Mathematics P2 Questions and Answers - Mangu High School Trial Mock Exams 2021/2022

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INSTRUCTIONS TO CANDIDATES

  1. Write your name and index number in the spaces provided above.
  2. Sign and write date of examination in the spaces provided above.
  3. This paper consists of two sections; Section I and Section II.
  4. Answer All questions in Section I and only Five questions from section II
  5. All answers and working must be written on the question paper in the spaces provided below each question.
  6. Show all the steps in your calculations giving answers at each stage in the spaces provided below each question.
  7. Marks may be given for correct working even if the answer is wrong.
  8. Non-programmable silent electronic calculators and KNEC Mathematical tables may be used except where stated otherwise.
  9. Candidates should answer questions in English.

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  ________



QUESTIONS

SECTION 1 (50 MARKS)
Attempt all questions.

  1. Factorise x2 – y2, hence evaluate 32822 - 32722 (3mks)
  2. Find cos x – Sin x, if tan x= ¾ and 90º≤ x ≤360º (3mks)
  3. Expand [1-2x]6 up to the fourth term. Hence use your expansion to evaluate (1.02)6 to four decimal places. (4mks)
  4. The average of the first and fourth terms of a GP is 140. Given that the first term is 64. Find the common ratio. (3mks)
  5. Make b the subject of the formula. (3mks)
    A = √   bd  
            b2 - d
  6. Two variables P and Q are such that P varies partly as Q and partly as the square root of Q. Determine the equation connecting P and Q. When Q=16, P=500 and when Q = 25, P = 800 (4mks)
  7. Calculate the interest on sh 10,000 invested for 1 ½ years at 12 % p.a. Compounded semi-annually. (3 mks)
  8. Given that x=2i+j-2k, y= -3i+4j-k and z =5i + 3j+2k and that P= 3x-y+2z, find the magnitude of vector p to 3 significant figure (4mks)
  9. Eighteen labourers dig a ditch 80m long in 5 days. How long will it take 24 labourers to dig a ditch 64 m long? (3mks).
  10. The expression 1+ x/2 is taken as an approximation for √1+x. Find the percentage error in doing so if x = 0.44 (3mks)
  11. The matrices A = [3 0] and B = [a  b]  are such that AB = A + B Find a, b, and c. (3mks)
                                [0 4]              [0  c]
  12. Simplify (3mks)
    2x2 – x-1
       x2 – 1
  13. On map of scale 1:25000 a forest has an area of 20cm2. What is the actual area in Km2 (3mks)
  14. In the figure below, DC = 6cm, AB = 5cm. Determine BC if DC is a tangent. (3mks).
    14 YATFTYDA
  15. Evaluate without using logarithm tables.
    3 log10 2 + log10 750 – log10 6 (3mks)
  16. A bag contains 10 balls of which 3 are red, 5 are white and 2 green. Another bag contains 12 balls of which 4 are red, 3 are white and 5 are green. A bag is chosen at random and a ball picked at random from the bag. Find the probability that the ball so chosen is red. (4mks).

SECTION II (50 MARKS)
Answer any five questions in this section.

