2016/2017 NECO FREE MATHEMATICS ANSWERS HERE

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1=10 deacdacbbe
11=20 cbaeebdacc
21=30 deedacbbce
31=40 abdcebaeec
41=50 caadedbaea
51=60 edeacbacdd
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(1a)
Tabulate: Draw both vertical and horizontal lines across
the digits. I drew only verticals cos it's txt
messaging.
x|1|2|3|4
1|1|2|3|4
2|2|4|0|2
3|3|0|3|0
4|4|2|0|4
(1b)
S.I=PRT
P=N15,000, R=10%, T=3yrs
S.I=#15,000, R=10%, T=3yrss
S.I=15,000* 10/100 *3
S.I=N4,500
A=P+S.I
A=N15,000+N4,500
A=#19,500
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(4a)
Convert 30000 litres to metres
= 30000/1000 metres
= 30metres
Depth of fuel = h * 7.5*4.2 = 30m^3
= 31.5hm^3 = 30m^3
= h = 30m^3/31.5m
= h = 0.9m
(4b)
Depth of the tank = l*b*h
where l= 7.5, b = 4.2, h = 1.2
=(7.5 * 4.2 * 1.2)m^3
=37.8m^3
Convert metres to litres = 37.8*
1000litres
= 37800litres
Litres of fuel needed to fill the
tank=37800litres/30800litres
=7800litres
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(5a)
sector for building project
=48000/144000*360=120degree
sector for education =
32,000/144000*360=80degree
sector for saving = 19200/144000*360=48degree
sector for maintenance =
12000/144000*360=30degree
sector for miscellaneous =
7200/144000*360=18degree
sector for food items = 360-(120+80+48+30+18)
=360-296
=64degree
THEN DRAW A PIE CHART WITH THE ANSWERS YOU
GOT ABOVE.
(5b)
Food = 48000 + 32000 + 19200 +
1200 +7200 + x = 144000
Food => 118400 + x = 144000
Food => x = 144000 — 118400
Food = x = #25,600
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(7a)
3^2n+1 - 4(3^n+1)+9=0
3^2-3 - 4(3^n -3)+9=0
(3^n)^2-3 - 4(3^n -3)+9=0
let 3^n = p
p^2 -3 - 4(p-3)+9=0
3p^2/3 - 12p/3 + 9/3 = 0
p^2 - 4p + 3 = 0
p^2 - 3p - p + 3 = 0
p^2p(p-3) - 1(p-3) = 0
(p-1)(p-3) = 0
p-1 = 0 or p-3 = 0
p = 1 or 3
Recall 3^n = p
when p=1
3^n = 3^0
n = 0
when p = 3
3^n = 3^1
n = 1
(7b)
log(x^2+4) = 2+logx - log^20
log(x^2+4) = log^100 = log^x - log^20
(x^2+4) = log(xx)
x^2+4 = 5x
x^2-5x+4 = 0
x^2-4x - x +4 = 0
x(x-4) - 1(x-4) = 0
(x-1)(x-4) = 0
x-1 = 0 or x-4 = 0
x = 1 or 4
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(9a)
Let the lens digit x and unit digit be y,
therefore x-y=5 ---(1)
3xy-(10x+y)=14 ----(2)
3xy-10x-y=14 -----(3)
frm eq(1); x=5+y --- (4)
therefore, 5(5+y)(y)-10(5+y)-y=14
(15+3y)y-50-10y-y=14
3y^2+4y-50-14=0
3y^2+4y-64=0
3y^2 -12y + 16y-64=0
3y(y-4)+16(y-4)=0
(3y+16)(y-4)=0
y=-16/3 or 4
therefore from eqn(1);
x+4=5
x=5+4=9
the number is 94
(9b)
(3-2x)/ 4 + (2x-3)/3
(3(3-2x)+4(2x-3))/12
(9-6x+8x-12)/12
=(2x-3)/12

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