INSTRUCTIONS TO THE CANDIDATES:-
- Write your name and index number in the spaces provided above.
- Sign and write the date in the space provided above.
- This paper consists two sections: Section I and Section II.
- Answer all the questions in Section 1 and any two questions from Section II.
- All working and answers must be written on the question paper in the spaces provided below each question.
- Show all the steps in your calculations, giving your answers at each stage in the spaces provided below each question.
- Non-programmable silent electronic calculators and KNEC mathematical tables may be used, except where stated otherwise.
- Marks may be given for correct working even if the answer is wrong.
- Candidates should check the question paper for error and omissions.
For Examiners’ Use Only.
Section I
Questions |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
Total |
Marks |
Section II
Questions |
17 |
18 |
19 |
20 |
21 |
22 |
Total |
Marks |
QUESTIONS
SECTION A (50 MARKS)
Answer all questions in this section
- Simplify (3mks)
(3 ¹/₅ - 2 ¹/₂) ÷ 6 ¹/₂
3 ¹/₃ - Solve the following simultaneous equation 3mks
x+2y=7
x-y=1 - Use logarithms to evaluate the following √30.07893 (4mks)
37.8 × 43.31 - Awinja is 100m from the foot of a tower and the angle of elevation of the tower from her position is 49º. Find the height of the tower (2mks)
- What is the total value of digit six in the solution to 1725 -125 + 6591 ÷ 39 x 5 of 90? (3mks)
- A square ABCD is such that A(-3,4) C(2,3). Equation of line AB is 3y-2x=18 and equation of line CD is 3y-2x=5. Determine
- Equation of line BC in the form y=mx+c (2mks)
- Equation of line AD in the form y=mx+c (2mks)
- A Kenyan bank buys and sells foreign currencies using the rates shown below.
Buying Selling
(Ksh) (Ksh)
1 Euro 86.25 86.97
100 Japanese Yen 66.51 67.26
A Japanese travelling from France arrives in Kenya with 5000 Euros, which he converts to Kenya shillings at the bank. While in Kenya he spent a total of Ksh.289,850 and then converted the remaining Kenya shillings to Japanese Yen at the bank. Calculate the amount of Japanese Yen that he received. (3marks) - A square brass plate is 2mm thick and has a mass of 1.05kg. The density of brass is 8.4g/cm^{3}. Calculate the length of the plate in centimeters. (3mks)
- Solve the equation -3x+2=x+6 (3mks)
- The G.C.D of three numbers is 30 and their L.C.M IS 900. If two of the numbers are 150 and 60, what are other three possible third numbers? (3 marks)
- Juma, Ali and Hassan share the profit of their business in the ratio 3: 7: 9 respectively. If Juma receives kshs 60, 000. How much profit did the Hassan get? (3 marks)
- Use tables of logarithms to evaluate 0.3 + √0.4983 (4 marks)
0.0351 - John who runs a clothing shop bought a shirt at Ksh. 500 and marked it at Ksh. 600. A customer bought it at Ksh. 550 after engaging John in a lengthy negotiation process. What was the customer’s percentage discount. (3mks)
- Line L passes through P(8,6) and perpendicular to the line 3y + 2x + 6 = 0. Find the equation of line L and write it in the form y = mx + c. (3 marks)
- Evaluate without using a calculator. (3 marks)
-2(+5+3)-9 ÷3+5
-3×-5-2×4 - Express 0.73 ̇ as fraction. (3 marks)
SECTION II (20 MARKS)
Answer any two questions in this section in the spaces provided
- A trader sold an item at sh. 10,625 after allowing his customers 15% discount on the marked price of the item. In so doing he made a profit of 25%
- Calculate the marked price of the item. (3marks)
- Calculate the price at which the trader had bought the item (2 mark)
- If the trader had sold the item without giving a discount, calculate the percentage profit he would have made. (3 marks)
- To clear his stock the trader decides to sell the remaining items at a loss of 10%. Calculate the price at which he sold each item. (2 marks)
- The table below shows measurements of a farm in a fields book. XY=2000m
Y
1800 G 100
F 200 1600
1200 E 300
900 D 100
C 150 600
300 B 200
A 200 100
X- Using a scale 1cm rep 100m. Sketch the map of the farm (2mks)
- Calculate the area of the farm in hectares. (8mks)
- Four towns R,T,K and G are such that T is 84km directly to the north of R and K is on bearing of 295º from R at a distance of 60km. G is on a bearing of 340º from K and at a distance of 30km.
- Using the scale of 1cm to represent 10km make an accurate scale drawing to show the relative positions of the towns.(3mks)
- Find:-
- The distance and the bearing of T from K (3mks)
- The distance and the bearing of G from T. (3mks)
- The bearing of R from G (1mk)
- The points A(2,6) B(1,1) C(3,4) and D (5,3) are the vertices of a quadrilateral ABCD.
