Cole had
110 as many coins as Zane. Zane had twice as many coins as Flynn. If Cole had 192 coins fewer than Flynn,
- how many coins did Cole have?
- how many coins did they have altogether?
(a)
Number of coins that Cole had = 1 u
Number of coins that Zane had = 10 u
Number of coins that Flynn had = 10 u ÷ 2 = 5 u
Number of coins that Cole had fewer than Flynn
= 5 u - 1 u
= 4 u
4 u = 192
1 u = 192 ÷ 4 = 48
Number of coins that Cole had
= 1 u
= 48
(b)
Total number of coins that they had
= 1 u + 10 u + 5 u
= 16 u
= 16 x 48
= 768
Answer(s): (a) 48; (b) 768
