Mathematics Paper 1 Questions and Answers - Mumias West Pre Mocks 2022

QUESTIONS

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

Without using a calculator evaluate(3mks)
5x⁶ + (-76)÷4+27÷3
(-5)÷3 x (-4)
1. Express 2268 in terms of its prime factors (1mk)
2. Hence determine the smallest positive number x such that 2268x is a perfect square. (2mks)
Elvis arrived in Kenya with 5000 sterling pound, he exchanged it to Kenya Shilling and spent sh. 267 100. Before jetting out of the country, he exchanged the balance into Euros. Using the exchange rates below, calculate the amount he obtained in Euros in Kenya shillings. (3mks)
Currency Buying Selling
1 Sterling pound 114.20 114.50
1Euro 101.20 101.30
Simplify the expression (3mks)
2x²+3x-2
x³- 4x
When two wires of length 179m and 234m are divided into pieces of equal lengths a remainder of 3m is left in each case. Find the least number of pieces that can be obtained. (3mks)
Without using calculator, solve for n in the equation 1 - (¹/₃)ⁿ = ²⁴²/₂₄₃ (3mks)
Solve for y in the equation 7 - y - 9 - 2y = ½ (3mks)
4 3
Two similar solids have surface area of 48cm² and 108cm²respectively. Find the volume of the smaller solid if the bigger solid has a volume of 162 cm³ (3 mks)
Use reciprocal table only to evaluate ¹/_0.325 (3mks)
Hence, evaluate √0.25 to 1.d.p
0.325
A plot measuring 1.2m by 19.1 m is surrounded by a path 0.5m wide. Find the area of the path in square metres. (3mks)
The interior angle of a regular polygon is 600 more than its exterior angle, find the number of sides of the polygon. (3mks)
Complete the following solid given that ABC is its cross-section (3mks)
If tan x = ¹/_√3 find without using tables or calculator the value of Sin (90-x)+ cos (90-x) leaving your answer in simplified surd form (3mks)
A line perpendicular to the line 3y-2x=2 passes through the point (-3,2). Determine the equation of the line and write it in the form ax +by = c where a, b, and c are constant. (3mks)
Given that A = (⁴_-1 ³_-2) and C = (³_-1 ⁷_-2) (4mks)
Find B such that A² + B = C^-1
Ali travelled a distance of 5km from village A to village B in direction of N60ºE. He then changed direction and travelled a distance of 4km in the direction of 135º to village C.
Using a scale of 1cm to represent 1.0 km represent the information on an accurate diagram. (2mks)
1. Using scale drawing (a) above determine
2. distance between A and C (1mk
3. bearing of A from C (1mk)

SECTION B (50 MARKS):
Answer all the questions in this section in the spaces provided.

The figure below shows a frustrum. The top and bottom radii are 5cm and 10cm respectively, while the vertical height of the frustrum is 12cm.

Find the:-
1. Slant height of the frustum. (3marks)
2. Curved area of the frustum. (3marks)
3. Volume of the frustum. (4marks)
Bumala is a market centre 600km from Kisumu town.A bus starts from Kisumu for Bumala at 7.00am at an average speed of 80 km/h. At 8.30 am a car started from Kisumu to Bumala and moved at an average speed of 120 km/hr. Calculate
1. The distance bus covered before the car started moving. (3marks)
2. The relative speed for the two vehicles. (2marks)
3. The time the car overtook the bus. (1 mark)
4. Distance covered by the car before overtaking the bus. (2marks)
5. Distance from Bumala to the car at the time the car was overtaking the bus. (2marks)
The height of 36 students in a class was recorded to the nearest centimeter as follows:-
148 159 158 163 166 155 155 179 158
161 160 157 165 165 175 173 172 178
147 168 157 172 165 154 170 157 167
155 159 173 171 168 160 172 156 167
1. Make a frequency distribution table using a class interval of 5 and starting with the class 145 – 149. (2marks)
2. From the table above
  1. Calculate the mean mark (3marks)
  2. Calculate the median (3marks)
3. Draw a frequency polygon using the table in (a) above. (2 marks)
Bujumba Boys Secondary School. Intends to buy a certain number of chairs For Ksh. 16,200. The supplier agreed to offer a discount of Ksh. 60 per chair which will enable the school to get 3 chairs more.
Taking y as the originally intended number of chairs:-
1. Write an expression in terms of y for
  1. Original price per chair. (1mark)
  2. Price per chair after discount. (1mark)
2. Determine
  1. The number of chair the school originally intended to buy. (4marks)
  2. Price per chair after discount. (2marks)
  3. The amount of money the school would have saved per chair of it got the intended number of chairs at a discount of 15%. (2marks)
1. Without using a protractor, construct triangle ABC such that angle ABC = 60º, BC = 8cm and AC = 9cm.Measure AB. (3marks)
2. Drop a perpendicular from A to BC and measure its length. (2marks)
3. Hence calculate the area of triangle ABC. (2marks)
4. Locate a point D on BC such that the area of triangle ABC is three times that of triangle ABD. (3marks)
In triangle ABC, shown below, AB = a AC = b point M lies on AB such that AM: MB = 2:3 and point N lies on AC such that AN: NC = 5:1 line BN intersects line MC at X.
1. Express the following in terms of a and b
  1. BN (1 mark)
  2. CM (1 mark)
2. Given that BX = kBN and CX = rCM where k and r are scalars
  1. Write two different expressions for AX in term of a, b, k and r (4marks)
  2. Find the values of k and r (4 marks)
A triangle ABC has vertices A(2,1), B(5,2) and C(0,4).
1. On the grid provided plot the triangle ABC. (2 marks)
2. A¹B¹C¹ is the image of ABC under a translation (²_-5). Plot A¹B¹C¹ and state its coordinates. (2 marks)
3. Plot A¹¹B¹¹C¹¹ the image of A¹B¹C¹after a rotation about the origin through a negative quarter turn. State its coordinates. (3 marks)
4. A¹¹¹B¹¹¹C¹¹¹ is the image of A¹¹B¹¹C¹¹ after a reflection on the line y = 0.
  Plot A¹¹¹B¹¹¹C¹¹¹ and state its coordinates.(3 marks)
The table below shows income tax for a certain year

