# Mathematics Paper 1 Questions and answers - Sukellemo Joint Pre Mock Exams 2022

QUESTIONS
SECTION 1 (50 MARKS)

Answer all the questions in the space provided below each question

1. Find the equation of a straight line passing through the points A (1,-3) and B (-2, 5).Express your answer in the form ax + by = c where a, b and c are integers. (3marks)
2. Evaluate without using mathematical tables or calculator -10÷2+6×4-8×5 (3marks)
-5+(-12)÷3×2
3. Solve for x in the equation Cos(2x-30)°= tan45° (3marks)
Sin(3x+10)°
4. Two taps P and Q together can fill a water tank in 6 minutes. Tap P alone takes 5 minutes longer than tap Q. How many minutes does it take tap P alone to fill the tank? (3marks)
5. Given that, 275x-2y=243 and 812x-y=3, Calculate the values of x and y. (3marks)
6. A point P is mapped onto P’ by a negative quarter turn about the origin. P’ is mapped onto P’’ by a translation represented by the vector (-23) . If P’’ has coordinates (11,-5) determine the coordinates of p. (3marks)

7. A metallic pipe which is 21 meters long has an internal radius of 13 cm and an external radius of 15 cm. if the density of the metal is 8620 kg/ m3, find its mass. (3marks)
8. Using logarithms evaluate ∛(82.73×0.29432)(3marks)
613.5
9. A proper fraction is such that the denominator exceeds the numerator by 3. If 2 is subtracted from both the numerator and denominator, the fraction formed is 1/8 less. Determine the original fraction. (3 marks)
10. Given that OM = 2i +3j -6k and ON = -3i + 5j +k.Find the magnitude of MN to 2 decimal places. (3marks)
11. Find the range of the integral values of x in the inequality 10<3(x+2)<35 , giving your answer in the form a≤x≤b (3marks)
12. Simplify completely    2-2x    ÷  x-1  (3marks)
6x2-x-12    2x-3
13. The marked price of a recliner sofa set in a furniture store was ksh 400,000.A customer bought the recliner at 10% discount. The dealer still made a profit of 20%, Calculate the amount of money the dealer paid for the recliner. (3marks)
14. Draw a line AB of length 9 cm. On one side of line AB construct the locus of a point P such that the area of triangle ABC is 13.5 cm2.On this locus locate two positions of a point P1 and P2 such that
15. Given that the area of an image is four times the area of the object under a transformation whose matrix is , find the possible value of x .(3 marks)
16. Construct a triangle ABC in which AB = 5cm and AC = 8cm and ∠ABC=105°. Using line AC, locate point x on AB produced such that AX: XB =3: -2. (4marks)

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

1. The table below shows the weekly salary (k£) paid to workers in a school.
 Salary (k£) 50≤x≤100 100≤x≤150 150≤x≤250 250≤x≤350 350≤x≤500 No. of Workers 25 27 30 26 24
1. Calculate the differences between the mean and the median. (6 marks)
2. Draw a frequency polygon to illustrate the above information. (4marks)
2.
1. Complete the table of values for the equation, y=-2x^(2 )+x+8. (2marks)
 x -3 -2 -1 0 1 2 3 4 y
2. Use the values above to draw the graph of y=-2x2+x+8 . (3marks)
3. Using the graph drawn above Solve the equations:-
1. 2x2=x+8 (2marks)
2. -2x2+4x+12=0 (3marks)
3. Three towns P, Q and R are such that Q is 16 km north of P and the distance of R is 12 km from P and on a bearing of 60º from Q.
1. Using a scale of 1cm to represent 4km, Make a scale drawing showing the relative positions of the three towns. (3marks)
2. Using the scale drawing above, find the
1. Distance of R from Q. (1mark)
2. Bearing of P from Q. (1mark)
3. How far town R is east of Q (1mark)
3. A Passenger in an aero plane after take-off from town R spotted town P at an angle of depression of 48º, by means of a scale drawing determine the vertical height of the plane at town R. (3marks)
4.
1. The equation of a straight line L1 is of the form 3y+2x=5.L1 is perpendicular to L2 and meets it at the point where X=-2, determine the equation of L2 in the form y = mx+c where m and c are constants. (5marks)
2. L3 is parallel to the line L2 and passes through the point (-3,2).,find the equation of L3, leaving your answer in its double intercept form. (3marks)
3. Determine the angle of inclination of L2 to the Y-axis. (2marks)
5. The points P, Q, R and S, have position vectors 2p, 3p, r and 3r respectively, relative to an origin O. A point T divides PS internally in the ratio 1:6.
1. Find, in its simplest form OT, QT and TR in terms of p and r. (6 marks)
2. Show that the points Q, T and R, are collinear. (3marks)
3. Determine the ratio in which T divides QR. (1mark)
6. In the figure below, O1 and O2 are the centers of the circles whose radii are 5 cm and 8 cm respectively. The circles intersect at A and B and angle AO1O2 = 64˚.

