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Candidate’s Examination Number ................................
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THE PRESIDENT’S OFFICE
REGIONAL ADMINISTRATION AND LOCAL GOVERNMENT
MOSHI MUNICIPAL COUNCIL
SECONDARY EDUCATION EXAMINATIONS SYNDICATE
041 BASIC MATHEMATICS
PRE-CERTIFICATE FOR SECONDARY EDUCATION EXAMINATION
Monday 23rd August 2021am Time: 3:00Hours
INSTRUCTIONS:
1) This paper consists of section A and B with a total of 14 questions
2) Answer all questions in section A and B
3) Each question in section A carries six (06) marks while each question in
section B carries ten.
4) Non programmable calculator, NECTA mathematical tables and graph papers
may be used.
5) All the working and answers for each section must be shown clearly.
6) You may use the following constant: π=3.14, R=64000km
SECTION A
1. (a) Express 0.567567567...................... in a form of
a
b
where a and b are
integers and b≠ 0
(b) Four bells ring at intervals of 8 seconds, 15 seconds, 12 second and 9
seconds. If they start together, how long will it take before they ring at the
same time again?
2. (a) Solve for x, given that (
1
5
)
x−1
(25)
x−2 =
1
625
(b) Determine the value of x in the following log5 (x+1) = log5 (x-3) + 1
3. (a) In a class of students,17participate in English debate, 12participated in
English and sports. If every student is required to participate at least one of
those two events, find the number of students who participate in: -
(i) English debate only
(ii) Sports only
(b) A bag contains 6 red balls and 8 blue balls. If one ball is drawn at random
from the bag and replacing before the second ball is drawn. Find the
probability of drawing: -
(i) Balls of different colour
(ii) Balls of the same colour
4. (a) Find equation of the line passing through point (3,4) and the mid-point of the
line segment joining the two points (5,8) and (11,-4) Give your answer in the
form of the ax+by+c =0
(b) Given that: ⏟a = 2i+5j and ⏟b=i-j find
i) z = a+b
ii) The unit vector of z
5. (a)The area of a regular 6-sides plot of land inscribed in a circular track is
720m2. Find the area of the track.
(b) Prove that AB̂C=AD̂C
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Candidate’s Examination Number ................................
Page 2 of 3
D
A C
B
6. (a)Three quantities x, y,and z are such that z is directly proportional to x. Given
that z=10 when x=5 and y=2. Find the value of x when y=12 and z=200
(b) The distance from Arusha to Dar es Salaam on the map is 6.8cm. If the scale
of the map is 1:500000 find the actual distance from Arusha to Dar es
Salaam
7. (a) In 2019 Mr. Massawe bought a computer which costs him Tshs.500,000/= he
used his computer for five years and decided to sell his computer as a second- hand computer to Mr. Faridi with a loss of 30% of its price when it was new
brand computer. What is the price of this second-hand computer?
(b) From the given information below
Opening stock .......... Ths34,430/=
Closing stock .......... Tshs26,720/=
Net purchases .......... Tshs212,290/=
Gross profit markup ...... 50%
Find (i) Cost of goods sold
(ii) Average stock
(iii) Sales
8. (a) Given the series -2+4-8+,...128. Find the tenth term of this series.
(b) The sum of the third term and sixth term of arithmetic progression is 38. If
the fourth term of this series is 18: -
(i) find the first term and common difference
(ii) find the sum of the first term of the series.
9. (a) Find the smallest angle in triangle below
(b) A ladder 24cm is learns against the wall of building. The top of the ladder is
17m above the ground. How far away is the foot of the ladder from the base of
the building?
10. (a) Solve the following quadratic equation by the method of completing square,
2x2 +7x+6 = 0
(b) For what value of N, the equation; 4x2 +
16
3
x + N is a perfect square?
SECTION B: (40 MARKS)
11. (a) Given the marks obtained by 20 students in mathematics test done by St.
Gasper Secondary School:
85, 87, 90, 87, 89, 90, 88, 89, 90, 87,
6cm
5cm 3cm
A
C
B
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Candidate’s Examination Number ................................
