Jane has 15 coins which include $2-coins and $5-coins only. If the total value of $2-coins is less than that of $5-coins by $33, how many $5-coins does Jane have?

Printable View

- May 5th 2009, 06:41 AMeconloverNumber and alegbra
Jane has 15 coins which include $2-coins and $5-coins only. If the total value of $2-coins is less than that of $5-coins by $33, how many $5-coins does Jane have?

- May 5th 2009, 06:47 AMRapha
Hello

let x be the amount of $2-coins

y the amount of $5-coins

Jane has 15 coins, therefor

x+y = 15

Suppose you have x=6 $2-coins, then its worth is 6*2 =$12, and you have y=9 coins (actually this is the solution) then you have 9*5 = $45. You know the difference between the worth of all your $2 coins and 5$-coins is 33$.

Hence

2x = 5y-33

(2 and 5 because of the worth of x and y)

You need to solve the equations

x+y = 15

2x = 5y-33

Regards,

Rapha - May 5th 2009, 06:55 AMeconlover
Thanks for the above explanation Rapha. I do have the same equations with yours but I am stuck on how to take the next step through these equations

x+y=15

2x+33=5y - May 5th 2009, 06:59 AMRapha
You're welcome

this is the first equation: x+y=15, solving for x => x = 15-y

now we know x, therefor:

2(15-y)+33 = 5y

and calculate this

30 - 2y +33 = 5y // +2y

63 = 7y // :7

63/7 = y

9 = y

Now our x depends on y, because we got x = 15-y. We know y exactly and thus

x = 15-9 = 6

Done.

Rapha - May 5th 2009, 07:11 AMeconlover
Ah...thanks Rapha!(Clapping) I finally got it !