Riordan started saving part of his pocket money by putting 2 coins in a money box every day. Each coin was either a 10¢ or 20¢ coin. His mother also put on a $1 coin in the box every 7 days. The total value of the coins after 91 days was $39.80.
- How many coins were there altogether?
- How many of the coins were 20¢ coins?
Total nuber of coins that his mother put in
= 91 ÷ 7
= 13
Total number of coins that Riordan saved
= 2 x 91 = 182 Total number of coins
= 13 + 182
= 195
(b)
Amount contributed by his mother
= 13 x 1
= $13
Amount that Riordan saved
= 39.80 - 13
= $26.80
$1 = 100¢
$26.80 = 2680¢
Number of 20¢ coins |
Total value of 20¢ coins |
Number of 10¢ coins |
Total value of 10¢ coins |
Total value |
182
|
182 x 20 = 3640¢ |
0
|
0
|
3640¢
|
181
|
181 x 20 = 3620¢ |
1
|
1 x 10 = 10¢ |
3630¢
|
86 |
86 x 20 = 1720¢ |
96 |
96 x 10 = 960¢ |
2680¢
|
Amount that Riordan saved if all he saved were 20¢ coins
= 182 x 20
= 3640
Big difference between the total values of 20¢ coins and 10¢ coins
= 3640 - 2680
= 960
Small difference between the value of 20¢ coins and 10¢ coins
= 20 - 10
= 10
Number of 10¢ coins
= 960 ÷ 10
= 96
Number of 20¢ coins
= 182 - 96
= 86
Answer(s): (a) 195; (b) 86