©WAEC RONSE™ | VERIFIED ESSAY AND O B J ANSWERS FOR MATHEMATICS

Maths O B J
1-10.DCBCABBADA
11-20.CCBCABABBA
21-30.AACCDCDABC
31-40.BCDCDCBABA
41-50.ACADBBDADD
MATHS- ANSWERS
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6a )
Sn= n /2 (2 a +( n- 1 ) d )
A =1 , d= 2 , n = n
Sn= n /2 (2 * 1 + (n - 1) 2)
N /2 (2 + 2, - 2 )
=n / 2* 2 =n ^ 2
6b )
n( E )=95 , n( BnT )=7
N (U )=95 , n (BnTnC ')=7 n (B 'nTnC )=3 , n (BnTnC )=8
n( BnT 'nC ')=x , n( BnT 'nC ) , n ( b)=47 , n (T )=30
i )
Draw d venn diagram
ii)
To find x ,
X +x + 7+ 8 =47
2x = 47 - 15 = 32
X =32 /2
=16
iii)
No of those who travelled by at least two means
7+ 3 + 16 + 8= 34
(6b)
IMG: IMG-20160421-WA016.jpg(93.37 KB)
13 a )
(x 1 , y 1) = (2 , - 3)
for 2x + y = 6
y = - 2x + 6
compare y = mx +6
m2 = - 2
for parallel lines , m1 = m2
: y - y 1 / x - x 1 = m
y - - 3/ x - 2 = - 2 / 1
y + 3 = - 2 (x - 2 )
y + 3 = - 2 x + 4
y = - 2 x + 4 - 3
y = - 2 x + 1
(11a )
3p + 4 q/ 3p - 4q * 2 /1 find p: q
1( rp+ 4q )= 2 ( 3p - 4q )
3p + 4 q= 6 p- 8 q
Collect like terms
3p - 6p = - 8 q - 4 q
- 3 p =- 12 q
Ther 4 p : q = - 3 p/ - 3 = - 12 q /- 3
P =4 q , p = 4, q =1
P : q = 4: 1
11 b )
Ts = pq - (2 + 2)
Ts = pq - 4
Circumference of Ts = 2pie (TS )
11 bi.
Ts = pq - (2 + 2)
Ts = pq - 4
Circumference of Ts = 2pie (TS )
=2 pie (pq - 4 )
Perimeter= pq + 4+ 4 + 2+ 2 +2 pie (pq - 4 )
34 = pq + 12 + 2 pie ( pq - 4 )
34 - 12 = pq + 2 piePQ- 8 pie
22 = PQ( 1 +2 pie ) - 8pie
22 = pq ( 1( 2* 22 / 7 ) - 8* 22 / 7
22 = pq ( 1+ 44 / 7 ) - 176/7
22 = PQ( 7 +44 /7 ) - 176/7
22 + 176/7 = pq ( 51 / 7)
154+176/ 7 =Pq (51 /7 )
330/7 = pq (51 /7 )
330/51 =pq
Pq = 6 . 4706
11 bii .
