Rodney has fifty coins, including at least one quarter, that are worth $1.00. If he loses one coin, what is the probability that it was a dime?

Question

calculusxy · Answer

@phi

phi · Answer

let me know when you figure it out.

calculusxy · Answer

I have no idea on how to figure this out.

phi · Answer

you have to examine the possibilities.
they say you have at least one quarter.
say you had 2 quarters  and 48 other coins (assumed to be pennies, nickels or dimes)
and they add up to 100 cents

if we subtract the 2 quarters off the 100 cents, the 48 coins have to add up to 50 cents

Let p, n and d be the number of pennies nickels and dimes
for this case, you have
p+n+d = 48  (48 coins)
from which we get  p= 48-n-d

also, the value of the coins is
p +5n +10d = 50 cents

replace p with 48-n-d and simplify
48 +4n + 9d = 50
4n + 9n = 2

if we have even 1 nickel or 1 dime, we will get a number bigger than 2
all of this means we cannot have 2 or more quarters.

anonymous · Answer

the only way to make $1 with 50 coins is, 1Q, 2D, 2N, 45P. So 2/50

phi · Answer

because there is at least 1 quarter, we have exactly 1 quarter, and 49 other coins

we have
p+n+d= 49  which means p= 49 - n - d

p+ 5n+10d= 75 cents (the quarter is worth the remainder 25 cents)


replace p with 49-n - d and simplify
4n + 9 d = 26

phi · Answer

now you have
4n + 9 d = 26

if you have 3 dimes, you get 27... which is too big.
so you know you have 0, 1 or 2 dimes.

test the 3 choices and see if you can get an *integer* number of nickels