NYAMIRA SUB-COUNTY JOINT EVALUATION EXAMS
MATHENATICS 121/1
MARKING SCHEME
JULY/AUGUST 2014
1.
NO
Log
4.283
0.6317
0.009478 3.9767
+
x 2
5.9534
4.6851
Log 9.814 1.9964
4.5887
2
2.2944
M1
M1
M1
10-2x ant 2944
10-2x1.9697
A1
0.019697
2.
V .S .F 
4752

1408 8
l.s. f  3 27

M1
27
8
3
2
2
9
 3
A.S.F=   
4
2 
l arg ercylinder 
9
 352
4
 792cm 2
3.
A1
03
1 22
x
x 4.2 x 4.2 x9.7  179.256
3 7
1 22
x 4.15 x 4.15 x9.65  174.1112738
Minimum volume = x
3 7
1 22
x 4.25 x 4.25 x9.75  184.4955357
Maximum volume = x
3 7
1
Absolute error = 184.4955357  174.1112738
2
5.192130957
x100
% error=
179.256
Actaul volume =
=2.896%
4.
M1
M1
M1
M1
A1
Ax  3 y  6  0
3 y  Ax


3
3
3
6
y  A x2
3
g1  
A
3
3
A
g2 
5x  7 y  k  0
(4,3)
5 x 4  7 x3  k  0
20  21  k  0
41  k  0
k  41
5 x  7 y  41  0
7y  5
41

x
7
7
7
 5 x 41
y

7
7
5 3

7
A
g2 
5.
 5 A 21

5
5
21
A
5
2
10(9 x  4 y 2 )
M1
10(3x  2 y )(3x  2 y )
A1
02
P
6
P
13cm
112.62o
30o
R
13
37.38o
θ
5
cos  
10
5
R
S
5
13
cos   0.3846
  cos 1 0.3846
  67.38
13
sin 

sin 30 sin 112.62
M1
7.
sin  
13 sin 112.62
sin 30
sin  
11.999
sin 30
M1
Sin θ=23.99997
Sin θ=24
A1
(21x  12 y )  29) x 4...........(i )
(12 x  5 y  19) x7.................(ii )
84 x  48 y  116
84 x  35 y  133
M1
 83 y  249
 83 y  249
y3
substitutey  3in (i )
A1
21x  12(3)  29
21x  29  36
21x  7
x
8
B1
1
3
Let R=resistance in Newton
V=speed in Kmh-1
M1
R=a+KV2 where a and k are both constants
a+1600k=530
And a+3600k=730
200=2000k
k
1
10
M1
Substituting in (I)
530  a  1600 x
1
 a  160
10
M1
a=370
R= 370 
1 2
v
10
When V=70
1
R  370  x 4900
10
=370+490
Resistance=860N
A1
04
√ formation
of
simultaneo
us equation
9.
10
60 x 24 x9 x6
9 x15 x160 x10
M1
60 x 24 x9 x6
9 x15 x160 x10
M1
9
25
A1
03
150 workers x6hr=900hr
30 days =900x30hrs
=27000hrs
1
 27000hrs
4
1  27000 x 4  108000hrs
M1
Number of hrs reguired =10800-27000
=81000
Remaining days=80-30=50 days
M1
81000
50 x 6
Number of workers=270
Number of workres =270-150
=120days
11.
dis tan ce 
1
x80 x(16  24)
2
dis tan ce 
1
 80 x(16  24)  1600m
2
 1600m
decelaration 
20  80
24  20
deceleration  80
4
A1
3
B1
M1
A1
03
 20m / s
12.
13
x-4<3x+2<2(x+5)
x-43x+2
-3x
3x+2<2(x+5)
X<8
-3x<8
Selling price=
87
xsh.800
100
=sh.696
Buying price=
sh.100
x696
120
M1
A1
2
M1
=sh.580
A1
2
14.
(a) CDF=110o-60=50o
(Two interior angles add up to one opposite exterior)
(b)BDE=
B1
B1
50
 25 0
2
B1
ABD=25o(alternate angles are equal)
15.
Let J represent James and P represent Peter
(J-3)=3(P-3)
J-3=3p-9
J-3P=-6…………..(i)
(J+5)+(P+5)=76
J+P=66…….(ii)
√equati
ons
B1
J+P=66
J-3P=-6
4P=72
P=18
J+18=66
J=48
16.
M1
A1
3
AB=B-A
 2  1
   
  3  2 
= 
M1
 1 

  5
= 
/AB/= 12   5
M1
= 26
=5.10
A1
2
17.
Tap
Rate of filling
03
cold
hot
drain
1
3
1
5
4
15
Rate of filling
1 1 4 534 4
 


