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
Users' Answers & CommentsNumber 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
Users' Answers & CommentsIf a pie chart is used to depict the data, the angle corresponding to 4 is?
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.
Users' Answers & CommentsE1 = {x : x is a multiple of 3}
E2 = {x : x is a multiple of 4}
and an integer is picked at random from U, find the probability that it is not in E2
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
Users' Answers & CommentsHint: 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
Users' Answers & CommentsRange = Highest - lowest number => 10-4 = 6
Mode is the number with highest occurrence => Mode = 10
Sum = 6 + 10 = 16
Users' Answers & CommentsNo 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
Users' Answers & CommentsHints:
- Integrate the given first derivative of f(x) at the boundaries, (0,0)
Then solve accordingly to get f(x) = y = (3x2/)2 + 2x
Users' Answers & CommentsAt 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
Users' Answers & Comments5-3log52 x 22log23
i Let -3log52 = p => log52-3 = p
∴2-3 = 5p
∴5-3log52 = 5log52-3 = 5p
ii 22log23 = q => log532 = q
∴32 = 2q
∴222log23 = 2q
= 32 = 9
i x ii = 2-3 x 9 = 1/23 x 9
= 1/8 x 9
= 9/8 = 11/8
Users' Answers & Comments