.(a) In a class of 80 students,\(\frac{3}{4}\) study Biology and \(\frac{3}{5}\) study Physics. If each student studies at least one of the subjects: (i) dr...

Assessment: WAEC SSCE - General Mathematics - 2021 (Essay) Subject: General Mathematics

Question 1 Report

.(a) In a class of 80 students,\(\frac{3}{4}\) study Biology and \(\frac{3}{5}\) study Physics.

If each student studies at least one of the subjects: (i) draw a Venn diagram to represent this information

(ii) how many students study both subjects

(iii) find the fraction of the class that study Biology but not Physics.

(b) Johnson and Jocatol Ltd. owned a business office with floor measuring 15m by 8 m which was to be carpeted.

The cost of carpeting was Gh¢ 890.00 per square metre. If a total of GH 216,120.00 was spent on painting and carpeting, how much was the cost of painting? 

Answer Details

(a) In a class of \(n=80\) students, let \(B\) be the set who study Biology and \(P\) the set who study Physics.

Number studying Biology \(=\tfrac{3}{4}\times80=60\).

Number studying Physics \(=\tfrac{3}{5}\times80=48\).

Since every student studies at least one subject, no one lies outside the two sets, i.e. \(n(B\cup P)=80\).

(i) Venn diagram. Let \(x\) be the number who study both subjects. Then Biology-only \(=60-x\) and Physics-only \(=48-x\):

figure
Venn diagram: Biology-only = 60 - x, both = x, Physics-only = 48 - x, with the solved values 32, 28 and 20 shown in brackets; none outside the circles.

(ii) How many study both subjects. Using the total,

\[ (60-x)+x+(48-x)=80 \] \[ 108-x=80\;\Rightarrow\;x=108-80=28. \]

So 28 students study both Biology and Physics. Hence Biology-only \(=60-28=32\) and Physics-only \(=48-28=20\). Check: \(32+28+20=80\). \(\checkmark\)

(iii) Fraction that study Biology but not Physics. These are the Biology-only students:

\[ \frac{60-28}{80}=\frac{32}{80}=\frac{2}{5}. \]

The fraction of the class that study Biology but not Physics is \(\dfrac{2}{5}\).

(b) Area of the office floor \(=15\text{ m}\times8\text{ m}=120\text{ m}^{2}\).

Cost of carpeting \(=120\text{ m}^{2}\times\text{GH}\cent890.00=\text{GH}\cent106{,}800.00\).

Total spent on painting and carpeting \(=\text{GH}\cent216{,}120.00\). Therefore

\[ \text{Cost of painting}=216{,}120.00-106{,}800.00=\text{GH}\cent109{,}320.00. \]

The cost of painting was GH\(\cent\)109,320.00.

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