A coins problem, I can never get these right!
- Thread starter ledhed318
- Start date
Okay, I came across a coins problem, which I never understand. It goes like this:
40 coins (all dimes, nickels, and quarters) are worth $6.20. There are four more dimes than quarters. How many of each coin are there?
1--You have n nickels, d dimes and q quarters translating to n + d + q = 40
2--Being worth $6.20, 5n + 10d + 25q = 620.
3--Multiplying (1) by 5 yields 5n + 5d + 5q = 200
4--Subtracting (3) from (2) yields 5d + 20q = 420.
5--Knowing that d = q + 4, substitute and solve for q, then d than n.
40 coins (all dimes, nickels, and quarters) are worth $6.20. There are four more dimes than quarters. How many of each coin are there?
1--You have n nickels, d dimes and q quarters translating to n + d + q = 40
2--Being worth $6.20, 5n + 10d + 25q = 620.
3--Multiplying (1) by 5 yields 5n + 5d + 5q = 200
4--Subtracting (3) from (2) yields 5d + 20q = 420.
5--Knowing that d = q + 4, substitute and solve for q, then d than n.
