1990 - JAMB Mathematics Past Questions and Answers - page 4

\(\frac{h_1}{h_2}\) = \(\frac{2}{3}\)

h₂ = \(\frac{2h_1}{3}\)

\(\frac{r_1}{r_2}\) = \(\frac{9}{8}\)

r₂ = \(\frac{9r_1}{8}\)

v₁ = \(\pi\)(\(\frac{9r_1}{8}\))₂(\(\frac{2h_1}{3}\))

= \(\pi\)r₁ 2h₁ x \(\frac{27}{32}\)

v = \(\frac{\pi r_1 2h_1 \times \frac{27}{32}}{\pi r_1 2h_1}\) = \(\frac{27}{32}\)

v₂ : v₁ = 27 : 32

The requires locus is angle bisector of the two lines

Given that 4, 16, 30, 20, 10, 14 and 26

Adding up = 120

nos \(\geq\) 16 are 16 + 30 + 20 + 26 = 92

The requires sum of angles = \(\frac{92}{120}\) x \(\frac{360^o}{1}\)

= 276^o

Mean of 10 numbers = 16

The total sum of numbers = 16 x 10 = 160

Mean of 11 numbers = 18

Total sum of numbers = 11 x 18

= 198

The 11th no. = 198 - 160

= 38

\(\begin{array}{c|c} Scores & 1 & 2 & 3 & 4 & 5 & 6 \ \hline \text{No. of students} & 1 & 4 & 5 & 6 & x & 2\end{array}\)
Average = 3.5

3.5 = \(\frac{(1 \times 1) + (2 \times 4) + (3 \times 5) + (4 \times 6) + 5x + (6 \times 2)}{1 + 4 + 5 + 6 + x + 2}\)

\(\frac{3.5}{1}\) = \(\frac{1 + 8 + 15 + 24 + 5x + 12}{18 + x}\)

\(\frac{3.5}{1}\) = \(\frac{60 + 5x}{18 + x}\)

60 + 5x = 3.5(18 \(\div\) x)

60 + 5x = = 63 + 1.5x

5x - 1.5x = 63 - 60

1.5x = 3

x = \(\frac{3}{1.5}\)

\(\frac{30}{15}\) = 2

\(\begin{array}{c|c} 1 & 2 & 3 & 4\\hline 1(1, 1) & (1, 2) & (1, 3) & (1, 4)\ \hline 2(2, 1) & (2 , 2) & (2, 3) & (2, 4) \ \hline 3(3, 1) & (3, 2) & (3, 3) & (3, 4)\ \hline 4(4, 1) & (4, 2) & (4, 3) & (4, 4)\end{array}\)

sample space = 16

sum of nos. removed are (2), 3, (4), 5

3, (4), 5, (6)

(4), 5, (6), 7

(5), 6, 7, (8)

Even nos. = 8 of them

Pr(even sum) = \(\frac{8}{16}\)

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

Given from 41 to 56

41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56

The nos multiple of 9 are: 45, 54

P(multiple of 9) = \(\frac{2}{16}\)

= \(\frac{1}{8}\)

The required equation is y = x² - x - 2

i.e. B where the graph touches the graph touches the x-axis y = 0

x² - x - 2 = 0 = (x + 1)(x - 2) = 0

Hence roots of the equation are -1 and 2 as shown in the graph

Since PS = QS = RS

S is the centre of circle passing through P, Q, R /PS/ = /RS/ = radius

QPR \(\pm\) \(\frac{100^o}{2}\) = 50^o