AAU Ekpoma Post UTME Past Questions and Answers
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AAU Post UTME Sample Questions and Answers 2020/2021
AMBROSE ALLI UNIVERSITY, EKPOMA
MANAGEMENT SCIENCES/MANAGEMENT SCIENCE
POST JAMB SCREENING EXERCISE
Time: 10 minutes
Instruction: Answer ALL Questions
A market situation where exist only one seller and many buyers is called what?
Monopoly is the only market structure where there is only one seller but many buyers e.g. PHCN in Nigeria.
What is the full meaning of OMO?
OMO stands for Open Market Operation. It is one of the techniques used by the Apex Bank (Central Bank) to control commercial banks. In order to reduce the volume of money in circulation.
3 A government by the people, for the people and of the people is called what?
4 What is the full meaning of NAFDAC and who is the Director of NAFDAC?
NAFDAC stands for National Agency for food and drugs administration and control.
In a legislative process, a bill is referred to as …………..
In a legislative process, a bill is an accepted motion in the floor of the house
6 Who was the first executive president of Nigeria?
The first executive president of Nigeria was Alhaji Shedu Shagari (1979).
How many different committees of 4 can be selected from 10 people?
Committee of 4 from 10 people
= 10C4 = 10!
10! = 10 x 9 x 8 x 7 x 6!
6!4! 6! X 4 x 3 x 2 x 1
Note: nCr = n!
If 5, x, y 40 are in geometric progression, find x and y respectively.
5, x, y, 40 are in G.P
= u, = 5 … … … … … … …, (i)
U4, = ar3 = 40 … … … (iii)
Dividing (ii) by (i) gives:
U4 = ar3 = 40
U1 a 5
\ r3 = 8, r = 8 r = 2
\ the first term, a = 5
common ratio, r = 2
x = U2 = ar = 5 x 2 = 10
y = U3 = ar Xr = 10 x 2 = 20
x = 10 and y = 20
note: for a G.P, Un = arn-1
Calculate the compound interest on N20,000 for 2 year at 10% per annum.
For compound interest
First yr, 1 = PRT = 20,000 X 10 X 1
\Principal of 2nd yr = N20000 + N2000
\The compound interest = N2,200
10 The width of a rectangle is 6m shorter than its length. If the area of the rectangle is 216cm2, find the perimeter of the rectangle.
With (w) = Length (L) – 6
W = 1-6
Area of sect angle = W x
\I (I – 6) = 216
L2 – 6l – 216 = 0
(I – 18) (l + 12) = 0
I – 18 = 0 and I = -12
Since length cannot be negative
\I = 18m
Perimeter = 2 (l + w) = 2(19 + 12)
Note: if I = 18m, w = 18-6 = 12m
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