# A family in Singapore part 33 (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 presenting the Math Concept, Math Questions with solutions that teaches in Singapore Primary Schools. You or the kids will able to acquire knowledge of dealing with Math Modeling, Math Problem Solving and 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 Fraction and Problem Sum.

Read also the posting on Sarin learns the concept of Fraction (mathematics concept) to understand the concept of Fraction.

Challenge yourself with the question before look for the given solution‼!

Upper primary school mathematics question UPQ321

Sarin and Hairu had some money. Sarin’s money was \$80 more than Hairu. After Sarin spent 1/4 of his money and Hairu spent 1/3 of his money, Sarin still had \$83 more than Hairu. How much money did Hairu have at first?

Solution:

Sarin had \$80 more than Hairu at first

The initial Math Model

Saroin spent 1/4 of his money and Hairu spent 1/3 of his money

Redraw the model by portion Sarin’s money to 4 equal portions and Hairu’s money to 3 equal portions

Subdivide the blue unit blocks to 4 × 3 = 12 equal units

After both of them spend the amount, Sarin still has \$83 more than Hairu

The model can be rearrange with the amount spent included as shown

Per the model

1 unit + 60 = 83

1 unit = 83 – 60 = \$23

The amount of money Hairu had at first = 12 × 23 = \$276

Alternative Solution(1):

By equations,

Set S as money that Sarin had at first

Set H as money that Hairu had at first

Sarin had \$80 more than Hairu at first

S = H + 80

After Spending, Sarin more than Hairu by \$83

Sarin spent 1/4 fraction of his money and left with 3/4 fraction of money

Hairu spent 1/3 fraction of his money and left with 2/3 fraction of money

3/4 S = 2/3 H + 83

S = 8/9 H + 83 × 4/3

H + 80 = 8/9 H + 83 × 4/3

1/9 H = 83 × 4/3 – 80

H = 9 × 83 × 4/3 – 9 × 80 = \$276

The amount of money Hairu had at first was \$276

Alternative Solution(2):

Set H as Hairu’s money at first

Sarin’s money = H + 80

Sarin spent 1/4 fraction of his money

Sarin left = 3/4 fraction of his money = 3/4 × (H + 80)

Hairu spent 1/3 fraction of his money

Hairu left = 2/3 × H

After spending, Sarin had \$83 more than Hairu

3/4 × (H + 80) – 2/3 × H = 83

3/4 H – 2/3 H = 83 – 3/4 × 80

1/12 H = 83 – 60 = 23

H = 12 × 23 = 276

The amount of money Hairu had at first = 12 × 23 = \$276