WAEC GCE Mathematics Questions and Answers for 2019/2020 Exam (Step-by-Step)

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WAEC GCE Mathematics Questions and Answers for 2019/2020 Exam (Step-by-Step) 1

WAEC GCE Mathematics Questions and Answers for 2019/2020 Exam

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EXAMPLE OF 2019 WAEC GCE MATHEMATICS QUESTIONS AND ANSWERS THAT WAS SERVED IN 2018.

The answers was past waec gce answers. pls don’t use it. 

(11a)
Loga(y + 2) = 1 + LogaX
=> Log^y a + Log^2 a = Log^a a + Log^x a

Loga^(y + 2) = Loga^(ax)

Y + 2 = ax
Hence y+2/a = ax/a
X = y+2/a

(11bi)
Bibiani = 600
Amenji = 700
Oda = 1800
Wawso = 1500
Sankose=2400
Total = 7200

Bibiani = 600/7200 × 360/1 = 30°
Amenji = 700/7200 × 360/1 = 45°
Oda = 1800/7200× 360/1 = 90°
Wawso = 1500/7200× 360 = 75°
Sankose = 2400/7200× 360/1 = 120°
Total = 30°+45°+90°+75°+120° = 360°

(11bii)
% of timber produced from Amenji = 900/7200 × 100/1 = 12.5%

(11biii)
Revenue received by Bibiani = 600×$560 = $336,000
Revenue received by Oda = 1800×560 = $1,008,000
Oda will receive(1,008,000 – 336000) = $672,000 more than Bibiani

==========================

(4a)
Rate = 2/100 * N0.02 per month
Rate per annum = 0.02 * 12 = 0.24 per annum

(4b) Draw the Diagram

=======================

No 1
1/4 * 9 1/7 + 2/5 [2/3 + 3/4] / (2/5 – 1/4)

(1/4 * 64/7 + 2/5)[17/12)] /8-20/20]

16/7 +2/5(17/2) *[20/3

(16/7 +1/5 *17/6)*20/3

(16/7+17/30)*20/3

(16*30+17*7 /210)*20/3

(480+119/210)*20/3 599/210 *20/3

599*2/63

1198/69
=19^1/63

1b)
Sin 48=x/250
X=250 sin 48 degrees
X= 250 * 0.7431
X=185.7775m
=186m

======================

2)
Let musa’s age=x.
Manya’s age=y.
x-y=3———(1)
Also x=3+y——(2)

7years ago
Musa’s age=x-7
Manya’s age=y-7
x-7=2(y-7)
x-7=2y-14
x-2y=-14+7
x-2y=-7——-eqn(3)

Put eqn(2) into eqn(3)
3+y-2y=-7
-y=-7-3
-y=-10
Y=10

But x=3+10=====>x=13
Also therefore Musa’s age is x =13,
And Manya’s age is y=10

====================

(3a)
[Diagram]
Distance covered by an athlete = Perimeter of A + Perimeter of rectangle CDEF + perimeter of B
Perimeter of A = 2πr/2 = π =22/7, r = d/2 = 120/2 = 60m
= 22/7 × 60 = 1320/7 = 188.57m

Perimeter of B = perimeter of A = 188.57m
Perimeter of rectangle CDEF= 2(L + B)
L = 120m; B = 60m
Perimeter = 2(120+60) = 2(180)
=360m
Distance covered by an athlete = 188.57 + 360 + 188.57
=737.14m
If the athlete runs the track two times = 2 × 737.14
= 1474.28m

(3b)
If the athlete spends 200seconds for the race
Speed = distance/time
Distance = 1474.28m
Time = 200second
Distance = 1474.28m = 1.47428km
Time= 200seconds = 3.3333hrs
Speed = 1.47428/3.3333 = 0.44kmhr-1

===================

(7a)
Reduction in the first sales = 40%
Reduction in the second sales = 30%
Price sold Ghc 3500 = 70% ie (100 – 30)%
GHc y = 100% second reduction sale
35 × 100 = 70y
35 × 100/70 = 70/70
Y = 350/7 = 50
Hence price after first sale = GHc50
But GHc50 = 60% ie (100-40)%
Therefore GHcx = 100% first reduction sale
100 × 50/60 = 60x/60
X=> 500/60 = GHc83.33
=>GHc83.3
Hence price before the first sales = GHc83.33

(7b)
Initil price of article = GHc = 180.00
In the first sales, reduction = 40%
i.e 100% – GHc 18.00
40% – GHc x
100x/100 = 40*180/100
.:. x = 4*18 = GHc 72.00
Since reduction in the first sale is GHc 72.00
Then reduction in the second = 30%
100% = GHc 108
30% = y
100y/100 = 30*108/100 = 324/10 = GHc 32.4

(i) Hence reduction in the price due to the two sales = (72+32.4)GHc = GHc 104.4

(ii) % reduction = Reduction/Original price * 100/1
=104.4/180 * 100/1 = 58%

======================

(9a)
Let the distance which the donckey can reach = x
Hence 17² – 15² = x²
289 – 225 = x²
64 = x²
√64 = √x²
8 = x
But since the donkey can move both sides, the lenght of the fence = 8*2 = 16m

(9bi)
Draw the diagram

(9bii)
(I)
(RV)²=(OR)²+(OV)²
==>32²=(10.6065)²+(OV)²
1024=112.4978+(OV)²
1024-112.4978=(OV)²
(OV)²=911.50215
OV= √911.50215
OV=height=30.1911
Height = 30.2cm

(II)
Volume=⅓×base area×height
=⅓×15×15×30•1911
=2,264•33≈
2,264cm³
======================
(13a)

1/3 [3] + [6] + [P] = [3]

-6    -2     q     4
[1] + [6] + [P] = [3]

-2    -2     q     4
1+6+p = 3

7 + p = 3

P = -4
-2 -2 + q = 4

-4 + q = 4

q = 4+4 = 8

T =    [-4]

[ 8]

(13b)

2[3] + n[1] = [8]

m      2     4
[6] +  [n] = [8]

2m      2     4
Hence 6 +n = 8 => n = 8-6

n = 2

Also

2m + 2n = 4

2m + 2(2) = 4

2m = 4 – 4

2m = 0

m = 0/2 = 0

Hence m = 0 , n = 2

(13c)

(Y+4)*y = 12

(Y+4+1)*y = 12

(Y+5)*y = 12

Y+5+Y+1 = 12

2y+6 = 12

2y = 12-6

2y = 6

Y = 6/2

Y = 3

Updated: July 21, 2019 — 2:58 pm

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