Twenty-five blue and twenty-five yellow marbles are placed in a jar. Thirty-four marbles are removed from the jar and put in a second jar. What is the positive difference between the number of blue marbles in the second jar and the number of yellow marbles remaining in the first jar?

a) Let b represent the number of blue marbles placed in the second jar. Then 34-b is the number of yellow marbles in the second jar, and 25 - (34-b) is the number of yellow marbles remaining in the first jar. That value simplifies to b-9. The positive difference between b and (b-9) is 9.

b) Let n, d, p represent the numbers of nickels, dimes, and pennies, respectively.

... (5n + 10d + p) / (n + d + p) = 7 ... the average value of the collection is 7 cents per coin

... (5 (n-1) + 10d + (p+5)) / ((n-1) + d + (p+5)) = 6 ... replacing one nickel with pennies changes the average to 6 cents per coin

Multiplying by the denominator and subtracting the right side of each resulting equation gives

... 10d + 5n + p = 7d + 7n + 7p

... 10d + 5n + p = 6d + 6n + 6p + 24

Subtracting the second equation from the first gives

... 0 = d + n + p - 24 ... tells us the total number of coins is 24. This tells us the total value is $1.68 = 24*$0.07.

This sum will consist of at least 3 pennies. If that is all the pennies, then the remaining 21 nickels and dimes must add to $1.65. If all were nickels, the value of the 21 coins would be $1.05. We have $0.60 more than that. For each nickel traded for a dime, we gain $.05, so 12 of the 21 nickels must be traded for dimes to raise the total coin value to $1.65.

Alternatively, you can write two equations in the two unknowns:

... d + n = 21

... 10d + 5n = 165

Subtracting 5 times the first equation from the second gives

... 5d = 60 ... which gives the same answer as reasoned above.

There are 3 pennies, 9 nickels, 12 dimes. (If there are 8 pennies to start, then the number of nickels is zero, so it is impossible to trade one for 5 pennies. The solution shown is the only one.)