Pierre started saving part of his pocket money by putting 2 coins in a money box every day. Each coin was either a 20¢ or 50¢ coin. His father also put on a $1 coin in the box every 7 days. The total value of the coins after 56 days was $51.40.
- How many coins were there altogether?
- How many of the coins were 50¢ coins?
Total nuber of coins that his father put in
= 56 ÷ 7
= 8
Total number of coins that Pierre saved
= 2 x 56 = 112 Total number of coins
= 8 + 112
= 120
(b)
Amount contributed by his father
= 8 x 1
= $8
Amount that Pierre saved
= 51.40 - 8
= $43.40
$1 = 100¢
$43.40 = 4340¢
Number of 50¢ coins |
Total value of 50¢ coins |
Number of 20¢ coins |
Total value of 20¢ coins |
Total value |
112
|
112 x 50 = 5600¢ |
0
|
0
|
5600¢
|
111
|
111 x 50 = 5550¢ |
1
|
1 x 20 = 20¢ |
5570¢
|
70 |
70 x 50 = 3500¢ |
42 |
42 x 20 = 840¢ |
4340¢
|
Amount that Pierre saved if all he saved were 50¢ coins
= 112 x 50
= 5600
Big difference between the total values of 50¢ coins and 20¢ coins
= 5600 - 4340
= 1260
Small difference between the value of 50¢ coins and 20¢ coins
= 50 - 20
= 30
Number of 20¢ coins
= 1260 ÷ 30
= 42
Number of 50¢ coins
= 112 - 42
= 70
Answer(s): (a) 120; (b) 70