Monthly Income in Kenya Shillings (Ksh) Tax Rate

0 - 10164 10%

10165 - 19740 15%

19741 - 29316 20%

29317 - 38892 25%

Over 38892 30%

A tax relief of Ksh 1162 per month was allowed. In a certain month of the year, an employee’s taxable income in the fifth band was Ksh.2108.
1. Calculate
  1. Employees total taxable income in that month (2 Marks)
  2. The tax payable by the employee in that month (5 Marks)
2. The employee’s income include a house allowance of Ksh 15,000 per month. The employee contributed of the 5% basic salary to a co-operative society. Calculate the employee’s net pay for that month (3 Marks)

MARKING SCHEME

5x⁶ +(-76) ÷ 4 +27÷3
-15÷ 3 x (-4)
= 30 + (-19) + 9
-5 x-4
= ²⁰/₂₀ = 1
2268= 2² x 3⁴ x 7
x = 2² x 34 x 7²= 7
2² x 3⁴ x 7
Amt in Ksh = 5000 x 114.2
= 571000
Remainder = 571000-276100
= 303900
Amt in Euros = 303900
101.30
= 3000 Euros M1
2x² + 3x -2 = (2x -1)( x+ 2)
X³-4x x(x² -4)
= (2x -1)( x+ 2)
x( x -2)( x + 2)
= 2x - 1
x( x- 2)
179 - 3 = 176
234 – 3 = 231
176 = 2⁴ x 11
234 = 3 x 7 x 11
GCD = 11
Number of pieces = ¹⁷⁶/₁₁ + ²³¹/₁₁
= 16 + 21
= 37
1(¹/₃)ⁿ =²⁴²/₂₄₃(¹/₃)ⁿ= 1 - ²⁴²/₂₄₃
= ¹/₂₄₃
(¹/₃)ⁿ = (¹/₃)⁵
∴ n =5
4 - y - 9 - 2y= ½
4 3
3(4-y) – 4(9-2y) =6
12-3y -36 +8y = 6
5y =30
Y=6
For locus (i)
For locus (ii)
For locus (iii)
For the region
1 = 1
0.325 3.25×10^-1
= 0.3077×10^-1
= 3.077
^√0.25/_0.325 = 0.5×3.077
= 1.5385
= 1.5
External area = 20.1 × 2.2 = 44.22
Internal area = 19.1 × 1.2 = 22.92
Area of path = 44.22 – 22.92
= 21.3 m²
x + x +60 = 180°
2x = 120˚
x = 60˚
Exterior =60
No. of sides = ³⁶⁰/₆₀
= 6 sides
Sides
Broken lines
Shape of solid
Total = 42×24 = 1008
Total with abs = 1008-65
= 943
Average = ⁹⁴³/₂₃ =41
γ = ^2x/₃ + ²/₃
Gradient = ^-3/₂
γ - 2 = ^-3/₂
x--3
2(γ-2) = -3(x+3)
2γ – 4 = -3x – 9
3x + 2γ = -5
B = A² + C^-1
= (⁴_-1 ³₂)(⁴_-1 ³₂) + (^-2₁ ^-7₃)
= (¹³_-6 ¹⁸₁) + (^-2₁ ^-7₃)
= (¹¹_-5 ¹¹₄)
1. AC = 7.2km
2. Bearing 273º

