Today's examination papers
Wednesday 22nd June.
Paper III: Objective – General Mathematics 10:00am – 11:45am
Paper II: Essay – General Mathematics 12:00noon – 2:30pm
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1 a )
Tabulate
x - 1 , 2 , 3 , 4
1 - 1 , 2 , 3 , 4
2 - 2 _ , 4, 0 _ , 2 _
3 - 3 , 0 _ , 3 , 0 _
4 - 4 _ , 2_ , 0 _ , 4
1 b )
I = PRT/ 100 , p =N 15000 R = 10 % and
I = 3 years
A = P+ I
where I = 15000* 10 * 3 /100 = N4500
A = 4500 + 15000 = N19500
NO 1 a , b IMAGE
================
2 a )
using sine rule
b / sin20 = 6/ sin30
bsin30 = 6 sin120
b 6 sin120/ sin30
b = 6 x 0 . 2511 /0 . 4540
b = 5 . 7063 / 0. 4540
b = 12 . 57 ≠ 12 . 6 cm
2 bi)
the diagram is euivalent triangles .
where
| AX |/ |BC| = | BY |/ |AC | = | XY |/ |YC |
XY = 9 , BY = 7
YC = 18 - 7 = 11
9 / 11 = 7 / |AC |
9 | AC | = 77
| AC | = 77 /9
| AC | = 8 cm
2 bii )
XY / AB = BY /AC
9 / |AB | = 7 / 8 . 6
| AB | = 9 x 8 . 6 / 7
| AB | = 11 cm
=================
3 )
let the son age be x
man = 5 x
son = x
4 yrs ago; the man age = 5 x - 4
the son age = x - 4
the product of their ages
( 5 x - 4 ) ( x - 4 )
= 448
=================
7 a )
3 ^ 2 n + 1 - 4 ( 3 ^ n + 1 ) + 9= 0
3 ^ 2 - 3 - 4 ( 3 ^ n - 3 ) + 9 = 0
( 3 ^n )^ 2 - 3 - 4 ( 3 ^n - 3 ) + 9 = 0
let 3 ^ n = p
p ^ 2 - 3 - 4 ( p - 3 ) + 9 = 0
3 p ^ 2 /3 - 12 p / 3 + 9 /3 = 0
p ^ 2 - 4 p + 3 = 0
p ^ 2 - 3 p - p + 3 = 0
p ^ 2 p ( p - 3 ) - 1 ( p - 3 ) = 0
( p - 1 ) ( p - 3 ) = 0
p - 1 = 0 or p - 3 = 0
p = 1 or 3
Recall 3 ^ n = p
when p =1
3 ^ n = 3^ 0
n = 0
when p = 3
3 ^ n = 3^ 1
n = 1
7 b )
log( x ^ 2 + 4) = 2 + logx - log^ 20
log( x ^ 2 + 4) = log^ 100 = log^ x - log^ 20
( x ^2 + 4 ) = log( xx )
x ^ 2 + 4 = 5 x
x ^ 2 - 5 x + 4 = 0
x ^ 2 - 4 x - x + 4 = 0
x ( x - 4 ) - 1 ( x - 4 ) = 0
( x - 1 ) ( x - 4 ) = 0
x - 1 = 0 or x - 4 = 0
x = 1 or 4
4a )
volume of fuel = cross - sectional area of X depth of fuel
rectangular tank
30 , 000litres = 7 . 5* 4 . 2 * d m ^ 3
but ; 1000 litres = 1m ^ 3
therefore ;30 ( M ^ 3 ) = 7 . 5 * 4 . 2 * d( M ^ 3 )
30 = 31 . 5d
====> d = 30 /31 . 5 = 0 . 95 (2 d. p)
4b )
to fill the tank / volume of fuel needed
= 7. 5 * 4 . 2 * 1. 2
= 37 . 8m ^ 3
= 37 , 800 litres
addition fuel = 37 , 800- 30 , 000
= 7, 800 litres
therefore , 7 , 800 more litres would be needed
==================================
5a )
sector for building project = 48000 / 144000* 360= 120degree
