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Neco 2016 Mathematics free Answers

Neco 2016 Mathematics free Answers

VERIFIED NECO 2016 MATHEMATICS QUESTIONS AND ANSWERS

Wednesday 22nd June.
Paper III: Objective – General Mathematics 10:00am – 11:45am

Paper II: Essay – General Mathematics 12:00noon – 2:30pm

1-10 deacdacbbe
11-20 cbaeebdacc
21-30 deedacbbce
31-40 abdcebaeec
41-50 caadedbaea
51-60 edeacbacdd

1a)
Tabulate
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)
I = PRT/100,
p=N15000
R=10% and I=3years
A = P I where I = 15000*10*3/100=N4500
A=4500 15000 =N19500
2a)
using sine rule
b/sin20 = 6/sin30
bsin30 = 6sin120
b 6sin120/sin30
b = 6×0.2511/0.4540
b = 5.7063/0.4540
b = 12.57 ≠ 12.6cm
2bi)
the diagram is euivalent triangles.
where
|AX|/|BC| = |BY|/|AC| = |XY|/|YC|
XY = 9, BY = 7
YC = 18-7=11
9/11 = 7/|AC|
9|AC| = 77
|AC| = 77/9
|AC| = 8cm
2bii)
XY/AB = BY/AC
9/|AB| = 7/8.6
|AB| = 9×8.6/7
|AB| = 11cm
3)
let the son age be x
man=5x
son=x
4yrs ago;the man age = 5x – 4
the son age = x – 4
the product of their ages
(5x – 4)(x – 4) =448
4a)
volume of fuel = cross-sectional area of X depth of fuel rectangular tank
30,000litres = 7.5*4.2*d m^3
but; 1000litres =1m^3
therefore;30(M^3) = 7.5*4.2*d(M^3)
30=31.5d
====> d = 30/31.5 = 0.95(2d.p)
4b)
to fill the tank/volume of fuel needed = 7.5*4.2*1.2 = 37.8m^3 = 37,800
litres addition fuel = 37,800-30,000 = 7,800
litres therefore, 7,800
more litres would be needed 
================ 
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
5b)
amount spent=144000- [48,000 32000 19200 12000 7200] =144000-118400 =N25600 
===============
7a)
3²ⁿ ¹ — 4(3ⁿ ¹) 9 = 0
3²ⁿ × 3 — 4(3ⁿ× 3¹) 9 = 0
(3ⁿ)² × 3 — 4(30ⁿ× ) 9 = 0
Let 3ⁿ= x
3x² — 4 × 3 × x 9 = 0
3x² — 12x 9 = 0
Divide all by 3
3x²/3 — 12x/3 9/3 = 0
x² — 4x 3 = 0
x² — 3x — x 3 = 0
x(x—3) -1(x—3) = 0
(x—3)(x—3) = 0
x—3 = 0 or x—1 = 0
x = 3, x = 1
Substitute x = 3
3ⁿ = 3 or 3ⁿ = 1
3ⁿ = 3¹ or 3ⁿ= 3°
n = 1 or n= 0
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

8)
|BC|² = |BD| + |CD|²
13² = BD² + 5²
169 = BD² + 25
√BD² = √144
BD= 12m
Both OB = OD and OB + OD
= 2OB/2 = 12cm/2
OB = 6cm

8b) Circumference = 2πr
r = 6cm, π = 22/7
= 22 × 2 × 6/7cm
= 264/7cm
= 37.7cm to 1 decimal places
=================

9a) Let the digits be y
10(5 + y)+y =3y(5 +y) —14
50 + 10y + y = 3y(5+y) —14
50 + 11y = 15y + 3y² — 14
3y² + 15y — 11y — 50 — 14 = 0
3y² + 4y — 64 = 0
(3y² — 12y) + (16y — 64) = 0
3y(y — 4)(3y + 16) = 0
y —4 = 0 or 3y + 16 = 0
y = 4 or —16/3

9b)
3—2x/4 + 2x—3″3
= 3(3—2x) + 4(2x—3)/12
= 9—6x + 8x—12/12
= 2x—3/12

10a)
y=(2x^2 + 3)^5
let U=2x^2 + 3
Y=u^5
du/dx = 4x
dy/du = 5u^4
dy/du = (2x^2 + 3)^4
dy/dx = du/dx dy/du
dy/dx = 4x.5(2x^2 + 3)^4
dy/dx = 20x(2x^2 + 3)^4

10b)
y=3x^2 + 2x +5
dy/dx =6x + 2
dy/dx =6(3) +2
dy/dx =18+2
dy/dx =20

10c)
R-W=Wv^2/gx
Wv^2=gx(R-W)
Wv^2=gRx-Wgx
Wv^2+Wgx=gRx
W(v^2 + gx) =gRx
W=gRx/V^2 + gx
R=2, g=10, x=3/2, V=3
W= 10*2*3/2/3^2 + 10*3/5
W=30/9+15
W=30/24
W=5/4

Posted by on June 22, 2016.

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