David J.

asked • 02/03/20

found $8.25 in pennies, nickels, dimes, and quarters. When deposited money, she noticed that she had twice as many nickels as pennies, 1 fewer dime than nickels, and I more quarter than nickels.

How many quarters did Lucky find that week?

Sid M.

tutor
I'm a bit rusty, but if I did the math correctly, I think that there might be a problem with the problem.... Let's count the numbers of each coin (pennies, nickels, dimes, and quarters) in terms of pennies. We do this because if we work in terms of nickels, the number of pennies becomes a fraction (# nickels / 2), and that might prove to be troublesome. # pennies = p # nickels (n) = 2 * # pennies, or 2 * p; so n = 2p # dimes (d) = # nickels - 1, or n - 1, or 2p - 1; so d = 2p - 1 # quarters (q) = # nickels + 1, or n + 1, or 2p + 1; so q = 2p + 1 So, with the above, we get the following: $8.25 = $0.01(p) + $0.05(2p) + $0.10(2p - 1) + $0.25(2p + 1) Rather than count count value in dollars, let's do it in pennies. I'm showing n pennies as np. (The cent sign, ¢, is a bit hard to see without confusion.) $8.25 = 825p = 1p(p) + 5p(2p) + 10p(2p - 1) + 25p(2p + 1) 825p = 1pp + 10pp + (20pp + 10p) + (50pp + 25p) 825p = 81pp + 35p 825p - 35p = 81pp 790p = 81pp 790p / 81p = p Unfortunately, 790 / 81 (9.753086419753086) is not a whole number, and we can't have fractions of a penny. If we set p = 10, then: value = 1p(p) + 5p(2p) + 10p(2p - 1) + 25p(2p + 1) = 1p(10) + 5p(2*10) + 10p(2*10 - 1) + 25p(2*10 + 1) = 10p + 100p + 200p - 10p + 500p + 25p = 925p = $9.25 A dollar too much! If we set p = 9, then: value = 1p(p) + 5p(2p) + 10p(2p - 1) + 25p(2p + 1) = 1p(9) + 5p(2*9) + 10p(2*9 - 1) + 25p(2*9 + 1) = 9p + 90p + 180p - 10p + 450p + 25p = 744p = $7.44 Too low, by $0.81! So, I suspect that there is something wrong with our problem. David J, can you recheck the problem and correct it if necessary?
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02/03/20

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