Oliver has some twenty-cent coins and fifty-cent coins which amount to more than $30 but less than $40. The number of twenty-cent coins is
19 of all the coins he has. When he exchanges some fifty-cent coins for twenty-cent coins, the ratio of the number of twenty-cent coins to fifty-cent coins he now has become 8 : 1.
- What is the largest possible amount of money Oliver has?
- What is the total value of fifty-cent coins that has been exchanged for twenty-cent coins?
|
Twenty-cent coins (1) |
Fifty-cent coins (2) |
Make b the same (2)x8 = (3) |
Before |
1 u |
8 u |
64 u |
Change |
+ 5 p |
- 2 p |
- 16 p |
After |
8 b |
1 b |
8 b
|
(a)
Total value of coins per set
= 1 x 20 + 8 x 50
= 20 + 400
= 420¢
$1 = 100¢
$40 = 4000¢
Number of sets of 420¢
= 4000 ÷ 420
= 9 r 220
Largest possible amount that Oliver has
= 9 x 420
= 3780¢
= $37.80
(b)
Twenty-cent coins : Fifty-cent coins = 1 : 8
1 u + 5 p =
8 b --- (1)
8 u - 2 p = 1 b --- (2)
Make b the same.
(2)
x8 = (3)
64 u - 16 p =
8 b --- (3)
(3) = (1)
64 u - 16 p = 1 u + 5 p
64 u - 1 u = 5 p + 16 p
63 u = 21 p
21 p = 63 u
1 p = 63 u ÷ 21 = 3 u
Since the number of sets of 420¢ is 9, 1 u is equal to 9.
Number of fifty-cent coins that have been exchanged for twenty-cent coins
= 2 p
= 2 x 3 u
= 6 u
= 6 x 9
= 54
Total value of fifty-cent coins that have been exchanged for twenty-cent coins
= 54 x 50
= 2700¢
= $27
Answer(s): (a) $37.80; (b) $27