JAMB Mathematics Past Questions & Answers - Page 197

$\frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})}$

$\log_{3} 27 = \log_{3} 3^{3} = 3\log_{3} 3 = 3$

$\log_{\frac{1}{4}} 64 = \log_{\frac{1}{4}} (\frac{1}{4})^{-3} = -3$

$\log_{3} (\frac{1}{81}) = \log_{3} 3^{-4} = -4$

$\therefore \frac{\log_{3} 27 - \log_{\frac{1}{4}} 64}{\log_{3} (\frac{1}{81})} = \frac{3 - (-3)}{-4}$

= $\frac{6}{-4} = \frac{-3}{2}$

Numbers divisible by 4 between 1 and 300 include 4, 8, 12, 16, 20 e.t.c.To get the number of figures divisible by 4, We solve by method of A.P

Let x represent numbers divisible by 4, nth term = a + (n - 1)d

a = 4, d = 4

Last term = 4 + (n - 1)4

296 = 4 + 4n - 4

= $\frac{296}{4}$

= 74
rn(Note: 296 is the last Number divisible by 4 between 1 and 300)

Prob. of x = $\frac{74}{298}$

= $\frac{37}{149}$

Total cost of production = N120.00

Labour Cost = $\frac{70}{120}$ x $\frac{360^o}{1}$

= 210o

Number of boy's = 30

Number of girls = x

Mean of boys' score = 6, for girls = 8

Total score for boys = 30 x 6

Total score for both boys and girls = 468

180 + 8x = 468

x = $\frac{288}{8}$

= 36

Total yield = N3,700

Total amount invested = N 50,000

Let x be the amount invested at 6% interest and let y be the amount invested at 8% interest

then yield on x = $\frac{6}{100}$ x and yield on y = $\frac{8}{100}$y

Hence, $\frac{6}{100}$x + $\frac{8}{100}$y

= 3,700.........(i)

x + y = 50,000........(ii)

6x + 8y = 370,000 x 1

x + y = 50,000 x 6

6x + 8y = 370,000.........(iii)

6x + 6y = 300,000........(iv)

eqn (iii) - eqn(2)

2y = 70,000

y = $\frac{70,000}{2]$

= 35,000

Money invested at 8% is N35,000