Coin Problem 2 Solution

1st of all on all these problems convert the total amount of money from dollar to cents-- $100 is equal to 10000 cents (just add 2 zeros), $7 is equal to 700 cents, and you already have 37 cents, so altogether you have 10000 + 600 + 87 = 10737 cents.

From the problem you can see that everything is relative to pennies so let that be 1(x) or just x

1 more than twice as many nickels as pennies is (2x + 1) and and since its nickels 5(2x +1) or 10x + 5

twice as many dimes as nickels 2(2x +1) = 4x + 2 and since its dimes 10(4x + 2) or 40x + 20

1 more than twice as many 50 cents pieces than dimes (this is tricky wording since you may expect the denomination to go to a quarter) which is 2(4x + 2) + 1 or 8x + 5 and since its 50 cent pieces 50(8x + 5) or 400x + 250

and twice as many quarters as 50 cent pieces or 2(8x + 5) or 16x + 10 and since its quarters 25(16x + 10) or 400x + 250

so algebraically we have
x + 10x + 5 + 40x + 20 + 400x + 250 + 400x + 250 = 10737 =
851x + 525 = 10737 =
851x = 10737 - 525 =
851x = 10212 =
x = 10212/851

x = 12

So you have 12 pennies, 2(12) + 1 or 25 nickels, 4(12) + 2 or 50 dimes, 8(12) + 5 or 101 Fifty cent pieces, and 16(12) + 10 or 202 quarters....

Yeah, a little tedious arithmetic wise...