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 $5y + 4x ≤ 20$