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Worked solution

Meat prices: use the printed table consistently

Four housewives bought three different types of meat from the same meat seller. The mass of meat and the total each housewife paid are shown in the table below.

Mrs LeeMrs NasriMrs VicneshMrs Parker
Chicken (kg)0.30.50.60.8
Mutton (kg)0.400.5aa
Duck (kg)0.60.401
Total amount ($)10.76.5718

Find the value of aa. marks

[2013/YJC/I/3]

CJC · Review Problems · Question 2

Answer

a=0.5a=0.5 kg.

Price equations

Let x,y,zx,y,z be the prices per kilogram of chicken, mutton and duck in dollars. The printed table gives 0.3x+0.4y+0.6z=10.7,0.5x+0.4z=6.5,0.6x+0.5y=7.0.3x+0.4y+0.6z=10.7,\qquad0.5x+0.4z=6.5,\qquad0.6x+0.5y=7. Zero entries mean the corresponding meat contributes no cost.

How did this part go?

Solve and check the source slip

Solving gives x=5,y=8,z=10x=5,y=8,z=10, matching the source’s displayed GC result. These values give 0.6(5)+0.5(8)=70.6(5)+0.5(8)=7, the table total. The source working prints 0.6x+0.5y=80.6x+0.5y=8 instead; that line is inconsistent with both its table and its displayed solution. The original 8 is retained in the native transcription and separately qualified here.

How did this part go?

Find the unknown mass

Mrs Parker’s total gives 0.8(5)+8a+1(10)=180.8(5)+8a+1(10)=18. Hence 8a=48a=4, so a=0.5a=0.5 kg, as in the printed answer.

How did this part go?

Common mistakes

A zero quantity is an actual coefficient zero, not missing data.

Solution notes

Source p 2 prints 7 in the table and 8 in one equation; the table and final answer agree on 7. Neither source value is silently overwritten in the retained transcription.

Also printed as Review 2 in both CJC Linear Equations lecture copies: question PDF page 13 / printed 173; answer PDF 15 / Student printed 175, Teacher visible footer 15 (extracted 175). Compared in evidence/review-cjc-linear-equations-lecture.md; the repeated asks, tables and marks retain this existing identity.

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