Calculate the area of the:-
1. Sector
1. AO1B (2marks)
2. AO2B (3 marks)
2. Intersecting region. (3marks)
3. The shaded region. (2marks)
7.
1. Find the x –intercept of the curve y = (x+2) (x-1)2. (1mark).
2. Find the gradient function of the curve y = (x+2) (x-1)2 (2marks)
3. Find the co-ordinates of the turning point. Hence sketch the curve y= (x+2) (x-1)2. (4 marks)
4. Calculate the exact area enclosed by the curve and the x - axis (3marks)
8. P and Q are two points on latitude 40°N.Their longitudes are 30°E and 150°W respectively. Find to one decimal place :( Take the radius of the earth = 6370km andπ=22/7)
1. The distance in km between P and Q along the parallel of latitudes. (2marks)
2. The shortest distance along the earth’s surface between P and Q in km. (3marks)
3. A weather forecaster reports that the center of a cyclone at (40°N, 60°W) is moving due north at 24 knots. How long will it take to reach a point (45°N, 60°W). (2marks)
4. A plane leaves P at 2.15 pm at a speed of 350 knots to town R (40°N, 65°E). Determine the time at R when the plane arrived. (3marks)

## MARKING SCHEME

1. Gradient = 5/-2 = 3/-1 = -8/3
y - 5-8/3
x + 2
3(y - 5) = -8(x + 2)
3y - 15 = -8 - 16
3y  + 8x = -1
2. Numerator = -5 + 24 - 40
-21
den. = -5 - 8
= -13
-21/-13
= 18/13
3. cos(2x - 30)º = sin(3x + 10)º
cos(2x - 30)º = cos 90 -(3x + 10)
2x - 30 = 90 - (3x + 10)
2x - 30 + (3x + 10) = 90
5x - 20 = 90
5x = 110
x = 22
4. 1/x+5 + 1/x = 1/6
6(x + x + 5) = x(x + 5)
6x + 6x + 30 = x2 + 5x
-7x - 30 = 0
- 10x + 3x - 30 = 0
x(x - 10) + 3(x - 10) = 0
(x + 3)(x - 10) = 0
x = -3 or 10
Top P = 5 + 10
=15mins
5. 33(5x - 2y) = 35
3(5x - 2y)= 5
15x - 6y = 5
34(2x - y) = 31
4(2x - y) = 1
(15x - 6y = 5)4
(8x - 4y = 1)6
60x - 24y = 20
48x - 24y = 6
12x = 14
x = 11/6
8x7/6 - 4y = 1
56/6 - 4y = 1
81/3 = 4y
21/12 = y

6. y - 2 = 11
y = 13
x + 3 = -5
x = -8
P(-8, 13)
7. External volume - Internal volume
22/7 x 152 x 2100 - 22/7 x 132 x 2100
22/7 x 2100 (225 - 169)
22/7 x 2100 x 56
369,600cm3
0.3696m3
Mass = 8620 x 0.3696
= 8620 x 3696
1000
3185.952
= 3186kg

8.  No Log 0.2943282.73613.50.2269 T.4688 x 22.93761.9177 + 0.85532.7878-2.06752.0675 = 3/3 + 1.0675 = T + 0.355833T.3558
9. x/y
y - x = 3
y = x = 3
y = 3 + x
x - 2x/y - 1/8
y - 2
x - 2  =  x   1/8
3+x-2    3+x
(x - 2) (x + 3 )8 = x (x + 1)8-((x+1)(x + 3))
(x2 + x - 6)8 = 8x2 + 8x - (x2 + 4x + 3)
8x2 + 8x - 48 = 8x2 + 8x - x- 4x - 3
0 = -y-4x + 45
x+ 4x - 45 = 0
x+ 9x - 5x - 45 - 0
x(x + 9) -5(x + 9) = 0
(x - 5)(x + 9) = 0
x = 5 or -9
10. MN = √(-3-2)2 + (5-3)2 + (1-6)2
√25+ 4 + 49
√78
8.832
8.83
11. 10<33(x + 2)
10<3x + 6
4<3x
11/3 < x
3x + 6 < 35
3x < 29
x < 92/3

2≤x≤9
12.         2(1-x)
6x2 - 9x + 8x - 12
-2(x - 1)
3x(2x-3)+4(2x-3)
-2(x-1)
(3x+4)(2x-3)
=  -2
3x+4
13. 100% = 400000
90% = 90 x 400000
100
= 360000
120% = 360000
100% = 100 x 360000
120
= 300000

14. ½ x 9 x h = 13.5
h = 3cm
15. x2 - ((x-4)(x+8)) = 4
x- (+4x - 32) = 4
x2- x2 - 4x + 32 = 4
- 4x = -28
x = 7
16.
17.

