1990 - JAMB Mathematics Past Questions and Answers - page 1

4\(\frac{3}{4}\) - 6\(\frac{1}{4}\)

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

\(\frac{21}{5}\) - \(\frac{5}{4}\).............(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 \(\frac{5}{20}\)

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

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

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

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

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

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

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

a²bx + ab²x; a²b - b²

abx(a + b); b(a² - b²)

b(a + b)(a + b)

∴ H.C.F. = (a + b)

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

(3 x 10³)²............(B)

= 3²x

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

(Note that x² = \(\frac{1}{x^3}\))

= 24.34 x 3² x \(\frac{1}{10^6}\)

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

= 0.00217206

= 0.002172(4 s.f)

Interest I = \(\frac{PRT}{100}\)

∴ R = \(\frac{100 \times 1}{100 \times 5}\)

= \(\frac{100 \times 7.50}{500 \times 5}\)

= \(\frac{750}{500}\)

= \(\frac{3}{2}\)

= 1.5%

You can use any whole numbers (eg. 1. 2. 3) to represent all the mangoes in the basket.

If the first child takes \(\frac{1}{4}\) it will remain 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)

Next, the second child takes \(\frac{3}{4}\) of the remainder

which is \(\frac{3}{4}\) i.e. find \(\frac{3}{4}\) of \(\frac{3}{4}\)

= \(\frac{3}{4}\) x \(\frac{3}{4}\)

= \(\frac{9}{16}\)

the fraction remaining now = \(\frac{3}{4}\) - \(\frac{9}{16}\)

= \(\frac{12 - 9}{16}\)

= \(\frac{3}{16}\)

\(\frac{0.00275 \times 0.0064}{0.025 \times 0.08}\)

Removing the decimals = \(\frac{275 \times 64}{2500 \times 800}\)

= \(\frac{88}{10^4}\)

88 x 10^-4 = 88 x 10^-1 x 10^-4

= 8.8 x 10^-3

use "T" to represent the total profit. The first receives \(\frac{1}{3}\) T

remaining, 1 - \(\frac{1}{3}\)

= \(\frac{2}{3}\)T

The seconds receives the remaining, which is \(\frac{2}{3}\) also

\(\frac{2}{3}\) x \(\frac{2}{3}\) x \(\frac{4}{9}\)

The third receives the left over, which is \(\frac{2}{3}\)T - \(\frac{4}{9}\)T = (\(\frac{6 - 4}{9}\))T

= \(\frac{2}{9}\)T

The third receives \(\frac{2}{9}\)T which is equivalent to N12000

If \(\frac{2}{9}\)T = N12, 000

T = \(\frac{12 000}{\frac{2}{9}}\)

= N54, 000

\(\sqrt{160r^2 + 71r^4 + 100r^8}\)

Simplifying from the innermost radical and progressing outwards we have the given expression

\(\sqrt{160r^2 + 71r^4 + 100r^8}\) = \(\sqrt{160r^2 + 81r^4}\)

\(\sqrt{160r^2 + 9r^2}\) = \(\sqrt{169r^2}\)

= 13r

3 log₆9 + log₆12 + log₆64 - log₆72

= log₆9³ + log₆12 + log₆64 - log₆72

log₆729 + log₆12 + log₆64 - log₆72

log₆(729 x 12 x 64) = log₆776

= log₆6₅ = 5 log₆6 = 5

N.B: log₆6 = 1