Jar L contains only 50-cent coins and Jar M contains only 5-cent coins. There are 21 more coins in Jar M than in Jar L. The amount of money in Jar L is $1.20 more than the amount of money in Jar M. Find the total amount of money in both jars.
| |
Jar L |
Jar M |
| Number |
1 u |
1 u + 21 |
| Value |
50¢ |
5¢ |
| Total value |
50 u |
5 u + 105 |
$1 = 100¢
$1.20 = 120¢
Total value of 50¢ coins in Jar L
= 50 x 1 u
= 50 u
Total value of 5¢ coins in Jar M
= 5(1 u + 21)
= 5 u + 105
The amount of money in Jar L is $1.20 more than the amount of money in Jar M. If another 120¢ is added into Jar M, the total value of coins in Jar L and Jar M will be the same.
50 u = 5 u + 105 + 120
50 u - 5 u = 225
45 u = 225
1 u = 225 ÷ 45 = 5
Total amount in both jars
= 50 u + (5 u + 105)
= 55 u + 105
= (55 x 5) + 105
= 275 + 105
= 380¢
380¢ = $3.80
Answer(s): $3.80
