.(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...

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 \(=48-x\):

(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.