sector for education = 32 , 000/144000* 360=80 degree
sector for saving = 19200 /144000* 360=48 degree
sector for maintenance = 12000 / 144000* 360= 30 degree
sector for miscellaneous = 7200 / 144000* 360= 18 degree
sector for food items = 360- (120+ 80 + 48 + 30 +18 )
=360- 296
=64 degree
5b )
amount spent = 144000-
[ 48 , 000+ 32000 + 19200 + 12000 + 7200 ]
=144000- 118400
=N 25600
6 )
tabulate
No
41 . 02 sqr 0 . 7124
42 . 81
0 . 207
0 . 0404
antilog ( 0 . 6616 )
= 4 . 587
log
1 . 6130 ===> 1 . 6130
T . 8527 / 2
= 2 ^ - + 1 . 8527 / 2 +
T + 0 . 9264 ===> T . 9264 1. 5394
1 . 6321
T . 3160 +
2 ^ - . 6064
T . 5545 ===> T . 5545
1 . 9849 ===> 1 . 9849 / 3
= 0. 6616
=========
4 b ) to fill the tank/ volume of fuel needed = 7. 5 * 4. 2 * 1. 2
= 37 . 8 m^ 3
= 37 , 800 litres
addition funaijacomms.comel = 37 , 800 - 30 , 000 = 7, 800 litres
therefore , 7, 800 more litres would be
needed
5 a ) sector for building project
= 48000/ 144000 * 360 =
120degree
sector for education =copied from naijacomms.com
32 , 000 / 144000 * 360= 80 degree sector for saving =
19200 /144000 * 360 = 48 degree sector for maintenance =
12000 /144000 * 360 = 30 degree sector for miscellaneous =
7200 /144000 *360 = 18 degree sector for food items = 360 -
( 120+ 80 + 48 + 30 + 18 )
= 360 - 296
= 64 degree
5 b )copied from naijacomms.com
amount spent = 144000 -
[ 48 , 000 + 32000+ 19200+ 12000+ 7200 ]
= 144000 - 118400
= N25600
1a ) Tabulate
x - 1 , 2 , 3 , 4
1- 1 , 2 , 3 , 4
2- 2 _ , 4 , 0 _ , 2 _
3- 3 , 0 _ , 3 , 0_
4- 4 _ , 2 _ , 0 _ , 4
7 b )
log( x ^ 2 + 4) = 2 + logx - log^ 20
log( x ^ 2 + 4) = log^ 100 = log^ x - log^ 20
( x ^2 + 4 ) = log( xx )
x ^ 2 + 4 = 5 x
x ^ 2 - 5 x + 4 = 0
x ^ 2 - 4 x - x + 4 = 0
x ( x - 4 ) - 1 ( x - 4 ) = 0
( x - 1 ) ( x - 4 ) = 0
x - 1 = 0 or x - 4 = 0
x = 1 or 4
4a )
volume of fuel = cross - sectional area of X depth of fuel
rectangular tank
30 , 000litres = 7 . 5* 4 . 2 * d m ^ 3
but ; 1000 litres = 1m ^ 3
therefore ;30 ( M ^ 3 ) = 7 . 5 * 4 . 2 * d( M ^ 3 )
30 = 31 . 5d
====> d = 30 /31 . 5 = 0 . 95 (2 d. p)
4b )
to fill the tank / volume of fuel needed
= 7. 5 * 4 . 2 * 1. 2
= 37 . 8m ^ 3
= 37 , 800 litres
addition fuel = 37 , 800- 30 , 000
= 7, 800 litres
therefore , 7 , 800 more litres would be needed
==================================
5a )
