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 ((43) + (75) + (82) + (117) + (132) + (81))/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