Mathematics Paper 1 Questions and Answers - Lanjet Joint Mock Exams 2020

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INSTRUCTIONS TO CANDIDATES

  • Write your name and Admission number in the spaces provided at the top of this page.
  • This paper consists of two sections: Section I and Section II.
  • Answer ALL questions in section 1 and ONLY FIVE questions from section II
  • All answers and workings must be written on the question paper in the spaces provided below each question.
  • Show all the steps in your calculation, giving your answer at each stage in the spaces below each question.
  • Non – Programmable silent electronic calculators and KNEC mathematical tables may be used, except where stated otherwise.

SECTION I (50 marks)
Answer all the questions in this section in the spaces provided.

  1. Without using mathematical tables or calculators, evaluate1leaving your answer as a simplified fraction (3mks)
  2. Two similar solids have surface areas 48cm2 and 108cm2respectively. Find the volume of the smaller solid if the bigger one has a volume of 162cm3. (3mks)
  3. A triangle flower garden has an area of 28m2. Two of its edges are 14 metres and 8 metres. Find the angle between the two edges. (2mks)
  4. A watch which looses a half a minute every hour.It was set read the correct time at 0445hr on Monday. Determine in twelve hour system the time the watch will show on Friday at 1845hr the same week. (3mks)
  5. Find the least whole number by which 25×54×73 must be multiplied with to get a perfect cube. What is the cube root of the resulting number? (3mks )
  6. A woman went on a journey by walking, bus and matatu. She went by bus 4/of the distance, then by matatu for 2/3 of the rest of the distance. The distance by bus was 55km more than the distance walked. Find the total distance. (3mks).
  7. Simplify the expression:    (9t2- 25a2)     (3mks).
                                      (6t2+ 19at+15a2)                  
  8. Solve the simultaneous equations
    X y = 4 and x + y = 5 (4mks)
  9. The size of an interior angle of regular polygon is 3xº. While its exterior angle is (x – 20)º. Find the number of sides of the polygon. (3mks)
  10. A Kenya company received US Dollars M. The money was converted into Kenya Shillings in a bank which buys and sells foreign currencies.
      Buying (in Ksh) Selling (in (Ksh)
    1 Sterling Pound   125.78   126.64
    1 Us Dollar   75.66  75.86
    1. If the company received Ksh.15, 132,000, calculate the amount, M received in US Dollar. (2mks)
    2. The company exchanged the above Kenya shillings into Sterling pounds to buy a car in Britain. Calculate the cost of the car to the nearest Sterling pound. (2mks)
  11. A plot in a shape of rectangle measurers 608m by 264m. Equidistance fencing posts are placed along its length and breadth as far apart as possible. Determine
    1. The maximum distance between the posts. (1mk)
    2. The number of posts used. (2mks)
  12. Given that sin (x – 30)º - Cos (4x) 0. Find the tan (2x+30)º (3mks)
  13. A trader sold a dress for Ksh 7200 allowing a discount of 10% on the marked price. If the discount had not been allowed the trader would have made a profit of 25% on the sale of the suit. Calculate the price at which the trader bought the dress. (3mks)
  14. In august, Joyce donated 1/6 th of her salary to a children’s home while Chui donated 1/5 th of his salary to the same children’s home. Their total donation for August was Kshs 14820. In September, Joyce donated 1/8 th of her salary to the children’s home while Chui donated 1/12 th of his salary to the children’s home. The total donation for September was Kshs 8675. Calculate Chui’s monthly salary. (4mks)
  15. Simplify completely 3n + 3 - 3n + 1 (3mks)
                                    4 x 3n + 2  
  16. In what ratio should grade A tea costing Sh. 180 per kg be mixed with grade B tea costing Sh. 300 per kg to produce Nganomu Tea which when sold at Kshs 270 a profit of 20% is realized? (3mks)

    SECTION II (50 MARKS)