3 5 15
15
14
M1
M1
It two minutes amount 1 the tank
2x
4
8

15 15
8
7

(b) Volume remaining = 1 
15 15
1 1 8
Rate of filling =  
3 5 15
A1
M1
A1
7
15
7 15 7
Time = 15 
x  min utes
8 15 8 8
M1
7
8
Amount of water delivered by the cold water tap in 2 min utes
A1
23
x60
8
 34500cm 3
200 x
Let the volume be equal to x
Water delivered by the cold water tap
2
7
x
x  34500
3
24
M1
23
x  34500
24
34500 x 24
x
23
=34500
X=36000cm3
X=36 litres
A1
10
36000
 120cm / s
5 x60
18
B1-A
B1-B
B1-C
B1-D
B1
B1
(b)(i)029o1o
(ii) 5.2km0.1
(iii)5.5km0.1
(c)CD=5.2km+
M1
DA=3km
ADC=5.2+3=8.2
Total time =
30  20
50 5
hrs
 hrs
60
60 6
Speed = 8.2 
8.2+
M1
A1
10
5
6
6
5
 9.84km/ hr
19
Let the length of the train be equal x
Distance covered =5+x
Relative speed= 30-25km/hr
=
M1
25
m/s
18
M1
M1
Time = 90 seconds
25
x90
18
x  120m
48
(b) time = 4
/ 60  0.08hrs
60
5 x 
A1
M1
Distance =5x0.08x1000
=400m
(c) speed=30km/hr
M1
45
Distance= 30x
60
A1
=22.5kh/hr
M
45
 18.75km
Distance = 25 x
60
M1
22.5-18.75=3.75
A1
3.75
Time =
25
10
=0.15x60
=9minutes
20.
J
2/3
3/10
1
7/10
5
4/5
ONJ……….all correct
1/3
J
1/3
ON1J
J
2/3
3/10
7/10
ONJ1
J1
2/3
1/3
2/3
B1
1
J
ON1J1
J1
O1N1J
J
1/3
J1
(b)(i) P(ONJ)
1 3 2
x x
5 10 3
6
3
1
or or
=
150 75 25
O1N1J1….all correct
B1
=
(ii) P(ON1J1) or P(O1NJ1) or P(O1N1J)
M1
A1
1 7 1  4 3 1  4 7 2
x    x x    x x  for any correct two
 5 10 3   5 10 3   5 10 3 
= x
M1
M1
7
12
56
for adding


150 150 150
75
25
5
1
=
or or or
150 50 10 2
=
A1
(iii) P(ONJ) or P(ONJ1) or P(ON1J) or P(O1NJ)
1 3 2 1 3 1 1 7 2  4 3 2
 x x  x x  x x  x x 
 5 10 3   5 10 3   5 10 3   5 10 3 
M1
M1
6
3
14
24



150 150 150 150
21.
A1
10
47
150
Vo AB 12


y
EF 6
y6 2

y
1
2y  y  6
y6
Volume of frustrum =
6
E
8
h
8
A 3
h 2  8 2  2 .5
12
2
h  7.599cm
1
10  57.559 x 2
2
 113.4cm 2
A
1
 1

 12 x10 x12   6 x5x6
3
 3

=88-60
=28cm3
F
3
B
5
E
(b)
8
A 2.5
H
h 2  82  3
8
10
Total area=133.5+113.4+30+120
=396.9cm2
22.
F
2.5 B
2
h  7.416cm
1
12  67.1416 x 2
2
 133.5cm 2
A
23
24.
M1
√exp. For
mid point
M1
√exp. For
mid point
y  3 .5  5

Eqn of to AB:=
x  2 .5
1
1
y  x3
5
A1
√coordinates
for both
y=-5x+29
Grad. Of AC=-5
Grad of to AC 1/5
M1
 3  8 1 2 
,

2 
 2
(a) Mid point of AB= 
Midpoint of AC=(5,5 1.5
 3  2 1 6 
,


2 
 2
=(2.5,3.5)
(b) Grad of AB= 1/5
Grad. Of to AB=-5
Both
eaquation
y  3 .5 1

Eqn of  to AC:
x  2 .5 5
y
1
x3
5
(c)
1
x3
5
y  5 x  29
y
M1
A1
Atleast
one. Exp.
For centre
of reading
1
x  3  5 x  29
5
5.2 x  26 x  5
X=5
A1
1
y  (5)  3  4
5
centre(5,4)
 3  5 
M1
A1
  2

r=       
1  4   3
/r/=
 2
2
 
 3
2
= 13units
Eqn: (x-5)2+(y-4)2=(√13)2
x 2  10 x  25  y 2  8 y  16  13  0
x 2  1x  y 2  8 y  28  0
√center and
radius
(both)
√exp for
eqn
√simplifica
tion of eqn