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Sarin and his coin bank part 6 (Math Question)

You can read posting Sarin and his coin bank on the story of coin bank. Read also postings on Sarin and the million numbers to understand concept of dollars and cents and Sarin learns concept of ratio in school on concept of ratio.

Upper primary school mathematics question UPQ44

Sarin had some 50cent coins and Hairu had some 20cent coins in their own coin bank. The number of coins that Sarin had was 2/3 of the number of coins that Hairu had at first. When Hairu gave Sarin $80, he changed all the $80 to 50cent coins. Then ratio of the number of coins that Sarin had to the number of coins that Sarin had was 10 : 11. How many coins did Hairu have at the end?

Solution:

One dollar = 100 cents

$80 = 80 × 100 = 8000 cents

$80 in number of 50cent coins = 8000 ÷ 50 = 160

$80 in number of 20cent coins = 8000 ÷ 20 = 400

Math model:

Before,

The blue portion = N = 1 unit based on after

4 unit = 160 + 400 + 80 = 640

1 unit (N) = 640 ÷ 4 = 160

At the end, Hairu had  = 160 × 11 = 1760 coins

Alternative solution:

            The initial ratio of Sarin’s coins to Hairu’s coins = 2 : 3

            Hairu given $80 to Sarin

            The number of coins Sarin recieved = 80 ÷ 0.5 = 160

            The number of coins that Hairu gave = 80 ÷ 0.2 = 400

            Finally, the coin ratio Sarin to Hairu = (2U + 160) : (3U – 400)

            Given the final ratio

            Sarin’s coins : Hairu’s coins = 10 : 11 = (2U + 160) : (3U – 400)

            We have

            11 × (2U + 160) = 10 × (3U – 400)

            22U + 1760 = 30U – 4000

            8U = 5760

            1U = 720

            The number of coins hairu has at the end = 3 × 720 – 400 = 1760

Sarin and his coin bank part 6(Math Question)

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