Mathematics Questions and Answers - Form 1 Opener Exam Term 3 2022

MATHEMATICS
FORM 1 OPENER EXAM
TERM 3 2022

INSTRUCTIONS TO CANDIDATES

These papers consist of two sections; Section A and section B.
Answer all questions in section A and Section B.
Marks are awarded for steps which are correctly worked.
Calculators must not be used.

SECTION A 50 (MARKS)
ANSWER ALL QUESTIONS IN THIS SECTION

Ayub travelled part of the journey by train and the rest of the journey by bus. The total fare was Ksh 4000. On return, the fare of the bus was hiked by a half of what he had paid and the total fare of the return journey hiked to Ksh 4800. Find the train fare. (3mks).
When 19346 is divided by a number, the quotient is 841 and the remainder is 3. What is the number? (3mks).
A square toilet is covered by a number of whole rectangular tiles of sides 60cm by 48cm. Calculate the least possible area of the room in square metres. (3mks).
Evaluate using prime factors and leave your answer in prime factors form. (4mks).
Evaluate (4mks).
Use number line to solve 6 – (- 4) (2mks).
Given that X= 4 and y = -6, evaluate the following. (3mks).
6x – 3xy
Simplify (3mks)
a + b - 2a - b
2 3
The ratio of boys and girls in a mixed school is 2:3. If there are 160 boys, how many girls are there? (3mks).
Find the ratio a:c, given that;
a:b = 7: 1, b:d = 1: 2, d: c = 2: e (4mks).
The perimeter of a semicircular protractor is 14.28cm. Find its radius. (3mks)
The area of the sector of a circle is 38.5cm². Find the radius of the circle if the angle subtended at the centre is 900. (Take π= 22/7). (3mks).
A cylindrical column of fat has diameter 17.5cm and height 10cm. Calculate the density in g/cm³ of fat if the column has a mass of 2kg. (3mks)
Express the following measurements in 3 s.f (3mks)
1. 36.7892
2. 0.09854
3. 345204
Find the square root of 0.001952 using mathematical table. (3mks)
Arrange the following numbers in an ascending order. (3mks)
2 ⁷/₈, ¹⁴/₅, 3, ¹⁷/₉

SECTION B 50 MARKS
ANSWER ALL QUESTIONS IN THIS SECTION

A rectangle measures 18cm by 12cm.
1. If each dimension is reduced by 2cm, by what percentage is:
  1. The perimeter reduced. (3mks)
  2. The area of the rectangle reduced. (3mks)
2. If each dimension is reduced by 2%, by what percentage is the area of the rectangle reduced. (4mks)
1. Study the equations below and complete the tables.
  7x + 2y = 16
  
  x 0 1 2
  
  y
  
  2x-y = 3
  
  x 0 1 2
  
  y
2. On the same axis, draw the graphs of y = 2x – 3 and y = 8 - ⁷/₂x (4mks)
3. Use the graph to solve the simultaneous equations; (2mks)
  y = 2x – 3
  y = 8 - ⁷/₂x
1. Draw triangle ABC in which AB = 11cm, AC = 8cm and BC = 5.6cm. (2mks)
2. Construct the bisectors of any two angles of the triangle and let the bisectors meet at R. (2mks)
3. Draw the perpendicular from R to AB so that it cuts AB at M. (2mks)
4. With centre R and the radius RM, Draw a circle. (2mks)
5. Calculate the area of the circle. (2mks)
A carpenter constructed a closed wooden box with internal measurements 1.5m long by 0.8m wide and 0.4m high. The wood used in constructing the box was 1.0cm thick and had a density of 0.6 g/cm³.
1. Determine the:
  1. Volume in cm³ of the wood used in constructing the box. (3mks)
  2. Mass of the box in kg correct to 1dp. (2mks)
2. Identical cylindrical tins of diameter 10cm and height 20cm with a mass of 120g each were packed in the box. Calculate;
  1. Maximum number of tins that were packed in the box. (3mks)
  2. Total mass of the box with the tins. (2mks)
Find the size of the angles marked with letters.
1. (1mk)
2. (5mks)
3. (4mks)
1. The GCD of two numbers and their LCM is 360. If one of the numbers is 72, what is the other number? (3mks).
2. When a number u is divided by 34 or 24 or 40, the remainder is always 5. Find the least value of u. (3mks)
3. Three bells ring at intervals of 12minutes, 15minutes and 18minutes. If they sound together at 10:00 am;
  1. After how long will they next sound together? (3mks).
  2. What time will this be? (1mks)

