EXAM TIME:
OBJECTIVES – 10.00 am – 11.45 am
ESSAY – 12.00noon – 2.30 pm
Mathematics OBJ:
1-10:CCDEAEBDCE
11-20:AAACEECACE
21-30:CBBAACEBAD
31-40:CDABEEBCAD
41-50:ECCECEDEAC
51-60:EBABCCCCCB
(12)
TABULATE:
Marks | frq | c.f | boundary
11-20 | 8 | 8 | 10.5 – 20.5
21-30 | 6 | 14 | 20.5 – 30. 5
31-40 | 10 | 24 | 30.5 – 40.5
41-50 | 12 | 36 | 40.5 – 50.5
51-60 | 8 | 44 | 50.5-60.5
61-70 | 6 | 50 | 60.5 – 70.5
(12b)
DRAW THE CUMULATIVE FREQUENCE CURVE: CLICK HERE FOR THE IMAGE
(12ci)
Medium mark = N + ½
= 50 + 1/2 = 51/2
= 25. 5
= 48
(12cii)
Lower quartile Q1
= ¼ x 50
= 12.5
= 28
(12ciii)
Upper quartile Q2
= ¾ x 50
= 37.5
=53
(12civ)
Both percentile
70/100 x 50
= 35
= 50
Hence we have A+B+C = 180
50+80+C = 180
130+C = 180
C = 180 – 130
C = 50
10ai)
The bearing of B from C = 90 – 50 = 40°
(10ii)
Bearing of A from B
90+50(alternate angle to A)
= 140°
(10iii)
Distance between B and C
Using sine rule
C/sinC = a/SinA
40/sin50 = a/sin50
Cross multiply
asin50 = 40sin50
a = 40km
Hence distance between B and C = 40km
(10iv)
Using cosine rule
b²= a²+c² – 2acCosB
where a = 40, c= 40 and B=80
b² = 40²+40² -2*40*40cos80
b² = 1600+1600 – 3200(0.1736)
b² = 3200 – 555.52
b² = 2644.48
b = √2644.48
b = 51.42km
(10v)
Height of ΔABC
Draw the triangle: CLICK HERE FOR THE IMAGE
Area of the triangle
=1/2acSinB
=1/2*40*40sin80
=800(0.9848)
=787.84km²
Hence 1/2bh=787.84
bh=2*787.84
h=2*787.84/b
h=2*787.84/51.42
h=30.64
=31
(6b)
Area of triangle = ½ x b x h
Let the Acheal Area = x
Area = ½ x b x h
Base = x – 9x/100 = 93/100
Height = 9x/100 + x = 107x/100
Area = ½ x 93x/100 x 107x/100
=963x/2000
% error = actual Area – wrong/actual area x 100
= x – 963x/2000 /x 100
= 20000x – 963x/20000 x 100
= 19037/20000 x 100
= 95.185%
(6c)
p/100 + 2p + 7 = 11.02 x 100
p + 200p + 700 = 1102
201p = 1102 – 700
201p = 402
P = 402/201
P = N2
P = 200K
(8bi)
T = 2π square l/g
Dividing both side by 2π
T/2π = square root of L/g
By squaring both side
(T/2 π)² = (L/g)²
T²/4 π² = l/g
Cross multiplication
gT²/T² = 4π²L/T²
g = 4π²L/T²
(8bii)
T = (0.4)1/2 = square root of 0.4
L = 0.04 T1 = 3.14
G = 4π²L/T2
= 4 x (3.14)² x 0.04/(0.4)²
= 4 x 9.8596 x 0.04/0.4
= 1.578/0.4
g = 3.94
(9b)
z^2 -25/z^2-9z+20
If Z is undefined
Z^2-9z+20=0
Using factorization method
-4z and -5z
Z^2 -4z-5z+20=0
Z(z-4) – 5(z-4)=0
(z-4)(z-5)=0
Z-4=0 or Z-5=0
Z=4 or Z=5
Z=4 or 5
Z is undefined when it is equal to 4 or 5
Pls objective
It’s good
obj /theory
good
good job