a) value of 5 $1 coins = 5 x $1 = $5

$5 = 500¢
500¢ ÷ 20¢ = 25

So 5 $1 coins were exchanged for 25 20¢ coins, which Louise put into Box A.



b)

Number of 20¢ coins in Box A at first
= 40 - 25 = 15

Since ratio of 20¢ coins to $1 coins in Box A at first = 1 : 2 ,

number of $1 coins in Box A at first
= 15 x 2 = 30


Realise that total value of coins in Box A before the exchange and after the exchange is the same. There is no change in value.


Total value of coins in Box A
= $1 x 30 + 15 x 20¢
= $30 + 300¢
= $30 + $3
= $33


Since ratio of value of coins in Box A to Box B = 5 : 4,

$33 = 5 units
4 units = $33 x 4/5 = $26.40

So total value of coins in Box B
= $26.40


$26.40 = $26 + 40¢

As there's only $1 and 20¢ coins,
the 40¢ must be made up of 2 20¢ coins.

To get the smallest number of coins,
the other $26 should be made up of all $1 coins
(The higher the value of the coin, the fewer coins you need to get a certain value)

$26 ÷ $1 = 26


So smallest possible number of coins
= 26 + 2 = 28