sector for building project = 48000 / 144000* 360= 120degree
sector for education = 32 , 000/144000* 360=80 degree
sector for saving = 19200 /144000* 360=48 degree
sector for maintenance = 12000 / 144000* 360= 30 degree
sector for miscellaneous = 7200 / 144000* 360= 18 degree
sector for food items = 360- (120+ 80 + 48 + 30 +18 )
=360- 296
=64 degree
5b )
amount spent = 144000-
[ 48 , 000+ 32000 + 19200 + 12000 + 7200 ]
=144000- 118400
=N 25600
6 )
tabulate
No
41 . 02 sqr 0 . 7124
42 . 81
0 . 207
0 . 0404
antilog ( 0 . 6616 )
= 4 . 587
log
1 . 6130 ===> 1 . 6130
T . 8527 / 2
= 2 ^ - + 1 . 8527 / 2 +
T + 0 . 9264 ===> T . 9264 1. 5394
1 . 6321
T . 3160 +
2 ^ - . 6064
T . 5545 ===> T . 5545
1 . 9849 ===> 1 . 9849 / 3
= 0. 6616
=========
( 9 a )
Let the lens digit x and unit digit be y ,
therefore x - y = 5 - - - ( 1 )
3 xy - ( 10 x + y )=14 - - - - ( 2 )
3 xy - 10 x - y = 14 - - - - - ( 3 )
frm eq ( 1); x = 5 + y - - - ( 4 )
therefore , 5( 5 + y ) ( y ) - 10 ( 5 + y ) - y = 14
( 15 + 3 y ) y - 50 - 10 y - y = 14
3 y ^ 2 + 4 y - 50 - 14 = 0
3 y ^ 2 + 4 y - 64 = 0
3 y ^ 2 - 12 y + 16 y - 64 = 0
3 y ( y - 4 ) + 16 ( y - 4 )= 0
( 3 y + 16 ) ( y - 4 )=0
y = - 16 / 3 or 4
therefore from eqn ( 1 );
x + 4 = 5
x = 5 + 4 = 9
the number is 94
( 9 b )
( 3 - 2 x ) / 4 + ( 2 x - 3 ) / 3
( 3 ( 3 - 2 x ) + 4 ( 2 x - 3 )) / 12
( 9 - 6 x + 8 x - 12 ) /12
= ( 2 x - 3 ) /12
8a) |AD|^2=13^2-5^2
|AD|^2=169-25
|AD|^2=144
AD=sqr144
AD=12CM
|AD|=12-r
r^2=(12-r)^2 - 5^2
r^2=(12-r)(12-r)+25
r^2=144-24r+25
r^2=169-24r
r^2+24r-169=0
r^2+24r=169
r^2+24r+14^2=169+14^2
(r+14)^2=169+196
(r+14)^2=365
(r+14=sqr365
r+14=19.105
r=19.105-14
r=5.105
r=5.1cm
8aii) circumfrenece of a circle=2pie R
C=2x22/2*(5.1)^2
C=1144.44/7
C=163.4914cm
C=163.5cm
8b) y2-y1/x2-x1=y-y1/x-x1
6-2/2-(-1)=y-2/x-(-1)
4/2+1 = y-2/x+1
4/3=y-2/x+1
3(y-2)=4(x+1)
3y-6=4x+4
3y-4x=4+6
3y-4x=10
y=4x/3+10/3
( 9 a )
Let the lens digit x and unit digit be y ,
therefore x - y = 5 - - - ( 1 )
3 xy - ( 10 x + y )=14 - - - - ( 2 )
3 xy - 10 x - y = 14 - - - - - ( 3 )
frm eq ( 1); x = 5 + y - - - ( 4 )
therefore , 5( 5 + y ) ( y ) - 10 ( 5 + y ) - y = 14
( 15 + 3 y ) y - 50 - 10 y - y = 14
3 y ^ 2 + 4 y - 50 - 14 = 0
3 y ^ 2 + 4 y - 64 = 0
3 y ^ 2 - 12 y + 16 y - 64 = 0
3 y ( y - 4 ) + 16 ( y - 4 )= 0
( 3 y + 16 ) ( y - 4 )=0
y = - 16 / 3 or 4
therefore from eqn ( 1 );
x + 4 = 5
