Mathematics Questions and Answers

If the majority are women implies that:

3 Women and 2 Men Or 4 Women and 1 Man

= $^4C_3 \times^6C_2+^4C_4\times^6C_1$

= $4 \times 15 + 1 \times 6$

= 60 + 6 = 66
 

Mean = $\frac{\sum fx}{\sum f} = \frac {820.5}{107}$ = 7.7 (1 d.p)

$lim_{x\to2} \frac {x^2 + 4x - 12}{x^2 - 2x}$ = $lim_{x\to2} \frac {(x - 2)(x + 6)}{x(x - 2)}$

$lim_{x\to2} \frac {x + 6}{x}$

$\frac {2 + 6}{2} = \frac {8}{2}$ = 4

Amount (A) = Principal (P) + Interest (I) i.e. A = P + I

1 = $\frac {PTR}{100}$

A = 3P; T = 10 years (Given)

Since A = P + I

$\implies 3P = P + \frac {P\times 10 \times R}{100}$

$\implies 3P = P + \frac {10PR}{100}$

$\implies 3P = P + \frac {PR}{10}$

$\implies 3P - P = \frac {PR}{10}$

$\implies 2P = \frac {PR}{10}$

$\implies \frac {2P}{1} = \frac {PR}{10}$

$\implies$ PR = 20P

$\therefore$ R = 20%

Since the rate stays the same

A = 5P; R = 20%;T =?; A  =P + I

$\implies 5P = P + \frac {p \times T \times 20}{100}$

$\implies 5P - P = \frac {2PT}{10}$

$\implies 4P = \frac {2PT}{10}$

$\implies 2P = \frac {PT}{10}$

$\implies$ PT = 20P

$\therefore$ T = 20 years

Let the given points be:

P(-3, -14) = (x$_1$, y$_1$)

Q(t, -5) = (x$_2$, y$_2$)

PQ = 9 units (given)

Using the distance formula,

d = √ [ $(x_2 - x_1)^2 + (y_2 - y_1)^2$]

PQ = √ [ $(t - (-3))^2 + (-5 + 14)^2$]

$\implies$ √ [ $(t + 3)^2 + 81$] = 9

Squaring on both sides,

⇒ (t + 3)$^2$ + 81 = 81

⇒ (t + 3)$^2$  = 0

⇒ t + 3 = 0

∴ t = -3