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90, 87, 87, 85, 86, 86, 86, 87, 89, 90.
Find (i) Mean mark. (ii) Mode (iii) Median (iv) Range.
(b) Prove that the angle subtended by a diameter in a semi-circle is a right angle.
(c) AB̅̅̅̅ is a diameter of a circle having a centre O and C is a point on the
circumference.
AB̅̅̅̅ = 29m and BC̅̅̅̅ = 21m. Find (i) AC̅̅̅̅ (ii)∠ ABC .
12. (a) Given two points: A(250N, 400E) and C(700N, 400E), find the angle subtended
by an arc AĈ and hence find its distance.
(b) Given two towns A and B, and all are found on the equator. If the longitudes
of A and B are 350E and 640Wrespectively, find the angle subtended by an arc
AB̂ at the center of the earth and hence determine its distance in kilometers.
(c) In the following cuboid, AB = 5 cm, BC = 12 cm and BG = 10 cm. Calculate:
(i) The length of AH (give your answer correct to one decimal place).
(ii) The angle CAH.
13. (a) Given B = (
2 x 5
−4 3 y
) and D = (
u 1 5
−4 3 9
), if B = D, find the values of
x, y, and u.
(b) (i) When a matrix is said to have no inverse?
(ii) If A =(
4 x
4 + x 3
) has no Inverse, find the value(s) of x.
(c) If T maps (3, 0) to (−3, 3) and (1, −3) to (−7, −11), and T is a matrix of
transformation, find the matrix T.
14. (a) A farmer can buy two types of food in cubic meters. Food I contain 20 units
of Vitamin A, 30 units of Vitamin B and 5 units of Vitamin C. Food II contains
10 units of Vitamin A, 30 units of Vitamin B and 10 units of Vitamin C. The
costs needed per Vitamins A, B and C are 460, 960, 220 shillings respectively.
A cost per each type is:
10,000/= and 15,000/= of Food I and II respectively.
(i) Formulate linear programming problem inequalities.
(ii) Find the minimum cost for each Type of Food.
(b) If f(x) =
x
3 − 2x2 +5
x
2 + 4x − 5
, when the function f(x) is not defined?
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Page 1 of 9
THE PRESIDENT’S OFFICE
REGIONAL ADMINISTRATION AND LOCAL GOVERNMENT
MOSHI MUNICIPAL COUNCIL
SECONDARY EDUCATION EXAMINATIONS SYNDICATE
041 BASIC MATHEMATICS MARKING SCHEME
PRE-CERTIFICATE FOR SECONDARY EDUCATION EXAMINATION
1. Let x = 0.567
1000x = 567.567.........................................................01mark
1000x-x = 567.567-0.567
999x = 567..........................................................01mark
x=
567
999
x=
21
37
.................................................................01mark
(b) Let’s find the LCM of 8,15,12 and 9
...............01mark
LCM= 2×2×2×3×3×5=360.........................................................01mark
360 c equals to 6 min they will ring together after 6 minutes..................01mark
2. (a) (
1
5
)
x−1
(25)
x−2 =
1
625
5-(x+1) X 52x-4 =5-4
52x-4(x+1) =5-4.........................................................01mark
Equate the powers,
2x-4-(x+1) =-4................................................01mark
2x-x-4-1 =-4
x-5=-4
x=1........................................................01mark
(b) log5 (x+1) = log5 (x-3) +1
Log 5 (x+1) –log5 (x-3) =1................................01mark
log5 (
x−1
x−3
) = 1 ........................................01mark
In exponent form
51 = x + 1
X -3
5 ( x- 3)= x+1
5x -15=x+1
4x=16
X= 4........................................................01mark
2 8 15 12 9
2 4 15 6 9
2 2 15 3 9
3 1 15 3 9
3 1 5 1 3
5 1 5 1 1
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3. (a) Let n(E) = number of who takes English, and
(S) = number of who takes sports
n (EՍS)= n(E) +n(S) –n(E∩S).....................................................01mark
30 = 17+ n(S) -12
30 = 5+n(S)