MARKING SCHEME

4800 – 4000 = 800
Let the bus fare be x.
1/2 x = 800
X = 800 x 2 = 1600
Train fare 4000 – 1600 = 2400
Let the number be x
19346 = 841 rem 3
x
19346 = x(841) + 3
19346 = 841x + 3
19346 – 3 = 841x
19343 = 841x
X = 23
2 48 60

2 24 30

2 12 15

2 6 15

3 3 15

5 1 5

L.C.M= 2⁴ x 3 x5
Area = 240 x 240 = 57600
57600 = 5.76 m²
10000
2 1470

3 735

5 245

7 49

7 7

1

2 7056

2 3528

2 1764

2 882

3 441

3 147

7 49

7 7

1

1470² = ( 2 x 3 x 5 x 7²)²
7056= 2⁴ x 3² x 7²
( 2 x 3 x 5 x 7²)²
√2⁴ x 3² x 7²
2² x 3² x 5² x 7⁴
2² x 3 x 7
3 x 5² x 7³
Numerator
¹/₂ x ⁷/₂ + ³/₂ (⁵/₂ - ²/₃)
⁷/₄ + ³/₂(¹¹/₆)
⁷/₄ + ¹¹/₄ = ¹⁸/₄ = ⁹/₂

Denomenator
³/₄ of 2¹/₂ ÷ ¹/₂
³/₄ x ⁵/₂ x ²/₁ =¹⁵/₄
⁹/₂÷¹⁵/₄
⁹/₂ x ⁴/₁₅
= 1¹/5
6 – (- 4) = 10
X = 4 y = -6
6x – 3xy
6(4) – 3 (4) (-6)
24 + 72
= 96
a+ b - 2a - b
2 3
3(a+ b) – 2(2a – b)
6
3a+ 3b – 4a +2b
6
5b – a
6
B : G: T
2 : 3 :5
2/5 ⇒160
3/5 ⇒?
3 x – 160
2
= 3 x 80
= 240 girls
a : b = 7 : 1
b : d = 1 : 2
d : c = 2 : e

a : b : d : c = 7 : 1
1 : 2
2 : e
a : b : d : c = 7 : 1 : 2 : e
a : c = 7 : e
C = 2 πr
2πr + 2r = 14.28
2
πr + 2r = 14.28
r(π + 2) = 14.28
r(3.142 + 2) = 14.28
r(5.142) = 14.28
r = 14.28
5.142
r = 2.777cm
A = 38.5 cm²
ϴ = 90⁰
A = ϴ x πr
360
38.5 = 90 x 22 x r²
360 7
r² = 38.5 x 360 x 7
90 x 22
r² = 49
r = 7
V = πr²h
V = 22 x 17.5 x 17.5 x 10
7 2 2
V= 2406.25 cm³
m = 2kg
m = 2 x 1000 = 2000g
ρ = m = 2000
v 2406.25
ρ = 0.8312 g/cm³
1. 36.7892
  = 36.8
2. 0. 0 985 4
  = 0.0985
3. 345204 = 345000
√0.001952
√0.001952 = √19.52 x10^-4
√19.52 x10-4 = √19.52 x √10^-4
= 4.4159
+23
4.4182
= 4.4182 x 10^-2
= 4.8182 x ¹/₁₀₀
= 0.044182
2 ⁷/₈, ¹⁴/₅, 3, ¹⁷/₉
²³/₈,¹⁴/₅ 3 , ¹⁷/₉L:C:M = 8 x 5 x 9 = 360
²³/₈ x 360 = 1035
¹⁴/₅ x 360 = 1008
3 x 360 = 1080
¹⁷/₉ x 360 = 680
¹⁷/₉, ¹⁴/₅, ²⁷/₈, 3
1. Reduced by 2cm
  Length = 18 - 2 = 16cm
  Width = 12 – 2 = 10cm
  1. Perimeter = 2(18 + 12)
    = 60cm
    New Perimeter = 2 (16 + 10) = 52
    % = (60 – 52) x 100
    60
    = 8 x 100 = 13.33%
    10
  2. Original area = 2 (18x 12)
    = 2 x 216
    = 432cm²New = 2(16 x 10) = 320cm²% change = (432 – 320) x 100
    432
    = 25.93%
2. Length = 98 x 18 = 17.64 cm
  100
  Width = 98 x 12 = 11.76 cm
  100
  Original area = 18 x 12 = 216cm²New area = 17.64 x 11.76 = 207.4464 cm²% change = (216 – 207.4464) x 100
  216
  = 3.96%
1. 2x – y = 3
  y= 2x – 3
  