  1. Income tax is charged on annual income at the rates shown below.
    Taxable Income K£            Rate (shs per K£)
    1 – 1500                                       2
    1501 – 3000                                 3
    3001 – 4500                                 5
    4501 – 6000                                 7
    6001 – 7500                                 9
    7501 – 9000                                10
    9001 – 12000                              12
    Over 12000                                 13
    A certain headmaster earns a monthly salary of Ksh. 8570.. He is entitled to tax relief of Kshs. 150 per month.
    1. How much tax does he pay in a year. ( 6 mks)
    2. From the headmaster’s salary the following deductions are also made every month;
      W.C.P.S 2% of gross salary
      N.H.I.F Kshs. 1200
      House rent, water and furniture charges Kshs. 246 per month.
      Calculate the headmaster’s net salary. (4 mks)
  2.                                
    1.                                
      1. Taking the radius of the earth, R = 6370 km and π = 22/7 calculate the shorter distance between the two cities P (60ºN , 29ºW) and Q (60ºN, 31ºE) along the parallel of latitude. (3mks)
      2. If it is 1200Hrs at P, what is the local time at Q. (3mks)
    2. An aeroplane flew due South from a point A (60ºN, 45ºE) to a point B. The distance covered by the aeroplane was 800km. Determine the position of B. (4mks).
  3. Triangle PQR whose vertices are p(2,2), Q(5,3) and R(4,1) is mapped onto triangle P'Q'R' by a transformation whose matrix is
    1 -1
    -2 1
    1. On the grid draw PQR and P'Q'R'. (4mks)
    2. The triangle P'Q'R' is mapped onto triangle P''Q''R'' whose vertices are P''(-2,-2), Q''(-5,-3) and R'' (-4,-1)
      1. Find the matrix of transformation which maps triangle P'Q'R' onto P''Q''R''.(2mks)
      2. Draw the image P''Q''R'' on the same grid and describe the transformation that maps PQR onto P''Q''R''. (2mks)
    3. Find a single matrix of transformation which will map PQR on to P''Q''R''.(2mks)
  4.                                    
    1. Complete the table for y = Sin x + 2 Cos x. (2mks)

      X

      0

      30

      60

      90

      120

      150

      180

      210

      240

      270

      300

      Sinx

      0

         

      1.0

       

      0.5

       

      -0.5

         

      -0.87

      2 cos x

      2

         

      0

       

      -1.73

       

      -1.73

         

      1.0

      Y

      2

         

      1.0

       

      -1.23

       

      -2.23

         

      0.13

    2. Draw the graph of y = Sin x + 2 cos x. (3mks)
    3. Solve sinx + 2 cos x = 0 using the graph. (2mks)
    4. Find the range of values of x for which y < -0.5 (3mks).
  5. A bag contains 3 red, 5 white and 4 blue balls. Two balls are picked without replacement. Determine the probability of picking.
    1. 2 red balls 2mks
    2. Only one red ball 2mks
    3. At least a white ball 2mks
    4. Balls of same colour. 2mks
    5. Two white balls 2mks
  6.                              
    1. Draw the graph of the function 2mks
      y = 10+3x – x2 for –2≤x≤5
    2. use of the trapezoidal rule with 5 stripes, find the area under the curve from x = -1 to x = 4. 4mks
    3. Find the actual area under the curve from x = -1 to x = 4. 2mks
    4. Find the percentage error introduced by the approximation. 2mks
  7. The figure below is a cuboid ABCDEFGH such that AB = 8cm, BC = 6cm and CF 5cm.
    Determine
    1. the length
      1. AC (2mks)
      2. AF (2mks)
        23 jgauygda
    2. The angle AF makes with the plane ABCD. (3mks)
    3. The angle AEFB makes with the base ABCD. (3mks)
  8. A manager wishes to hire two types of machine. He considers the following facts.
                                                                 Machine A              Machine B
    Floor space                                              2m2                      3m2
    Number of men required to operate          4                          3
    He has a maximum of 24m2 of floor space and a maximum of 36 men available. In addition he is not allowed to hire more machines of type B than of type A.
    1. If he hires x machines of type A and y machines of type B, write down all the inequalities that satisfy the above conditions. 3mks
    2. Represent the inequalities on the grid and shade the unwanted region. 3mks
    3. If the profit from machine A is sh. 4 per hour and that from using B is kshs8 per hour. What number of machines of each type should the manager choose to give the maximum profit? (4mks)


MARKING SCHEME

1.

(x- y) ( x+y)
( 3282 – 3272) ( 3282 + 3272)
65540

M1

M1

A1

 

 

 

3

 

2.

Tan x = is positive 3rd quadrant
Then sin x = -3
                     5
hypotenese uygz
h =  √42 + 32 = √25 = 5
Sin  x = -3
              5
Cos x – sin x = -4   - -3   = -3
                         5      5       5
 = -1
     5

 

 

 

B1

 

 

M1

 

 

A1

 

 

 

Identification the hypotenuse

 

 

 

 

Cao
accept (-0.2)

 

 

3

 

3.

