Cole has 40% more coins than Riordan. Albert has 80% the number of coins Cole has.
- What is the ratio of the number of Riordan's coins to the number of Albert's coins?
- If the total number of coins of Cole's and Albert's is 304 more than Riordan's, how many coins does Cole have?
Cole |
Riordan |
Albert |
7x5 |
5x5 |
|
5x7 |
|
4x7 |
35 u |
25 u |
28 u |
100% + 40% = 140%
140% =
140100 =
75 Cole's coins : Riordan's coins = 35 : 25
80% =
80100 =
45 Albert's coins : Cole's coins = 28 : 35
Riordan's coins: Albert's coins = 25 : 28 (a)
Cole's coins + Albert's coins = 35 u + 28 u = 63 u
Riordan's coins = 25 u
63 u - 25 u = 304
38 u = 304
1 u = 304 ÷ 38 = 8
Number of Cole's coins
= 35 u
= 35 x 8
= 280 (b)
Answers: (a) 25 : 28; (b) 280