Reggie has some ten-cent coins and fifty-cent coins which amount to more than $85 but less than $90. The number of ten-cent coins is
14 of all the coins he has. When he exchanges some fifty-cent coins for ten-cent coins, the ratio of the number of ten-cent coins to fifty-cent coins he now has become 11 : 1.
- What is the largest possible amount of money Reggie has?
- What is the total value of fifty-cent coins that has been exchanged for ten-cent coins?
|
Ten-cent coins (1) |
Fifty-cent coins (2) |
Make b the same (2)x11 = (3) |
Before |
1 u |
3 u |
33 u |
Change |
+ 5 p |
- 1 p |
- 11 p |
After |
11 b |
1 b |
11 b
|
(a)
Total value of coins per set
= 1 x 10 + 3 x 50
= 10 + 150
= 160¢
$1 = 100¢
$90 = 9000¢
Number of sets of 160¢
= 9000 ÷ 160
= 56 r 40
Largest possible amount that Reggie has
= 56 x 160
= 8960¢
= $89.60
(b)
Ten-cent coins : Fifty-cent coins = 1 : 3
1 u + 5 p =
11 b --- (1)
3 u - 1 p = 1 b --- (2)
Make b the same.
(2)
x11 = (3)
33 u - 11 p =
11 b --- (3)
(3) = (1)
33 u - 11 p = 1 u + 5 p
33 u - 1 u = 5 p + 11 p
32 u = 16 p
16 p = 32 u
1 p = 32 u ÷ 16 = 2 u
Since the number of sets of 160¢ is 56, 1 u is equal to 56.
Number of fifty-cent coins that have been exchanged for ten-cent coins
= 1 p
= 1 x 2 u
= 2 u
= 2 x 56
= 112
Total value of fifty-cent coins that have been exchanged for ten-cent coins
= 112 x 50
= 5600¢
= $56
Answer(s): (a) $89.60; (b) $56