Form 3 Maths paper 1 Exams and Marking Schemes Free

NAME…………………………………………………………………………

 

ADM NO………………………………………………………………………

 

SCHOOL………………………………….………………………………………..

 

DATE……………………………

121/1

MATHEMATICS

PAPER 1

2 HOURS

FORM THREE

121/1

MATHEMATICS

PAPER1

INSTRUCTIONS TO CANDIDATES

1. Write your name and index number in the spaces provided at the top of this page.
2. This paper consists of two sections: Section I and Section II.
3. Answer all questions in section I and any five questions from Section II.
4. Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
5. Marks may be given for correct working even if the answer is wrong.

 

Non- programmable silent electronic calculators and KNEC Mathematical tables may be used

FOR EXAMINER’S USE ONLY

SECTION I

1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	Total

 

SECTION II

17	18	19	20	21	22	23	24	Total

GRAND

             TOTAL

This paper consists of 13 pages.  Candidates should check the question paper to ensure that all the pages are printed as indicated and no questions are missing.

 

 

 

SECTION A (50 MARKS)

Attempt all the questions

 

1. Without using mathematical tables or calculators, evaluate the following leaving your answer as a fraction in its simplest form.             (3mks)

 

 

 

2. Two boys and a girl shared some money .The elder boy got of it, the younger boy got of the remainder and the girl got the rest. Find the percentage share of the younger boy to the girl’s share. (4mks)

 

 

 

 

 

3. Solve for t in the equation

9^t+1 +3^2t=30     (3mks)

 

 

 

4. The exterior angle of a regular polygon is (c – 50)° and the interior angle is (2c + 20)°. Find the number of sides of the polygon.                                (3 mks)

 

 

 

5. A salesman is paid a salary of Sh. 10,000 per month. He is also paid a commission on sales above Sh. 100,000. In one month he sold goods worth Sh. 500,000. If his total earning that month was Sh. 56,000. Calculate the rate of commission. (3 mks)

 

 

 

 

 

 

 

 

6. In a book store, books packed in cartons are arranged in rows such that there are 50 cartons in the first row, 48 cartons in the next row, 46 in the next and so on.

• How many cartons will there be in the 8^th row? (2 mks)

 

 

 

 

 

 

 

• If there are 20 rows in total, find the total number of cartons in the book store

(2 mks)

 

 

 

 

 

 

7. A rectangle whose area is 96m² is such that its length is 4metres longer than its width.

Find

• It dimensions (2 mks)

 

 

 

 

 

 

• Its perimeter (1 mk)

 

 

 

 

8. (a) Find the gradient of a straight line joining the points P(2,3) and Q(8,-6) (1mk)

 

 

 

 

 

 

1. b) Hence find the equation of the line through P perpendicular to line PQ. [3mks]

 

 

 

 

 

 

 

 

9. Given that OP ⃗= 2i + 3j and 𝑂𝑄⃗= 3i – 2j. Find the magnitude of PQ correct to three decimal spaces.

(3mks)

 

 

 

 

 

10. Solve the following inequality and state the integral solutions.             (2mks)

 

 

 

11. Given that is a perfect square. Find the value of k.       (3 mks)

 

 

 

 

 

 

 

 

12. Two matrices A and B are such that A= and B =  given that the determinant of AB = 10, find the value of k.                                                 (3mks)

 

 

 

 

 

13. Omwando borrows sh. 90,000 for 5 years at 6 ½ % simple interest p.a.What amount does he have to pay at the end of that time? (3mks)

 

 

 

 

14. Given that log a = 0.30 and log b = 0.48 find the value of . (3mks)

 

 

 

 

15. Find the value of x in the equation.

Cos(3x -180⁰) =    For the range  x 180⁰         (3 mks)

2

 

 

 

 

 

 

 

 

16. Given that the dimensions of a rectangle are 20.0cm and 25.0. Find the percentage error in calculating  the                                                                     (3 marks)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

SECTION B (50 MARKS)

Attempt  5 questions only

 

• Water flows through a circular pipe of cross-sectional area of 6.16cm² at a uniform speed of 10cm per second. At 6.00 a.m. water starts flowing through the pipe into an empty tank of base area are 3m².

