NABTEB Mathematics Answer - May/June 2017 Expo
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NABTEB General Mathematics Obj And Theory/Essay Questions and Answer - May/June 2017 Expo Runz English Answer is ready now. Start subscribe...

NABTEB General Mathematics Obj And Theory/Essay Questions and Answer - May/June 2017 Expo Runz

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*2017 NABTEB VERIFIED MATHS ANSWERS*

*=====================*

*MATHS OBJ *

*=====================*

*VERIFIED NABTEB MATHS*

*OBJ:*

*1-10=DBDCDABBCD*

*11-20=DCDBDCDAAB*

*21-30=BCDAADCACD*

*31-40=DABABCBDCD*

*41-50=ABBBAABDAD*

*COMPLETED*

*=======================*

*VERIFIED-MATHS-THEORY ANSWERS*

*=======================*

*INSTRUCTION:ANSWER questions 1 to 5 and any other four questions *

*MATHS THEORY*

*======================*

*SECTION A(ANSWER ALL QUESTION)*

*QUESTION 1 -5)*

*======================*

*1a)*

*1 4/5 × 2 1/3 / 3 3/4 – 4/5 × 2/3*

*=9/5 × 7/3 / 18/5 – 4/5 × 2/3*

*=21/5 / 18 – 4/5 × 2/3*

*=(21/5 ÷ 14/5) × 2/3*

*=21/5 × 5/14 × 2/3*

*=1*

*1b)*

*x^2 +5x – 6= 0*

*(x^2 + 6x) – ( x– 6) = 0*

*x(x+6) –1 ( x+6)= 0*

*(x–1) (x +6)= 0*

*x – 1 =0 or x+6 = 0*

* x=1 or x =– 6*

*=====================*

*2a)*

*U ={2,3 ,4 ,5, 6, 7,8 ,9}*

*A ={2,3 ,5 , 7}*

*B ={3,6,9 }*

*i)AUB ={2,3,5,6,7,9}*

*ii)*

*A'nB'*

*A' ={4,6, 8,9,}*

*B' ={2,4,5,7,8,}*

*A'nB' {4,8}*

*2b)*

*161n = 32less down 5*

*1×n^2+6×n^1+1×n° = 3× 5^1+2×5°*

*n^2 + 6n + 1 = 15 + 2*

*n^2 +6n + 1 =17*

*n^2 + 6n – 16 = 0*

*(n^2 +8n) – (2n – 16) = 0*

*n(n + 8) –2 (n + 8) = 0*

*(n – 2)(n + 8) = 0*

*n – 2 = 0 or n + 8 =0*

*n = 2 or n = –8*

*hence,*

*n = 2*

*=====================*

*3a)*

*1/2logy^8 = 2*

*logy^√81 =2*

*logy^9 =2*

*y^2 = 9*

*y = √9=3*

*y = 3*

*3b)*

*0.016 × 0.048 / 0.64*

*=16 ×10^-3 / 64 × 10^-2*

*=16 × 48 × 10^-6 / 64 × 10^-2*

*=768 / 64 × 10^-6 × 10^2*

*=12 × 10^-4*

*=======================*

*4)*

*h/8 = h + 20/12*

*12h = 8h + 160*

*4h =160*

*h = 160/4*

*h =40cm*

*Hence ,H =h +20=40 +20 =60cm*

*H=60cm*

*Volume of bucket =*

*1/3Ï€R^2H – 1/3Ï€r^2h*

*=1/3Ï€(R^2H – r^2h)*

*=1/3(3.142)( (12)^2(60) – (8)^2(40))*

*=1/3(3.142)(8640 – 2560)*

*=1/3 × 3.142 × 6080*

*volume is =6367.7cm^3*

*since ,*

*1litre = 1000cm^3*

*Capacity =6367.7 / 1000*

*=6.3677litre*

*=6.4litre*

*=====================*

*5a)*

*2/3(x – 2) – x –1/x–2*

*=2–3(x –1)/3(x–2)*

*=2 – 3x + 3/5x – 6*

*=5–3x/3x–6*

*5b)*

*x^2 + 3x + 2/ x^2 – 4*

*=(x^2 + x) + (2x + 2)/(x+2)+(x –2)*

*=x(x+1) +2(x+1)/(x+2) +(x – 2)*

*= (x+2) (x – 2)/(x+2)(x –2)*

*= x+1/x –2*

*====================*

*SECTION B*

*ANSWER ONLY FOUR QUESIONS*

*QUESTION 8,9,10,11*

*POSTED*

* ===================*