x = 5 + 4 = 9
the number is 94
( 9 b )
( 3 - 2 x ) / 4 + ( 2 x - 3 ) / 3
( 3 ( 3 - 2 x ) + 4 ( 2 x - 3 )) / 12
( 9 - 6 x + 8 x - 12 ) /12
= ( 2 x - 3 ) /12
10 a )
y = ( 2 x ^ 2 + 3 )^5
let U =2 x ^ 2 + 3
Y = u ^ 5
du / dx = 4 x
dy / du = 5 u ^ 4
dy / du = ( 2 x ^2 + 3 )^4
dy / dx = du / dx dy / du
dy / dx = 4 x . 5( 2 x ^ 2 + 3 )^4
dy / dx = 20 x ( 2x ^ 2 + 3 )^4
10 b )
y = 3 x ^ 2 + 2 x + 5
dy / dx = 6 x + 2
dy / dx = 6 ( 3 ) + 2
dy / dx = 18 + 2
dy / dx = 20
10 c )
R - W= Wv^ 2 / gx
Wv^ 2 = gx ( R - W)
Wv^ 2 = gRx - Wgx
Wv^ 2 + Wgx= gRx
W ( v ^ 2 + gx ) = gRx
W = gRx/ V^ 2 + gx
R = 2 , g = 10 , x = 3 / 2 , V=3
W = 10 *2 * 3 / 2 /3 ^ 2 + 10 * 3 / 5
W = 30 / 9 + 15
W = 30 / 24
W = 5/ 4
1b ) I = PRT /100, p =N 15000 R =10 % and
I =3 years
A = P + I
where I = 15000 * 10 * 3 /100= N 4500
A =4500 + 15000 = N 19500
========================= 2a ) using sine rule
b/ sin 20 = 6 /sin 30
bsin 30 = 6sin 120
b 6sin 120/ sin30
b = 6x 0 . 2511 /0 . 4540
b = 5. 7063 /0 . 4540
b = 12 . 57 ≠ 12 . 6 cm
2bi ) the diagram is euivalent triangles . where
|AX |/| BC | = |BY |/ |AC | = |XY |/
|YC|
XY = 9, BY = 7
YC = 18 - 7 = 11
9/ 11 = 7/ |AC |
9| AC | = 77
|AC | = 77 /9
|AC | = 8 cm
2bii ) XY/ AB = BY/ AC
9/ |AB | = 7 / 8. 6
|AB | = 9 x 8. 6/ 7
|AB | = 11 cm
======================
3) let the son age be x
man= 5 x
son =x
4yrs ago; the man age = 5x - 4
the son age = x - 4
the product of their ages
(5 x - 4) (x - 4) = 448
======================
7a )
3²ⁿ +¹ — 4 (3 ⁿ +¹ ) + 9 = 0
3²ⁿ × 3 — 4 (3 ⁿ × 3¹) + 9 = 0
(3 ⁿ ) ² × 3 — 4 (30 ⁿ × ) + 9 = 0
Let 3ⁿ = x
3x ² — 4 × 3 × x + 9 = 0
3x ² — 12 x + 9 = 0
Divide all by 3
3x ² /3 — 12 x /3 + 9 /3 = 0
x ² — 4x + 3 = 0
x ² — 3x — x + 3 = 0
x ( x — 3) - 1 (x — 3 ) = 0
(x — 3 )(x — 3 ) = 0
x — 3 = 0 or x — 1 = 0
x = 3, x = 1
Substitute x = 3
3ⁿ = 3 or 3ⁿ = 1
3ⁿ = 3¹ or 3 ⁿ = 3 °
n = 1 or n = 0
.
7 a )
3 ^ 2 n + 1 - 4 ( 3 ^ n + 1 ) + 9= 0
3 ^ 2 - 3 - 4 ( 3 ^ n - 3 ) + 9 = 0
( 3 ^n )^ 2 - 3 - 4 ( 3 ^n - 3 ) + 9 = 0
let 3 ^ n = p
p ^ 2 - 3 - 4 ( p - 3 ) + 9 = 0
3 p ^ 2 /3 - 12 p / 3 + 9 /3 = 0
p ^ 2 - 4 p + 3 = 0
p ^ 2 - 3 p - p + 3 = 0
p ^ 2 p ( p - 3 ) - 1 ( p - 3 ) = 0
( p - 1 ) ( p - 3 ) = 0
p - 1 = 0 or p - 3 = 0
p = 1 or 3
Recall 3 ^ n = p
when p =1
3 ^ n = 3^ 0
n = 0
when p = 3
3 ^ n = 3^ 1
n = 1
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Wednesday 22nd June.
