Benjamin has a total of two hundred fifty-eight coins. He has three times as many nickels as pennies and one-half as many pennies as quarters. How much money does he have?
p+n+q = 258
n = 3p
q = 2p
so
p + 3p + 2p = 258
he has 43 pennies
now finish it off
let x be the no. of pennies
let y be the no. of nickels
let x be the no. of quarters
x + y + z = 258 —> Eq. 1
y = 3x —> Eq. 2
x = 1/2z —> Eq 3.
Solution: substitute y = 3x to Eq. 1
x + y + z = 258
x + 3x + z = 258
4x + z = 258
Substitute x = 1/2z to
4(1/2z) + z = 248
2z + z = 258
3z = 258 ; z = 86 quarters
x = 1/2 (86)
x = 43 pennies
y = 3 (43)
y = 129 nickels
Total Money = 0.01x + 0.05y + 0.25z
= 0.01 (43) + 0.05 (129) + 0.25 (86)
Total Money = $28.38
To find out how much money Benjamin has, we need to determine the number of each type of coin he has.
Let's call the number of pennies P, nickels N, and quarters Q.
From the problem, we can gather the following information:
1. Benjamin has a total of 258 coins: P + N + Q = 258.
2. Benjamin has three times as many nickels as pennies: N = 3P.
3. Benjamin has one-half as many pennies as quarters: P = Q/2.
Now, we can use these equations to solve for the number of each type of coin.
First, replace the value of N in terms of P from equation 2 into equation 1:
P + 3P + Q = 258.
We can simplify this to: 4P + Q = 258.
Next, substitute the value of P from equation 3 in terms of Q:
Q/2 + 3(Q/2) + Q = 258.
We can simplify this to: 4Q/2 + 3Q/2 + Q = 258.
Simplifying further: 2Q + 3Q + 2Q = 516.
Now, combine like terms: 7Q = 516.
Divide both sides by 7: Q = 516/7.
So, the number of quarters, Q, is approximately 73.714.
Since we can't have fractions of a coin, we know that Q must be a whole number. Given that P = Q/2, we can determine that P = 73/2 = 36.5, which is not a whole number either.
This means there must be an error in the problem because we can't have half coins. Please double-check the problem details, especially in relation to the number of pennies and quarters, as it seems there might be a mistake.