# Hairu shares his stickers part 63 (Math Question)

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The blog postings are about the Singapore Math. The readers can learn from the postings about Solving the 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 will learn the skills of dealing with the Math Problem Solving, the Math Modeling 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 Percentage, AlgebraRatio and Problem Sum.

You can read the posting on Sarin learns the concept of Ratio (mathematics concept) to understand the concept of Ratio.

You should also read the posting on Sarin learns Average Number, Percentage and Charting in school (Math Concept) to understand Percentage.

Read the posting on 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 UPQ501

Hairu and Sarin have a sum of stickers. The number of stickers Hairu had was 30 more than Sarin at first. After Sarin lost 25% of his stickers and then Hairu got 40% more of the number of stickers he has, the ratio of the number of stickers that Hairu has to the number of stickers that Sarin has become 8 : 3. Find the number of stickers that Hairu has finally.

Solution:

Draw the model

Final ratio, Hairu : Sarin = 8 : 3

8 parts ==> 40U + 16 U + 30 + 12 = 56U + 42

3 parts ==> 30U

1 parts ==> 10U

8 parts ==> 80U

80U = 56U + 42

24U = 42

1U = 42/24

The number of stickers Hairu has finally = 80 × 42/24 = 140

Or,

Per the model

24U = \$42

1U = 14/8

The number of stickers Hairu has at the end = 80 × 14/8 = 140

Alternative Solution(1):

The initial ratio,

Hairu : Sarin = (1U + 30) : 1U

Hairu gain 40% and Sarin lost 25%

The final ratio,

Hairu : Sarin = (1U + 30) × 1.4 : 1U ×0 .75 = (1.4U + 42) : 0.75U

The final ratio given, Hairu : Sarin = 8 : 3

3 × (1.4U + 42) = 8 × 0.75U

4.2U + 126 = 6U

1.8U = 126

1U = 70

The number of stickers Hairu has at first = 70 + 30 = 100

The number of stickers Hairu has finally = 100 × 1.4 = 140

Alternative Solution(2):

Using algebra expression

Set S as the number of stickers Sarin had at first

After Sarin lost 25% of his stickers

Sarin left = 0.75S

The number of sticker Hairu has at first = S + 30

After Hairu gain 40% of stickers

The number of stickers Hairu has = 1.4(S + 30) = (1.4S + 42)

The final ratio, Hairu : Sarin = 8 : 3

Normalised to 1 unit

1 unit = 0.75S/3

1 unit = (1.4S + 42)/8

0.75S/3 = (1.4S + 42)/8

6S = 4.2S + 126

1.8S = 126

S = 70

The number of stickers Hairu has finally = 1.4 (70 + 30) = 140