2019 - JAMB Mathematics Past Questions and Answers - page 6

Let the number of days worked by the assistant = d

\(\therefore\) The bricklayer worked (d + 10) days.

1500(d + 10) + 500(d) = N 95,000

1500d + 15,000 + 500d= N 95,000

2000d = N 95,000 - N 15,000

2000t = N 80,000

d = 40 days

\(\therefore\), the assistant worked for 40 days and received N (500 x 40)

= N 20,000

Given that A(x, y) is the point of equidistance between B and C, 

AB = AC

(AB)\(^2\) = (AC)\(^2\)

Hence, 

(x - 0)\(^2\) + (y - 2)\(^2\) = (x - 2)\(^2\) + (y - 1)\(^2\)

x\(^2\) + y\(^2\) - 4y + 4 = x\(^2\) - 4x + 4 + y\(^2\) - 2y + 1

x\(^2\) - x\(^2\) + y\(^2\) - y\(^2\) + 4x - 4y + 2y = 5 - 4

4x - 2y = 1

Education => \(\frac{15}{100} \times N 50,000\) = N 7,500

Food => N 13,600

Electricity => \(\frac{3}{100} \times N 50,000\) = N 1,500

Leftover => N (50,000 - (7,500 + 13,600 + 1,500))

= N 27,400

a Δ b = a + 3b + 2

(3 Δ 2) Δ 5 = (3 + 3(2) + 2) Δ 5

= 11 Δ 5

= 11 + 3(5) + 2

= 28

\(M \propto N \) ; \(M \propto \frac{1}{\sqrt{P}}\).

\(\therefore M \propto \frac{N}{\sqrt{P}}\)

\(M = \frac{k N}{\sqrt{P}}\)

when M = 3, N = 5 and P = 25;

\(3 = \frac{5k}{\sqrt{25}}\)

\(k = 3\)

\(M = \frac{3N}{\sqrt{P}}\)

when M = 2 and N = 6,

\(2 = \frac{3(6)}{\sqrt{P}} \implies \sqrt{P} = \frac{18}{2}\)

\(\sqrt{P} = 9 \implies P = 9^2\)

P = 81

Mode = score that has the highest frequency

Mode = 5

the mean = \(\frac{\sum fx}{\sum f}\)

= \(\frac{156}{40}\)

= 3.9

M.D = \(\frac{\sum f|x - \bar{x}|}{\sum f}\)

= \(\frac{60.4}{40}\)

= 1.51

\(\int (4x^{-3} - 7x^2 + 5x - 6) \mathrm d x\)

= \(\frac{4x^{-3 + 1}}{-3 + 1} - \frac{7x^{2 + 1}}{2 + 1} + \frac{5x^{1 + 1}}{1 + 1} - 6x\)

= \(-2x^{-2} - \frac{7}{3} x^3 + \frac{5}{2} x^2 - 6x\)

Set, S = {21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34}

n(S) = 14

multiples of 3 = {21, 24, 27, 30, 33}

n(multiples of 3) = 5

Probability of picking a multiple of 3 = 5/14

\(T_n = ar^{n - 1}\) (terms of a G.P)

\(T_3 = ar^2 = 9\) ... (i)

\(T_7 = ar^6 = \frac{1}{9}\) ... (ii)

Divide (i) by (ii);

\(\frac{ar^6}{ar^2} = \frac{\frac{1}{9}}{9}\)

\(r^4 = \frac{1}{81}\)

\(r^4 = (\frac{1}{3})^4\)

\(r = \frac{1}{3}\)