2004 - JAMB Mathematics Past Questions and Answers - page 3

Gold medals = 30
silver medals = 22
Bronze medals = 18/70
P(Gold medals) = 30/70 = 3/7
∴ P(not Gold medal) = 1 - 3/7
= 4/7

\(^{3}C_{1} \times ^{9}C_{5}=\frac{3!}{(3-1)!1!}\times \frac{9!}{(9-5)!5!}\=\frac{3!}{2!1!}\times \frac{9!}{4!5!}\=\frac{3 \times 2!}{2!\times 1}\times \frac{9\times8\times7\times6\times5!}{4\times3\times2\times1\times5!}\=3\times9\times2\times7\=378 \hspace{1mm}ways\)

Range = 18 - 15
= 3

y ∝ 1/x
y = K/x
K = xy
K = (1/2) * 4 ( when y = 4 and x = 1/2)
K = 2
∴y = ²/_x
10 = 2/x (when y = 10)
10x = 2
x = 2/10
x = 1/5

-1 < 3 -2x and 3 - 2x \(\leq\) 5
2x < 3 + 1 and -2x \(\leq\) 5 - 3
2x < 4 and -2x \(\leq\) 2
x < 2 and x \(\geq\) 2/-2 = x \(\geq\) -1
If -1 \(\leq\) x < 2
it implies that x = -1, 0, 1

\(a=6, U_4 = 162\U_4 = ar^{4-1}\U_4 = ar^3\162 = 6(r)^3\r^3 = 27\r=\sqrt[3]{27}\r=3\S_4 = \frac{a(r^{n}-1)}{r-1}\=\frac{6(3^{3}-1)}{3-1}\=\frac{6(27-1)}{2}\=3\times26\=78\)

\(P\times Q=\sqrt{PQ}\4\times(8\times32)=4\times\sqrt{8\times32}\=4\times\sqrt{256}\=4\times16\=\sqrt{4\times16}\=\sqrt{64}\=8\)

x = -3
substitute x = -3 in 3x³ + 5x² - 11x + 4
3(-3)³ + 5(-3)² - 11(-3) + 4
-81 + 45 + 33 + 4
-81 + 82
= 1

Q_n = 3 * 2^{n - 2}
U_m = 3 * 2^{2m - 3}
Q₂ * U₂ = (3 * 2^{2 - 2}) * (3 * 2^{2(2) - 3})
= 3 * 2⁰ * 3 * 2¹
= 3 * 1 * 3 * 2
= 18