either X, Y or Z) equals 60. We can rule out Z=60, because in equation [1], when Z=60, 25Z = 25*60 = 1250. This alone already exceeds 500 (the left hand side of the equation). As we know, we X and Y represent the number of coins, both cannot be negative. So, there is no way that we can make [1] hold true if Z=60. This leaves us with two remaining possibilities. Either Y=60 OR Z=60.

Let's consider the first scenario. If Y=60, [1] becomes 500 = X + 10*60 + 25Z. After subtracting 600 from both sides, we require -100 = X + 25Z ....[1a] This is an impossible situation, because both X and Z must be non-negative as they represent the number of pennies and dimes we have. If we ignore this and continue, we will reach a contradiction.
From [2], when Y=60, we must have 40 = X+Z. Rearranging this to make X the subject, X = 40-Z. Substituting this into [1a], we get
-100 = (40-Z) + 25Z
-140 = 24Z
This forces Z to be a negative, non-integer solution which makes no sense.

So, the only alternative is for X=60. Let us proceed and see what happens. When X=60, [1] becomes 500 = 60 + 10Y + 25Z. This simplifies to 440 = 10Y + 25Z...let's call this [1b]. From [2], when X=60, Y=40-Z. Substituting this into [1b], we get 440 = 10*(40-Z) + 25Z, 40 = 15Z. This does not produce an integer solution. Something is wrong...It works out alright if part of the question is changed -- from "100 coins" to "101 coins".