TS = PQ - 4 = 6. 4706- 4 =2 . 4706 m
Area of semi circle TS =pieD
=7 . 7626m^ 2
Area of rectangle PqRU=| pu |*| Pq |
4** 6. 4706= 25 . 8824
Area of the cross section = area of rectangle- area of semi
circle
=25 . 8824 - 7. 7626
=18 . 1198 m^ 2
(No9)
IMG: IMG-20160421-WA022.jpg(0 B)
9a )
x 62 63 64 65 66 67 68
tally 1 iiiii iiiii iiiii ii iiii iiii iiii iiiii iiiii iiiii i ii
freq 1 5 12 14 10 6 2 total = 50
fx 62 315 768 910 660 402 136 tot = 3253
x - x - - 3 . 06 - 2. 06 - 1 . 06 0 . 06 0. 94 1 . 94 2 . 94
(x - x - ) 9. 3636 4. 2436 1. 1236 0 . 0036 0 . 8836 3 . 7636
8. 6436
f ( x - x ) 2 9 . 3636 21 . 218 13 . 4832 0 . 0504 8 . 836 22 . 5816
17 . 2872 tot = 92 . 82
bi) mean = EFX/EF = 3253 /50 = 65 . 06
bii ) Standard deviation = Square root of EF (x - x ) 2/ Ef
= Square root of 92 . 82 /50
= square root of 1. 8564
= 1. 36
7a )
x - 2 /4 : x + 2 /2 x
Cross multiply
2x ( x - 2 )=4( x + 2 )
Opening the bracket
2x ^ 2- 4 x = 4x + 8
Collect like terms together
2x ^ 2- 4 x - 4 x - 8 =0
2x ^ 2- 8 x - 8 =0
Solving with completing the square method
2x ^ 2- 8 x = 8
Divide through with the coefficient of x ^ 2
2x ^ 2/ 2 - 8 x /2 = 8/ 2
X ^ 2 - 4 x = 4
Half of coefficient of x
(- 4 x 1/ 2 )^ 2 = (- 2)^ 2
X ^ 2 - 4 x + (- 2)^ 2 = 4+ (- 2 )^ 2
X ^ 2 - 4 x + (- 2)^ 2 = 4+ 4
(x - 2 )^ 2 = 8
X - 2 : square root 8
X =2 + or - squar root 8
X =2 + square root 8 or x =2 - square root 8
X =2 + 2. 828=4 . 828
X =2 - 2. 828= - 0 . 828
X =4 : 83 or - 0 . 83 .
8a )
< KGF = 110
2x = r = y
if < KGF = 110, < MGH = 180 - 110
< MGH = 70 degree
< HMG = Y
< MHG = 180 - Y - 70
< MHG = 110 - Y
< LHJ = 180 - (110 - Y )
=180 - 110 + Y = 70 + Y
nOW : X + r + 70 + y = 180
2x = r + - y
if 2x = r
x = r / 2 =y
: x + 2 x + 70 + 2x = 180
5x = 180 - 70 = 110
x = 110/5 = 22 degree
8b )
let A collection of boy be X and that of girl be y
Tot collected by boy
600 + Y = x
Tot collected by girl
x - 600 = y
x - y = 600 - - (i )
Ave collection for boys
x / 10 = 100+y / 12 (multipled by 120)
x / 10 * 120 = 100 * 120 + y / 12 * 120
12 x = 12000 + 10 y
12 x - 10 y = 12000 - - - ( ii) /x - y = 600 * 12
12 x - 10 y = 12000 * 1
12 x - 12 y = 7200 - - - - ( iii)
12 x - 10 y = 12000
= - 2 y = - 4800
y = 4800/ 2 = 2400
y = 4800/ 2 = 2400
subt 2400 for y in eq(i )
600 + y = x
600 + 2400 = x
x = 3000,
boys collection = # 3000
girls collection = # 24000
total collected = # 54000
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5a )
Area of triangle= 1/ 2 bh where h =6
: . 1 /2 * 6 / 1* b /1 =36
6b / 2= 36 / 1
6b = 72
B =72 /6
Base of angle PSR =12 cm
Since |TS | / /PQ
QR= PR - PQ
QR= 12 - 8
QR= 4 cm
5b )
Draw d triangle
From angle ABC, to get | BC |
Tan 60 / 1= 10 cm / |BC |
|BC |= 10 . 65 /tan 60
|BC |= 10 . 65 /1 . 7321
=6 . 1486 =6 . 15 m
Hence | BC | =| DE|
From angle AED , to get | AE |
Draw d diagram
|AB |= 6. 15 tan 45
6. 15 * 1
=6 . 15 m
There 4 , th height of the tree = 10 . 65 - 6 . 15
=4 . 50 m
TYPING……………………