## WAEC GCE Mathematics Obj And Essay/Theory Solution Questions and Answer – NOV/DEC 2018 Expo Runz.

##

**MATHS OBJ:**

1-10=ACCAACDBBB

11-20=BCABABCABB

21-30=DCCCBCBCCA

31-40=BDADBBDBAA

41-50=DABCDBADAC

COMPLETED

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

<ZWY = <XWY = 180 degree(opp angles of cyclic quad. are supplimantary)

<ZWY = 100 = 180

<ZWY = 180-100

<ZWY = 100degree

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

6a)

Tanx = 5/12

Using the diagram

Sinx = 5/13

Cosx = 12/13

Sinx/(sinx)² + cosx = 5/13/(5/13)² + 12/13

= 5/13all over 25/169 + 12/13

= 5/13/25+156/169

=5/13/181/169

= 5/13 × 169/181 = 65/18

6b)

9b

(PR)²=(PS)²+(SR)²

(PR)²=15²+15²

(PR)²=225+225

(PR)²=450

PR=sqr root 225×2

PR=15root2cm

But OR=PR÷2 = 15root 2÷2

=7.5×1.4142

=10.6065

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%

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

2b)

Let the time be y

( x + y) + (x + 3 + y) = 45

(10 + y) + (10 + 3 +y) = 45

10+10+3+2y = 45

23+2y = 45

2y = 45-23

2y = 22

Y = 22/2

Y = 11years

The sum of their ages will be 45 after 11 years

**MATHS OBJ:**

1-10=ACCAACDBBB

11-20=BCABABCABB

21-30=DCCCBCBCCA

31-40=BDADBBDBAA

41-50=DABCDBADAC

COMPLETED

1-10=ACCAACDBBB

11-20=BCABABCABB

21-30=DCCCBCBCCA

31-40=BDADBBDBAA

41-50=DABCDBADAC

COMPLETED

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

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

<ZWY = <XWY = 180 degree(opp angles of cyclic quad. are supplimantary)

<ZWY = 100 = 180

<ZWY = 180-100

<ZWY = 100degree

4a)

Rate = 2/100 * N0.02 per month

Rate per annum = 0.02 * 12 = 0.24 per annum

4b) Draw the Diagram

<ZWY = <XWY = 180 degree(opp angles of cyclic quad. are supplimantary)

<ZWY = 100 = 180

<ZWY = 180-100

<ZWY = 100degree

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

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

6a)

Tanx = 5/12

Using the diagram

Sinx = 5/13

Cosx = 12/13

Sinx/(sinx)² + cosx = 5/13/(5/13)² + 12/13

= 5/13all over 25/169 + 12/13

= 5/13/25+156/169

=5/13/181/169

= 5/13 × 169/181 = 65/18

6b)

6a)

Tanx = 5/12

Using the diagram

Sinx = 5/13

Cosx = 12/13

Sinx/(sinx)² + cosx = 5/13/(5/13)² + 12/13

= 5/13all over 25/169 + 12/13

= 5/13/25+156/169

=5/13/181/169

= 5/13 × 169/181 = 65/18

6b)

9b

(PR)²=(PS)²+(SR)²

(PR)²=15²+15²

(PR)²=225+225

(PR)²=450

PR=sqr root 225×2

PR=15root2cm

But OR=PR÷2 = 15root 2÷2

=7.5×1.4142

=10.6065

9b

(PR)²=(PS)²+(SR)²

(PR)²=15²+15²

(PR)²=225+225

(PR)²=450

PR=sqr root 225×2

PR=15root2cm

But OR=PR÷2 = 15root 2÷2

=7.5×1.4142

=10.6065

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%

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%

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

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

2b)

Let the time be y

( x + y) + (x + 3 + y) = 45

(10 + y) + (10 + 3 +y) = 45

10+10+3+2y = 45

23+2y = 45

2y = 45-23

2y = 22

Y = 22/2

Y = 11years

The sum of their ages will be 45 after 11 years

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

2b)

Let the time be y

( x + y) + (x + 3 + y) = 45

(10 + y) + (10 + 3 +y) = 45

10+10+3+2y = 45

23+2y = 45

2y = 45-23

2y = 22

Y = 22/2

Y = 11years

The sum of their ages will be 45 after 11 years

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

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Verified WAEC GCE 2018 Jan/Feb Mathematics OBJ and Essay Answer and Solution to the questions.

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