16 + 6(- ½  x ) + 15(- ½  x )2 + 20(- ½  x )3
 = 1 – 3x + 15x2  - 5 x3
                    4
X = -0.04
1-3 ( -0.04) + 15  (-0.04)2  5 ( -0.04) 3
                       4                 4
= 1 + 0.12 + 0.006 + 0.000616
= 1.12616
= 1.1262

M1

M1

M1

 

A1

For simplification

 

 

 

For substitution of x

 

 

4

 

4.

 a + ar3 = 140
     2
64 + 64 r3 = 140
     2
64 + 64 r3  = 280      64 r3  = 280 – 64
  64 r3  = 216        r  =   √216
                                        64 
r = 3
     2

M1

 

 

M1

 

A1

 

 

 

3

 

5. a2 =    b2d2  

       b2 –d
a2b2 - a2d  = b2d2
a2b2  - b2d2  = a2d
b2(a2 - d2) = a2d
b2 =    a2 d
          a2 - d2
=   √ a2 d
       a2 d2

M1

M1

A1
sq on both sides
    3  
6 P = aQ +  √Q
P = 16a + 4b

( 500 = 16a + 4b)
(800 = 25a + 5b)

2500 = 80a + 20b
3200 = 100a + 20b
-700 = -20a
35 = a

Then b = -15
Equation connecting P and Q
p = 35Q – 15 √Q

M1

 

M1

 

M1

A1

 

For equation

For formation of simultaneous equations 

For  values of both a  and   b

    4  
7. 1000 [1 + 6]3
              100
1000 x 1.063
ksh 11910.16
11910
M1

A1
 
    2  
8 4.562 x 0.38 = 1.73356
4 √1.73356      = 1.14745 ÷ 0.82
 = 1.3993
= 1.4

M1

M1

A1

 
    3  
9. 18 x 64 x 5
  24 x 80
6 x 64
8x 16
3 days

M1

M1

A1

For simplification
    3  
10 True value = √1 + n = 1.44 = 1.2
Approx. value
1 + n = 1 + 44 = 1.22
      2           2
= 1.22 – 1.2 
 = 0.02
0.02    x 100 =
1.2
= 1.67 %

M1

M1

A1

 
    3  
11 [3  0]      [a   b]   =   [3 + a    b]
[0  4]      [0   c]        [0    4 + c]
3a + 0 = 3  + a
3b  + 0  = b
3a = 3 + a     →      a = 3
                                     2
3b + 0 = b
2b = 0

B = 0

0 + 4c = 4 + c
3c = 4

C= 4
     3

M1

M1

A1

For matrix equation 

For forming of simultaneous equation

For values of a, b and c ( correct)

    3  
12. 2x2 – 2x  + x -1
( x + 1 ( x – 1)

2x ( x – 1 ) + 1 ( x- 1)
  ( x + 1 ) (n- 1 )
= ( 2x + 1)
    ( x + 1 ) 
= 2x + 1
    x + 1

M1

M1

A1

 
    3  
13.

1              cm = 25000cm
2              1cm = 250m
3              1cm = 0.25

1cm2 = 0.0625
20cm2  = 20 x 0.0625
             = 1.25/ cm2

M1

M1

A1

 
    3  
14

AB . BC = DC -2
     5: BC = 36
         BC = 36
                    5
             = 7.2 cm

M1

M1

A1

 
    3  
15.

Log108 + Log10750  - Log106
Log10  Log10 ( 8 x 750)
                          6
             = Log101000
            = 3

M1

M1

A1

 
    3  
16.

