# Sarin participates in school anniversary event part 28 (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 to deal 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.

Read the posting on Sarin participates in school anniversary event part 1 for story on school anniversary.

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 UPQ579

Fatimah were given equal number funfair coupons to sell. They did not manage to completely sell all their share of coupons. Sarin sold 7 times as many coupons as Hairu and was left with 15 unsold coupons . 1/2 of the coupons sold by Fatimah was 5 more than those by Hairu. The 3 of them sold a total of 260 funfair coupons.

a.         How many coupons did Fatimah sell?

b.         How many coupons did each of them receive?

Solution:

a.         The total number of coupons sold = 260

From the model,

10 units + 10 = 260

10 units = 250

1 units = 25 coupons

The number that Fatimah sold = 2 × 25 + 10 = 60 coupons

b.         The number of coupons each of them has = 7 × 25 + 15 = 190 coupons

Alternative solution:

Half of Fatimah sold was 5 more than Hairu sold,

Fatimah : Hairu = (2U+10) : 1U

Sarin sold 7 times as many as Hairu sold,

Sarin : Hairu = 7U : 1U

The ratio of number of coupons sold,

Sarin : Hairu : Fatimah = 7U : 1U : (2U+10)

In total,  7U + 1U + 2U + 10 = 10U + 10

Given the total number of ticket sold = 260

Hence,

10U = 250

1U = 25

a.         The number of coupons sold by Fatimah = 2 × 25 + 10 = 60

b.         The number of coupons each of them has at first

= 7 × 25 + 15 = 190