2015 - JAMB Mathematics Past Questions and Answers - page 3

tan x = \(\frac{2}{3}\)(given), is illustrated in a right-angled \(\Delta\)

thus m² = 2² + 3²

= 4 + 9 = 13

m = \(\sqrt{13}\)

Hence, 2sin x = 2 x \(\frac{2}{m}\)

2 x\(\frac{2}{\sqrt{13}}\)

= \(\frac{4}{\sqrt{13}}\)

= \(\frac{4}{\sqrt{13}} = \frac{\sqrt{13}}{\sqrt{13}}\)

= \(\frac{4\sqrt{13}}{13}\)

Percentage birthrate = 1.0%

1¼ ÷ [ 2 ÷ ¼] of 28

Apply BODMAS rules

\( \frac{5}{4}\) ÷ [ 2 ÷ ¼ × 28 ]

\( \frac{5}{4} ÷ \frac{2}{7} \)


\( \frac{5}{4} × \frac{7}{2} \)

\( \frac{35}{8}\)

= \(4 \frac{3}{8}\)

\( x^2 + 9x + 20 \)

Find the two numbers whose product is 20 and its sum is 9}

\( 5x \times 4x = 20x^2\)


\(5x + 4x = 9x\)

\((x^2 + 5x) + (4x + 20)\)
\(

x(x + 5)+ 4(x + 5)

= (x + 5)(x + 4)\)

n(X ∩ Y) = nx + ny − n(xX ∩ Y)

= 15 + 12 − 7

∴ n(X ∩ Y) = 20

Starting from 3x − z \(\leq\) 9

x = 0, substitute the value of x (i.e. x = 0) into the equation 3x − z \(\leq\) 9

3(0) − z \(\leq\) − 9

z \(\leq\) − 9

If z = 0

Then, 3x − 0 \(\leq\) 9

3x \(\leq\) 9

x \(\leq\) 3

Intercept of x and z i.e (x,z)

= (3, -9)

log₁₀1.5 + 3 log₁₀2 − log₁₀0.3

log₁₀1.5 + log₁₀23 − log₁₀0.3

[log₁₀(1.5 × 23) ÷ log₁₀0.3]

[log₁₀(15/10 × 8 ) ÷ log₁₀( 3/10)]

log₁₀(15/10 × 8 × 10/3 )

log₁₀40

The total surface area of a cylinder = 2πrl + 2πr2

= 2πr(l + r)

= 2 × 3.14(7+%=5)

2 × 3.14 × 12

= 75.4cm (1DP)