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.
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s = ones , f = fives , t = tens , w = twent
s + f + t + w = 26
s + 5f + 10t + 20w = 120
s = 3f , t = f - 1
the total value is divisible by 5, so the number of ones must be also
... the number of ones must also be divisible by 3
didn't see earlier answers, otherwise i would not have responded
Thank you all very much for your help. it's much appreciated.
X=FIVES
3X=ONES
X-1=TENS
27-5X=TWENTIES (26-(5X-1))
5X+3x+10(x-1)+20(27-5x)=120
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To solve this problem, we can break it down into equations and solve them step by step.
Let's assume that the cashier has:
- x number of fives,
- 3x number of ones, (since there are three times as many ones as fives)
- (x-1) number of tens, (since there is one less ten than fives)
- The rest, which are twenties.
Now, let's add up the values of these bills to form an equation.
The value of the fives is 5x.
The value of the ones is 3x.
The value of the tens is 10(x-1).
The value of the twenties is the remaining difference from the total value of $120.
So we can write the equation:
5x + 3x + 10(x-1) + 20(26 - x - 3x - (x-1)) = 120
Now, let's solve the equation step by step.
5x + 3x + 10(x-1) + 20(26 - x - 3x - (x-1)) = 120
5x + 3x + 10x - 10 + 20(26 - x - 3x - (x-1)) = 120
5x + 3x + 10x - 10 + 20(26 - 4x - (x-1)) = 120
5x + 3x + 10x - 10 + 20(25 - 5x) = 120
5x + 3x + 10x - 10 + 500 - 100x = 120
18x + 490 - 100x = 120
-82x + 490 = 120
-82x = -370
x = 4.51
Since the number of bills cannot be fractional, there seems to be an error in the problem statement or the calculations.
Please recheck the problem statement and confirm the values or any missing information to proceed with solving it accurately.