- Plot points A,B,C and D on the graph provided and join them to form quadrilateral ABCD(2mks)
- Locate and write down the coordinates of A', B', C' and D' and to the image of ABCD under a rotation of positive 90o centre (0,0) on the same grid (3mks)
- Reflect A'B'C'D' on the x-axis and draw the image of quadrilateral A'' B'' C'' D'' (3mks)
- Draw the mirror line MM for the reflection of ABCD whose image is A'' B'' C'' D'' (2mks)
MARKING SCHEME
SECTION A (50 MARKS)
Answer all questions in this section
- Simplify (3mks)
(3 ¹/₅ - 2 ¹/₂) ÷ 6 ¹/₂
3 ¹/₃
BODMAS"
2 ¹/₂ = ⁵/₂ × ³/₁₀
3 ¹/₃
= ³/₄
3¹/₅ - ³/₄ =
¹⁶/₅ - ³/ = 64 - 15
20
= 49
20
49 ÷ 6¹/₂
20
49 × 2 = 49
20 13 130 - Solve the following simultaneous equation 3mks
x+2y=7 ..........................(i)
x-y=1 .............................(ii)
x = 1 + y
Replacing in eqt (i)
(1 + y) + 2y = 7
1 + 3y = 7
3y = 6
y = 2
x = 1 + y
= x = 1 + 2
= 3
x = 3 ; y = 2 - Use logarithms to evaluate the following √30.07893 (4mks)
37.8 × 43.31
No Std Form Log 30.07893 3.008 × 10^{1} 1.4781 37.8 3.78 × 10^{1} 1.5775 43.81 4.381 × 10^{1} 1.6416 3.2191 10^{-2} × ant .2590
←
3.2191
2.2590 - Awinja is 100m from the foot of a tower and the angle of elevation of the tower from her position is 49º. Find the height of the tower (2mks)
SOHCAHTOA
Tan 49 = h
100
h = 100 Tan 49º
= 115m - What is the total value of digit six in the solution to 1725 -125 + 6591 ÷ 39 x 5 of 90? (3mks)
= 1725 - 125 + 7650
= 77650
⇒ 6 × 100 = 600 - A square ABCD is such that A(-3,4) C(2,3). Equation of line AB is 3y-2x=18 and equation of line CD is 3y-2x=5. Determine
- Equation of line BC in the form y=mx+c (2mks)
BC CD
Eqtn CD
3y = 2x + 5
y = ²/₃x + ⁵/₃
m_{1}m_{2} = -1
²/₃xm_{2} = -1
m2 = - ³/₂
(x,y) (2,3)
y - 3 = 3
x - 2 -2
-2(y - 3) = 3x - 6
-2(y - 3) = 3x - 6
y = ^{3}/_{2}x + 6 - Equation of line AD in the form y=mx+c (2mks)
AD II BC
m = -^{3}/_{2}
A (-3,4) (x,y)
y - 4 = 3
x + 3 -2
-2(y - 4) = 3x + 9
-2y + 8 = 3x + 9
y = -^{3}/_{2}x - ^{1}/_{2}
- Equation of line BC in the form y=mx+c (2mks)
- A Kenyan bank buys and sells foreign currencies using the rates shown below.
Buying Selling
(Ksh) (Ksh)
1 Euro 86.25 86.97
100 Japanese Yen 66.51 67.26
A Japanese travelling from France arrives in Kenya with 5000 Euros, which he converts to Kenya shillings at the bank. While in Kenya he spent a total of Ksh.289,850 and then converted the remaining Kenya shillings to Japanese Yen at the bank. Calculate the amount of Japanese Yen that he received. (3marks)
1 Euro ⇒ 5000 Euro
sh 86.25
sh 431 250 - 289850
= sh 141 400
100 J.Y = sh 67.26
x = 141 400
⇒ sh 210 228.962 - A square brass plate is 2mm thick and has a mass of 1.05kg. The density of brass is 8.4g/cm^{3}. Calculate the length of the plate in centimeters. (3mks)
D = m/v
v = m/D = 1050/8.4 g
= 125 cm^{3}
x^{2} × 0.2 =125
x^{2} = 1250 = 625
2
x = √625 = ± 25
x = 25 cm - Solve the equation -3x+2=x+6 (3mks)
-3x + 2 = x + 6
-3x - x = 6 - 2
-4x = 4
x = -1 - The G.C.D of three numbers is 30 and their L.C.M IS 900. If two of the numbers are 150 and 60, what are other three possible third numbers? (3 marks)
150 = 2 × 3 × 5^{2}
60 = 2^{2} × 3 × 5
G.C.D = 2 × 3 × 5
LCM = 2^{2} × 3^{2} × 5^{2}
other numbers
2 × 3^{2} × 5 = 90
2 × 3 × 5^{2} = 150
2 × 3^{2} × 5^{2} = 450 - Juma, Ali and Hassan share the profit of their business in the ratio 3: 7: 9 respectively. If Juma receives kshs 60, 000. How much profit did the Hassan get? (3 marks)
3/19 = 60000
9/19 = x
3x = 60000 × 9
x = 60000 × 9
3
x = 180,000
squares, squareroot & receiprocals - Use tables of logarithms to evaluate 0.3 + √0.4983 (4 marks)
0.0351
