# Mathematics Questions and Answers - Form 1 End Term 3 Exams 2023

INSTRUCTIONS TO CANDIDATES

• This paper consists of TWO sections. Section A and Section B
• Answer ALL the questions in section A and only FIVE questions from Section B
• All answers and working 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 below each question.
• Marks may be given for correct working even if the answer is wrong
• Non- programmable silent calculators and KNEC mathematical tables may be used except where stated otherwise.
• Write the answers in English.

SECTION A (50 MARKS)

1. Simplify without using a calculator, giving your answer as a fraction in its simplest form.
21/3 – 12/3 ÷ 5/9       (3marks)
4/7 of 21/3 – 22/7
2. Evaluate without tables or calculator.   (3 marks)
–12 ÷ (–3) x 4 – (–20)
– 6 x 6 ÷ 3 + –6
3. An airbus left town N at 1945hrs and arrived in town L at 0320hrs. It stayed for 11/2 hours for rest and refreshment of passengers and crew. It then headed for town W and took 11/2 hours.
1. How long did the journey from town N to town L take in hours and minutes? (2 marks)
2. At what time did it arrive in town W? (1 mark)
4. Mzee is three times as old as his son now. If ten years ago the sun of their ages was 44, how old was Mzee when his son was born? (3 marks)
5. Given that  2x + 3y  = 5 find the ratio y:x   (3 marks)
3x + 4y
6. Express  as improper fraction. (3 marks)
7. An Italian tourist, on arrival in Kenya, converted 6000 Euros into Kenya shillings. During her three day’s stay the country, she spent Kshs 260,000. She converted the remaining amount into US dollars. Using the exchange rates table below calculate how many US dollars she got?   (3 marks)
1 US dollar \$          96.20               96.90
1 Euro C                112.32             112.83
8. Find the area in hectares of a field book measurement is recorded in metres as follows.
 To E 50 To F 80 D 170 140 110 100  30 A 60 to C 50 to B
9. The perimeter of a triangular field is 120 m. Two of the sides are 21m and 40m. Find the largest angle of the field. (4 marks)
10. A paint dealer mixes three types of paint A, B and C in the ratios A: B =3:4 and B:C=1:2. The mixture is to contain 168litres of C.
1. Find the ratio A: B:C (2 marks)
2. Find the required number of litres of B. (1 marks)
11. Annette has some money in two denominations only. Fifty shillings notes and twenty shilling coins. She has three times as many fifty shilling notes as twenty shilling coins. If altogether she has sh. 3,400, find the number of fifty shilling notes and 20-shilling coin. (3 marks)
12. The mean of five numbers is 20. The mean of the first three numbers is 16. The fifth number is greater than the fourth by 8. Find the fifth number. (3 marks)
13. A farmer has a piece of land measuring 840m by 396m. He divides it into square plots of equal size. Find the maximum area of one plot. (3 marks)
14. Each interior angle of a regular polygon is 1200 larger than the exterior angle. How many sides have the polygon? (3 marks)
15. A photograph is reduced in the ratio 3:5 for a newspaper and further reduced in the ratio 4:5 for a text book. Find the ratio of the photograph size to the text book size. (3 marks)
16. The net of a solid is as shown below.
1.  Sketch the solid if ABCD is the base. (2 marks)
2. Calculate the volume of the solid sketched in (a). (2 marks)

SECTION B (50 MARKS)

Attempt ONLY FIVE questions in this section.

