2016 NECO Mathematics Objective and theory Answers – june/July Solutions

2016 NECO Mathematics Objective and theory Answers – june/July Solutions

Mathematics Questions and Answers

2018 NECO SSCE QUESTIONS and  Answers – june/July Expo

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

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