n (S) = 30 -5 .............................................................01mark
n (S) = 25
؞ Students who do English only are 17-12 = 5 .........................0.5marks
Students who do sports only are 25-12 =13............................0.5marks
(b) Let R represent red balls and
B represent blue balls
Using tree diagram,
8
14
B
8
14
B 6
14
B
Start
8
14
B ........................ 01mark
6
14
R
6
14
R
i) Probability of different colour = P(Band R) or P(R and B)
=
8
14
×
6
14
+
6
14
×
8
14
=
48
196
+
48
196
=
12
49
+
12
49
=
24
49
...................................01mark
؞ The probability of drawing two balls of different colour is 24
49
ii) Probability of the same colour = P(B and B) or P(R and R)
= P(BꓵB) + P(RꓵR)
= P(B) x P(B) + P(R) x P(R)
=
8
14
×
8
14
+
6
14
×
6
14
=
64
196
+
36
196
=
100
196
؞ The probability of the same colour = 25
49........................01mark
4. (a) Midpoint = (
x1+x2
2
,
y1+y2
2
) given the points (5, 8) and (11, -4)
Midpoint = (
5+11
2
,
8+ −4
2
)
= (16
2
,
4
2
)
= (8, 2).........................................................01mark
The equation of the point (3,4) and (8, 2)
Slope M = ∆y
∆x
=
2−4
8−3
=
−2
5
then from equation y = m(x + x1
) + y0
y =
−2
5
(x + 3) + 4 ..................................................................01mark
5y = −2x − 6 + 20
Equation of a line 2x + 5y − 14 = 0..........................01mark
(b) i) Z = a+b
Z= (2i+5j)+(i-j)
Z = 3i + 4j ...........................................................1.5marks
ii)The unit vector of Z = Z
|z|
=
3i+4j
√3
2+4
2
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Page 3 of 9
=
3i+4j
√25
Ẑ =
3i+4j
5
......................................1.5marks
5. (a) Given number of sides n= 6
Area = 720 m2 and radius r = ?
A = 1
2
nr
2
sin 360°
n
720 = 1
2
× 6 × r
2
sin 360°
n
720 = 3× r
2
sin 360°
n
r
2 =
720
3×0.8660
r = 16. 65 m
(b) To prove that AB̂C = AD̂C
D
C A
B
Required to prove that ΔADC≡ABC ..................................1mark
Proof: Argument Reason
(i) AB = AD Given
(ii) BC = DC Given
(iii) Ac = AC Common ..............................................01mark
Hence ADC≡ABC SSS theory
؞�A� ̂C = AD̂C by definition .........................................01mark
6. (a) zα
y
x
the z =
ky
x
Given that z=10, y=2 and z=5
Hence 10 = 2×k
5
, k=25
If Z = 200, y= 12, x =?
x =
ky
z
→
25×12
200
=
3
2
= �� ؞
3
2.....................................................03marks
(b) Given: scale 1: 500,000
Distance on a map = 6.8cm
Scale = Measurement on map
Actual measurement
.................................................01mark
1
500,000
=
6.8 cm
x
..............................................01mark
x = 500,000 × 6.8
؞ The actual distance from Arusha to Dar es Salaam= 34 km..........01mark
7. (a) Cost of computer = 500,000
% loss =30%
Loss made = 30 x 50000
= 150,000
Selling price = Cost price – Loss made
= 500,000 – 150,000
= 350,000/=
The price of second hand computer = 350,000/= ...................03marks
(b) (i) Cost of goods sold = Opening stock + Net purchase – Closing stock
= 34,430 + 212,290 – 26,720
= 246,720 – 26,720
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24m 17m
A
B C
Cost of goods sold = 220,000/=...............................01mark
(ii) Average stock = Opening stock + Closing stock
2
=
61159
2
Average stock = 30,575/=................................................01mark
(iii)Sales = Cost of goods sold + Gross profit but gross profit = 50
1000
× cost of sales
Sales = 220,000+ 50
100
× 220,000
= 220,000+ 110,000
Sales = 330,000/= ....................................................01mark
8. (a) -2 + 4 – 8 + ........ -128
WhereG1 = −2, r = −2, n = 10,G10 =?