8. What will be the depth of the water at 8.30 a.m.?                                  (5 marks)

 

 

 

 

 

 

 

 

 

 

1. If the tank is 1.2m high and a hole at the bottom through which water leaks at a rate of 11.6cm³ per second. Determine the time at which the tank will be filled.                                                                                                                               (5 marks)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

• Using a ruler and a pair of compass only, construct a triangle ABC such that AB = 8cm, BC = 6cm and <ABC = 30⁰ (2mks)

 

 

 

 

 

 

 

 

 

 

1. Measure the length AC (1mk)

 

 

1. Construct a circle that touches sides AB, BC and AC (2mks)

 

 

 

 

 

 

 

 

 

1. Measure the radius of the circle. (1mk)

 

 

 

 

1. Hence or otherwise calculate the area of the triangle not in the circle. (4mks)

 

 

 

 

 

• A field was surveyed and its measurements recorded in a field book as shown below.

E 40   C 40

F 100 80 60 40 20 A

D 50   B 30

 

(a)Using a scale of 1cm to represent 10m, draw a map of the field.                        (4mks)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(b)       Calculate the area of the field.

(i) In square metres.                                                                           (4mks)

 

 

 

 

 

 

 

 

 

 

 

 

(ii) In hectares.                                                                                   (2mks)

 

 

 

20. . The table below shows the income tax rates in a certain year.

Total income in k£per annum	Rate in shs per pound
1-3900 3901-7800 7801-11,700 11701-15600 15601-19500 Over 19500	2 3 4 5 7 7.5

 

Mrs.Masau earned a basic salary of ksh18600 per month and allowances amounting to ksh 7800 per month. She claimed a personal relief of ksh 1080 per month.

Calculate:

1. Total taxable income in k£ p.a                      (2 marks)

 

 

 

 

 

1. i) the tax payable in ksh per month without relief (4marks)

 

 

 

 

ii)the tax payable in ksh per month after  relief                                                     (2marks)

 

 

 

 

1. Musau’s net monthly income                                                                   (2marks)

 

 

21 An arithmetic progression of 41 terms is such that the sum of the first five terms is 560 and the sum of the last five terms is -250. Find:

(a) The first term and the common difference                                              (5mks)

 

 

 

 

 

 

 

 

(b) The last term                                                                                                   (2mks)

 

 

 

(c) The sum of the progression                                                                      (3mks)

 

 

 

• Three towns P, Q and R are such that P is on a bearing of 120⁰ and 20 km from Q, town R is on a bearing of 220⁰ and 12km from P.
• Using a scale of 1cm to represent 2km draw and locate the position of the three towns (3 mks)

 

 

 

 

 

 

 

 

 

 

(b) Measure

1. The distances between Q and R  in kilometres                                                 (2mks)

 

 

 

 

1. The bearing of P from R (1mk)

 

 

• The bearing of R from Q (2mks)

 

 

(c) Calculate the area of the figure bounded by PQR                       (2mks)

 

 

23. 23.

500

8cm

2.82cm

7cm

B

C

D

A

6cm

In the figure below (not drawn to scale). AB = 8cm, AC = 6cm, AD = 7cm, CD = 2.82cm and angle CAB = 50⁰

 

 

 

 

Calculate (to 2 decimal places)

(a) The length BC,                                                                                         (2mks)

 

 

 

(b) The size of angle ABC                                                                             (3mks)

 

 

 

(c) The size of angle CAD                                                                             (3mks)

 

(d) The area of triangle ACD.                                                                              (2mks)

 

 

 

 

 

• The figure below shows a tumbler with diameters 6cm and 10cm and height 15cm.

 

 

 

 

 

 

 

 

• If it is filled with water, what area is in contact with water? (7 mks)

 

 

 

 

 

 

 

 

 

 

 

 

1. Find the volume of the tumbler. (3 mks)

 

 

__________________________________________________________________

F3 MATHS PP1 TERM 3 MARKING SCHEME

 

2. Share of elder boy =

Share of younger boy=

=

Girls share =

% share of younger boy to girls share

 

 

 

3. 3^2(t+1)+3^2t=30

3^2t . 3²+3^2t =30

3^2t(3²+1)=30

3^2t×10=30

3^2t=3¹

2t=1

t= ½

 

 

 

4. 2x + 20 + x-50=180

3x – 30 = 180

3x = 210

3        3

x = 70

Each exterior angle = 70 – 50

= 20⁰

No. of sides = 360

20

= 18

 

5. Commission = 56,000 – 10 000 = Ksh. 46,000   Sales above 100 000 = 500 000 – 100 000 =  Ksh. 400 000   Rate of commission = 46000 x 100% 400 000   = 11.5%

M1     M1           A1

 

 

 

 

6. 50,48,46,……………

= 50 + 7x (-2)

