# 100% REAL GCE WAEC GENERAL MATHEMATICS RUNZ ANSWERS

## 100% REAL GCE WAEC GENERAL MATHEMATICS RUNZ ANSWERS

100% REAL GCE WAEC GENERAL MATHEMATICS RUNZ ANSWERS | SIMPLIFIED WAEC GCE MATHEMATICS EXPO ANSWERS & RUNZ | 100% REAL GCE WAEC GENERAL MATHEMATICS RUNZ ANSWERSThis post give current answers on waec gce, 2017 waec gce expo waec gce answers, 2017 gce waec runz, 2017 gce waec exam answers 2017 2018 and beyond – SIMPLIFIED WAEC GCE MATHEMATICS EXPO ANSWERS & RUNZ – TODAY’S BIOLOGY PRACTICAL ANSWERS Neco answers.Waec Gce 2017 General Mathematics Obj And Theory Answers – Nov/Dec Expo Now Available.100% REAL GCE WAEC GENERAL MATHEMATICS RUNZ ANSWERS.

“); }10a)
YX/XZ=XM/MZ
W/10=8/15
15W=10*8 W=5.33cm.

10bi)
q^2=p^2+r^2-2prcos tita
q^2=20^2+15-2*20*15 c0s 90
q^2=400+225-0
q^2=625 q=sqroot625
q=25km.

10bii)
p/sinP=q/sinQ=r/sinR
25/sin90=15/sinR
sinR=15*1/25 sinR=0.6
R=sin^-1(0.6)
R=36.86degrees
The bearing of p from R
=90+90+90+alpha
alpha=45-36.86 =90+90+90+8.14
=278.14
=278degrees
The bearing of p from R=278degrees.

## 100% REAL GCE WAEC GENERAL MATHEMATICS RUNZ ANSWERS

==================================
9a)
CBA=180-(128-x) sum of angle on a straight line
CBA=52+x
=52+x
Also BCD=180-x(angle on a straight line)
DCQ=180-(180-x) DCQ=180-180+x
DCQ=x
x+52+x+76=180
2x=180-52-76
2x/2=52/2
x=26degrees.
=====================================
8a)
diagram
(8+x)^2 = x^2+32
64+16x+x^2= x^2 + 1024
16x=1024-64=960
therefore 960/16= 60 x=960/16
=60
=68cm.
=====================================
8b)
diagram
(i)volume of a pyramid
=1/3 AH
2601= 1/3 * A * 27
A=7803/27
=289cm^3
Area of square =289 t^2= 289
t= sqr rut(289)
l=17cm.

## 100% REAL GCE WAEC GENERAL MATHEMATICS RUNZ ANSWERS

8bii)
AC^2 = 17^2 +17^2
AC^2 = 289 +289 Ac^2 =578
AC =sqr root (578)
AC=24.04cm
for the triangele COP
CO= 1/2 AC
=1/2 * 24.04 VC^2= 27^2 + 12.07
VC^2= 929 +144.49
VC= SQR root (873.48)
=29.55cm
cos x= 8.5/29.55 cos x=0.2877
x=cos^-1 0.2877
=73.66 degree.
==========================
1a)
1/2log 25/4- 2log10 4/5+log10 320/125
Firstly Log(25/4)^1/2-log(4/5)^2+log10 320/125
Log10 sqrt25/4-log16/25+log10 320/125
Log10 5/2-log10 16/25+log10 320/125
Furthermore – Log10 5/2+log10 320/125-log10 16/25
Log10 [5/2*320/125÷16/25]
Log10 [5/2*320/125*25/16]
Lastly – Log10 10=1.

1b)
% income= 20%
Grant per land =GHC€15.00
Total population from
2003-2007
=1.2*1.2*1.2*1.2*3000=6220.8
=6220.8*15
=ghc€93312
Total grant=GHC€93312.

## 100% REAL GCE WAEC GENERAL MATHEMATICS RUNZ ANSWERS

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
2a)
1/x+1/x+3=1/2
L.C.M=x(x+3)
X+3+x/x(x+3)=1/2
2(2x+3)/x(x+3)=1/2
2(2x+3)=x(x+3)
4x+6=x^2+3x
X^2+3x=4x+6
X^2+3x-4x-6=0
X^2-x-6=0
(X^2-3x)+(2x-6)=0
X(x-3)+2(x-3)=0
(X+2)(x-3)=0
X=-2 or x=3.

2b)
Let d bag of rice be X
Let d bag of beans be Y
X+Y=17—>(I)
2250x+2400y=39600—>(2)
From eqtn (I)
X=17-y
Using elimination method to eliminate Y
2250x+2250y=38250—>(3)
2250x+2400y=39600—>(4)
Eqtn 4- eqtn 3
2400x-2250x = 39600-38250
150x=1350
X=1350/150
X=9
D trader bought 9bags of beans.

## 100% REAL GCE WAEC GENERAL MATHEMATICS RUNZ ANSWERS

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
(3)
=Draw The Diagram=
Area of path –> 2[1/2 (x+2)] + 2(x+1) = 17
x + 2 + 2x = 17
3x + 2 = 17
3x = 17 – 2
also 3x = 15
x = 15/3
x = 5.

