2000 - JAMB Mathematics Past Questions and Answers - page 4

Mean = {0 + (x+2) + (3x+6) + (4x+8)}/4 = 4
=> 0 + (x+2) + (3x+6) + (4x+8) = 16
8x + 16 = 16
x = 0

Now prepare a table showing the deviation of each of 0, (x+2), (3x+6) and (4x+8), adding the deviations will give 12.

Thus M.D = 12/4 = 3

Number of letter in MATHEMATICS = 11
Number of letter M = 2
Number of letter E = 2
Number of letter A = 2

Arrangement = 11!/(2! 2! 2!) ways

Angle corresponding to 4 is 40/240 x 360/1 = 60°

Note that total angle in a circle is 360°.
Note also that sum of the frequencies given is 240.

U = {1, 2, 3, 4, 5,..., 20}
E₁ = {3, 6, 9, 12, 15, 18}
E₂ = {4, 8, 12, 16, 20}
P(E₁) = 5/20
P(not E₁) = 1 - (5/20) = 15/20 = 3/4

Hint: prepare a three-columned table, one for (x), another for the (deviation), and the last for the (squared-deviation).

Sum of (x) = 15x, algebraic sum of (deviation) = zero (0)
Sum of (squared-deviation) = 10x²

Variance = ∑(squared-deviation)/n = 10x²/5 = 2x²

Range = Highest - lowest number => 10-4 = 6
Mode is the number with highest occurrence => Mode = 10

Sum = 6 + 10 = 16

No of ways of choosing 1 man, 2 women = 5C1 x 3C2
No of ways of choosing 2 men, 1 woman = 5C2 x 3C1
Summing, => (5C1 x 3C2) + (5C2 x 3C1) = 15 + 30 = 45

Hints:
1. Integrate the given first derivative of f(x) at the boundaries, (0,0)

Then solve accordingly to get f(x) = y = (3x²/)2 + 2x

At x = 1, substituting x = 1 in the equation: ax² + bx + c = 5;
f(1) => a + b + c = 5 .....(1)

Taking the first derivative of f(x) in the original equation gives dy/dx = 2ax + b = 2x + 1 (given)....(2)

From (2),=> b = 1, and 2ax = 2x, => a = 1.

substituting into (1) 1 + 1 + c = 5, => c = 5 - 2 = 3

Thus a = 1, b = 1 and c = 3

5^-3log₅2 x 2^2log₂3
i Let -3log₅2 = p => log₅2^-3 = p
∴2^-3 = 5^p
∴5^-3log₅2 = 5log₅2^-3 = 5^p
ii 2^2log₂3 = q => log₅3² = q
∴3² = 2^q
∴2^22log₂3 = 2^q
= 3² = 9
i x ii = 2^-3 x 9 = ¹/_2³ x 9
= ¹/₈ x 9
= ⁹/₈ = 1¹/₈