Mathematics Questions and Answers

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

$PR^2 = PQ^2 + RQ^2 - 2(PQ)(RQ)cos Q$

$\implies cos Q = \frac{PQ^2 + RQ^2 - PR^2}{2(PQ)(RQ)}$

$\implies cos Q = \frac{8^2 + 5^2 - 7^2}{2\times8\times5}$

$\implies cos Q = \frac{64 + 25 - 49}{80}$

$\implies cos Q = \frac{40}{80} = 0.5$

$\implies Q = cos^{-1} (0.5) = 60^0$

$\therefore x = 60^0$

$T_2 = \frac{-2}{3};S_\infty \frac {3}{2}$

$T_n = ar^n - 1$

∴ $T_2 = ar = \frac{-2}{3}$---eqn.(i)

$S_\infty = \frac{a}{1 - r} = \frac{3}{2}$---eqn.(ii)

= 2a = 3(1 - r)

= 2a = 3 - 3r

∴ a = $\frac{3 - 3r}{2}$

Substitute $\frac{3 - 3r}{2}$ for a in eqn.(i)

= $\frac{3 - 3r}{2} \times r = \frac{-2}{3}$

= $\frac{3r - 3r^2}{2} = \frac{-2}{3}$

= 3(3r - 3r$^2$) = -4

= 9r - 9r$^2$ = -4

= 9r$^2$ - 9r - 4 = 0

= 9r$^2$ - 12r + 3r - 4 = 0

= 3r(3r - 4) + 1(3r - 4) = 0

= (3r - 4)(3r + 1) = 0

∴ r = $\frac{4}{3} or - \frac{1}{3}$

For a geometric series to go to infinity, the absolute value of its common ratio must be less than 1 i.e. |r| < 1.

∴ r = -$^1/_3$ (since |-$^1/_3$| < 1)

Let the length of the longer side = $x$ cm

∴ The length of the shorter side = ($x$ - 6) cm

If we increase each side's length by 2 cm, it becomes

($x$ + 2) cm and ($x$ - 4) cm respectively

Area of a rectangle = L x B

$A_1 = x(x - 6) = x^2 - 6x$

$A_2 = (x + 2)(x - 4) = x^2 - 4x + 2x - 8 = x^2 - 2x - 8$

$A_1 + 68 = A_2$ (Given)

⇒ $x^2 - 6x + 68 = x^2 - 2x - 8$

⇒ $x^2 - x^2 - 6x + 2x$ = -8 - 68

⇒ -4$x$ = -76

⇒ $x$ = $\frac{-76}{-4}$ = 19cm

∴ The length of the shorter side = $x$ - 6 = 19 - 6 = 13 cm

$\frac{5}{8} - \frac{3}{4} ÷ \frac{5}{12} x \frac{1}{4}$

⇒ $\frac{5}{8} - (\frac{3}{4} ÷ \frac{5}{12}) \times \frac{1}{4}$

⇒ $\frac{5}{8} - (\frac{3}{4} \times \frac{12}{5}) \times \frac{1}{4}$

⇒ $\frac{5}{8} - (\frac{9}{5} \times \frac{1}{4})$

⇒ $\frac{5}{8} - \frac{9}{20}$

∴ $\frac{7}{40}$