I would solve the above question as follows :

Siti started a savings plan by putting a coin in a money box everyday. Each coin is either a 20-cent coin or a 50-cent coin. Her mother also put in a $1 coin in the box every 5 days. The total value of the coins after 68 days is $34.70. (a) How many coins were there altogether? (b) How many of the coins were 20-cent coins?

(a)

68/5 = 13R3 (Her mother put in 13 $1 coins)

13 + 68 = 81

(b)

$34.70 – $13 = $21.70 (amount Siti put in after 68 days.) 

68 x 0.20 = 13.60

21.70 – 13.60 = 8.10

0.50 – 0.20 = 0.30

8.10/0.30 = 27 (number of 50-cent coins)

68 – 27 = 41 (number of 20-cent coins)

Ans : (a)81 coins; (b) 41 coins.

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This is a Quantity Each Amount question. 

a)    68 ÷ 5 = 13

        68 + 13 = 81 coins

b) 34.70 – 13 = 21.70

34 – 0.3u = 21.70

34 – 21.7 = 0.3u

12.3 = 0.3u

1u = 41       (20 cent coins )

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(a)

68 ÷ 5 = 13 R 3

Her mother put in 13 $1 coins

There were 68 + 13 = 81 coins altogether

(b)

Her mother put in $13

The value of the 20¢ and 50¢ coins was $34.70 – $13 = $21.70

The total number of the 20¢ and 50¢ coins was 68

Now we have a “Guess and Check” question

Assume all the 68 coins were 50¢ coins

The value would be 68 x $0.50 = $34

The excess in value would be $34 – $21.70 = $12.30

The excess is reduced by $0.50 – $0.20 = $0.30 per 20¢ coin

There were $12.30 ÷ $0.30 = 41 20¢ coins

Hope it helps 🙂

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