2001 - WAEC Mathematics Past Questions and Answers - page 1

48 = 2 x 2 x 2 x 2 x 3

$({20}_{three}{)}^2-({11}_{three}{)}^2$

= $({20}_3-{11}_3)({20}_3+{11}_3)$

= $(2_3)({101}_3)$

= ${202}_3$ 

(log_{10}5 + log_{10}20<br /> log_{10}[5\times 20]<br /> log_{10}100<br /> log_{10}10^2<br /> 2log_{10}10=2)

Let c = no of students that offered both subjects

$\therefore$ No of students offering Economics = 65 - c

No of students offering Geography = 50 - c

65 - c + c + 50 - c = 80

115 - c = 80

c = 35

35 students offer both Economics and Geography.

N ∝ M
∴N = kM where k is a constant
N = 8 when M = 20
∴ 8 = k x 20
(k = \frac{8}{20} = \frac{2}{5}<br /> ∴ N = \frac{2}{5}M<br /> If N = 7<br /> ∴ 7 = \frac{2}{5}M<br /> M = \frac{35}{2} = 17\frac{1}{2})

Ratio of good ones to bad ones is 5:4; If 36 is bad;

∴ the good ones = $\frac{5\times 36}{4}=45$ oranges. 

The total number of oranges is 36 + 45 = 81.

The percentage error is (\frac{error}{actual}\times \frac{100}{1}%<br /> =\frac{200-190}{200}\times \frac{100}{1}<br /> =\frac{10}{200}\times \frac{100}{1}% = 5%)

(5\frac{2}{5}\times \left(\frac{2}{3}\right)^2 \left(1\frac{1}{2}\right)^{-1}<br /> \frac{27}{5}\times \frac{4}{9} \times \frac{3}{2}=3\frac{3}{5}
)

$T_n=2^{2n-1}$

$2^{2n-1}=2^9$

$2n-1=9⟹2n=9+1$

$2n=10⟹n=5$

The 5th term = 2${}^9$