Tin Can A contains 50-cent coins and Tin Can B contains 10-cent coins. There are 32 more coins in Tin Can B than in Tin Can A. The amount of money in Tin Can A is 80¢ more than the amount of money in Tin Can B. How many total coins are there?
| |
Tin Can A |
Tin Can B |
| Number |
1 u |
1 u + 32 |
| Value |
50¢ |
10¢ |
| Total value |
50 u |
10 u + 320 |
Total value of 50¢ coins in Tin Can A
= 50 x 1 u
= 50 u
Total value of 10¢ coins in Tin Can B
= 10 x (1 u + 32)
= 10 u + 320
The amount in Tin Can A is 80¢ more than Tin Can B. If another 80¢ is added into Tin Can B, both Tin Can A and Tin Can B will have the same amounts of money.
50 u = 10 u + 320 + 80
50 u - 10 u = 320 + 80
40 u = 400
1 u = 400 ÷ 40 = 10
Total number of coins
= 1 u + (1 u + 32)
= 2 u + 32
= 2 x 10 + 32
= 20 + 32
= 52
Answer(s): 52
