A cashier has 26 bills consisting of three times as many ones as fives, one less ten than fives, and the rest twenties. If the value is $120, find how many of each she has.
x fives
3x ones
(x-1) tens
sum so far = (4x-1)
[26 - (4x-1)] twenties = (27-4x)
3x + 5x + 10(x-1) + 20(27-x) = 120
proceed :)
Damon, I have the answers but don't know how to get the equation. Ones - 15; Fives - 5; Tens - 4; Twenties - 2. I don't get that answer with your equation. Do you mind looking at it again?
3x+5x+10(x-1)+20(27-x)=120
Thank you for your help. It's much appreciated.
3f is the number of ones
f-1 is the number of ten
rest is 20s
total is 120
break down the problem and write what you know.
120=3f+f-1, the rest is 20s.
120=4f rest is 20s
4 times five is 20
20+ the rest is twenties=120
-20 -20
the rest are 20s=100
100/20=5 there are five 20s. You do the rest.
Well, this cashier certainly has an interesting mix of bills! Let's break it down:
Let's suppose that the number of fives is "x". Since there are three times as many ones as fives, we can say that there are 3x ones.
According to the information given, there is also one less ten than fives. So, we have (x-1) tens.
The rest of the bills are twenties. Since we know there are 26 bills in total, we can subtract the number of fives, ones, and tens from 26 to find the number of twenties: 26 - (x + 3x + x - 1) = 26 -5x + 1 = 27 - 5x
Now, let's calculate the total value:
5(x) + 1(3x) + 10(x-1) + 20(27-5x) = 120
5x + 3x + 10x - 10 + 540 - 100x = 120
-92x + 530 = 120
-92x = -410
x = 4.46
Uh oh! It seems like our calculations didn't give us a whole number for the number of fives. Since you can't have a fraction of a bill, I'm afraid I can't give you an exact answer here. Perhaps there was an error in the information provided?
But don't worry, a cashier with fractional bills could surely join the circus as the mathematical magician of funny money!
To solve this problem, let's assign variables to each type of bill.
Let:
x = the number of ones
y = the number of fives
z = the number of tens
We are given three pieces of information:
1) "A cashier has 26 bills"
This information implies that the total number of bills should be equal to 26:
x + y + z = 26
2) "Three times as many ones as fives"
This means the number of ones is three times the number of fives:
x = 3y
3) "One less ten than fives"
This means the number of tens is one less than the number of fives:
z = y - 1
Now, let's express the value of the bills in terms of their denominations:
The value of ones = 1 * x
The value of fives = 5 * y
The value of tens = 10 * z
The value of twenties = 20 * (26 - x - y - z), since the remaining bills are in twenties.
According to the problem, the total value of the bills is $120:
1 * x + 5 * y + 10 * z + 20 * (26 - x - y - z) = 120
Now, we have a system of equations:
x + y + z = 26
x = 3y
z = y - 1
1 * x + 5 * y + 10 * z + 20 * (26 - x - y - z) = 120
By substituting the values from equations (2) and (3) into equation (4), we can solve for the unknowns.
Following these steps, we can find the values of x, y, and z.