Tin Can R contains only 50-cent coins and Tin Can S contains only 10-cent coins. There are 21 more coins in Tin Can S than in Tin Can R. The amount of money in Tin Can R is $1.90 more than the amount of money in Tin Can S. Find the total amount of money in both tin cans.
| |
Tin Can R |
Tin Can S |
| Number |
1 u |
1 u + 21 |
| Value |
50¢ |
10¢ |
| Total value |
50 u |
10 u + 210 |
$1 = 100¢
$1.90 = 190¢
Total value of 50¢ coins in Tin Can R
= 50 x 1 u
= 50 u
Total value of 10¢ coins in Tin Can S
= 10(1 u + 21)
= 10 u + 210
The amount of money in Tin Can R is $1.90 more than the amount of money in Tin Can S. If another 190¢ is added into Tin Can S, the total value of coins in Tin Can R and Tin Can S will be the same.
50 u = 10 u + 210 + 190
50 u - 10 u = 400
40 u = 400
1 u = 400 ÷ 40 = 10
Total amount in both tin cans
= 50 u + (10 u + 210)
= 60 u + 210
= (60 x 10) + 210
= 600 + 210
= 810¢
810¢ = $8.10
Answer(s): $8.10
