JAMB Mathematics Past Questions & Answers - Page 3

∠PQR + ∠PSR = 180o (opp. angles of cyclic quad. are supplementary)

⇒ 5\(x\) - \(y\) + 10 + (-2\(x\) + 3\(y\) + 145) = 180

⇒ 5\(x\) - \(y\) + 10 - 2\(x\) + 3\(y\) + 145 = 180

⇒ 3\(x\) + 2\(y\) + 155 = 180

⇒ 3\(x\) + 2\(y\) = 180 - 155

⇒ 3\(x\) + 2\(y\) = 25 ----- (i)

∠QPS + ∠QRS = 180o (opp. angles of cyclic quad. are supplementary)

⇒ -4\(x\) - 7\(y\) + 150 + (2\(x\) + 8\(y\) + 105) = 180

⇒ -4\(x\) - 7\(y\) + 75 + 2\(x\) + 8\(y\) + 180 = 180

⇒ -2\(x\) + \(y\) + 255 = 180

⇒ -2\(x\) + y = 180 - 255

⇒ -2\(x\) + \(y\) = -75 ------- (ii)

⇒ \(y\) = -75 + 2\(x\) -------- (iii)

Substitute (-75 + 2\(x\)) for \(y\) in equation (i)

⇒ 3\(x\) + 2(-75 + 2\(x\)) = 25

⇒ 3\(x\) - 150 + 4\(x\) = 25

⇒ 7\(x\) = 25 + 150

⇒ 7\(x\) = 175

⇒ \(x = \frac{175}{7} = 25\)

From equation (iii)

⇒ \(y\) = -75 + 2(25) = -75 + 50

⇒ \(y\) = -25

∴ \(x\) + \(y\) = 25 + (-25) = 0

Area of trapezium = \(\frac{1}{2}(a + b) h\)

⇒ \(\frac{1}{2} (a + b)\times 16 = 200\)

⇒ 8(a + b) = 200

⇒ a + b = \(\frac{200}{8}\) = 25 -----(i)

⇒ a : b = 2 : 3

⇒ \(\frac{a}{b} = \frac{2}{3}\)

⇒ 3a = 2b

⇒ a = \(\frac{2b}{3}\) -------(ii)

Substitute \(\frac{2b}{3}\) for a in equation (i)

⇒ \(\frac{2b}{3}\) + b = 25

 \(\frac{5b}{3}\) = 25

⇒ b = 25 ÷ \(\frac{5}{3} = 25\times\frac{3}{5} = 15cm\)

From equation (ii)

⇒ a = \(\frac{2 \times 15}{3} = 2\times5 = 10cm\)

∴ Lengths of each parallel sides are 10cm and 15cm

Total number of students that scored at most 50 marks = 100 + 80 + 60 + 40 + 80 = 360

Volume of a cone = \(\frac{1}{3}\pi r^2 h\)

r = 5 cm

l = 13 cm

Using Pythagoras theorem

⇒ \(13^2 = 5^2 + h^2\)

⇒ \(169 = 25 + h^2\)

⇒ \(169 - 25 = h^2\)

⇒ \(h^2 = 144\)

⇒ \(h = \sqrt144 = 12 cm\)

∴ Volume of the cone = \(\frac{1}{3} \times\pi\times 5^2 x 12 = 100\pi\) cm\(^3\)

First, we find the equation of the boundary line using the two intercepts.The slope is

m = \(\frac{4 - 0}{0 - 5} = {4}{5}\)

The y-intercept is 4

The slope-intercept form of the equation is therefore

y = -\(\frac{4}{5} x + 4\)

\(\implies y + \frac{4}{5} x = 4\)

Multiply both sides by \(\frac{5}{4}\)

\(\implies\frac{5}{4}(y +\frac{4}{5} x) = 4\times\frac{5}{4}\)

\(\implies\frac{5}{4} y + x = 5\)

Multiply both sides by 4

⇒ \(5y + 4x = 20\)

The inequality is therefore either \(5y + 4x ≤ 20\) or \(5y + 4x ≥ 20.\)

Using the test point (0, 0) -The origin

⇒ 5(0) + 4(0) ≤ 20

⇒ 0 ≤ 20 (True)

∴ The inequality is \(5y + 4x ≤ 20\)