JAMB Mathematics Past Questions & Answers - Page 11

$\frac {N + 1}{2} = \frac {107 + 1}{2}$ th = 54th value i.e the median age is in the interval "7 - 8" (29 + 25 = 54)

Median = $1_m + (\frac {^{\sum f} /_2 - f_b}{f_m})$c

$l_m$=lower class boundary of the median age = 6.5

$\sum f$ = 107

$f_b$ = sum of all frequencies before the median age = 29

$f_m$=frequency of the median age = 40

c = class width = 8.5 - 6.5 = 2

Median = 6.5 + $(\frac {^{107}/_2 - 29}{40})$ x 2

= 6.5 + $(\frac {53.5 - 29}{40})$ x 2

= 6.5 + $(\frac {24.5}{40})$ x 2

= 6.5 + 1.225

$\therefore$ median age = 7.725

Let the number of white balls be n

The number of red balls = 8

Now, the probability of drawing a white ball = $\frac {n}{n + 8}$

The probability of drawing a red ball = $\frac {8}{n + 8}$

Since Pr(White ball) = $\frac{1}{2}$ x Pr(Red ball)

$\therefore \frac {n}{n + 8} = \frac {1}{2}\times \frac {8}{n + 8}$

= $\frac {n}{n + 8} = \frac {4}{n + 8}$

$\therefore n = 4$

Number of white balls = 4

The possible number of outcomes = n + 8 = 4 + 8 = 12

Pr(Red ball and White ball) = Pr(Red ball) x Pr(White ball)

Pr(Red ball) = $\frac {8}{12}$

Pr(Red ball) = $\frac {4}{11}$

Pr(Red ball and white ball) = $\frac {8}{12} \times \frac {4}{11}$

$\therefore$Pr(Red ball and white ball) = $\frac {8}{33}$

A = { 1, 2, 3, 4, 5, 6}

B = { 2, 4, 6, 8 }

A - B = { 1, 3, 5 }

B - A = { 8 }

∴ (A - B) ⋃ (B - A) = {1, 3, 5, 8}

16$x^4 - y^4$

= 2$^4x^4 - y^4$

= $(2x)^4 - y^4$

= $((2x)^2)^2 - (y^2)^2$

Using a$^2 - b^2$ = (a - b)(a + b) identity

= ((2x)$^2 - y^2)((2x)^2 + y^2)$

Using the identity one more time

= $(2x - y)(2x + y)((2x)^2 + y^2)$

∴ $(2x - y)(2x + y)(4x^2 + y^2)$

Let F be the set of people who can speak French and E be the set of people who can speak English. Then,

n(F) = 400

n(E) = 350

n(F ∪ E) = 500

We have to find n(F ∩ E).

Now, n(F ∪ E) = n(F) + n(E) – n(F ∩ E)

⇒ 500 = 400 + 350 – n(F ∩ E)

⇒ n(F ∩ E) = 750 – 500 = 250.

∴ 250 people can speak both languages.