*8a)*

*if (x – 6), 2x and (8x – 20) are consecutive terms of G.P*

*the common ratio is *

*r = 2x / x– 2 --------(1)*

*r = 8x + 20 / 2x --------(2)*

*Equating (1) and (2)*

*2x/x – 6 = 8x + 20/2x*

*4x^2 = (x – 6) (8x + 20)*

*4x^2 = 8x^2 + 20x – 48x – 120*

*4x^2 – 28x – 120 = 0*

*x^2 – 7x – 30 = 0*

*Solving quadratically,*

*(x^2 + 3x) – (10x – 30) = 0*

*x(x + 3) –10 (x + 3) = 0*

*(x - 20) (x + 3) = 0*

*x – 10 =0 or x + 3 = 0*

*x = 10 or x = -3*

*8b)*

*√72 × 3√18 × 14√6 / 2√24 ×√12 √36×2 × √9×3 × 24√6 /2√4×6 × √4×3*

*6√2 x 9√2 × 14√16 / 4√6 × 2√3*

*6 × 9 × 2 × 14 √6 / 4√6 × 2√3*

*3 × 9 × 9 / √3*

*by rationalizing the denomenator*

*= 189/√3 × √3/√3*

*=189√3 / 3*

*=63√3*

*===================*

*9a)*

*x =30° <base angle of issoceless triangle is equal >*

*y =Î¸ = 180 – 60*

*Î¸=120° , r = 5cm*

*the length of the chord AC*

*L =2rsinÎ¸/2*

*=2 × 5 × sin120/2*

*=10sin60*

*=10 × 0.866*

*Length of chord is =8.66cm*

*9aii)*

*Area of shaded segment = area of sector – Area ot triangel*

*Î¸/360 × Ï€r^2 – 1/2 (5)^2sinÎ¸*

*=120/360 × 3.142 × (5)^2sin120*

*=120 × 3.142 × 25 /360 – 25 × sin120/2*

*=26.18 – 10.825*

*=15.358cm^2*

*9b)*

*Speed = Distance / time*

*Distance = 500 × 2*

*=1000km*

*x^2 =(1000)^2 + (450)^2 -2(1000)(450)cos120*

*=1000000 + 202500 +450,000*

*x^2 =1652500*

*x =√1652500*

*x = 1285.5km*

*The bearing the airport (Î¸)*

*450/sinÎ¸ = 1285.5/sin120*

*sinÎ¸ = 450 × sin120 / 1285.5*

*sinÎ¸ =0.3032*

*Î¸ = sin(0.3032)= 17.6°*

*=18°*

*10a)*

*blue marble = 3*

*white marble = 2*

*Red marble = 4/9*

*i)*

*Pr (both of them will be red) *

*first drawn = 4/9*

*second drawn = 3/8*

*Pr (both red) =4/9 × 3/8 *

*=1/8*

*ii)*

*Pr (the two are of the same color)*

*= RR or BB or WW*

*=(4/9 × 3/8)+ (3/9 × 2/8)+ (2/9×1/8)*

*=12/72 + 6/72 + 2/72*

*=12+6+2 / 72*

*=20/72*

*=5/18*

*10b)*

*sine(2Î¸ – 30°) and (3Î¸ – 45)are supplementary that there sum is 180°*

*2Î¸ – 30 + 3Î¸–45=180*

*5Î¸ –75 =180*

*5Î¸=180 + 75*

*5Î¸ =225*

*Î¸=51°*

*====================*

*11a)*

*hence, volume of hemispherical portion is half of the volume of the cone*

*Volume of hemisphere=2/3Ï€r^3*

*Volume of cone =1/3Ï€r^3*

*2/3Ï€r^3 = 1/2(1/3Ï€r^2 h)*

*2/3Ï€r^3 = 1/6 Ï€r2 h*

*h = 2 × 6 × Ï€r^3 / 3Ï€r^2 = 4r*

*h = 4r = 4 × 4 = 16cm*

*hence the vertical angle is*

*tanÎ¸ = r/h*

*tanÎ¸ = 4/6 = 1/4*

*Î¸ =tan(0.25) = 14.036°*

*Î¸ =14°(correct to nearest degree)*

*11b)*

*Total volume of solid*

*=1/3Ï€r^2h (h + 2r)*

*=1/3 × 22/7 × (4)^2 (16 + 8)*

*=1/3 × 22/7 × 16 × 24*

*= 22 × 16 × 24 / 21*

*= 402.285cm^3*

*=====================*

*COMPLETED*

*====================*

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