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Solution Centre for All Examinations both WAEC/WASSCE/SSCE, NECO, NABTEB, JAMB (UTME), Post-UTME, WAEC GCE, NECO GCE, NABTEB GCE etc

**Mathematics OBJ: 1-10: CABDBADCBC 11-20: BCBCBBAACD 21-30: DCCBCCABBA 31-40: AABDDCDDCC 41-50: BBCDCCBCCD**

Section A:

Section B:

(1a)

110ₓ = 40₅

change all to base ten.

110ₓ = 40₅

1x²+1x ¹+0x⁰ = 45¹+0*5⁰

x² + x + 0 = 20

x² + x – 20 = 0

Solve quadratically

x² + 5x – 4x – 20 = 0

x(x+5) – 4(x+5) = 0

(x-4)(x+5) = 0

Either x-4 = 0 or x + 5 = 0

x = -5 or 4

therefore, x= 4

(1b)

15/√75 + √108 + √432

= 15/√253 + √363 + √144*3

= 15/5√3 + 6√3 + 12√3

= 3/√3 + 18√3

= 3√3/3 = 18√3

= √3 + 18√3

=19√3

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(2a)

y2-y1/x2-x1 = y-y1/x-x1

Given : A(-2,7) and B(2,-3)

-3-7/2+2 = y-7/x+2

-10/4=y-7/x+2

4(y-7)= -10(x+2)

4y-28= -10x-20

4y= -10x+28-20

4y= -10x+8

2y= -5x+4

OR

5x+2y=4

(2b)

5b-a/8b+3a=1/5

(5b-a)/b/(8b+3a)/b =1/5

5-(a/b)/8+3(a/b) =1/5

8+3(a/b)= 5 [ 5-(a/b)]

8+3(a/b)=25-5(a/b)

3(a/b) + 5(a/b)=25-8

8(a/b)=17

a/b=17/8

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(3a)

Total ratio =3+5+8=16

Ali’s share =3/16 * 420,000/1

=210,000

Son’s of Ali and Yusuf’s share = N78,750+ N210,000

= N288,750

(3b)

2(1/8)^x=32^x-1

2^1(2^-3)^x=2^5(x-1)

2^1 * 2^-3x=2^5(x-1)

2^-3x+1=2^5(x-1)

-3x+1=5(x-1)

-3x+1=5x-5

-5x-3x= -5-1

-8x/-8= -6/-8

x=3/4

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(4)

Since <PQR = <PRS = 90°

Using Pythagoras theorem

|PR|² = |PQ|² + |QR|²

|PR|² = 3² + 4²

|PR|² = 9 + 16

|PR|² = 25 PR = √25

|PR| = 5cm

Considering PRS

|PS|² = |PR|²+|SR|²

13² = 5² + |SR|²

169 = 25 + |SR|²

|SR|² = 169 – 25

|SR|² = 144

|SR| = √144 = 12cm

Hence the area of the quadrilateral = Area of triangle PQR + area of PRS

= 1/2bh + 1/2bh

= 1/2×4×3 + 1/2×12×5

= 6+30 = 36cm

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(5a)

No of red balls = 3

No of green balls = 5

No of blue balls = x

Prob.(red ball) = no of total outcome/no of possible outcome

Pr(red) = 3/3+5+x = 1/6

3/8+x = 1/6

6(3) = 1(8+x)

18 = 8 + x

X = 18 – 8 = 10

Therefore the no of blue ball = 10

(5b)

Probability of picking a green ball

P(g) = no of green balls/no of possible outcome

P(g) = 5/3+5+10 = 5/18

=5/18

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(6ai)

F α M1M2/d²

F = KM1M2/d²

Given F = 20N, M1= 25kg, M2 = 10kg and d = 5m

20 = k(25)(10)/5²

250k = 500

k = 500/250 = 2

Expression is

F = 2M1M2/d²

(6aii)

Making d subject

d = √2M1M2/F

d = √2 ×7.5×4/30

d = √60/30 = √2

d = √2m or 1.41m

(6b)