  7x + 2y = 16
  2 y = 16 – 7 x
  2 2 2
2. x 0 1 2
  
  y -3 -1 1
  
  x 0 2 4
  
  y 8 1 -6
  
  B₁B₁
3. B1B1
  Graph
  
  Point of intersection = (2,1)
  X= 2, y = 1
e. Area of the circle = 3.142 x 1.7 x 1.7
A= 9.08038cm²
A= 9.080cm²
1. 1. Internal volume = 150 x 80 x 40 = 480000 cm³
    External Volume = 152 x 82 x 42 = 523, 488 cm³Volume of wood = 523,488 – 480,000 = = 43488cm³
  2. Mass of the box
    m = ρ x v
    m = 0.6 x 4388
    = 26092.8g
    26092.8 = 26.0928kg
    10000
    = 26.1kg (1dp)
2. 1. V = πr2h
    3.142 x 5 x 5 x 20
    V = 1571cm³
    Number of tins
    = 480000
    1571
    = 305.54
    = 305 tins
  2. Total mass of the box with tins
    = 26.0928 + (120 x 305)
    1000
    = 26.0928 + 36.6
    = 62.6928kg
1. n = 5
  (2n - 4)90
  (2(5) - 4)90
  (10 – 4)90
  6 x 90 = 540
  2(180 – x) + 100 + 130 = 540
  360 – 2x + 330 = 540
  690 – 2x = 540
  - 2x = -150
  X = 75⁰
2. y = 180 – 60 = 1200
  4x + 60 = 180
  4x = 120⁰
  x = 30⁰
  2x = 2(30) = 60⁰
  3x = 3(30) = 90⁰
  4x = 4(30) = 120⁰
3. y = 87
  w = 180 – 87 = 93⁰
  z = 180 – (87+45) = 48⁰
  x = 180 – (87 +45) = 45⁰
1. G.C.D = 8 = 23
  L.C.M = 360 = 2³ x 3² x 5
  X = 72 = 2³ x 3²y = ? = 2³ x 5
  y = 40
2. 2 24 34 40
  
  2 12 17 20
  
  2 6 17 10
  
  3 3 17 5
  
  5 1 17 5
  
  17 1 17 1
  
  1 1 1
  
  L.C.M = 2³ x 3 x 5 x 17 = 2040
  Value of u= 240 + 5
  u= 245
3. 1. 2 12 15 18
    
    2 6 15 9
    
    3 3 15 9
    
    3 1 5 3
    
    5 1 5 1
    
    1 1 1
  2. L.C.M = 22 x 32 x5 = 180 mins = 3hrs
    10: 00
    + 3. 00
    13:00
    - 12:00
    1:00 pm