MARKING SCHEME FOR JSS1 MATHEMATICS THIRD TERM EXAMINATION

SECTION A: OBJECTIVE TEST

ALLOCATION OF MARKS: ½ mark each = 30 marks

SECTION B: ESSAY

ALLOCATION OF MARKS: 30 marks

PART I

1.(a) \(24 = 2 × 2 × 2 × 3\)

\(48 = 2 × 2 × 2 × 2 × 3\)  (1 mark)

HCF \(= 2 × 2 × 2 × 3\)

HCF \(= 24\)  (1 mark)

Also

(1 mark for table. Accept any other correct method.)

LCM \(= 120\)  (1 mark)

LCM of \(24\) and \(30\) – HCF of \(24\) and \(48\)

\(120 -24\)

\(= 96\) (1 mark)

(b) \(111_2 +  101_2    + 111_2 = 10011_2\)  (1 mark)

(c) Let the budget be represented by \(x\). The amount spent on

Anti-malarial drugs \(= \frac{1}{3}\) of \(x = \frac{x}{3}\) (1 mark)

Remainder \(= x -\frac{x}{3} = \frac{2x}{3}\)  (1 mark)

Amount spent on immunizations

\(= \frac{3}{5}\) of \(\frac{2x}{3} = \frac{2x}{5}\)  (1 mark)

The budget left for other things

\(= \frac{2x}{3} -\frac{2x}{5}\)

\(= \frac{4x}{15}\)

Hence, the fraction of the budget left for other things \(= \frac{4}{15}\)  (1 mark)

TOTAL = 10 MARKS

2.(a) Original cost of the plot of land \(= \#238,000\)

With yearly rise by \(10\%\)

In \(2015\), the rise will be \(10\%\) of \(\#238,000\)

\(\frac{10}{100} × \frac{\#238,000}{1}\)

\(= \#23,800\)  (1 mark)

Therefore, the value of one plot of land in \(2015\) is \(\#238,000  +  \#23,800\)

\(= \#261,800\).