Jar K contains 20-cent coins and Jar L contains 10-cent coins. There are 11 more coins in Jar L than in Jar K. The amount of money in Jar K is 110¢ more than the amount of money in Jar L. How many total coins are there?
| |
Jar K |
Jar L |
| Number |
1 u |
1 u + 11 |
| Value |
20¢ |
10¢ |
| Total value |
20 u |
10 u + 110 |
Total value of 20¢ coins in Jar K
= 20 x 1 u
= 20 u
Total value of 10¢ coins in Jar L
= 10 x (1 u + 11)
= 10 u + 110
The amount in Jar K is 110¢ more than Jar L. If another 110¢ is added into Jar L, both Jar K and Jar L will have the same amounts of money.
20 u = 10 u + 110 + 110
20 u - 10 u = 110 + 110
10 u = 220
1 u = 220 ÷ 10 = 22
Total number of coins
= 1 u + (1 u + 11)
= 2 u + 11
= 2 x 22 + 11
= 44 + 11
= 55
Answer(s): 55
