2003 - JAMB Mathematics Past Questions and Answers - page 4

(\begin{array}{c|c} \text{Age in years} & 10 & 20 & 30 & 40 & 50 \hline \text{frequency} & 3 & 4 & 5 & 6 & 2\end{array})

(\sum f ) = number of people involved in the pollution

(\sum f) = 3 + 4 + 5 + 6 + 2 = 20

Note that if we set (log_{\sqrt{2}}4 = x_{1}, \hspace{1mm}solving\hspace{1mm}gives\hspace{1mm}x_{1}=4\if\hspace{1mm}log_{\frac{1}{2}}16 = x_{2}, \Rightarrow x_{2} = -4\Also \hspace{1mm}for\hspace{1mm}log_{4}32 = x_{3} = 2.5\Combining\hspace{1mm}results,\hspace{1mm}x_{1}+x_{2}+x_{3} = 4+(-4)-2.5=-2.5)

Note that the solution has been separated to simplify solving difficulty due to different bases.

Using a venn diagram, let x = number who offer both Maths and Physics. So that (32-x) offer Maths and another (24-x) offer Physics.

(32-x) + (24-x) + (x) + (4 who offer neither) = 40

=> 60 - x = 40

=> -x = -20

Therefore x = 20.

C.P = N1800

S.P = (270/5) x 400 = N2160.00

Profit = S.P - C.P = N2160 - N1800 = N360.00

(√98 - √50)/√32 = √(49x2) - √(25x2) √(16x2)

=> (7√2 - 5√2)/4√2 = 1/2

25/2% + 95/2% = 70%

100% - 70% = 30% (i.e women)

=> 30% = 84

1% = 84/30

95/2% = (84/30 x 95/2) = 133.

Cross-multiplying, 9^2x-1 = 1 x 27^x+1

3^2(2x-1) = 3^3(x+1)

Solving by equating index: 2(2x-1) = 3(x+1), => x = 5