### 100% REAL GCE WAEC GENERAL MATHEMATICS RUNZ ANSWERS | SIMPLIFIED WAEC GCE MATHEMATICS EXPO ANSWERS & RUNZ | 100% REAL GCE WAEC GENERAL MATHEMATICS RUNZ ANSWERSThis post give current answers on waec gce, 2017 waec gce expo waec gce answers, 2017 gce waec runz, 2017 gce waec exam answers 2017 2018 and beyond – SIMPLIFIED WAEC GCE MATHEMATICS EXPO ANSWERS & RUNZ – TODAY’S BIOLOGY PRACTICAL ANSWERS Neco answers.Waec Gce 2017 General Mathematics Obj And Theory Answers – Nov/Dec Expo Now Available.100% REAL GCE WAEC GENERAL MATHEMATICS RUNZ ANSWERS.

(a) Perimeter of garden
= 4x
= 4 x 5
= 20m.

(3b)
=Draw The Diagram=
Area Covered By Both Garden And path
= (x+1) (x+2)
= (5+1) (5+2)
= 6×7
= 42m^2.

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
(4a)
Given: Sinx = 3/5
i.e
=Draw The Diagram=
Using pythogoras triple 3,4,5
CosX + TanX/SinX
=4/5 + 3/4 ÷ 3/5
= 16+15/20 ÷ 3/5
= 31/20 ÷ 3/5
= 31/20 x 5/3
lastly = 31/12.

(4b)
=Draw The Diagram=
Reflex P’Q’T + R’Q’T + P’Q’R = 360 (sum at a point)
200 + 32 + P’Q’R = 360
232 + P’Q’R = 360
P’Q’R = 360 – 232
P’Q’R = 128°
From the Diagram
Q’R’U = P’Q’R = 128° (alternate angles)
Firstly X = Q’R’U + S’R’U
X = 128° + 180
X = 308°.

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
5a)
=Draw The Diagram=
1. | 2. | 3. | 4. | 5. | 6. |
——————————————
1|1,1|1,2 |1,3 |1,4 |1,5 |1,6 |
——————————————-
2|2,1|2,2 |2,3 |2,4 |2,5|2,6 |
——————————————–
3|3,1|3,2 |3,3 |3,4 |3,5|3,6 |
———————————————
4|4,1|4,2 |4,3 |4,4 |4,5|4,6 |
———————————————
5|5,1|5,2 |5,3 |5,4 |5,5|5,6 |
———————————————–
6|6,1|6,2 |6,3 |6,4 |6,5|6,6 |
————————————————

### B) pr(sum of outcome is 8) = 4/36 = 1/9 Bii) pr(product of outcom <10) = 30/36 = 5/6 Biii) pr(outcom contain atleast a 3) = 24/36 = 4/6 = 2/3. °°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°° (6a) 2 X (37)x = 75x 2 X (3x + 7)10 = (7x + 5)10 Finally 2 (3x + 7) = 7x + 5 6x + 14 = 7x + 5 14 – 5 = 7x – 6x 9 = x :- x = 9.

(6b)
Let The Number Of boys be X
Number Of girls in class = x + 5
GIVEN: x+5/+2 = 5/4
5(x+2) = 4(x+5)
5x + 10 = 4x + 20
5x – 4x = 20 – 10
x = 10
(i)Numbers of girsl in class
= x + 5
= 10 + 5
= 15
(ii)Total number of students
= x + x + 5
= 10 + 10 + 5
= 25
(iii)Probability of selecting a boy as class prefect
= 10/24
= 2/5.

°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
7a)
|PQ|^2 = |PB|^2 + |BQ|^2
|PQ|^2 = (5 – x )^2 + X^2
Furthermore |PQ|^2 = 25 – 10x + x^2 + x^2 [note ^ means Raise to power]
|PQ|^2 = 2x^2 – 10x + 25
NB: PQ = QR (side of a square)
Area of PQRS = PQ x QR
= PQ x PQ = PQ^2
Area of PQRS = (2x^2 – 10x + 25)m square
GIVEN: 2x^2 – 10 + 25 = 3/5 of 25
= 2x^2 – 10x + 10 = 0
= x^2 – 5 + 5 = 0.

Using General Formular mthod
X = -(-5) ± Square Root (-5)^2 – 4(1)(5)/ (2 x 1)
X = 5 ± Square Root 25 – 20/ 2
Furthermore X = 5 ± Square Root 5 / 2
X = 5 + Square Root 5 / 2 OR 5 – Square Root 5 / 2
X = 3.62 or 1.68.

7b )
L + a / n – 1 = d ———— (I)
2s = n (a + 1) —————- (2)
From eq (1)
L + a = d (n – 1) ————– (3)
Put eq (3 into 2)
2s = nd (n-1)
2s = d (n^2 – n)
S = d ( n^2 – n ) / 2 or
S = dn(n-1)/2.
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°.

#### 100% REAL GCE WAEC GENERAL MATHEMATICS RUNZ ANSWERS

8a)
By using pythagoras
r^2 = (r – 8)^2 + 32^2
r^2 = r^2 – 16r + 64 + 1024
16r = 64 + 1024
r = 1088/16
r = 68
r=68cm
8bi)
{st/pt/
q2=27^2 +12^2
q=sqrt27^2 + 12^2
q=sqrt729+144
q=sqrt873
q=29.5cm.
================================
Good Luck

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