    Answer any five questions from this section in the spaces provided
  17. Atambo poured spirit into a test tube which has hemispherical bottom of inner radius 1.5cm. He noted that the spirit is 8cm high.
    1. What is the area of surface in contact with spirit? (4mks)
    2. Calculate volume of spirit in the test tube. (4mks)
    3. If Atembo obtained the mass of the spirit as 10g. Calculate the density of the spirit. (2mks).
  18. A bus left Nairobi at 7.00 am and traveled towards Eldoret at an average speed of 80Km/hr. At 7.45am a car left Eldoret towards Nairobi at an average speed of 120Km/hr. The distance between Nairobi and Eldoret is 300 km. Calculate:
    1. The time the bus arrived at Eldoret. (2mks)
    2. The time of the day the two met. (4mks)
    3. The distance of the bus from Eldoret when the car arrived in Nairobi. (2mks)
    4. The distance from Nairobi when the two met. (2mks)
  19. The figure below C is a point on AB such that AC: CB=3:1 and D is the mid –point of OA. OC and BD intersect at X.
    2
    Given that OA = a and OB = b
    1. Write the vectors below in terms of a andb.
      1. AB (1mk)
      2. OC (2mks)
      3. BD (1mk)
    2. If BX = h BD, express OX in terms of a, b, and h. (1mk)
    3. If OX = KOL, find h and k. (4mks)
    4. Hence express OX in terms of a andb only. (1mk).
  20.  
    1. Using a ruler and a pair of compasses only, draw a triangle ABC such that AB = 5cm,BC = 8cm and <ABC = 60º. Measure AC and <CAB.(4mks)
    2. Find a point O in ΔABC such that OA = OB = OC. (2mks)
    3. Construct a perpendicular from A to BC to meet BC at D. Measure AD. Hence calculate the area of the ΔABC (4mks)
  21. A boy started walking due East from a dormitory 100m South of a bore-hole. He walked to the school library from which the bearing of the bore-hole is 315º. He then walked on a bearing of 030º to the water tank. From the water tank he went west to the bore-hole.
    1. Using a scale of 1cm to represent 20m, construct a diagram to show the positions of the tank, borehole, dormitory and library. (5mks).
    2. Find the distance and bearing of the bore-hole from the water tank. (3mks)
    3. Calculate the total distance covered by the boy. (2mks).
  22. The table below shows the amount in shillings of pocket money given to students in a particular school.
    Pocket Money (Ksh)  210 – 219  220-229  230-239   240-249  250-259 260-269    270-279  280-289 290-299
    No. of Students  13 23 32 26 20 15 12 4
    1. State the modal class. (1mk)
    2. Calculate the mean amount of pocket money given to these students to the nearest shilling. (4mks).
    3. Use the same axes to draw a histogram and a frequency polygon on the grid provided(5mks)
      3
  23.  
    1. Given that y = 7 + 3χ - χ², complete the table below. (2mks)
      χ -3 -2 -1 0 1 2 3 4 5 6
      y -11      7           -11
    2. On the grid provided and using a suitable scale draw the graph of y = 7 + 3χ - χ².(3mks)
      3
      On the same grid draw the straight line and use your graph to solve the equation χ² - 4χ– 3 = 0. (3mks)
    3. Determine the coordinates of the turning point of the curve. (2mks)
  24. A straight line L1 has a gradient ˉ½ and passes through point P (-1, 3). Another line L2 passes through the points Q (1, -3) and R (4, 5). Find.
    1. The equation of L1. (2mks)
    2. The gradient of L2. (1mk)
    3. The equation of L2. (2mks)
    4. The equation of a line passing through a point S (0, 5) and is perpendicular to L2. (3mks)
    5. The equation of a line through R parallel to L1.(2mks)


MARKING SCHEME

SNO: WORKING MARKS
1

1408X594X12
   605X125X100

128X54X12
   5X125X100

√[128X9X2X3X3X2
          100X625]

