# Thread: Algebra Help urgently needed

1. ## Algebra Help urgently needed

Dear anyone.
Could anyone please show me the easiest most simplest way to sork out this solution please ?

Aaron has am assortment of coins and notes in his money box. He has 9 times as many $1 coins as$5 notes, while there are 8 times as many $2 coins as$5 notes. After counting his money, Aaron discovered that he had exactly $90. How many$2 coins did Aaron have ?

Thanks a billion.

2. Originally Posted by chhoeuk
Dear anyone.
Could anyone please show me the easiest most simplest way to sork out this solution please ?

Aaron has am assortment of coins and notes in his money box. He has 9 times as many $1 coins as$5 notes, while there are 8 times as many $2 coins as$5 notes. After counting his money, Aaron discovered that he had exactly $90. How many$2 coins did Aaron have ?

Thanks a billion.
Let $\displaystyle a$ be the number of $1 coins Aaron has Let$\displaystyle b$be the number of$2 coins Aaron has
Let $\displaystyle c$ be the number of $5 notes Aaron has since we have 9 times as many$1 coins as $5 notes, we have:$\displaystyle 9c = a$.............(1) since we have 8 times as many$2 coins as $5 notes, we have$\displaystyle 8c = b$..............(2) the monetary value from the$1 coins will be $a dollars, the monetary value from the$2 coins will be $2b and the monetary value from the$5 notes will be $5c. Since Aaron has a total of$90, we have:

$\displaystyle a + 2b + 5c = 90$ ...................(3)

plug in the values for $\displaystyle a$ and $\displaystyle b$ from equations (1) and (2) respectively into equation (3), we get:

$\displaystyle 9c + 2(8c) + 5c = 90$

$\displaystyle \Rightarrow 30c = 90$

$\displaystyle \Rightarrow \boxed { c = 3 }$

But $\displaystyle a = 9c = 9(3) \implies \boxed { a = 27 }$

Also, $\displaystyle b = 8c = 8(3) \implies \boxed { b = 24 }$

So, Aaron has 27 $1 coins, 24$2 coins and 3 $5 notes. Adding these up, we get$90 total

EDIT: I seem to have misread the question, I found how many of each coin he had, oh well, it still gets the problem done, so i'll leave it. There is probably a way to solve for how many \$2 coins there are without finding how many of the others there are, but i'm too lazy to think about that right now. (wait, not much thinking is required here, perhaps, we could have solved for the other two in terms of b and write equation 3 in terms of b as opposed to c and solve for b right off the bat. you can try that if you wish).