1. Milk in a cooling factory is stored in a rectangular tank whose internal dimensions are 1.7m by 1.4m by 2.2m one day the tank was 75% full of milk.
1. Calculate the volume of milk in the tank in litres. (3 marks)
2. The milk is packed in small packets which are in the shape of a right pyramid on an equilateral triangle base of side 16cm. The height of each packet is 13.6cm. Each packet is sold at Sh.30. Calculate
1. The volume of milk in milliliters, contained in each packet to 2 significant figures. (4 marks)
2. The exact amount of money that was realized from the sale of all the packets of milk. (3 marks)
2.
1. Using a ruler and a pair of compasses only construct a rhombus A B C D such that AB = 6cm and <ABC = 135°. (4 marks)
2. Drop a perpendicular from C to AB extended to meet AB at N. Measure BN and CN. (3 marks)
3. Bisect <ABC and <DAB, let the two bisectors meet at M. Measure MA. (1 mark)
4. Determine the area of triangle ABM. (2 marks)
3.
1. A manufacturer sells goods to a shopkeeper at a profit of 15%. The shopkeeper sells them so as to make a profit of 25%. During a sale, the shopkeeper reduced his prices by 10%. Find, to the nearest shilling, the factory price of an article which was marked at sh. 450 during the sale. (3 marks)
2. Mrs. Kinuthia bought a car in USA for sterling pounds 10,000. She paid transportation charges from USA to Kenya amounting to 500 US dollars. On arrival to Kenya she paid import duty of 20% of the buying price and clearance tax of 0.5% of its total value in Kenya. If the exchange rate were as follows:
1 Sterling pound = 1.50 US dollars
1 Sterling pound = 120.00 Kenya shillings
1 US dollars = 80.00 Kenya shillings
1. Calculate the total amount paid for the car by Mrs. Kinuthia in Kshs. (5 marks)
2. If Kinuthia later on sold the lorry at a profit of 10% find how much profit he was paid. (2 marks)
4. The following table gives the population of a town at different times.
 YEAR 1990 1995 2000 2005 2010 2015 2020 POPULATION 5000 6100 7500 9100 11100 13600 16600
1. Draw a graph of population against year to represent the information above. (5 mark)
2. From the graph estimate the population of the town in:
1. 1998
2. 2011
3. 2017
3. At what time was the population double the figure in 1999.
5. Three villages A, B and C drawn on a map of scale 1: 100,000 are such that B and C are 8km apart while A and B are 6.3 km apart on the actual ground. C is on a bearing of 315° from B and A is due south of C,
1. Use scale drawing to show relative positions of A, B and C. (Also show two possible positions of A). (4 marks)
2. What is the longest actual distance between villages A and C? (2 marks)
3. A market serving the villages is placed in the most central position from each of the three villages. Using the longest position of A from C, locate the market with centre M, hence state the actual distance from M to vertex A. (4 marks)
6.
1. Three people Amina, Bundi and Chari formed a business partnership. Amina invested sh.80,000 for 2 years, Bundi sh.50,000 for 3 years and Chari invested his money for 4 years. They agreed that the profits should be shared in proportion to the amount invested and the time for which it was invested. How much did Chari invest if Amina’s share of profit of sh.129000 was sh.48000? (5 marks)
2. A lady buys a car for sh.40,000 paying sh.16000 and the remainder in instalments of sh.8000 paid at the end of each of the first three quarters together with a final payment at the end of the fourth quarter to clear the debt. Interest at 3% per quarter, reckoned on the amount owing at the beginning of each quarter is added at the end of each quarter.
Calculate the amount of the three quarters and also the final payment to clear the debt. (5 marks)
7. A garden measures 10m long and 8m wide. A path of uniform width is made all round the garden. The outer perimeter of the path is 52m.
1. Find the width and the area of the path. (4 marks)
2. The path is to be covered with square concrete slabs. Each corner of the path is covered with a slab whose side is equal to the width of the path. The rest of the path is covered with slabs of side 50cm. The cost of making each corner slab is sh.600 while the cost of making each smaller slab is sh.50.
Calculate :
1. the number of the smaller slabs used (3 marks)
2. the total cost of the slabs used to cover the whole path. (3 marks)
8.
1. Complete the table for the following equations
1. y−1 + x+3 = x−3
2        3        2
 x 0 1.5 3 4.5 y
2.
y+1 = x−2
3        2
 x 0 1 2 3 y
2. On the grid provided draw the graph of the above equations
3. Use the graphs above solve for the equations (i) and (ii)

### MARKING SCHEME

1. 7/3 − (5/3 × 9/5)
(4/7 × 7/3) − 16/7
7/3 − 3/1
4/3 − 16/
− 2/3
28 − 48
21
−2/3 ÷ −20/21
+2/3 × 21/20
= 7/10
2. 4 × 4 + 20
−6 × 2 − 6
16 + 20
−12−6
36
−18
= −2
3.
1. 0320hrs − 1945hrs
2720
1945 −
0735
= 7hours 35 minutes
2. Arrival time in w = 0320hrs + 1½hours + 1½hours
= 0320hrs + 3 hours
= 0620hrs
4. Let mthe age of the son be x
(3x − 10) +(x −10) = 44
3x − 10 + x − 10 = 44
4x − 20 = 44
4x = 64
x = 16
Mzee's present age = 3 × 16
= 48 years
48 − 16 = 32years
5. 2x + 3y = 15x + 20y
3y − 20y = 15x − 2x
− 17y = 13x
y = − 13/17x
y:x =−13:17
6.        r = 2.8383....
100r = 283.8383
99r = 281
r = 281
99
7. 6000 × 112.32 = Ksh 673,920
Balance in Ksh = Ksh 673,920 − 260,000
= Kshs 413, 920
Balance in US dollars = 413,920
96.90
= 4271.62 US dollars
8.