= 72 
  625 

M1

M1

A1

    03
2

ASF = 48 =
         108   9

LSF = √ 4/9 = 2/3

VSF= (2/3)3 = 8/27

Vol=8/27 x162

=48cm
M1

M1

A1

    03
3 0.5 x14x8 sin θ=28m2
Sin θ=0.5
θ=Sin-1 (0.5)
θ =30º
M1

A1

    02
4

Tuesday-Thursday=24x3=72hours
Monday=2400-0445=19hours 15minutes
Friday=18hours 45minutes
Total time=72+19.25+18.75=110hours
Time lost=0.5x110=55minutes
1845hrs-55minutes=1750hours
=5.50pm 

B1

M1

A1

    03
5

25X54X73X(2X52)
2X52=50
(26X56X73)1/3
=22X52X7
=700

M1

M1

A1

    03
6

Bus═> 4x 
           5
matatu═>2/3x1/5x=2/15x
walking═>1-(4/5+2/15)
=1/15X
4/5X=1/15X+55
11/15X=55
X=75Km 

M1

M1

A1

    03
7

N=9t2-25a2=(3t-5a)(3t+5a)

D=6t2+19at+15a2=6t2+9at+10at+15a2
=3t(2t+3a)+5a(2t+3a)
=(3t+5a)(2t+3a)
N=(3t-5a)(3t+5a)
D ( 3t+5a)(2t+3a)
= 3t-5a 
   2t+3a

M1

M1

A1

    03
8 x =
      y
+ y = 5
y
4 + y2 = 5y
y2 - 5y + 4 = 0
y2 - y - 4y + 4 = 0
y(y - 1) -4(y - 1) = 0
(y - 4)(y - 1) = 
y = 4 ; y = 1
x = 1   x = 4

M1

M1

M1

A1

    04
9

3x+x-20=180º

4x=200º

X=50º

n=360º=360º
     ext     30º

12sides

M1

M1

A1

    03

10(a)

 

 

(b)

M = 15,132,000
           75.66
= 200,000 USDollars

15,132,000
   126.64
11944 Sterling pounds

M1

A1

M1

A1

    04

11(a)

 

 

 

(b)

HCF of 608 and 264
608 = 2 x 2 x 2 x 2 x 2 x 19
264 = 2 x 2 x 2 x 3 x 11

HCF = 23 = 8

Maximum distance between the posts = 8m

Number of posts=2(608+264)÷8
=218posts

A1


M1A1

    03
12 x - 3 + 4x = 90º
5x = 120º
x = 24º
tan/92 x 24 + 30)º = tan 78º
= 4.705

M1

M1

A1

    03
13

Marked Price = 100  x 7200
                        90
= sh8000
Selling price = 100 x 8000
                      125

=sh6400

M1

M1

A1

    03
14

1/6 x + 1/5y = 14820
1/8 y + 1/12 y = 8 675
5x + 6y = 444 600
3x + 2y = 208 200

x= 45 000
y= 36 600

Chui’s salary =Kshs 36,600

M1

M1

A1

A1

    04
15 27(3n) -3(3n
        36(3n)
= 24(3n)
   36(3n)
2/3

M1

M1

A1

16 cost price = 270  x 100= 225
                  120
180x + 300y  = 225
     x + y
180x + 300y = 225 + 225y
45x = 75y
75 
y      45       3
ratio 5:3

M1

M1

A1

    03

17(a)

 

 

 

 

 

(b)

 

 

 

 

 

 

(c)

Surface area of hemisphere=2πr2

S.A= 2xπ x1.52
14.14cm2
Surface area of cylindrical part
2πrh
S.A=2xπ xrxh
22πx1.5x6.5=61.29cm2

T.S.A=14.14+61.29
=75.43cm2
Volume of hemispherical part=2/3 πr3
=2/3xπx1.53
=7.071cm3
Volume of cylindrical part=πr2 h
=2/3πx1.52 x6.5
=45.96cm3
T. volume=7.069+45.96
=53.04cm3

Density=M/v
=10/53.04
=0.1886g/cm3

 