Paper III: Objective – General Mathematics 10:00am – 11:45am
Paper II: Essay – General Mathematics 12:00noon – 2:30pm
Content will be available after 10 minutes, so re- visits this page
At Naijacomms.com
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Or Pay a one-time fees of N1500 for all subjects To Our bank Account below
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MATHEMATICS OBJ
MAths- Obj
1 ECADDDCEEE
11 CCDCAACECA
21 DCDEECCDED
31 EAADBEEADC
41 BBDCCDCECB
51 CADBAECDEC
1 a )
Tabulate
x - 1 , 2 , 3 , 4
1 - 1 , 2 , 3 , 4
2 - 2 _ , 4, 0 _ , 2 _
3 - 3 , 0 _ , 3 , 0 _
4 - 4 _ , 2_ , 0 _ , 4
1 b )
I = PRT/ 100 , p =N 15000 R = 10 % and
I = 3 years
A = P+ I
where I = 15000* 10 * 3 /100 = N4500
A = 4500 + 15000 = N19500
NO 1 a , b IMAGE
================
2 a )
using sine rule
b / sin20 = 6/ sin30
bsin30 = 6 sin120
b 6 sin120/ sin30
b = 6 x 0 . 2511 /0 . 4540
b = 5 . 7063 / 0. 4540
b = 12 . 57 ≠ 12 . 6 cm
2 bi)
the diagram is euivalent triangles .
where
| AX |/ |BC| = | BY |/ |AC | = | XY |/ |YC |
XY = 9 , BY = 7
YC = 18 - 7 = 11
9 / 11 = 7 / |AC |
9 | AC | = 77
| AC | = 77 /9
| AC | = 8 cm
2 bii )
XY / AB = BY /AC
9 / |AB | = 7 / 8 . 6
| AB | = 9 x 8 . 6 / 7
| AB | = 11 cm
=================
3 )
let the son age be x
man = 5 x
son = x
4 yrs ago; the man age = 5 x - 4
the son age = x - 4
the product of their ages
( 5 x - 4 ) ( x - 4 )
= 448
=================
7 a )
3 ^ 2 n + 1 - 4 ( 3 ^ n + 1 ) + 9= 0
3 ^ 2 - 3 - 4 ( 3 ^ n - 3 ) + 9 = 0
( 3 ^n )^ 2 - 3 - 4 ( 3 ^n - 3 ) + 9 = 0
let 3 ^ n = p
p ^ 2 - 3 - 4 ( p - 3 ) + 9 = 0
3 p ^ 2 /3 - 12 p / 3 + 9 /3 = 0
p ^ 2 - 4 p + 3 = 0
p ^ 2 - 3 p - p + 3 = 0
p ^ 2 p ( p - 3 ) - 1 ( p - 3 ) = 0
( p - 1 ) ( p - 3 ) = 0
p - 1 = 0 or p - 3 = 0
p = 1 or 3
Recall 3 ^ n = p
when p =1
3 ^ n = 3^ 0
n = 0
when p = 3
3 ^ n = 3^ 1
n = 1
7 b )
log( x ^ 2 + 4) = 2 + logx - log^ 20
log( x ^ 2 + 4) = log^ 100 = log^ x - log^ 20
( x ^2 + 4 ) = log( xx )
x ^ 2 + 4 = 5 x
x ^ 2 - 5 x + 4 = 0
x ^ 2 - 4 x - x + 4 = 0
x ( x - 4 ) - 1 ( x - 4 ) = 0
( x - 1 ) ( x - 4 ) = 0
x - 1 = 0 or x - 4 = 0
x = 1 or 4
4a )
volume of fuel = cross - sectional area of X depth of fuel
rectangular tank
30 , 000litres = 7 . 5* 4 . 2 * d m ^ 3
but ; 1000 litres = 1m ^ 3
therefore ;30 ( M ^ 3 ) = 7 . 5 * 4 . 2 * d( M ^ 3 )
30 = 31 . 5d
====> d = 30 /31 . 5 = 0 . 95 (2 d. p)
4b )
to fill the tank / volume of fuel needed
= 7. 5 * 4 . 2 * 1. 2
