# Sarin and his coin bank part 53 (Math Question)

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The blog postings are about the Singapore Math. The readers can learn from the postings about Solving Singapore Primary School Mathematics. The blog presents the Math Concept, the Math Questions with solutions that teaches in Singapore Primary Schools. You or the kids can learn the skills of dealing with the Math Modeling, the Math problem solving and the Problem Sum from Lower Primary School to Upper Primary School level after reading the blog postings. This posting is an upper primary school math question on Ratio and Problem Solving.

You can read the posting on Sarin and his coin bank for the story of coin bank.

Read the posting on Sarin learns concept of ratio in school to understand Ratio.

Challenge yourself with the Question before look out for the given solution‼!

Upper primary school mathematics question UPQ559

Sarin had some 50₵ coins in his coin bank. Hairu had some \$1 coins his coin bank. Sarin gave half of his coins to Hairu and Hairu gave half of his coins to Sarin. Hairu used some of the 50₵ coins to buy a mini fan that cost \$5. Sarin used some of his \$1 coins to buy a racket that cost \$16. After that the ratio of the number of 50₵ coin to \$1 coin Sarin had was 1 : 2 and the number of 50₵ coins to the number of \$1 coins Hairu had was 1 : 5.

How much did Sarin have at first in his coin bank?

Solution: To note:

1. Sarin and Hairu both did not use the coin they have at 1st but give half of their coins to each other.
2. Sarin spent \$16 of \$1 coins, so he used 16 of the \$1 coins given by Hairu.
3. Hairu spent \$5 of 50₵, so he used 5 ÷ 0.5 = 10 of the 50₵ coins given by Sarin.

Final coins ratio for Sarin in qty,

50₵ : \$1 = 1 : 2

Sarin used \$1 coins to pay for \$16, the number of coins used = 16

The initial coins ratio Sarin had

50₵ : \$1 = 1U : (2U + 16)

Final coins ratio for Hairu in qty,

50₵ : \$1 = 1 : 5

Hairu used 50₵ coins to pay for \$5, the number of coins used = 5 ÷ 0.5 = 10

50₵ : \$1 = (1P + 10) : 5P

Since both received half of the coins from each other, they have same number of 50₵ and \$1 coins initially.

1U = 1P + 10 ==> 2U = 2P + 20

2U + 16 = 5P ==> 2U = 5P – 16

2P + 20 = 5P – 16

3P = 36

1P = 12

1U = 12 + 10 = 22

The number of coins Sarin had at first = 2 × 22 = 44

The amount Sarin had at first = 44 × 0.5 = \$22