= 36

=  (2 x 50 + (20 – 1) (x – 2)

= 620

 

 

7. (a) x(x + 4) = 96

+ 4x – 96 = 0

(x-8) (x + 12) = 0

x = 8

Length = 12

Width = 8

 

(b) Perimeter = 2 (8 + 12) = 40m

 

 

8. M₁ = y

x

= 3 + 6

2 – 8

= 9 = -3

-6     2

M₁ x M₂ = -1

–  x m₂ = -1

M₂ =

Taking (x,y) and P(2,3)

y-3 = 2

x- 2   5

3y – 9 = 2x -4

3y = 2x + 5

y = 2x + 5

3       3

 

9. PQ = q – p

~   ~

= (3ɩ -2j)- (2ɩ +3j)

= 3ɩ – 2j – 2ɩ- 3j

= I – 5j

/PQ/=

=

= 5.099

 

 

10. 12 – 2x ˃ 18x – 8

= 20x ˃ – 20

x ˂ 1

18x – 8 ≥ -28 – 2x

20x ≥ – 20

X ≥ -1

-1 ≤ x ˂ 1

 

Integral solutions: 01, 0.

 

 

 

• b² = a.c

2

 

 

-20² = 25k

2

100 = 25 k

K = 100

25

= 4

 

 

12

 

AB= -2(K+12) -9 (2K-16)=10 -2K-24-18K+144=10 -20K = -110

∵K=5.5

 

 

13 I=

90,000 x 6.5 x 5

100 x 2

= sh.29,250

A =(90,000+29,250)

=SH. 119,250

 

14

= 2(0.48) – 0.30 0.96 – 0.30

= 0.66

 

15. 3x – 180 = 30 or 330

3x – 180 = 30

3x = 210

x = 700

 

OR

3x – 180 = 330

3x = 510

x = 170⁰

16.      Min Area = (19.95( (24.95) = 497.7525 Max. Area = (20.05)(25.05) = 502.2525   502.2525 – 497.7525 2   2.25  x 100

• = 0.45%

 

 

17. Time of = 2 ½ hrs Flow Volume in 2 ½ hrs = 6.16 x 10 x 2 ½ x 3600 = 554400 cm3   Volume of tank = 3h = 554400 10000   H =  554400 m 30000   = 18.48m	B1   M1     M1     M1     A1
Volume in per sec. = 6.16 x 10 – 11.6   = 61.6 – 11.6 = 50cm3   Volume of tank = 1.2 x 30000 x 100   Time = 3600000  sec 50 = 72000 3600 = 20 hrs	M1   A1   M1   M1       A1                             (10)

 

 

 

 

 

18

Triangle ABC

AC = 4.1cm

Bisecting <S

Circle

Radius = 1.2cm

Area = ½ x 8 x 6 sin 30⁰ – x 1.2²

= 4 x 6 x 0.5 – 4.5257

= 12 – 4.5257

 

= 7.4743

 

 

 

 

 

 

 

19

 

 

 

 

 

 

 

21

(a) Sum of arithmetic progression   Last five terms     term is a + 40d term is a + 39d term is a + 38d term is a + 37d term is a + 36d total   Solving (i) and (ii) simultenously;     (b) Last term is  a + 40d   (c)

M1                   M1           M1     A1       A1   M1   A1           M1       M1 A1

Formation of each equation                             Solving two equations simultaneously For common difference   For the first term

 

 

 

 

 

 

 

22.a)

 

10. b) i)10.8 x2 = 21.6 km

ii)040⁰+ 1

iii) 153⁰+ 1

1. c) A = ½ a b sin

= ½ x 12 x 20 sin 80⁰

= 118.18 km²

 

 

 

 

23

(a) cm (b) Let  be (c) Let  be (d) Area of ΔACD

M1 A1     M1     M1 A1   M1 M1 A1   M1 A1

Accept 47.940,47.960 depending on the method             22.890 is possible.

24

 

h    =     6

15 + h   10

 

10h = 90 + 6h

4h = 90

H = 22.5

 

H = h + 15

= 37.5

 

 

 

 

 

 

L = ² + 9

= .25

= 22.70

 

L = ² + 25

=

= 37.83

S.A = ( ²

 

= (3.142 x 5 x 37.83 – 3.142 x3 x 22.70) + (3.142.9)

= 380.3391 + 28.278

= 408.6111 cm²

1. b) Volume = AH – Ah

= (  x 3.142 x 25 x37.5)- (3.142 x 9 x 22.5)

= 981.875 – 212.085

= 769.79 cm³