# Sarin and his coin bank part 41 (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, Algebra 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.

Read also the posting Sarin learns Symbol, Algebra and Equation in school (Math Concept) to understand the concept of Algebra.

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

Upper primary school mathematics question UPQ412

Sarin and Hairu have some money in their coin banks. If Sarin gives \$20 to Hairu, he will have thrice as much money as Hairu. If Hairu gives \$16 to Sarin, he will have 1/9 of what Sarin has. How much money does Sarin have at first?

Solution:

Two conditions given and based on the conditions to put up the model

Start with Sarin’s money is 3 times Hairu’s money after he gives \$20 to Hairu

Determine the amount they have at first

Add the inform of Sarin’s money is 9 times Hairu’s money after he receives \$16 from Hairu

Determine the relationship of the two conditions in the model

From the model

9U = 3 × (1U + 20 + 16) + 20 + 16

9U = 3U + 144

6U = 144

1U = 24

The amount of money Sarin has at first = 9 × 24 – 16 = \$200

The amount of money Hairu has at first = 24 + 16 = \$40

To check:

Sarin gives Hairu \$20

Sarin’s money = 200 – 20 = \$180

Hairu’s money = 40 + 20 = \$60

180 ÷ 60 = 3

Sarin’s money is 3 times of Hairu’s money

Hairu gives Sarin \$16

Sarin’s money = 200 + 16 = \$216

Hairu’s money = 40 – 16 = \$24

216 ÷ 24 = 9

Sarin’s money is 9 times of Hairu’s money

Alternative Solution(1):

By using algebra and equations

Set S as the amount of money Sarin has at first

Set H as the amount of money Hairu has at first

Base on condition 1

Sarin gives \$20 to Hairu

Sarin’s money = (S – 20)

Hairu’s money = (H + 20)

Given that Sarin’s money is thrice of Hairu’s money

S – 20 = 3 × (H + 20) = 3H + 60

S = 3H + 80

Based on condition 2

Hairu gives \$16 to Sarin

Sarin’s money = (S + 16)

Hairu’s money = (H – 16)

Given that Hairu’s money is 1/9 fraction of Sarin’s money

S + 16 = 9 × (H – 16) = 9H – 144

S = 9H – 160

Solve the two equations

S = 3H + 80

S = 9H – 160

3H + 80 = 9H – 160

6H = 240

H = 40

S = 3 × 40 + 80 = \$200

The amount of money sarin has at first = \$200

Alternative Solution(2):

Using Ratio method

The final ratio  after Sarin gives \$20 to Hairu, Sarin’s is thrice as many as Hairu’s money

Sarin’s money : Hairu’s money  = S : H = 3 : 1

To tack back the \$20 to get the original ratio

S : H = (3U + 20) : (1U – 20)

If Hairu gives \$16 to Sarin,

S : H = (3U + 20 + 16) : (1U – 20 – 16) = (3U + 36) : (1U – 36)

Based on the raito of Sarin’s money to Hairu’s money is 9 : 1

3U + 36 = 9 × (1U – 36)

3U + 36 = 9U – 324

6U = 360

1U = 60

The amount of money Sarin has at first = 3 × 60 + 20 = \$200