0.3 × (^{1}/_{0.0351}) + (49.83 × 10^{-2})^{½}
0.3 × 10^{2} × 0.2849
8.547 + 7.059 × 10^{-1}
=9.2529 - John who runs a clothing shop bought a shirt at Ksh. 500 and marked it at Ksh. 600. A customer bought it at Ksh. 550 after engaging John in a lengthy negotiation process. What was the customer’s percentage discount. (3mks)
600 - 550 = 50
^{50}/_{600} × 100 = 8.333% - Line L passes through P(8,6) and perpendicular to the line 3y + 2x + 6 = 0. Find the equation of line L and write it in the form y = mx + c. (3 marks)
3y + 2x + 6
3y = -2x - 6
y = -2x + 2
3
m1m2 = -1
m1x - 2/3 = -1
m1 = 3/2
p(8,6) (x,y)
y - 6 = 3
x - 8 2
2y - 12 = 3x - 24
2y = 3x - 12
y = 3/2x - 6 - Evaluate without using a calculator. (3 marks)
-2(+5+3)-9 ÷3+5
-3×-5-2×4
BODMAS
-2(8) - 3 + 5
15 - 8
N = -14
D = 7
-14/7
= -2 - Express 0.73 ̇ as fraction. (3 marks)
let r = 0.73333
10r = 7.3333
100r = 73.3333
100r = 73.3333
10r = 7.3333
90r = 66
r = 66
90
=11
15
SECTION II (20 MARKS)
Answer any two questions in this section in the spaces provided
- A trader sold an item at sh. 10,625 after allowing his customers 15% discount on the marked price of the item. In so doing he made a profit of 25%
- Calculate the marked price of the item. (3marks)
10625 × 100 = 12500
85 - Calculate the price at which the trader had bought the item (2 mark)
10625 × 100 = 8500
1.25 - If the trader had sold the item without giving a discount, calculate the percentage profit he would have made. (3 marks)
12000 - 8000 = 4000
4000 × 100
8500
= 47.06% - To clear his stock the trader decides to sell the remaining items at a loss of 10%. Calculate the price at which he sold each item. (2 marks)
8500 × 0.9 = sh.7650
- Calculate the marked price of the item. (3marks)
- The table below shows measurements of a farm in a fields book. XY=2000m
Y
1800 G 100
F 200 1600
1200 E 300
900 D 100
C 150 600
300 B 200
A 200 100
X- Using a scale 1cm rep 100m. Sketch the map of the farm (2mks)
- Calculate the area of the farm in hectares. (8mks)
1 ⇒½ × 100 × 200 = 10000m^{2}
2 ⇒½ (200 + 150) × 500 = 87 500m^{2}
3 ⇒½ (150 + 200) × 1000 = 175 000
4 ⇒½ (400) × 200 = 40000
5 ⇒½ (300) × 200 = 30000
6 ⇒½ (100 + 200) × 600 = 90000
7 ⇒½ (100 + 300) × 300 = 60000
8 ⇒½ (300 + 100) × 600 = 120 000
9 ⇒½ (200) × 100 = 10000
622500
= 62.25 ha
- Using a scale 1cm rep 100m. Sketch the map of the farm (2mks)
- Four towns R,T,K and G are such that T is 84km directly to the north of R and K is on bearing of 295º from R at a distance of 60km. G is on a bearing of 340º from K and at a distance of 30km.
- Using the scale of 1cm to represent 10km make an accurate scale drawing to show the relative positions of the towns.(3mks)
- Find:-
- The distance and the bearing of T from K (3mks)
8.1 cm × 10
= 81 km - The distance and the bearing of G from T. (3mks)
7.3 cm ⇒ 73km
Bearing 245º - The bearing of R from G (1mk)
= 131º
- The distance and the bearing of T from K (3mks)
- Using the scale of 1cm to represent 10km make an accurate scale drawing to show the relative positions of the towns.(3mks)
- The points A(2,6) B(1,1) C(3,4) and D (5,3) are the vertices of a quadrilateral ABCD.
- Plot points A,B,C and D on the graph provided and join them to form quadrilateral ABCD(2mks)
- Locate and write down the coordinates of A', B', C' and D' and to the image of ABCD under a rotation of positive 90o centre (0,0) on the same grid (3mks)
- Reflect A'B'C'D' on the x-axis and draw the image of quadrilateral A'' B'' C'' D'' (3mks)
- Draw the mirror line MM for the reflection of ABCD whose image is A'' B'' C'' D'' (2mks)
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