Question 1203240: There are $925 in $5, $10, and $20 bills in a cash register. There is a total of 71 bills. The number of $20 bills is 7 less than the total number of $5 bills and $10 bills. How many bills of each denomination are in the cash register?
Found 3 solutions by math_tutor2020, greenestamps, ikleyn:
Answer by math_tutor2020(3835)   (Show Source):
You can put this solution on YOUR website!
Hints:
a = number of 5's
b = number of 10's
c = number of 20's
a+b+c = 71 is one equation to form since there are 71 bills overall.
c = a+b-7 because of the sentence "The number of $20 bills is 7 less than the total number of $5 bills and $10 bills"
Another equation to form is 5a+10b+20c = 925 because there's $925 total in the cash register.
5a = amount of money from the 5's only
10b = amount of money from the 10's only
20c = amount of money from the 20's only
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So we have this system of equations
a+b+c = 71
c = a+b-7
5a+10b+20c = 925
There are 3 equations and 3 unknowns.
Try to make an equivalent system such that there are 2 equations and 2 unknowns.
Use the substitution property to replace every copy of 'c' in equation (1) with a+b-7. This gets rid of c and we're left with 'a's and 'b's only. Repeat the same idea for equation (3) as well.
I'll let the student take over from here.
Let me know if these hints help or not.
Answer by greenestamps(13334)  (Show Source):
You can put this solution on YOUR website!
The problem is more easily solved if you do a bit of logical reasoning and mental arithmetic before plunging into solving the problem with a system of 3 equations.
The total number of bills is 71, and the number of $20 bills is 7 less than the total number of $5 and $10 bills. A bit of mental arithmetic (or formal algebra, if you want) tells us that the number of $20 bills is 32 and the total number of $5 and $10 bills is 39.
So now we have 32($20) = $640 in $20 bills and $925-$640 = $285 in $5 and $10 bills.
For me, the easiest way to finish solving the problem is to first consider the case where there are 39 $5 bills, then see how many of those need to be replaced with $10 bills to make the required total of $285.
39 $5 bills make a total of $195
The actual total is $285, which is $90 more than $195
Each time we replace a $5 bill with a $10 bill, the total increases by $5. So the number of times we need to do that is $90/$5 = 18. So the number of $10 bills is 18, leaving 39-18 = 21 $5 bills.
ANSWER: 32 $20 bills, 18 $10 bills, and 21 $5 bills
CHECK: 32($20)+18($10)+21($5)= $640+$180+$105 = $925
Answer by ikleyn(53765)  (Show Source):
You can put this solution on YOUR website! .
There are $925 in $5, $10, and $20 bills in a cash register. There is a total of 71 bills.
The number of $20 bills is 7 less than the total number of $5 bills and $10 bills.
How many bills of each denomination are in the cash register?
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I will show you how to solve the problem
using two unknowns (and two equations) only.
Let x be the number of the 5-dollar bills;
Let y be the number of the 10-dollar bills.
Then the number of the 20-dollar bills is (x+y-7).
Write an equation for the total number of bills
x + y + (x+y-7) = 71.
Simplify it
2x + 2y = 71 + 7 = 78,
or
x + y = 39.
Write second equation for the total money
5x + 10y + 20(x+y-7) = 925.
Simplify it
x + 2y + 4(x+y-7) = 185
x + 2y + 4x + 4y - 28 = 185
5x + 6y = 185 + 28 = 213
So, you have now the system of two equations
x + y = 39, (1)
5x + 6y = 213. (2)
To solve it, multiply equation (1) by 5 (both sides); keep equation (2) as is. You will get
5x + 5y = 195, (1')
5x + 6y = 213. (2')
Subtract equation (1') from equation (2'). You will get
6y - 5y = 213 - 195
y = 18.
Now from equation (1) x = 39-y = 39-18 = 21.
ANSWER. There are 21 of the 5-dollar bills, 18 of the 10-dollar bills, and (x+y-7) = (21+18-7) = 32 of the 20-dollar bills.
CHECK. 21*5 + 18*10 + 32*20 = 925 dollars total. ! correct !
Solved.
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