JAMB Mathematics Past Questions & Answers - Page 251

first work out the expression and then correct the answer to 4 s.f = 241.34..............(A)

(3 x 10-\(^3\))\(^2\)............(B)

= 3\(^2\)x\(^2\)

= \(\frac{1}{10^3}\) x \(\frac{1}{10^3}\)

(Note that x\(^2\) = \(\frac{1}{x^3}\))

= 24.34 x 3\(^2\) x \(\frac{1}{10^6}\)

= \(\frac{2172.06}{10^6}\)

= 0.00217206

= 0.002172(4 s.f)

a²bx + ab²x; a²b - b²

abx(a + b); b(a² - b²)

b(a + b)(a + b)

∴ H.C.F. = (a + b)

\(\frac{4\frac{3}{4} - 6\frac{1}{4}}{4\frac{1}{5} \text{ of } 1\frac{1}{4}}\)

\(\frac{19}{4}\) - \(\frac{25}{4}\)............(A)

\(\frac{21}{5}\) x \(\frac{5}{4}\).............(B)

Now work out the value of A and the value of B and then find the value \(\frac{A}{B}\)

A = \(\frac{19}{4}\) - \(\frac{25}{4}\)

= \(\frac{-6}{4}\)

B = \(\frac{21}{5}\) x \(\frac{5}{4}\)

= \(\frac{105}{20}\)

= \(\frac{21}{4}\)

But then \(\frac{A}{B}\) = \(\frac{-6}{4}\) \(\div\) \(\frac{21}{4}\)

= \(\frac{-6}{4}\) x \(\frac{4}{21}\)

= \(\frac{-24}{84}\)

= \(\frac{-2}{7}\)

Coca-cola = 10 bottles, Fanta = 8 bottles

Sprite = 6 bottles, Total = 24

P(cola-cola) = \(\frac{10}{24}\)

P(not coca-cola) = 1 - \(\frac{10}{24}\)

\(\frac{24 - 10}{24}\) = \(\frac{14}{24}\)

= \(\frac{7}{12}\)

% of boxes containing at least 5 defective bolts each

= \(\frac{7 + 17 + 10 + 8 + 6}{50}\)

= \(\frac{48}{50}\) x \(\frac{100}{1}\)

= 96%