Jar K contains only 50-cent coins and Jar L contains only 5-cent coins. There are 11 more coins in Jar L than in Jar K. The amount of money in Jar K is $1.70 more than the amount of money in Jar L. Find the total amount of money in both jars.
| |
Jar K |
Jar L |
| Number |
1 u |
1 u + 11 |
| Value |
50¢ |
5¢ |
| Total value |
50 u |
5 u + 55 |
$1 = 100¢
$1.70 = 170¢
Total value of 50¢ coins in Jar K
= 50 x 1 u
= 50 u
Total value of 5¢ coins in Jar L
= 5(1 u + 11)
= 5 u + 55
The amount of money in Jar K is $1.70 more than the amount of money in Jar L. If another 170¢ is added into Jar L, the total value of coins in Jar K and Jar L will be the same.
50 u = 5 u + 55 + 170
50 u - 5 u = 225
45 u = 225
1 u = 225 ÷ 45 = 5
Total amount in both jars
= 50 u + (5 u + 55)
= 55 u + 55
= (55 x 5) + 55
= 275 + 55
= 330¢
330¢ = $3.30
Answer(s): $3.30
