2001 - JAMB Mathematics Past Questions and Answers - page 5

Using the L.C.M of the fractions, convert them to uniform base.
L.C.M = 36, the new fraction becomes: 6/36, 12/36, 54/36, 24/36, 32/36 and 48/36.
Range = highest value - lowest value
Highest among the given fraction = 54/36
Lowest among the given fraction = 6/36.

Range = (54/36) - (6/36) = 48/36 = 4/3

Mode = 11
Square of the mode = 11²
=121

Mean ((4*3) + (7*5) + (8*2) + (11*7) + (13*2) + (8*1))/20
(12 + 35 + 16 + 77 + 26 + 8)/20
174/20 = 8.7

P(Games ends in draw)
This implies that Team P and Q wins
∴ P(P wins) = 1/2
P(Q wins) = 1/2
∴ P(game ends in a draw) = 1/2*1/2 = 1/4

Total number of beads
= 1+2+4+5+3
=15
Number of blue beads = 1
P(Blue beads) = 1/15
Numbers of white beads = 4
P(white beads) = 4/15
∴P(Blue of white beads)
= P(Blue) + P(White)
= 1/15 + 4/15
= 5/15
= 1/3

mean of X = x = (2+6+8+6+2+6)/6 = 30/6 = 5
X → 2,6,8,6,2,6
X - x = -3, 1, 3, 1, 3, 1
(X - x)² = 9, 1, 9, 1, 9, 1
∑(X-x) = 9+1+9+1+9+1 = 30
Variance = ∑(X-x)²/n = 30/6 = 5

Combination is the number (n) of ways of selecting a number of (m) of objects from n
\(^{12}C_{8}\frac{12!}{8!(12-8)!}\=\frac{12!}{8!4!}\\frac{(12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1)}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}\After\hspace{1mm} cancelling \hspace{1mm}out \hspace{1mm}we\hspace{1mm} have\11\times 5\times 9 = 495 \)

1/6, 1/3, 3/2, 2/3, 8/9 and 4/3
L.C.M of the denominators = 36
The fraction now becomes
6/36, 12/36, 54/36, 24/36, 32/36 and 48/36
Highest fraction = 54/36
Lowest fraction = 6/36
Range = Highest - Lowest
(54/36) - (6/36)
= 48/36
= 4/3

Δ SPT is the solution of the inequalities
2y - x - 2 ≤ 0
2y ≤ 2 + x
y ≤ ²/₂ + ^x/₂
y ≤ 1 + ¹/₂x
y + 2x + 2 ≥ 0,
y ≥ -2 -2x
∴ the solution of the inequalities
2y - x - 2 ≤ 0,
y + 2x + 2 ≥ 0
= -2 ≤ x ≤ -1