2024 NABTEB GCE MATHEMATICS: Nabteb Gce 2024. Maths answers (1533)
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OBJ
1-10: DABBABBBAD
11-20: CDBDCCCBBA
21-30: CDABDACCCC
31-40: BCBCABABBC
41-50: DACCCCDBDB
Welcome to official 2024 Mathematics NABTEB GCE answer page. We provide 2024 Mathematics NABTEB GCE Questions and Answers on Essay, Theory, OBJ midnight before the exam, this is verified & correct NABTEB GCE Maths Expo. NABTEB GCE Mathematics Questions and Answers 2024. NABTEB GCE Maths Expo for Theory & Objective (OBJ) PDF: verified & correct expo Solved Solutions, Nabteb Gce 2024. Maths answers . 2024 NABTEB GCE EXAM Mathematics Questions and Answers
*SECTION A*
*_(Answer ALL QUESTIONS From This Section)_*
(1a)
6^n+1*9^n*4^2n/18^n*2^n*12^2n
=(2*3)^n+1*(3*3)^n*(2*2)^2n/(2*3*3)^n*2^n*(2*2*3)^2n
=2^n+1*2^n+1*3^n*3^n*2^2n*2^2n/2^n*3^n*3^n*2^n*2^n*2^n*3^2n
= 2^n*2^1*2^n*2^1*3^n*3^n*2^n*2^n*2^n*2^n/2^n*3^n*3^n*2^n*2^n*2^n*2^n*2^n*3^n*3^n
= 2^6n*3^2n*2*2/2^6n*3^4n
= (2^6n / 2^6n) *(3^2n / 3^4n)*2*2
=2^6n - 6n * 3^2n - 4n * 4
= 2^0 * 3^-2n * 4
= 4*3^-2n
= 4 * 1/3^2n
= 4/3^2n
(1b)
Log 243/log 27
= log 3^5/log 3^3
= 5log3/3log3 = 5/3 = 1 whole no 2/3
============================================
(2a)
3t - 2p = 8Γ2 ...(I)
2t - 3p = 14Γ2...(II)
6t - 4p = 16 ...(III)
6t - 9p = 42 ...(IV)
9P - 4P = -42 + 16
5p = -26
P = -26/5 = -5 1/5
Put p = -26/5 into eqn (I)
3t - 2(-26/5) = 8
3t + 52/5 = 8
3t = 8 - 52/5
3t = 40 - 52/5 = -12/5
Divide both sides by 3
36/3 = -12/5 / 3
t = -12/5 Γ 1/3
= -12/15 = -4/5
(2b)
3/β3(2/β3 - β12/6)
2/β3 Γ 3/β3 - 3/β3 Γ β12/6
6/β9 - 3β12/6/β3
6/3 - 3β12/6β3
2 - β12/2β3
2 Γ 2β3 - β12/2β3
4β3 - (β4 Γ 3)/2β3
= 4β3 - 2β3/2β3
= 2β3/2β3 = 1
============================================
(3)
P(k) = 2/3
P(Y) = 5/8
P(I) = 3/4
P(k fail) = 1- 2/3 = 1/3
P(Y fail) 1 - 5/8 = 3/8
P(I fail) = 1 - 3/4 = 1/4
(a) 2/3*5/8*3/4 = 5/16
(b) 1/3*3/8*1/4 = 1/32
(c) 2/3*5/8*1*4 = 5/48
==============================================
(4)
Draw the triangle
<GFN = 180 - ( 68+90)
Sum of angles in a Ξ
= 180 - 158 = 22
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