100% VERIFIED WAEC GCE MATHEMATICS ESSAY AND OBJ ANSWER

8a) By using pythagoras r^2 = (r - 8)^2 + 32^2 r^2 = r^2 - 16r + 64 + 1024 16r = 64 + 1024 r = 1088/16 r = 68 r=68cm 8bi) {st/pt/ q2=27^2 +12^2 q=sqrt27^2 + 12^2 q=sqrt729+144 q=sqrt873 q=29.5cm ================================== 7a) |PQ|^2 = |PB|^2 + |BQ|^2 |PQ|^2 = (5 - x )^2 + X^2 |PQ|^2 = 25 - 10x + x^2 + x^2 [note ^ means Raise to power] |PQ|^2 = 2x^2 - 10x + 25 NB: PQ = QR (side of a square) Area of PQRS = PQ x QR = PQ x PQ = PQ^2 Area of PQRS = (2x^2 - 10x + 25)m square GIVEN: 2x^2 - 10 + 25 = 3/5 of 25 = 2x^2 - 10x + 10 = 0 = x^2 - 5 + 5 = 0 Using General Formular mthod X = -(-5) ± Square Root (-5)^2 - 4(1)(5)/ (2 x 1) X = 5 ± Square Root 25 - 20/ 2 X = 5 ± Square Root 5 / 2 X = 5 + Square Root 5 / 2 OR 5 - Square Root 5 / 2 X = 3.62 or 1.68 7b ) L + a / n - 1 = d ------------ (I) 2s = n (a + 1) ---------------- (2) From eq (1) L + a = d (n - 1) -------------- (3) Put eq (3 into 2) 2s = nd (n-1) 2s = d (n^2 - n) S = d ( n^2 - n ) / 2 or S = dn(n-1)/2 ================================== 6a) 2 X (37)x = 75x 2 X (3x + 7)10 = (7x + 5)10 2 (3x + 7) = 7x + 5 6x + 14 = 7x + 5 14 - 5 = 7x - 6x 9 = x :- x = 9 6b) Let The Number Of boys be X Number Of girls in class = x + 5 GIVEN: x+5/+2 = 5/4 5(x+2) = 4(x+5) 5x + 10 = 4x + 20 5x - 4x = 20 - 10 x = 10 i)Numbers of girsl in class = x + 5 = 10 + 5 = 15 ii)Total number of students = x + x + 5 = 10 + 10 + 5 = 25 iii)Probability of selecting a boy as class prefect = 10/24 = 2/5 (4a) Given: Sinx = 3/5 i.e Using pythogoras triple 3,4,5 CosX + TanX/SinX =4/5 + 3/4 ÷ 3/5 = 16+15/20 ÷ 3/5 = 31/20 ÷ 3/5 = 31/20 x 5/3 = 31/12 (4b) =Draw The Diagram= Reflex P'Q'T + R'Q'T + P'Q'R = 360 (sum at a point) 200 + 32 + P'Q'R = 360 232 + P'Q'R = 360 P'Q'R = 360 - 232 P'Q'R = 128° From the Diagram Q'R'U = P'Q'R = 128° (alternate angles) X = Q'R'U + S'R'U X = 128° + 180 X = 308° ================================== 5a) 1. | 2. | 3. | 4. | 5. | 6. | ------------------------------------------ 1|1,1|1,2 |1,3 |1,4 |1,5 |1,6 | ------------------------------------------- 2|2,1|2,2 |2,3 |2,4 |2,5|2,6 | -------------------------------------------- 3|3,1|3,2 |3,3 |3,4 |3,5|3,6 | --------------------------------------------- 4|4,1|4,2 |4,3 |4,4 |4,5|4,6 | --------------------------------------------- 5|5,1|5,2 |5,3 |5,4 |5,5|5,6 | ----------------------------------------------- 6|6,1|6,2 |6,3 |6,4 |6,5|6,6 | ------------------------------------------------ B) pr(sum of outcome is 8) = 4/36 = 1/9 Bii) pr(product of outcom <10) = 30/36 = 5/6 Biii) pr(outcom contain atleast a 3) = 24/36 = 4/6 = 2/3 ================================== (3) =Draw The Diagram= Area of path --> 2[1/2 (x+2)] + 2(x+1) = 17 x + 2 + 2x = 17 3x + 2 = 17 3x = 17 - 2 3x = 15 x = 15/3 x = 5 (3ai) Perimeter of garden = 4x = 4 x 5 = 20m (3aii) =Draw The Diagram= Area Covered By Both Garden And path = (x+1) (x+2) = (5+1) (5+2) = 6x7 = 42m^2 ================================== 2a) 1/x+1/x+3=1/2 L.C.M=x(x+3) X+3+x/x(x+3)=1/2 2(2x+3)/x(x+3)=1/2 2(2x+3)=x(x+3) 4x+6=x^2+3x X^2+3x=4x+6 X^2+3x-4x-6=0 X^2-x-6=0 (X^2-3x)+(2x-6)=0 X(x-3)+2(x-3)=0 (X+2)(x-3)=0 X=-2 or x=3 2b) Let d bag of rice be X Let d bag of beans be Y X+Y=17--->(I) 2250x+2400y=39600--->(2) From eqtn (I) X=17-y Using elimination method to eliminate Y 2250x+2250y=38250--->(3) 2250x+2400y=39600--->(4) Eqtn 4- eqtn 3 2400x-2250x = 39600-38250 150x=1350 X=1350/150 X=9 D trader bought 9bags of beans ================================== 1a) 1/2log 25/4- 2log10 4/5+log10 320/125 Log(25/4)^1/2-log(4/5)^2+log10 320/125 Log10 sqrt25/4-log16/25+log10 320/125 Log10 5/2-log10 16/25+log10 320/125 Log10 5/2+log10 320/125-log10 16/25 Log10 [5/2*320/125÷16/25] Log10 [5/2*320/125*25/16] Log10 10=1 1b) % income= 20% Grant per land =GHC€15.00 Total population from 2003-2007 =1.2*1.2*1.2*1.2*3000=6220.8 Total grant=population*grant per head =6220.8*15 =ghc€93312 Total grant=GHC€93312 =============keep refreshing this site