X = 100-78
= 22
Modal class 35-44

	x	f	cf	d	fd
15-24	19.5	6	6	-40	-240
25-34	29.5	14	20	-30	-420
35-44	39.5	24	44	-20	-480
45-54	49.5	14	58	-10	-140
55-64	59.5	22	80	0	0
65-74	69.5	10	90	10	100
75-84	79.5	6	96	20	120
85-94	89.5	4	100	30	120
		100			-940

Median = 44.5 + 50-44 × 10
14
= 44.5 + ⁶/₁₄ × 10
= 44.5 + 4.29
= 48.79

Mean = ^∑fd/_Σf + A
= 59.5 - ⁹⁴⁰/₁₀₀
= 50.1

1. x = ⁵/₁₀
  x + 12
  10x = 5x + 60
  5x = 60
  X = 12
  L = 24² + 10² = 26
  ¹/₂₆ = ⁵/₁₀
  L = 13
  Slant height 26 – 13
  = 13
2. Curved area of the frustum
  S.A = πRL - πrl
  = π(260 – 65)
  = 612.6
3. Volume of the frustrum
  ¹/₃π(R²H – r²h)
  =¹/₃π(2400 – 300)
  =¹/₃π(2100)
  = 2200 cm²
1. 1½ x 80
  = 120km
2. 120 – 50
  = 40
3. Time taken
  ¹²⁰/₄₀= 3hrs
  8.300 + 3
  11.30A.m
4. 3 x 120
  = 360km
5. 600 – 360 = 240km
  For 240
  Total marks 10

Class	Tally	Frequency	Mid point	Fx	Cf
145 – 149	11	2	147	294	2
150 – 154	1	1	152	152	3
155 – 159	1	11	157	1727	14
160 – 164		5	162	810	19
165 – 169	11	7	167	1169	26
170 – 174	11	7	172	1204	33
175 - 179	111	3	177	531	36
				Fx²5887

1. 5887
  36
  = 163.53
2. 159.5+(³⁶/₂-14)5
  11
  = 159.5 + 1. 8182
  = 161 . 32
Graph

Plotting
Curve
(No hanging curve. Not drawn by use of free hand)
Total marks 10

1. 1. 16200
    Y
  2. 16 200 or 16 200 – 60
    y + 3 y
2. 1. 16 200 – 16 200 = 60
    y y + 3
    60y² + 180y – 48 600 = 0
    Y² + 3y – 810 = 0
    (y + 30) ( y – 27) = 0
    = 27 or -30
    = y = 27
  2. 16200 = Ksh 540
    30
  3. 15 x 16 200
    100 27
    = Ksh 90
1. Constructed locating point A
  Dropping a
  Length the 8.2 + 0.1
  For a line at an angle to BC
  For joining C to the last point
  For locating point D
  (following through)
  Total marks 10
2. 1⁄2×8×8.2=32.8Cm²
1. 1. BN = BA + AN
    = -a + ⁵/₆ b
    = ⁵/₆b – a
  2. CM = -b + ²/₅a
    = b + r (²/₅a - b)
  3. AX = b + rCM
    = b – rb + ²/₅ra
    = (1 – r) b + ²/₅ar
    Ax = a + k ( ⁵/₆b – a)
    (1 – k)a + ⁵/₆kb
  4. (1 – r) b + ²/₅ar = ( 1 -k) a + ⁵/₆kb
    ⁵/₆K = 1 – r – (i)
    ²/₅r = 1 – k -- (ii)
    K = ⁶/₅ – ⁶/₅r
    ²/₅ = 1- (⁶/₅ –⁶/₅r) = 1 – ⁶/₅ +⁶/₅r
    -4r = -1 = 7 r = ¼
    ⁵/₆K = ¾
    K = ⁹/₁₀
1. Equating two values of AX.
  Two equations extracted
2. A¹(4,-4) B¹(7,-3) C¹(2,-1)
3. A¹¹(4,4) B¹¹(3,7) C¹¹(1,2)
4. A¹¹¹(4,-4) B¹¹¹(3,-7) C¹¹¹(1,-2)
  
  For plotting
  For ∆ABC
  For ∆A¹B¹C¹
  For construction or otherwise
  For ∆A¹¹B¹¹C¹¹
  For construction or otherwise
  For ∆A¹¹¹B¹¹¹C¹¹¹
1. S = t (t² – t – 2t + 2)
  = t³ – 3t² + 2t
  ^ds/_dt = 3t² – 6 (2) + 2
  V = 3 ( 4) - 6 ( 2) + 2
  V = 2m/s
2. ^dv/_dt = 0
  ^dv/_dt = 6t – 6
  6t – 6 = 0
  t = 1
  = 3(1)2 – 6(1) + 2
  = -1m/s
  3t² – 6t + 2 = 0
3. T = 6 + (-6)2 – 4 ( 3) ( 2)
  T = 6 + 3.464
  T = 1.577 or 0.4227
  a = 6 ( 3) – 6 = 12m/s²