Jar Y contains 20-cent coins and Jar Z contains 10-cent coins. There are 28 more coins in Jar Z than in Jar Y. The amount of money in Jar Y is 20¢ more than the amount of money in Jar Z. How many total coins are there?
| |
Jar Y |
Jar Z |
| Number |
1 u |
1 u + 28 |
| Value |
20¢ |
10¢ |
| Total value |
20 u |
10 u + 280 |
Total value of 20¢ coins in Jar Y
= 20 x 1 u
= 20 u
Total value of 10¢ coins in Jar Z
= 10 x (1 u + 28)
= 10 u + 280
The amount in Jar Y is 20¢ more than Jar Z. If another 20¢ is added into Jar Z, both Jar Y and Jar Z will have the same amounts of money.
20 u = 10 u + 280 + 20
20 u - 10 u = 280 + 20
10 u = 300
1 u = 300 ÷ 10 = 30
Total number of coins
= 1 u + (1 u + 28)
= 2 u + 28
= 2 x 30 + 28
= 60 + 28
= 88
Answer(s): 88
