JAMB Mathematics Past Questions & Answers - Page 4

Intersecting Chords Theorem states that If two chords intersect in a circle, then the products of the measures of the segments of the chords are equal.

⇒ AE * EB = CE * ED

⇒ 6 * \(x\) = 4 * (\(x\) + 5)

⇒ 6\(x\) = 4\(x\) + 20

⇒ 6\(x\) - 4\(x\) = 20

⇒ 2\(x\) = 20

∴ \(x = \frac{20}{2}\) = 10 units

Area of the composite figure = Area of semi circle + Area of rectangle + Area of triangle

Area of semi circle = \(\frac{1}{2}\pi r^2 = \frac{1}{2}\times\pi\times\frac{d^2}{4} = \frac{1}{2}\times\frac{22}{7}\times\frac{42^2}{4} = 693 m^2\)

Area of rectangle = l x b = 42 x 60  =2520 m\(^2\)

Area of triangle = \(\frac{1}{2}\times b \times h = \frac{1}{2}\times 36 \times 42 = 756 m^2\)

∴ Area of the composite figure = 693 + 2520 + 756 = 3969 m\(^2\)

\(log_2 (6 - x) = 3 - log_2 x\)

⇒ \(log_2 (6 - x) = 3 log_2 2 - log_2 x\) (since \(log_2\) 2 = 1)

⇒ \(log_2 (6 - x) = log_2 2^3 - log_2 x\) \((a log\) c = \(log\) c\(^a)\)

⇒ \(log_2 (6 - x) = log_2 8 - log_2 x\)

⇒\(log_2 (6 - x) = log_2 \frac{8}{x}\) (\(log\) a - \(log\) b = \(log \frac{a}{b})\)

⇒ \(6 - x = \frac{8}{x}\)

⇒ \(x (6 - x) = 8\)

⇒ \(6x - x^2 = 8\)

⇒ \(x^2 - 6x + 8 = 0\)

⇒ \(x^2 - 4x - 2x + 8 = 0\)

⇒ \(x (x - 4) - 2(x - 4) = 0\)

⇒ \((x - 4)(x - 2) = 0\)

⇒ \(x - 4 = 0 or x - 2 = 0\)

∴ x = 4 or 2

Let number of children's ticket at ₦250.00 each = \(x\)

∴ Number of adult tickets at ₦520.00 each = 5\(x\)

Then,

Total amount of money received from children's tickets = 250\(x\)

Total amount of money received from adult tickets = 520(5\(x\))

⇒ 250\(x\) + 520(5\(x\)) = 171,000

⇒ 250\(x\) + 2600\(x\) = 171,000

⇒ 2850\(x\) = 171,000

⇒ \(x = \frac{171,000}{2850} = 60\)

∴ 60 tickets were sold at ₦250.00 and 300 tickets were sold at ₦520.00

The line: \(3y + 6x\) = 48

Divide through by 3

⇒ y + 2\(x\) = 16

⇒ y = -2\(x\) + 16

∴ The gradient of the line = -2

The points: A(-2, k) and B (4, 8)

m =\(\frac{y2 - y1}{x2 - x1} = \frac{8 - k}{4 - (-2)}\)

⇒ m =\(\frac[8 - k}{4 + 2} = {8 - k}{6}\)

Since the line passes through the points

∴ -2 = \(\frac{8 - k}{6}\)

⇒ \(\frac{-2}[1} = \frac{8 - k]{6}\)

⇒ 8 - k = -12

⇒ k = 8 + 12

∴ k = 20