Waec 2018 Further Mathematics Answers / WASSCE 2018 Further Maths Obj And Theory Expo

Waec 2018 Further Mathematics Answers

Here Is The Waec 2018 Further Maths Questions And Answers Runz Waec Further Maths Obj And Theory Questions And Answers Expo, 2018/2019 Waec Further Maths Expo.

*FURTHER MATHS OBJ:*
*F/maths*
*1-10: BCADDDDDBB*
*11-20: DBABCCDDCC*
*21-30: BCABBAABAB*
*31-40: CBADBDDCAB*

*Completed•••••••••••••••••*

Further Maths Obj And Theory Monday 9th April, 2018

Waec 2018 Further Maths Obj And Theory Expo

*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

=-=-=-=-=-=-=-=-=

(2)
Log 3x – 3logx³+ 2 = 0
Log 3x – 3log3³/log3x + 2 = 0
Log3x – 3/log3x + 2 = 0
P – 3/p + 2 = 0 where P = log3x
P² – 3 + 2p = 0
P² + 2p – 3 = 0
(p + 3) ( p – 1) = 0
P = -3 OR p = 1
But log3x = p
when p = -3
Log3x = -3
X = 3-³ = 1/27
And when P = 1
Log3x = 1
X = 3¹ = 3
And X = 1/27 OR 3

=-=-=-=-=-=-=-=-=

(3a)

U= x-2 ; hence X = u+2
Therefore :
X^3+5/(x-2)^4 = (u+2)^3+5/ (u+2-2)^4
=(u+2)^3+5/u^4

(3b)

(u+2)^3+5/u^4 = (u+2)^3/u^4 + 5/u^4

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

No 4
Sn = n/2[2a+(n-1)d]
S12 = 12/6[2a + 11d]
S12= 6[2a + 11d] = 168
2a + 11d = 28——-(1)
Also
T3 = a + 2d = 7———(2)
Multiplying 1 by 2
2a + 4d = 14
Eqn 1 minus Eqn 3 gives
7d = 14
d = 2
Putting this in eqn 2
a + 2(2) = 7
a + (4) = 7
a = 7 – 4
a = 3
Common difference = 2
First term = 3

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

(5)
TABULATE
cooking oil type|a|b|c|d|e|f|g|h
x|8|5|1|7|2|6|3|4
y|6|3|4|8|5|7|1|8
d|2|2|-3|-1|-3|-1|2|2
d²|4|4|9|1|9|1|4|4

£d² = 4+4+9+1+9+1+4+4 = 36

r = 1- 6£d²/n(n²-1)

= 1 – 6×36/8(8²-1)
= 1 – 216/504
= 1 – 0.43
= 0.57

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=/=-=

(6a)

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

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-

(8)
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

(b)
In opposite direction
m1u1 – m2u2 = (m1 + u2)V
20×30 – 30×50 = (20+30)V
1600 – 1500 = 50V
100/50 = 50V/50
V = 2m/s

Thanks For Visit Our Website, The Number One Exam Expo Runz Online Website 2018, Waec 2018, We Also Do Neco, NABTEB, GCE Exam Runz.