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HELP ME OUT PLEASE SOLVE IT..............

Jeannie has some \$10 bills and some \$20 bills.if she has 273 bills worth a total of \$4370,how many of the bills are \$10 bills and how many are \$20 bills.?

4.9 4.9 (229 lesson ratings) (229)
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Hey Ma -- try a \$30 "half-and-half" pack ... 4370/30 ... 145 with a \$20 remainder ->
291 is too many bills ... need to drop 18 bills ... change 36 tens to 18 twentys ...
109 tens and 164 twentys -> 1090 + 3280 is 4370 ... Best wishes :)
Lisa K. | 4.0 graduate... Let me show you how to ACE your classes!4.0 graduate... Let me show you how to A...
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There are 164 twenties and 109 tens.

A reps \$10 bills
B reps \$20 bills

A+B=273 bills
A=273-B

value*numberOfTens + value*numberOfTwenties = \$4370
10A+20B=\$4370
replace A for one variable:
10(273-B)+20B=4370
2730-10B+20B=4370
10B=1640
B=164 twenties !!

A=273-B
A=273-164
A=109 tens !!

CHECK:
10(109)+20(164)=_
1090+3280=4370
Gabrielle P. | I love to teach just about everythingI love to teach just about everything
4.8 4.8 (15 lesson ratings) (15)
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take total amount divided by \$20 bills that will give you how many \$20's don't forget how to round. You can't have half of a \$20 and then subtract how much money you have in \$20 from your total to find how many \$10 you need. =) good luck!
Patricia S. | Math Tutoring for K-12 & CollegeMath Tutoring for K-12 & College
5.0 5.0 (39 lesson ratings) (39)
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This word problem describes a situation that is solved by a system of equations (in other words, you have more than one variable and so you need more than one equation to solve for all the variables).  In general the number of equations that you need to work with is the same as the number of variables that you have.

The best way to start these types of questions is to first determine how many variables are involved in the situation, then write out as many equations as are needed.  In this question, you can tell how many variables are in the situation by looking at the different kinds of bills that the question talks about (\$10 bills and \$20 bills).  The two equations, then, would deal with the total number of bills (273) and the total value of the bills (\$4370)
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Variable #1: Number of \$10 bills (I'm going to call this x)    \________  This is often called a
Variable #2: Number of \$20 bills (I'm going to call this y)  _/                 Let Statement and is written as
"Let x = # of \$10 bills
Let y = # of \$20 bills"
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Equation #1: x + y = 273             \_________   EQ #1 refers to the number of each type of bill, and so
Equation #2: 10x + 20y = 4370  _/                   the number of \$10 bills (x) plus the number of
\$20 bills (y) is equal to the total number of bills (273).
EQ #2 refers to the value of each type of bill, so the total value of the \$10 bills is \$10 times the number of \$10 bills you have (10x).  Likewise, the total value of the \$20 bills is \$20 times the number of \$20 bills (20x).  The sum of the total values of each type of bill equals the total value that Jeannie owns (\$4370).

From here, there are a few methods to solve this system of equations.  Since the question didn't specify how they wanted the question solved, I'm going to use the Substitution Method (in which you use one of the equations to solve for one of the variables in terms of the other variable; plug it into another equation; solve for the variable in that equation; plug the number value of the variable back into either equation and solve for second variable).

EQ #1: x + y = 273  ---------------> x + y = 273
EQ #2: 10x + 20y = 4370                   - y          - y
x = 273 - y  <----------  Plug this into EQ #2.

EQ #2: 10x + 20y = 4370
10 (273 - y) + 20y = 4370
10*273 - 10y + 20y = 4370
2730 + 10y = 4370
-2730           -2730
10y = 1640
10       10
y = 164   <----------  This is the number of \$20 bills.  I'm going to plug this number into
EQ #1 to solve for the number of \$10 bills.

EQ #1: x + y = 273
x + 164 = 273
-164   -164
x = 109   <----------  This is the number of \$10 bills.

Jeannie has 109 \$10 bills and 164 \$20 bills.   <------  Typically the final step to all system of equation