# WAEC GCE 2018 Mathematics Obj And Essay Answer – Jan/Feb Expo

## MATHS OBJ:1-10=ACCAACDBBB11-20=BCABABCABB21-30=DCCCBCBCCA31-40=BDADBBDBAA41-50=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 = axHence y+2/a = ax/aX = y+2/a 11bi)Bibiani = 600Amenji = 700Oda = 1800Wawso = 1500Sankose=2400Total = 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 = \$336,000Revenue received by Oda = 1800×560 = \$1,008,000Oda will receive(1,008,000 – 336000) = \$672,000 more than Bibiani 4a)Rate = 2/100 * N0.02 per monthRate per annum = 0.02 * 12 = 0.24 per annum 4b) Draw the Diagram 3a)[Diagram]Distance covered by an athlete = Perimeter of A + Perimeter of rectangle CDEF + perimeter of BPerimeter 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 = 188.57mPerimeter of rectangle CDEF= 2(L + B)L = 120m; B = 60mPerimeter = 2(120+60) = 2(180)=360mDistance covered by an athlete = 188.57 + 360 + 188.57=737.14mIf the athlete runs the track two times = 2 × 737.14= 1474.28m 3b)If the athlete spends 200seconds for the raceSpeed = distance/timeDistance = 1474.28mTime = 200secondDistance = 1474.28m = 1.47428kmTime= 200seconds = 3.3333hrsSpeed = 1.47428/3.3333 = 0.44kmhr-1 6a)Tanx = 5/12Using the diagramSinx = 5/13Cosx = 12/13Sinx/(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=450PR=sqr root 225×2PR=15root2cmBut 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 sale35 × 100 = 70y35 × 100/70 = 70/70Y = 350/7 = 50Hence price after first sale = GHc50But GHc50 = 60% ie (100-40)%Therefore GHcx = 100% first reduction sale100 × 50/60 = 60x/60X=> 500/60 = GHc83.33=>GHc83.3Hence price before the first sales = GHc83.33 7b)Initil price of article = GHc = 180.00In the first sales, reduction = 40%i.e 100% – GHc 18.0040% – GHc x100x/100 = 40*180/100.:. x = 4*18 = GHc 72.00Since reduction in the first sale is GHc 72.00Then reduction in the second = 30%100% = GHc 10830% = y100y/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 *20/3599*2/631198/69=19^1/63 1b)Sin 48=x/250X=250 sin 48 degreesX= 250 * 0.7431X=185.7775m=186m 2)Let musa’s age=x.Manya’s age=y.x-y=3———(1)Also x=3+y——(2) 7years agoMusa’s age=x-7Manya’s age=y-7x-7=2(y-7)x-7=2y-14x-2y=-14+7x-2y=-7——-eqn(3) Put eqn(2) into eqn(3)3+y-2y=-7-y=-7-3-y=-10Y=10 But x=3+10=====>x=13Also 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) = 4510+10+3+2y = 4523+2y = 452y = 45-232y = 22Y = 22/2Y = 11yearsThe sum of their ages will be 45 after 11 years

=========================
042tvseries.Com send all exam expo answer earlier than others
============================