From Gn = G1r
n−1
G10 = −2r − 2
10−1
= −2−2
9
= −2
10
The 10th term of the series = 1024..................................03marks
(b) (i) A3 + A6 = 38
A1+ 2d + A1+5d =38
2A1+7d =38 ......................................(i) .......................0.5marks
A4 = A1 + 3d = 18
A1 +3d = 18 ....................................(ii) ....................0.5marks
Solving the two equations simultaneously, (i) and (ii)
A1 =12 and d =2 ........................................0.5marks
(ii) Sn =
n
2
[(2A1 + (n − 1)d] ........................................0.5marks
S10 =
10
2
[2 × 12 + (10 − 1)2].....................................0.5marks
S10 = 5(24+ 9X2)
= 5(24 + 18)
= 5x 42
= 210 The sum of the first 10 terms = 210 ...................0.5marks
9. (a) From triangle ABC
The smallest angle is the angle opposite to smallest side, then; from the
triangle, the smallest angle is A
a
2 = b
2 + c
2 − 2bccosA
Cos A= b
2+c
2−a
2
2bc
Given that a = 3 cm, b = 6 cm and c = 5 cm
CosA =
6
2+5
2−3
2
2×6×5
=
52
60
CosA = 0.8660
Cos-1(0.8660)=A (03marks)
י 55 290 = A
The smallest angle of ΔABC = 290 55''
(b)
From the Pythagoras theorem
(BC)2= (AC)2 – (AB)2
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Page 5 of 9
= (24)2 – (17)2 = 576 – 289
(BC)2 = 287
BC = √287 (03marks)
BC = 16.94 m
The foot of the ladder from the base of the building = 16.94m
10.(a) 2x
2 + 7x + 6 = 0
2x
2 +
7
2
x + 3 = 0
x
2 +
7
2
x + (
7
4
)
2
= −3 + (
7
4
)
2
(x +
7
4
)
2 =
−48+49
16
=
1
16
(x +
7
4
) = √
1
16
=
1
4
then, (03marks)
x = ±
1
4
−
7
4
Therefore x =
−3
2
or x = 2
(b) 4x
2 +
16
3
x + N Where a = 4, b = 16
3
and c = N
For perfect square b2 = 4ac
(
16
3
)
2 = 4 × 4 × N
N =
256
9×16
The value of N = 4
9
11.Solution:
(a) Frequency Distribution table
Marks (x) frequency fx
85 2 170
86 3 258
87 6 522
88 1 88
89 3 267
90 5 450
∑f = 20 ∑fx = 1755
[01mark]
(i) Mean = ∑ fx
∑ f
x̅ =
1755
20
. [01mark]
(ii) Median = 87 .[01mark]
(iii) Mode =87 (number which occurs more frequently) .[01mark]
(iv) Range= highest mark – lowest mark
90-85=5. [01mark]
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Page 6 of 9
(b) Consider the figure below:-
C If O is the centre of a circle.
Then required to prove that ∠ ACB = 900
[0.5 mark]
A B From:< AOB = 2ACB(Angle at the centre)[01 mark]
But < AOB = 1800
(Angle formed on a straight line)
1800 = 2ACB, thus,∠ ACB = 900Proved. [01mark]
(c) C
A B [0.5 mark]
If AB = 29m,BC = 21m. Required to find
(i) AC. Using Pythagoras theorem. AC̅̅̅̅ = √(29)
2 − (21)
2 → AC̅̅̅̅ = 20m [01 mark]
(ii)Using trigonometric ratios ∠ ABC = cos−1
21
29
→ ∠ ABC = 43. D
0
[01 mark]
12. Solution
(a) Given that: A(250N, 400E)andC(700N, 400E).
