2018 WAEC MAY/JUNE FURTHER MATHEMATICS OBJ AND THEORY

 WAEC FURTHER MATHEMATICS ANSWERS.

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2018 WAEC MAY/JUNE FURTHER MATHS OBJ AND THEORY

Please Note the below symbols

^ means raise to power
* means multiplication
/ means division


(1)

| x-3  -4  3 |
| 5     2   2 |       = -24
| 2   -4 6-x |

(x-3)[2(6-x) +8]+4[5(6-x) -4]+3(-20-4)  =-24
(x-3)[-2x+20]+4[-5x+26]+3(-24)= -24
-2x^2+26x-60-20x+104-72= -24
-2x^2+6x-4=0
X^2-3x+2=0
X^2-2x-x+2=0
X(x-2)-1(x-2)=0
(x-1)(x-2)=0
X-1=0 OR x-2=0
X=1 OR x=2

12)
P:F=4:1 =4x+1x=100
5x=100
x=100/5
x=20
pass=20*4=80%
fail= 20*1=20%
p(pass)=80/100=0.8
p(fail)=20/100=0.2
n=7
12ai)
P(at least 3passed)
P=0.8
Q=0.2
P(x=r)=n(rP^rq^n-r
P(x>/3)=1-P(x<2) P(x<2)=P(x=0)+P(x=1)+P(x=2)
P(x=0)=7dgree (0.8)degree (0.2)^7 P
(x=0)=0.0000128 P(x=1)=^7( (0.8)^1 (0.2)^6
=0.0003584 P(x<2)=7^C2 (0.8)^2 (0.2)^5
=0.0043008 P
(X<2)=0.0000128+0.0003584+0.004300
=0.004672 P(x>3)=1-0.004672
=0.995321
=0.10(2d.p)
12aii)
P(between 3 and 6 failed)
P=0.2
q=0.8
P(36)
P(x=3) + P (x=4)+p(x=5)+P(x=6)
p(x=3) 7^C3 (0.2)^3 (0.8)^4
=0.114688
p(x=4)=7^C4 (0.2)^4 (0.8)^3
0.028672
P(x=5)=7^C5 (0.2)^5 (0.8)^2
=0.0043008
P(x=6)=7^C6 (0.2)^6 (0.8)^1
=0.0003584
p(36)
=0.114688+0.028672+0.0043008
+0.0003584
=0.1480192
=0.15(2d.p)
==================
4)
(x^2+5x+1)sqroot(2x^3+mx^2+nx+11)=(2x-5)
remainder:30x+16
(x^2+5x+1)(2x-5)
=2x^3+10x^2+2x-5x^2-25x-5
=2x^3+10x^2-5x^2-25x-5
=2x^3+5x^2-23x+30x+16-5
=2x^3+5x^2+7x+11
Therefore m=5, n=7
=================
5a)
pr(age)=4/5
pr(fully)=3/4
pr(must)=2/3
pr(age not admitted)=1-4/5
=1/5
pr(fully not admitted)=1-3/4
=1/4
pr(must not admitted)=1-2/3
=1/3
Therefore pr(none admitted)=1/5*1/4*1/3
=1/60
5b)
pr(only age and fully gained
admission)=4/5*3/4*1/3
=1/5
================
[10:29 AM, 4/9/2018] +234 814 459 8832: (6)
It follow that:
P(n) = 1/3 and p(k)¹ = 1 - 1/3 = 2/3
P(T) = 1/5 and P(T)¹ = 1- 1/5 = 4/5

Hence,
probability that only one if the them will be solve the questions will be:

= (1/3 × 4/5) + (1/5 × 2/3)
= 4/15 + 2/15
= 6/15
= 2/5

(6)
It follow that:
P(n) = 1/3 and p(k)¹ = 1 - 1/3 = 2/3
P(T) = 1/5 and P(T)¹ = 1- 1/5 = 4/5

Hence,
probability that only one if the them will be solve the questions will be:

= (1/3 × 4/5) + (1/5 × 2/3)
= 4/15 + 2/15
= 6/15
= 2/5


(7)
m = 3i - 2j ; n = 2i + 3j ; p = i + 6j

Therefore:
4(3i - 2j) +2(2i +3j) -3(-i + 6j)
12i - 8j + 4i + 6j + 3i - 18j
12i + 4i + 3i - 8j + 6j - 18j
19i = 20j

(8a)
m1 = 20kg
u1 = 8ms-1
m2 = 30kg
u2 = 50ms-1
(a)  in same direction
m1u1 + m2u2 = (m1+m2)v
20 × 80 + 30 × 50 = (20+30)v
1600+1500 = 50v
3100/50 = 50/50
V = 62ms-1

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