M1

M1


M1

A1

M1

A1

M1

A1


M1


A1

    10

18(a

 

 

(b)

 

 

 

 


(c)

 

 

(d) 

Time = = 300 
           S       80
=33/hrs
= 10.45am

Distance covered by bus at 7.45am
=80x 3/4 = 60km
Distance left = 300 - 60 = 240km =hrs
                                                     5
Relative speed =200km/h
Time taken = 240  =hrs
                    200      5
1hr 12min
Therefore 7.45 am + 0112 = 8.57am

Time taken by car =
Time of travel of bus 2 ½ hrs + ¾ hrs =3 ¼ hrs
Distance travelled = 80 x 13/4 = 260
Distance from Eldoret = 300 – 260 = 40km
Distance travelled by car in 6/5 hrs
= 120x 6/5
= 144km
Distance from Nairobi is 300 – 144
=156km

M1

A1

M1

M1

M1

A1

M1

A1


M1


A1

    10

19(a)
(i)

(ii)

 

(iii)

 

(b)

 

 

(c)

 

 

 


(d)

AB = b - a


OC = ¼ a+¾ b


BD =½ a - b

OX=b(1-h)+½ ah

OX=b(1-h)+½ ah

OX=¼ ak+¾ bk

½ah =¼ak

2h = k
¾ b k=b (1-h)
¾ k=1 - h
¾ (2 h)=1 - h
5/2h =1 ═> h = 2/5
K=2(2/5) = 4/5
K = 4/5

1/5 a+ 3/5 b

 


A1

A2


A1


A1

M1


M1


M1

A1

A1

    10
20 4
AB=5cm, BC=8cm

ABC=60º

AC=7.0 ± 0.1cm

BAC=83±1º

Bisectors of any 2 sides

centre o marked

Bisector of BC from A

AD measured 4.3±0.1

Area=½ x 8 x 4.3

=17.2cm2

 B1

B1


B1


B1


B1

B1


B1


B1

M1

A1

    10

21(a)


(b)


(c)

5
Dormitory located
Tank located
Borehole located
Library located
Correct scale

Distance=20x8

=160km

Bearing =270º± 2º

Total distance walked

= 100+120+160

=380km

B1
B1
B1
B1
S1


M1


A1

A1


B1


A1

    10

22(a)


(b)

Modal class=240-249

Amount (ksh) (X) mid point (f) frequency fx
 210-219  214.5  5  1072.5
 220-229  224.5  13  2918.5
 230-239  234.5  23  5393.5
 240-249  244.5  32  7824
 250-259  254.5  26  6617
 260-269  264.5  20  5290
 270-279  274.5  15  4117.5
 280-289  284.5   12  3414
 290-299  294.5  4  1178
     ∑f=150 ∑fx=37825

Mean=∑fx =37825 = ksh 252
∑f 150
6

B1


B1 for X
B1 for fx


M1
A1


S2


P2

L1

    10

23(a)

 

(b)

 


(c)

 


(d)

 X -3 -2 -1 0 1 2 3 4 5 6
 Y -11 -3 3 7 9 9 7 3 -3 -11

7
y = 7 + 3x - x2
0 = -3 - 4x + x2
X=-0.7 or 4.7

x 0 2
4 2

(1.5, 9.8)

B2

S1

C1

P1

M1

L1

A1

B2

    10

24(a)

 

 

(b)

 

 

(c)

 

 

 

(d)

 

 

(e)

y - 3= ½
x + 1
y - 3 = -½x - ½
y = -½ + 5/2

g = 5 - - 3
       4 - 1
8/3

y - 5 8/3
x - 4
y - 5 = 8/3x - 32/3
y = 8/3 x - 32/3 + 5 
y = 8/3x - 17/

y - 5  = -
   x          8
y =  -3  x + 5
       8
y - 5 = -½
x - 4
y = -½x + 7

M1

A1


A1

M1


A1


M1

M1
A1

M1

A1

    10

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