= 37 . 8m ^ 3
= 37 , 800 litres
addition fuel = 37 , 800- 30 , 000
= 7, 800 litres
therefore , 7 , 800 more litres would be needed
==================================
5a )
sector for building project = 48000 / 144000* 360= 120degree
sector for education = 32 , 000/144000* 360=80 degree
sector for saving = 19200 /144000* 360=48 degree
sector for maintenance = 12000 / 144000* 360= 30 degree
sector for miscellaneous = 7200 / 144000* 360= 18 degree
sector for food items = 360- (120+ 80 + 48 + 30 +18 )
=360- 296
=64 degree
5b )
amount spent = 144000-
[ 48 , 000+ 32000 + 19200 + 12000 + 7200 ]
=144000- 118400
=N 25600
6 )
tabulate
No
41 . 02 sqr 0 . 7124
42 . 81
0 . 207
0 . 0404
antilog ( 0 . 6616 )
= 4 . 587
log
1 . 6130 ===> 1 . 6130
T . 8527 / 2
= 2 ^ - + 1 . 8527 / 2 +
T + 0 . 9264 ===> T . 9264 1. 5394
1 . 6321
T . 3160 +
2 ^ - . 6064
T . 5545 ===> T . 5545
1 . 9849 ===> 1 . 9849 / 3
= 0. 6616
=========
4 b ) to fill the tank/ volume of fuel needed = 7. 5 * 4. 2 * 1. 2
= 37 . 8 m^ 3
= 37 , 800 litres
addition funaijacomms.comel = 37 , 800 - 30 , 000 = 7, 800 litres
therefore , 7, 800 more litres would be
needed
5 a ) sector for building project
= 48000/ 144000 * 360 =
120degree
sector for education =copied from naijacomms.com
32 , 000 / 144000 * 360= 80 degree sector for saving =
19200 /144000 * 360 = 48 degree sector for maintenance =
12000 /144000 * 360 = 30 degree sector for miscellaneous =
7200 /144000 *360 = 18 degree sector for food items = 360 -
( 120+ 80 + 48 + 30 + 18 )
= 360 - 296
= 64 degree
5 b )copied from naijacomms.com
amount spent = 144000 -
[ 48 , 000 + 32000+ 19200+ 12000+ 7200 ]
= 144000 - 118400
= N25600
1a ) Tabulate
x - 1 , 2 , 3 , 4
1- 1 , 2 , 3 , 4
2- 2 _ , 4 , 0 _ , 2 _
3- 3 , 0 _ , 3 , 0_
4- 4 _ , 2 _ , 0 _ , 4
Updated now
7 b )
log( x ^ 2 + 4) = 2 + logx - log^ 20
log( x ^ 2 + 4) = log^ 100 = log^ x - log^ 20
( x ^2 + 4 ) = log( xx )
x ^ 2 + 4 = 5 x
x ^ 2 - 5 x + 4 = 0
x ^ 2 - 4 x - x + 4 = 0
x ( x - 4 ) - 1 ( x - 4 ) = 0
( x - 1 ) ( x - 4 ) = 0
x - 1 = 0 or x - 4 = 0
x = 1 or 4
4a )
volume of fuel = cross - sectional area of X depth of fuel
rectangular tank
30 , 000litres = 7 . 5* 4 . 2 * d m ^ 3
but ; 1000 litres = 1m ^ 3
therefore ;30 ( M ^ 3 ) = 7 . 5 * 4 . 2 * d( M ^ 3 )
30 = 31 . 5d
====> d = 30 /31 . 5 = 0 . 95 (2 d. p)
4b )
to fill the tank / volume of fuel needed
= 7. 5 * 4 . 2 * 1. 2
= 37 . 8m ^ 3
= 37 , 800 litres
