1993 - JAMB Mathematics Past Questions and Answers - page 4

10² = 2x² - 2x² cos 120^o (Cosine rule)

100 = 24² - 2x² - 2x² x -\(\frac{1}{2}\)

100 = 3x² + x²

= 3x²

x = \(\sqrt{\frac{100}{3}}\)

= \(\frac{10}{\sqrt{3}}\) x \(\frac{3}{\sqrt{3}}\)

x = 10\(\frac{3}{3}\)cm

Angle corresponding to 7 in a pie chart will be \(\frac{7 \times 360}{\text{sum of items}}\)

= \(\frac{7 }{18}\) x 360

= 140^o

Number of students scoring at least 50 marks = Number of students scoring 50 and above

From the table 53, 70, 84, 59, 90, 60, 81, 73, 50, 37, 67, 68, 64, 52. Hence, 14 students scored at least 50 marks

Mode = a + \(\frac{(b - a)(F_m - F_b)}{2F_m - F_a - F_b}\)

= \(L_1 + \frac{\Delta_1 x^\text{c}}{\Delta_1 + \Delta_2}\)

= \(10 + \frac{(20 - 10)(27 - 10)}{2(27) - 10 - 19}\)

= 10 + \(\frac{170}{25}\)

= 10 + 6.8

= 16.8

Average age of 110 students = 16

∴ Total age = 16 x 10 = 160 years

Age of teachers = x, total number of people now = 11

mean age = 19

Total age of new group = 19 x 11 = 209

Age of teachers = x = (209 - 160) = 49 yrs

Median = L₁ + (\(\frac{Ef}{fm}\)) - fo

\(\frac{\sum f}{2}\)

= \(\frac{20}{2}\)

= 10, L₁ = 10.5, fo = 6, fm = 5

Median = 10.5 + \(\frac{(10 - 6)}{5}\)5

= 10.5 + 4

= 14.5

Possible outcomes are 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30. Prime numbers has only two factors

itself and 1

The prime numbers among the group are 23, 29. Probability of choosing a prime number

= \(\frac{\text{Number of prime}}{\text{No. of total Possible Outcomes}}\)

= \(\frac{2}{11}\)

\(\begin{array}{c|c} x & x - x & (x - x)^2\ \hline 7 & -3 & 9\8 & -2 & 4 \9 & -1 & 1\10 & 0 & 1\11 & 1 & 1\ 12 & 2 & 4\13 & 3 & 9\ \hline & & 28\end{array}\)

S.D = \(\sqrt{\frac{\sum(x - x)^2}{N}}\)

= \(\sqrt{\frac{\sum d^2}{N}}\)

= \(\sqrt{\frac{28}{7}}\)

= \(\sqrt{4}\)

= 2

Chance of x = \(\frac{1}{2}\)

Change of Y = \(\frac{2}{3}\)

Chance of Z = \(\frac{1}{4}\)

Chance of Y and Z only occurring

= Pr (Y ∩ Z ∩ X^c)

where X^c = 1 - Pr(X)

1 = \(\frac{1}{2}\) = 1\(\frac{1}{2}\)

= Pr(Y) x Pr(Z) x Pr(X^c)

= \(\frac{2}{3}\) x \(\frac{1}{4}\) x \(\frac{1}{2}\)

= \(\frac{1}{12}\)

The shaded part exists on x \(\cap\) z but not in y