JAMB Mathematics Past Questions & Answers - Page 239

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

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) = $\frac{8}{16}$

= $\frac{1}{2}$

$\begin{array}{c|c} Scores & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text{No. of students} & 1 & 4 & 5 & 6 & x & 2\end{array}$
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

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

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

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 $\frac{360^o}{1}$

= 276o