He has 13 dimes and 7 quarters.Let a=number of dimesLet b=number of quartersSo the amount of money he has in cents is: 10a + 25bGiven: a+b=20, then a = 20-band 10a + 25b = 25a + 10b – 90Sub in a = 20 – b gives 10(20-b) + 25b = 25(20-b)10b-90Multiply out: 200-10b25b = 500-25b+10b-90Reduce: 200+15b =410-15bSolve for b: 30b=210 b=7Thus a=13so he has $1.30 in dimes and $1.75 in quarters for $3.05If dimes were quarters and quarter were dimes, he has$3.25 in quarters and $0.70 in dimes for $3.95, or ninety cents more!
SDSillyCyclist about 8 years ago
Spoiler Alert: Answer in next message.
SDSillyCyclist about 8 years ago
He has 13 dimes and 7 quarters.Let a=number of dimesLet b=number of quartersSo the amount of money he has in cents is: 10a + 25bGiven: a+b=20, then a = 20-band 10a + 25b = 25a + 10b – 90Sub in a = 20 – b gives 10(20-b) + 25b = 25(20-b)10b-90Multiply out: 200-10b25b = 500-25b+10b-90Reduce: 200+15b =410-15bSolve for b: 30b=210 b=7Thus a=13so he has $1.30 in dimes and $1.75 in quarters for $3.05If dimes were quarters and quarter were dimes, he has$3.25 in quarters and $0.70 in dimes for $3.95, or ninety cents more!
thefoxapproves almost 6 years ago
me in math class all the time
markkahler52 4 months ago
He has what he has! What matter does it make? (Guess I’m more philisophical than mathematic…)