addition fuel = 37 , 800- 30 , 000
= 7, 800 litres
therefore , 7 , 800 more litres would be needed
==================================
5a )
sector for building project = 48000 / 144000* 360= 120degree
sector for education = 32 , 000/144000* 360=80 degree
sector for saving = 19200 /144000* 360=48 degree
sector for maintenance = 12000 / 144000* 360= 30 degree
sector for miscellaneous = 7200 / 144000* 360= 18 degree
sector for food items = 360- (120+ 80 + 48 + 30 +18 )
=360- 296
=64 degree
5b )
amount spent = 144000-
[ 48 , 000+ 32000 + 19200 + 12000 + 7200 ]
=144000- 118400
=N 25600
6 )
tabulate
No
41 . 02 sqr 0 . 7124
42 . 81
0 . 207
0 . 0404
antilog ( 0 . 6616 )
= 4 . 587
log
1 . 6130 ===> 1 . 6130
T . 8527 / 2
= 2 ^ - + 1 . 8527 / 2 +
T + 0 . 9264 ===> T . 9264 1. 5394
1 . 6321
T . 3160 +
2 ^ - . 6064
T . 5545 ===> T . 5545
1 . 9849 ===> 1 . 9849 / 3
= 0. 6616
=========
( 9 a )
Let the lens digit x and unit digit be y ,
therefore x - y = 5 - - - ( 1 )
3 xy - ( 10 x + y )=14 - - - - ( 2 )
3 xy - 10 x - y = 14 - - - - - ( 3 )
frm eq ( 1); x = 5 + y - - - ( 4 )
therefore , 5( 5 + y ) ( y ) - 10 ( 5 + y ) - y = 14
( 15 + 3 y ) y - 50 - 10 y - y = 14
3 y ^ 2 + 4 y - 50 - 14 = 0
3 y ^ 2 + 4 y - 64 = 0
3 y ^ 2 - 12 y + 16 y - 64 = 0
3 y ( y - 4 ) + 16 ( y - 4 )= 0
( 3 y + 16 ) ( y - 4 )=0
y = - 16 / 3 or 4
therefore from eqn ( 1 );
x + 4 = 5
x = 5 + 4 = 9
the number is 94
( 9 b )
( 3 - 2 x ) / 4 + ( 2 x - 3 ) / 3
( 3 ( 3 - 2 x ) + 4 ( 2 x - 3 )) / 12
( 9 - 6 x + 8 x - 12 ) /12
= ( 2 x - 3 ) /12
8a) |AD|^2=13^2-5^2
|AD|^2=169-25
|AD|^2=144
AD=sqr144
AD=12CM
|AD|=12-r
r^2=(12-r)^2 - 5^2
r^2=(12-r)(12-r)+25
r^2=144-24r+25
r^2=169-24r
r^2+24r-169=0
r^2+24r=169
r^2+24r+14^2=169+14^2
(r+14)^2=169+196
(r+14)^2=365
(r+14=sqr365
r+14=19.105
r=19.105-14
r=5.105
r=5.1cm
8aii) circumfrenece of a circle=2pie R
C=2x22/2*(5.1)^2
C=1144.44/7
C=163.4914cm
C=163.5cm
8b) y2-y1/x2-x1=y-y1/x-x1
6-2/2-(-1)=y-2/x-(-1)
4/2+1 = y-2/x+1
4/3=y-2/x+1
3(y-2)=4(x+1)
3y-6=4x+4
3y-4x=4+6
3y-4x=10
y=4x/3+10/3
( 9 a )
Let the lens digit x and unit digit be y ,
therefore x - y = 5 - - - ( 1 )
3 xy - ( 10 x + y )=14 - - - - ( 2 )
3 xy - 10 x - y = 14 - - - - - ( 3 )
frm eq ( 1); x = 5 + y - - - ( 4 )
therefore , 5( 5 + y ) ( y ) - 10 ( 5 + y ) - y = 14
( 15 + 3 y ) y - 50 - 10 y - y = 14
3 y ^ 2 + 4 y - 50 - 14 = 0
3 y ^ 2 + 4 y - 64 = 0
3 y ^ 2 - 12 y + 16 y - 64 = 0
3 y ( y - 4 ) + 16 ( y - 4 )= 0
