Thread: Algebra story problem. Need Help ASAP

1. Algebra story problem. Need Help ASAP

ok...you have some $1 bills,$5 bills, and $10 bills. There are 225 bills total and they are worth$596. If you have seven times as many $1 as you do$10, how many of each bill do you have?????

i have tried this problem over 20 times!!!!!!

Sincerly
armath

2. Originally Posted by armath
ok...you have some $1 bills,$5 bills, and $10 bills. There are 225 bills total and they are worth$596. If you have seven times as many $1 as you do$10, how many of each bill do you have?????

i have tried this problem over 20 times!!!!!!

Sincerly
armath
Let x = number of $1 bills, y = number of$5 bills and z = number of $10 bills. Then: x + y + z = 225 .... (1) x + 5y + 10z = 596 .... (2) x = 7z .... (3) I suggest substituting (3) into (1) and (2) to eliminate x. Now solve (1) and (2) simultaneously for z and y. Now sub the value of x into (3) to get x. 3. Originally Posted by armath ok...you have some$1 bills, $5 bills, and$10 bills. There are 225 bills total and they are worth $596. If you have seven times as many$1 as you do $10, how many of each bill do you have????? i have tried this problem over 20 times!!!!!! Sincerly armath You have 2 equations here, let "A" stand for the number of 1 dollar bills you have, "B" stand for the number of 5 dollar bills you have, and C stand for the number of 10 dollar bills you have. Then you know that the total number of bills is 225, so A + B + C = 225 And you know that each A is worth 1 dollar, each B is worth 5 dollars, and each C is worth 10 dollars, and that you have 596 dollars total. So A + 5B + 10C = 596 So set the equations next to eachother:$\displaystyle \begin{array}{ccccc}
A&+B&+C&=&225\\
A&+5B&+10C&=&596\\
\end{array}$Now, we have three variables, but only 2 equations. But we know that there are 7 times as many ones as tens. This means that the number of ones is equal to 7 times the number of tens. (Think if there was 1 ten, then there would need to be 7 ones) so A = 7C. This means we can substitute out A:$\displaystyle \begin{array}{ccccc}
7C&+B&+C&=&225\\
(7C)&+5B&+10C&=&596\\
\end{array}$Simplify$\displaystyle \begin{array}{ccccc}
&B&+8C&=&225\\
&5B&+17C&=&596\\
\end{array}$Now you can solve this any way you like, I'll do it with substitution. Since B + 8C = 225, it we can see that B = 225-8C So lets substitute that into our other equation:$\displaystyle 5(225-8C) +17C = 596\displaystyle 1125-40C +17C = 596\displaystyle -23C = -529\displaystyle C = 23$Now we can plug that back into our other equations:$\displaystyle \begin{array}{ccccc|cccccc}
A&+B&+C&=&225& \Longrightarrow& A&+B&+23&=&225\\
A&+5B&+10C&=&596&&A&+5B&+230&=&596\\
\end{array}$And simplify$\displaystyle \begin{array}{cccc}
A&+B&=&202\\
A&+5B&=&-366\\
\end{array}$Again, you can solve this any way you like, I'll use matrices this time:$\displaystyle \left(\begin{array}{cccc}
A&+B&=&202\\
A&+5B&=&366\\
\end{array}\right)$Subtract line 1 from line 2:$\displaystyle \left(\begin{array}{cccc}
A&+B&=&202\\
&4B&=&164\\
\end{array}\right)$Divide line 2 by 4$\displaystyle \left(\begin{array}{cccc}
A&+B&=&202\\
&B&=&41\\
\end{array}\right)$And substitute The value of B into the first equation:$\displaystyle \left(\begin{array}{cccc}
A&+41&=&202\\
&B&=&41\\
\end{array}\right)$Subtract 41 from line 1$\displaystyle \left(\begin{array}{cccc}
A&&=&161\\
&B&=&41\\
\end{array}\right)$So we have A=161, B=41, and C=23 We can verify our answer by plugging them back into our equations:$\displaystyle \begin{array}{ccccc|cccccc}
A&+B&+C&=&225&\Longrightarrow&161&+41&+23&=&225\ \
A&+5B&+10C&=&596&&161&+5(41)&+10(23)&=&596\\
\end{array}\$

You can use your calculator to verify that these answers are correct (they are, I checked them, but you should get into that practice).

And so we only need to verify that they also meet the last criteria that A=7C, so we plug the values in: 161=7(23) and we see that this is correct as well. So we know that our answer is correct.

Some parting advice: Make sure you set them up right, ie seven times as many ones as tens might make you want to say 7A=C, but plug in a value for A, and you will see that C is actually 7 times bigger than A, so it must be the other way around that A=7C. Second, when manipulating equations, use one to manipulate the other, you can never use an equation to manipulate itself, because you will end up with A=A or 0=0 or some other such useless equation. So if you solve for a variable in one equation, use that on the other equation.