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

`Real & Confirmed WAEC GCE Mathematics Questions and Expo Answer - Jan/Feb 2018`

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

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`MATHS OBJ:1-10: ADCAACCBBB11-20: BACBBBCABB21-30: ABDBBCBCCB31-40: BAADBDDBAA41-50: DABDDBBDAC`

`(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`

`1198/63=19^1/63`

`(1b)Sin 48 =x / 250X =250 sin 48 degreesX = 250 * 0 . 7431X =185 . 7775 m=186 m`

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`(2a)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=-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======================(3)CIRCUMFERENCE OF TWO SEMI CIRCLES* =PIEd22 / 7 X 120= * 377 . 142 *2 ( 377 . 142 + 60)=874 . 29 Km(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=======================(4a)Rate = 2/100 * N0.02 per monthRate per annum = 0.02 * 12 = 0.24 per annum`

`(4b) Draw the Diagram 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%========================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`

`9bii)(RV)²=(OR)²+(OV)²==>32²=(10.6065)²+(OV)²1024=112•4978+(OV)²1024-112•4978=(OV)²(OV)²=911•50215OV= √911•50215OV=height=30•1911≈Height = 30•2cm(ii)Volume=⅓×base area×height=⅓×15×15×30•1911=2,264•33≈2,264cm³===========================(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`

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`GOODLUCK!!!`

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