1.  0-50 f x fx cf fd 50-100 25 75 1875 25 0.5 100-150 27 125 3375 52 0.54 150-250 30 200 6000 82 0.3 250-350 26 300 7800 108 0.26 350-500 24 425 10200 132 0.16 500-650 Σf=132 Σfx= 29250
Mean = 29250
132
= 221.60
Median = 66th position
150 + 82
150 + (66 - 52) x 100
200
221.60 - 157
= 64.60
2.
18.
1.
 x -3 -2 -1 0 1 2 3 4 y -13 -2 5 8 7 2 -7 -20
2.
3.
1. 2x2 + x + 8 = 0
-2x2 + x + 8 = y
0 = y
x = -1.78 or x =2.2
2. -2x2 + x + 8 = y
-3x - 4 = y
x = -1.55
x = 3.67
19.
1.
2.
1. 5.3cm = 5.3 x 4 = 21.2km
21 ± 0.4
2. 215º
214 ± 1
3. 4.7cm = 4.7 x 4 = 18.8km
18.1 ± 0.4
3.
20.
1. 3y = -2x + 5
y = -2/3x + 5/3
Gr = 3/2
at x = -2
y = 4/3 + 5/39/3 = 3
(-2, 3)
y - 33/2
x + 2
2y - 6 = 3x + 6
2y = 3x + 12
y = 3/2x + 6
2.  y- 2 3/2
x + 3
3(x + 3) = 2(y - 2)
3x + 9 = 2y - 4
3x - 2y = -13
3x/-13 + 2y/13 = 1
x   +  y   = 1
-13/   13/
x    +  y  = 1
-41/3   6½

3. tanθ = 3/2
θ = 56.31º
90 - 56.31º = 33.69º
angle 33.69º or 146 .31º
21.

1. OT = 3/7r + 12/7p
QT = QO + OT
= -3p +3/7r + 12/7p
= 3/7r - 9/7p
TR = TO + OR
= -3/7r - 12/7p + r
= 4/7r -12/7p
2. QT = 3/7(r - 3p)
TR = 4/7(r - 3p)
4/7QT = 3/7TR
4QT = 3TR
QT//TR
3. QT/TR = ¾
QT:TR = 3:4

22.
1.
1. 128/360 x 22/7 x 25
27.94
2. 8/sin64 = 5/sinx
sin x = 5 x sin64
8
= 0.5617
= 34.18º
2x = 68.36º
68.3622/7 x 64
360
= 38.19
2. (27.94 - x 25 x sin128) + 38.19 - x 64 x sin68..36
27.94 - 9.850                        38.19 - 29.74
18.09 + 8.45
26.54
3. 9.850 + 29.74 - 26.54
39.59 - 26.54
13.05
23.
1. (x+2)(x-1) = 0
x = 1 or -2
2. y = (x+2)(x2 - 2x + 1)
y = x3 - 2x2 + 2x2 + x - 4x
y = x3 - 3x
dy/dx = 3x2 - 3
3. 3(x2 -1) = 0
3(x-1)(x+1) = 0
3(x-1)(x-1) = 0
x = 1 or -1
when x = 1, y = 0
(1,0)
d2y = 6x at x=1
dx2
= 6
(1,0)minimal
when x = -1
y = (-1+2)(-1-1)2
y = 4
(-1,4)
d2y = 6x at x = -1
dx2
22y = -6
dx

4. -1.25 - (-2)
0.75
24.

1. dif in longitude = 180
18022/7 x 2 6370 x cos40
360
= 15336.2

2. 100
22/7 x 2 x 6370
360
= 8272.7

3. differences in latitude = 45-40
= 50
= 5 x 60 = 300NM
Time = 300/24
= 12.5hrs

4. differences in longitude = 65-0
= 35º
24hrs = 360º
35 x 24
360
2hrs 20mins
Distance = 35 x 60 x c0s40
= 1,608.7NM
Time = 1608.7
350
= 4.596hrs
4hrs, 36 mins
= 8.11p.m

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Read 4729 times Last modified on Thursday, 28 July 2022 11:26

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