P (R )= ½ x 4
                  12
P (R )= ½ x 3
                  18
=    1    +   3
      6         20
=   20 + 18
       120
= 38    =     19
  120          60  

M1

M1

M1

A1

 
    4  
17. Taxable income 115 x 8570  = 9855.50
                          100
9855.50 x 12  p.a
                                 20
                         5913.30

                            Tax
1       – 1500  →   150
1501 – 3000   →  225
3001 – 4500  →   375
4501 – 5913.30→494.30
                            1244.30 -
                                90.00
 K £ 1154.30 pa . or
Ksh 1923.83 per month
Total Decuctions
 2    x   9855.5
100
197.11   ( wcps)
  +
20.00
246.00
  +
Tax per month  1923.83
                         2386.94
Net salary
9855.50 – 2386.94
Ksh 7468.65

M1 

A1

M1

M1

M1

A1 

M1

A1

M1

A1

 
    10  
18.
  1.    θ   ( 2πR cos θ)
     360
    60 x 2 x 22 x 6370 cos 60
        360         7
    = 1/3 x 22 x 910 x 0.5
    = 3336    7
  2. Time ( 4 x 60) hrs
                 60
         4 hrs.
    Local time 1200 + 4
                  = 1600hrs

b)  θ   x 2πR = 800
   360
= θ x 2 x π x 6370 = 800
 360 
θ  = 800 x 360
      2 x π x 6370
= 7.196°
∠(60 – 7.196) = 52.80°
( 52.8° N 45°E)

M1 

A1 

B1

M1 

M1

A1 

B1

 
    10  
19.      
    10  
20

X

30

60

120

180

240

270

Sin x

0.5

0.87

0.87

0

-0.87

-1.0

2 cos x

1.73

1.0

-1.0

-2

-1.0

0

             

Y

2.23

1.87

-0.13

-2

-1.87

-1.0

 

  1.                                            
  2.                                                    
    a2 graph uagd
  3. x = 114 ± 3º and x = 294 ± 3º
    line thro y = -1.5
  4.                   
 

B2 

S1 

P1 

C1 

B2 

L1

B2

 

For all 6 values of y
B1 for at least 4 
Appropriate scale use
plotting
curve
Points identified and stated
B1 only stated
line

Points identified and stated

B1 only stated

    10  
21.

tree diagram jygad

  1. P ( RR) = 3/12      x   2/11 
         1/22     
  2. P(IR) = RW or RB or WR or BR
    15/132   +    12/132    + 15/132    +    12/132
     9/22   
  3. p( At least white Ball ) =
    P(RW) + P(WR) + P(WW) + P ( WB) + P(B)
    15/132   +  9/132  +     20/132   +   20/132    +    20/132 
    =    84/132        or    7/11
  4. P(RR or WW or BB)
    = 6/ 132    +  20/132  + 12/132   
    = 19/66   
B2

M1

A1

M1

A1 

M1

A1

M1

A1

prob tree 

Or equivalent 0.04545 

Or equivalent 0.4091 

Or equivalent 0.6364 

Or equivalent 0.2424

    10  
22.

X

-2

-1

0

1

2

3

4

5

Y

0

6

10

12

12

10

6

0

kunta kinte graph uigad

½ ( 6 + 10 ) + ½ ( 12 + 12) + ½ ( 12 + 10 ) + ½ ( 10 + 6)
= 8 + 11 + 12 + 11 + 8  = 50cm2
u jkhgbaugda

B2

S1

P1

C1

M1 

A1

M1

M1

M1

A1

8 values 
B1 at least 6
Appropriate scale use
plotting
curve
    10  
23
  1.                        
    1. AC2 = 82 + 62 = 100
             AC  = 10cm
    2. AF2  = 102 + 52  = 125
                  AF = 11.18cm
  2. Tan x = 5/11 = 0.5
            x = 26.52°
  3. Tan x = 5/6 = 0.8333
    x = 39.7°

M1 

A1 

M1 

A1

B1

M1

A1

B1

M1 

A1

Sketch 

Sketch
    10  
24.

24 aujgdau

Inequalities
x ≥ O
y ≤ O
4x + 3y ≤ 36
4x + 3y = 24
y = n
For shading of x ≥ o and y ≥ o
2x + 3y ≤ 36
y ≤ n

P profit function object we function
P = 4x + 3y

Max profit at point (6,4)
P = 4(6) + 4,88)(4)
= 56

Hence he should here 6 medium of type A and 4 machine of type B

 

 

B1 

B1

B1

B1

B1
shading and line

B1
shading and line drawn
B1-for  
shading and line drawn

    10  

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