SOLUTION: Another story problem I can't seem to understand. Paul has $1.47 in change consisting of pennies, nickels and quarters. He has three more pennies than quarters and one more nic

Question 185512: Another story problem I can't seem to understand.
Paul has $1.47 in change consisting of pennies, nickels and quarters. He has three more pennies than quarters and one more nickel than pennies. How many of each coin does he have?
Found 2 solutions by jnaert08, Alan3354:
Answer by jnaert08(2) (Show Source): You can put this solution on YOUR website!
Okay, so you know he has 1.47, right? This consists of pennies, nickles and quarters. So keep in mind that this is .01, .05, and .25 amounts.
3+q=p (three more pennies than quarters)
1+p=n (one more nickel than pennies)
where p is the number of pennies (not the total cost of the pennies)
n is the number of nickels
and q is the number of quarters
The way I would start this out is plug and chug and then see what to do from there.
If you had 2 quarters, then that would be .50 cents already taken care of. And with the 3+q=p equation, you would find out that that would mean 3+2=p, so there are 5 pennies. 5*.01=5. So thats .55 cents total taken care of. And then 1+p=n, so 1+5=n, so n=6. so then you multiple 6*.05 and and thats .30, so that's .85 cents taken care of and you needed 1.47. This is much easier once you get the hang of it, and especially if you have a calculator.
4 quarters = 1.00
3+4=p, so 7 pennies. 7*.01= .07
1+7=n, so 8 nickels. 8*.05= .40
So then you add 1.00 plus .07 plus .40, which equals 1.47!
So that means there were
4 quarters
7 pennies
and 8 nickels!
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Another story problem I can't seem to understand.
Paul has $1.47 in change consisting of pennies, nickels and quarters. He has three more pennies than quarters and one more nickel than pennies. How many of each coin does he have?
---------------
p = q+3 (3 more pennies than quarters)
n = p+1
p + 5n + 25q = 147 (the total)
-------------Substitute for n and q
p + 5(p+1) + 25(p-3) = 147
p + 5p+5 + 25p-75 = 147
31p + 70 = 147
31p = 217
p = 7
You can do the other 2.
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