Draw the diagram

X+X+60+X+80+X+40+X+20 = 540(sum of angles in a Pentagon)

5x + 200 = 540

5x = 540 – 200

5x = 340

X = 340/5

X = 68

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(8a)

1/3x – 1/4(x+2)>_ 3x -1⅓

1/3x – 1/4(x+2)>_3x – 4/3

Multiply through by the L. C. M(12), we have

4x – 3(x + 2)>_36x – 16

4x – 3x – 6 >_ 36x – 16

-6+16 >_36x + 3x – 4x

10 >_ 35x

35x _< 10

X = 10/35

X = 2/7

(8bi)

Draw the triangle

|AB|/66 = sin35

|AB| = 66sin35 = 66×0.5736 = 37.8576

Draw the right angled triangle

|AD|/|AB| = Tan52

|AD| = 37.8576 × Tan52° = 37.8576 × 1.2799 = 48.45m

Height of tower = 48.45m

(8bii)

|AC|/66 = Cos35°

|AC| = 66 x cos35°

= 66 x 0.8192

= 54.0672

Tan = 41.86°

Angle of elevation of top of tower from c = 41.85°

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(10)

130kg of tomatoes for #52,000

Half of the tomatoes

130/2 = 65kg sold at 30%

Profit = #52,000/2 = 26,000

#26,000 = 100%

X = 130%

X = 26000 × 130/100

= #33,800

Then 65kg was then sold at reduction of 12% per kg

Recall that the initial cost price = 52000/130

=400kg

65kg sold at = 33,000/65

=#520/kg

Then for 12% reduction

520 × 88/100 = 457.6/kg

(a)

The new selling price per kg = #457.6/kg

(b) 65kg – 5kg = 60kg

(60kg×457.6kg)+33,800

= #61,256.00

#profit = selling price /cost price × 1000/1

=61256/52000×100/1= 117.8

= 17.8%

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(11ai)

ar² = 1/4 ……(1)

ar^5= 1/32 …..(2)

Divide eqn (2) by eqn(1)

ar^5/ar² = 1/32÷1/4

r³ = 1/32 × 4/1

r³= 1/8

r³ = 2-³

r = 2-¹

r = 1/2

Common ratio = 1/2

Put this into eqn (1)

a(1/2)² = 1/4

a(1/4) = 1/4

a = (1/4)/(1/4) = 1

First term, a = 1

(11aii)

Seventh term, T7 = ar^6

=(1)(1/2)^6

=1/64

(11b)

Given : X = 2 and X = -3

(X – 2)(X + 3) = 0

X² + 3x – 2x – 6 , 0

X² + x – 6 = 0

Comparing with ax²+bx+c = 0

a = 1

b = 1

C = -6

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(12a)

Given : siny = 8/17

Draw the right angle

From Pythagorean triple, third side is 15

Draw the right angle triangle

tan y = 8/15

tan y/1+2tany = 8/15/1+2(8/15) = 8/15/1+16/15

tany /1+2tan y = 8/31

(12b)

Amount shared = #300,000

Otobo’s share = #60,000

Ade’s share = 5/12 × #(300,000-60,000)

= 5/12 × #240,000

=#100,000

Adeobi’s share = #300,000 – (#60,000 + #100,000)

= 300,000 – 160,000

=#140,000

Ratio : Otobo : Ade : Adeola

60,000 : 100,000 : 140,000

60 : 100 : 140

6 : 10 : 14

3 : 5 : 7

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(13a)

RT^0=RS^0=90° (radius o targets)

T^0S=2TU^0 (angle at centre = 2 time angle at circumstance)

TOs=2*68°=136°

Now RT^0+RS^0+T^0S+SRT^0=360°(sum of angle in a quadrilateral)

90° + 90° + 136° + x = 360°

X+316°=360°

X=360°-316

X=44°

(13b)

Let tank B hold x litres

; Tank A hold (x+600)literally

3(x-100)=(x+600-100)

3x-300=x+500

3x-x=500+300

2x=800

X=800/2=400

Tank B holds 400 litres

Tank A holds (400+600)=1000litres

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