( 3 y + 16 ) ( y - 4 )=0
y = - 16 / 3 or 4
therefore from eqn ( 1 );
x + 4 = 5
x = 5 + 4 = 9
the number is 94
( 9 b )
( 3 - 2 x ) / 4 + ( 2 x - 3 ) / 3
( 3 ( 3 - 2 x ) + 4 ( 2 x - 3 )) / 12
( 9 - 6 x + 8 x - 12 ) /12
= ( 2 x - 3 ) /12
10 a )
y = ( 2 x ^ 2 + 3 )^5
let U =2 x ^ 2 + 3
Y = u ^ 5
du / dx = 4 x
dy / du = 5 u ^ 4
dy / du = ( 2 x ^2 + 3 )^4
dy / dx = du / dx dy / du
dy / dx = 4 x . 5( 2 x ^ 2 + 3 )^4
dy / dx = 20 x ( 2x ^ 2 + 3 )^4
10 b )
y = 3 x ^ 2 + 2 x + 5
dy / dx = 6 x + 2
dy / dx = 6 ( 3 ) + 2
dy / dx = 18 + 2
dy / dx = 20
10 c )
R - W= Wv^ 2 / gx
Wv^ 2 = gx ( R - W)
Wv^ 2 = gRx - Wgx
Wv^ 2 + Wgx= gRx
W ( v ^ 2 + gx ) = gRx
W = gRx/ V^ 2 + gx
R = 2 , g = 10 , x = 3 / 2 , V=3
W = 10 *2 * 3 / 2 /3 ^ 2 + 10 * 3 / 5
W = 30 / 9 + 15
W = 30 / 24
W = 5/ 4
1b ) I = PRT /100, p =N 15000 R =10 % and
I =3 years
A = P + I
where I = 15000 * 10 * 3 /100= N 4500
A =4500 + 15000 = N 19500
========================= 2a ) using sine rule
b/ sin 20 = 6 /sin 30
bsin 30 = 6sin 120
b 6sin 120/ sin30
b = 6x 0 . 2511 /0 . 4540
b = 5. 7063 /0 . 4540
b = 12 . 57 ≠ 12 . 6 cm
2bi ) the diagram is euivalent triangles . where
|AX |/| BC | = |BY |/ |AC | = |XY |/
|YC|
XY = 9, BY = 7
YC = 18 - 7 = 11
9/ 11 = 7/ |AC |
9| AC | = 77
|AC | = 77 /9
|AC | = 8 cm
2bii ) XY/ AB = BY/ AC
9/ |AB | = 7 / 8. 6
|AB | = 9 x 8. 6/ 7
|AB | = 11 cm
======================
3) let the son age be x
man= 5 x
son =x
4yrs ago; the man age = 5x - 4
the son age = x - 4
the product of their ages
(5 x - 4) (x - 4) = 448
======================
7a )
3²ⁿ +¹ — 4 (3 ⁿ +¹ ) + 9 = 0
3²ⁿ × 3 — 4 (3 ⁿ × 3¹) + 9 = 0
(3 ⁿ ) ² × 3 — 4 (30 ⁿ × ) + 9 = 0
Let 3ⁿ = x
3x ² — 4 × 3 × x + 9 = 0
3x ² — 12 x + 9 = 0
Divide all by 3
3x ² /3 — 12 x /3 + 9 /3 = 0
x ² — 4x + 3 = 0
x ² — 3x — x + 3 = 0
x ( x — 3) - 1 (x — 3 ) = 0
(x — 3 )(x — 3 ) = 0
x — 3 = 0 or x — 1 = 0
x = 3, x = 1
Substitute x = 3
3ⁿ = 3 or 3ⁿ = 1
3ⁿ = 3¹ or 3 ⁿ = 3 °
n = 1 or n = 0
.
7 a )
3 ^ 2 n + 1 - 4 ( 3 ^ n + 1 ) + 9= 0
3 ^ 2 - 3 - 4 ( 3 ^ n - 3 ) + 9 = 0
( 3 ^n )^ 2 - 3 - 4 ( 3 ^n - 3 ) + 9 = 0
let 3 ^ n = p
p ^ 2 - 3 - 4 ( p - 3 ) + 9 = 0
3 p ^ 2 /3 - 12 p / 3 + 9 /3 = 0
p ^ 2 - 4 p + 3 = 0
p ^ 2 - 3 p - p + 3 = 0
p ^ 2 p ( p - 3 ) - 1 ( p - 3 ) = 0
( p - 1 ) ( p - 3 ) = 0
p - 1 = 0 or p - 3 = 0
p = 1 or 3
Recall 3 ^ n = p
when p =1
3 ^ n = 3^ 0
n = 0
when p = 3
3 ^ n = 3^ 1
n = 1
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