Gene has less than 12 nickels,dimes,and quarters in his pocket.The probability of pulling out a nickel or a quarter is 3/4.The probabilty of pulling out a dime is 1/4.How many coins does Gene have in his pocket?How many of each does he have?
To solve this problem, we can use algebraic equations. Let's break it down step by step.
Let's assume the number of nickels Gene has is represented by "N," the number of dimes by "D," and the number of quarters by "Q."
1. We know that Gene has less than 12 nickels, dimes, and quarters combined, which we can write as: N + D + Q < 12.
2. We also know that the probability of pulling out a nickel or a quarter is 3/4, which means the combined total of nickels and quarters is 3/4 of the total coins Gene has. We can express this as: N + Q = (3/4)(N + D + Q).
3. Furthermore, we know that the probability of pulling out a dime is 1/4, so the number of dimes is 1/4 of the total coins Gene has, which can be written as: D = (1/4)(N + D + Q).
Now, we can solve these equations to find the values of N, D, and Q.
From equation 2, we can simplify it as follows:
N + Q = (3/4)(N + D + Q)
4N + 4Q = 3N + 3D + 3Q
N + Q = 3D
Substituting equation 3 into the above equation, we get:
3D = 3D
This means N + Q = 3D.
Now, let's go back to equation 1 and substitute N + Q with 3D:
3D + D < 12
4D < 12
D < 3
Since D represents the number of dimes, and it must be less than 3, we can conclude that Gene has either 0, 1, or 2 dimes.
Therefore, the possible combinations of coins Gene could have are as follows:
1. If Gene has 0 dimes (D = 0):
N + Q = 3D becomes N + Q = 0 (since D = 0)
N + Q < 12 still holds true.
Now, let's try to find the possible values of N and Q that satisfy this condition:
N + Q = 0
N < 12, Q < 12
Since N and Q both have to be less than 12, the only combination that works is N = 0 and Q = 0.
Therefore, if Gene has 0 dimes, he must have 0 nickels (N) and 0 quarters (Q).
2. If Gene has 1 dime (D = 1):
N + Q = 3D becomes N + Q = 3 (since D = 1)
N + Q < 12 still holds true.
Let's find the possible values of N and Q that satisfy this condition:
N + Q = 3
N < 12, Q < 12
There are multiple combinations that could work here, such as N = 2 and Q = 1 or N = 1 and Q = 2.
Therefore, if Gene has 1 dime, he could have either 2 nickels (N) and 1 quarter (Q) or 1 nickel (N) and 2 quarters (Q).
3. If Gene has 2 dimes (D = 2):
N + Q = 3D becomes N + Q = 6 (since D = 2)
N + Q < 12 still holds true.
Let's find the possible values of N and Q that satisfy this condition:
N + Q = 6
N < 12, Q < 12
Again, there are multiple combinations that could work here, such as N = 4 and Q = 2 or N = 3 and Q = 3.
Therefore, if Gene has 2 dimes, he could have either 4 nickels (N) and 2 quarters (Q) or 3 nickels (N) and 3 quarters (Q).
In conclusion, there are three possible scenarios:
1. Gene has 0 nickels, 0 dimes, and 0 quarters.
2. Gene has either 2 nickels and 1 quarter or 1 nickel and 2 quarters, along with 1 dime.
3. Gene has either 4 nickels and 2 quarters or 3 nickels and 3 quarters, along with 2 dimes.