Math Is Fun Forum

jxharmon

Member

Registered: 2007-01-08

Posts: 104

number problem

have $3.50 in nickels and dimes; total 50 coins how many of each

n + d =50                 
5n + 10d = 350                                       
5 d = 35
d= -5 + 35=30
n + 30 = 50
n=20

20 + 30 = 50

5(20) + 10(30)= 100 +  300 = $4.00
what do I have to change to get $3.50? they should have been just the opposite

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: number problem

You're lucky.  I was given the same problems, only it was 50 coins, pennies, nickels, dimes, and quarters totaling $3.50, and all 37 solutions I found by hand.

n + d =50                 
5n + 10d = 350

So then 5n - 5n + 10d - 5d = 350 - 5*50

Try to solve for d again.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

jxharmon

Member

Registered: 2007-01-08

Posts: 104

Re: number problem

thanks a million I could not figure out what I was doin wrong but I got it now
5n-5n+10d-5d=350-5*50; 5d=350-250;5d=100; d=20 so n+20=50; n=50-20=30; now I have n=30 and d=20 for a total of 50 coins and 5*30=150 and 10*20=200 for 350 in cash.