Angle subtended by an arc AC is700 − 250 = 450
[0.5 mark]
Distance =
π_α
1800
[0.5 mark]
Distance =
22
7
× 6,370km ×450
`
1800
[01 mark]
Distance= 5, 005km. [01 mark]
12.(b) speed =
distance
time ,then time =
distance
speed [0.5 mark]
time =
5005km
715km/hr [01 mark]
Time = 7hrs. [01 mark]
(c) (i) consider triangle ABC below
From Pythagoras theorem
AC2 = AB2 + BC2
(0.5 mark )
AC2 = 5
2 + 122
AC2 = 25 + 144
AC2 = 169 (0.5 mark )
AC = √169
AC = 13 cm (0.5 mark )
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Page 7 of 9
Consider triangle ACH below
From Pythagoras theorem AH2 = AC2 + CH2
(0.5 mark )
AH2 = 132 + 102
AH2 = 169 + 100
AH2 = 269 (0.5 mark )
AH = √269
AH = 16.4 cm (0.5 mark )
(ii) Consider triangle CAH
tan A =
opposite
adjacent
tan A =
10 cm
13 cm
(0.5 mark )
tan A = 0.769
A = tan−1
(0.769) (0.5 mark )
A = 37034′
(0.5 mark )
13. (a) If B = D,
B = (
2 x 5
−4 3 y
) and D = (
u 1 5
−4 3 9
) then by making comparison x = 1, y = 9
and u = 2 [01 mark] @
(b) (i) When a Matrix is said to have no inverse?
It is when determinant equals to zero [01 mark]
(ii) If A =(
4 x
4 + x 3
) has no Inverse, Find the value(s) of x.
|A| = |
4 x
4 + x 3
| = 0 (For a matrix to have no inverse). [01 mark]
12 − x(4 + x) = 0 [01 mark]
12 − 4x − x
2 = 0,then by solving x = 2 or − 6 [01 mark]
(c) If T maps (3, 0) to (−3, 3) and(1, −3) to (−7, −11), and T is a matrix of
transformation, find the matrix T.
let T = (
a c
b d
) [01 mark]
(
x
′
y
′) = (
a c
b d
) (
x
y
).
(
−3
3
) = (
a c
b d
) (
3
0
) . 3a = −3 a = −1. And 3b = 3 b = 1
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Page 8 of 9
(
−7
−11) = (
a c
b d
) (
1
−3
), a − 3c = −7, a = −1, c = 2, b − 3d = −11,d = 4
Then (
a c
b d
) = (
−1 2
1 4
) [02 mark]
14. (a) Let units for Food I be represented by x and for Food II be represented by y.[1 mark]
Table of summary:
Food contents in kg-units.
Vitamin A Vitamin B Vitamin C cost
Food type I 20m3 30m3 5m3 10,000/=
Food type II 10m3 30m3 10m3 15,000/=
Available units. 460m3 960m3 220m3 f(x, y)
Now, formulate constraints/inequalities.
20x + 10y ≥ 460
30x + 30y ≥ 960
5x + 10y ≥ 220 [1.5 mark]
Add non negative constraints.
x ≥ 0 and y ≥ 0.
Objective function:
f(x, y) = 10, 000x + 15, 000y. [0.5 mark]
Then convert/change the constraints to linear equations.
2x + y = 46(0,46) (23,0)
x + y = 32(0,32) (32,0)
x + 2y = 44(0,22) (44,0)
Graph for presenting L.P.P.
Y
46 A
32
FEASIBLE REGION
22 B [04 mark]
x = 0 C
D
0- 23 y = 032 44X
2x + y =46x + y = 32 x + 2y = 44
Page 12 of 12
Page 9 of 9
Food type I required is 20 m3 and food type II is 12m3 of all food contents
(Vitamins) which yields a minimum cost of Tsh. 380, 000/= [01 mark]
(b) f(x)is not defined whenthe denominator = 0 that is x
2 + 4x – 5 = 0 02 mark]
Corner point f(x, y) = 10, 000x + 15, 000y.
A(0,46) 690,000
B(14,18) 410,000
C(20,12) 380,000
D(44,0) 440,000