2017 - JAMB Mathematics Past Questions and Answers - page 2

3(^{n + 1}) × 27(\frac{n + 1}{81^n})

= 3(^{n + 1}) × 3 (\frac{3^{(n + 1)}}{3^{4n}})

= 3(^{n + 1 + 3n + 3 − 4n})

= 3(^{4n − 4n − 1 + 3})

= 3²

= 9

The locus of points at a fixed distance from the point P is a circle with the given P at its centre.

The locus of points at a fixed distance from the point Q is a circle with the given point Q at its centre

The locus of points equidistant from two points P and Q i.e line PQ is the perpendicular bisector of the segment determined by the points

Hence, The locus of a point which is equidistant from the line PQ forms a perpendicular line to PQ

Given T = {even numbers from 1 to 12}

= { 2, 4, 6, 8,10, 12}

N = {common factors of 6, 8 and 12}

= {2} Find T n N = {2}

If RST = 60^o

RXT = 2 × RST

(angle at the centre twice angle at the circumference)

RXT = 2 × 60

= 120^o

Range = Highest Number - Lowest Number

Mode is the number with highest occurrence

10, 9, 10, 9, 8, 7, 7, 10, 8, 4, 6,, 9, 10, 9, 7, 10, 6, 5

Range = 10 − 4 = 6

Mode = 10

Sum of range and mode = range + mode = 6 + 10

= 16

Sum to infinity

∑ = arn − 1

= (\frac{a}{1}) − r

a = (\frac{1}{4})

r = (\frac{1}{8}) ÷ (\frac{1}{4})

r = (\frac{1}{s}) × (\frac{4}{1})

= (\frac{1}{2})

S = (\frac{1 \div 4}{1}) − (\frac{1}{2})

= (\frac{1}{4}) ÷ (\frac{1}{2})

= (\frac{1}{4}) × (\frac{2}{1})

= (\frac{1}{2})

2x + x = 180^o

3x = 180^o

x = 60^o (exterior angle of the polygon)

angle = (\frac{\text{total angle}}{\text{number of sides}})

60 = (\frac{360}{n})

n = (\frac{360}{60})

n = 6 sides

Capacity = Volume = 3080m³

base diameter = 14m

radius = (\frac{\text{diameter}}{2})

= 7m

Volume of Cylidner = Capacity of cylinder

πr²h = 3080

(\frac{22}{7}) × 7 × 7 × h = 3080

h = (\frac{3080}{22 \times 7})

h = 20m