In May, Abi and Dylan had coins in the ratio 11 : 9. In June, each of them gave away the same number of coins. Abi then had four times as many coins as Dylan.
- Find the ratio of the number of coins Dylan had in May to the number of coins he had in June.
- Find the ratio of the total number of coins both had in May to the total number of coins both had in June.
| |
Abi |
Dylan |
Difference |
| Before |
11x3 =
33 u |
9x3 =
27 u |
2x3 =
6 u |
| Change |
- 25 u |
- 25 u |
|
| After |
4x2 =
8 u |
1x2 =
2 u |
3x2 =
6 u |
(a)
Since Abi and Dylan gave away the same number of coins, the difference in the number of coins between Abi and Dylan at first and in the end remains the same.
Make the difference in the number of coins at first and in the end the same. LCM of 2 and 3 is 6.
Dylan's coins in May : Dylan's coins in June
27 : 2
(b)
Total number of coins in May
= 33 u + 27 u
= 60 u
Total number of coins in June
= 8 u + 2 u
= 10 u
Total coins in May : Total coins in June
60 : 10
(÷10)6 : 1
Answer(s): (a) 27 : 2; (b) 6 : 1
