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)<br />

∴ 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