Total Area
= ½ × 30 × 50 + ½(60+50)80 + ½ × 60 × 60 + ½ × 50 × 30 + ½(50+80)40 + ½ × 80 × 100
750 + 4400 + 1800 + 750 + 2600 + 4000 = 14,300m2
Area in ha = 14300 = 1.43ha
10000
9. 120m = 21 + 40 + y
y = 59m
Let 1cm rep 10m

Largest angle = 148° ±1°
10.
1. A:B
3:4
B:C
1:2
A:B:C = 3:4:8
2. If 8 → 168litres
∴ 4 → 4 × 168
8
= 84 litres
11. Let the nhumber of 20 shilling coins be y
fifty shilling notes = 3y
3y(50) + y(20) = 3400
150y + 20y = 3400
170y = 3400
y = 20
fifty shillings notes
= 20 × 3
= 60
12. Sum of the 5 numbers = 20 × 5
= 100
Sum of the first 3 numbers = 16 × 3
= 48
Sum of the fourth & fifth numbers = 100 − 48
= 52
Let the 4th be a
a + a + 8 .= 52
2a = 44
a = 22
the 5th number
= 22+ 8
= 30
13.

G.C.D = 2 × 2 × 3
= 12
Maximum A = L2
= 122
= 144m2
14. n =         360
exterior angle
a + a + 120 = 180
2a  = 60
a = 30
Exterior angle = 30°
n = 360
30
= 12 sides
15. Newspaper = 3/5 photograph
Text book = 4/5 newspaper
= 4/5 × 3/5 photograph
Text book = 12/25 photograph
Photgraph : Textbook = 25:12
16.
1.
2. Volume = ½bh × L
= ½ × 4 × 3 × 3
= 18cm2
17.
1. Volume of milk = (1.7 × 1.4 × 2.2 × 75/100)m3
= 3.927m3
1m3 = 1000litres
∴ 3.927m3 = 3.927 × 1000
= 3927litres

2.
1. V = 1/3 Base area × Height
Base area = ½bh
= ½ × 16 × 13.86
=110.88cm2
h = √(162 − 82
= 13.86cm
V = 1/3 × 110.88 × 13.6
= 502.66cm3
1cm3 = 1 ml
∴ 502.66cm3
= 502.66cm3 × 1ml
1cm3
= 502.66ml
≅ 500ml (2s.f)
2. Total cost = 3927L  × 30
0.5L
= Sh. ,235,620
18.
1.

BN = 4.2cm
CN = 4.3cm
2.
3.
4.  AB = 6cm
Bm = 2.4cm
∠ABm = 67.5°
Area of ABM = ½ × 6 × 2.4 sin 67.5
= 6.652cm2
19.
1. The price before reduction by 10%
= 450 × 100%
90%
=500/=
125% = 500/=
∴ 100% = ? 100%× 500
125%
= Sh.400
115% = 400/=
∴ 100% = ? 100% × 400
115%
= Sh. 347.83
≅ Sh. 348
2.
1. Buying price = 10000 × 120
= Kshs .1,200,000
Transportation charges = 500 × 80
= Ksh. 40,000
Import duty = 20/100 × 1,200,000
= Kshs 240,000
Clearance tax = 0.5/100 × (1,200,000 + 40000 + 240,000)
= Sh. 1,480,000 × 0.5/100
= Sh. 7,400
Total amount paid = Sh(1,480,000 + 7,400)
= Kshs 1,487,400
2. Profit = 10/100 × 1,487,400
= Kshs. 148, 740
20.
1.
2.
1. 1998 - 7000
2. 2011 - 11600
3. 2017 - 14800
3. Population in 1999 = 7200
Double the figure = 7200 × 2
= 14,400
21.
1.
2. 87cm × 1km
1cm
= 8.7km
3. MA = 4.5cm
1cm rep 1km
∴ Actual distance MA = 4.5km
22.
1.   Amina    :    Bundi   : Chari      Total
2×80000   3×50000   4×y
160,000     150,000    4y         310,000 +4y
160000     × 129000 = 48,000
310000+4y
160000 × 129000 = 48000  (310000+4y)
48000              48000
10 × 43000 = 310000 + 4y
430000 − 310000 = 4y
4y = 120,000
y = 30,000
Chari invested Sh. 30,000/-
2. The initial balance to be paid
= Sh. 40,000,000 − Sh. 16,000,000
= Sh. 24,000,000/=
The amount of the 3 quarters = Sh. 24,000,000/=
Interest at the beginning of the 1st quarter = 3/100 × 24,000,000
= 720,000/=
Interest at the beginning of the 2nd quarter = 3/100 × 16,000,000
= 480,000/=
Interest at the beginning of the 3rd quarter = 3/100 × 16,000,000
= 240,000/=
The final payment = 720,000 + 480,000 + 240,000
= Sh. 1,440,000

23.
1. Let the width of the path be x
2(10 + 2x) + 2(8+2x) = 52
36 + 8x = 52
8x = 16
x = 2m
The length is 10 + 2(2m) = 14m
The width is 8+2(2) = 12m
Area of the path = 14 × 12 − 10 × 8
= 168 − 80
= 88m2
2.
1. Area co\vered by the small slabs = 10×2 + 10×2 + 8×2 + 8×2
= 20 + 20 + 16 + 16
= 72m2
=720,000cm2
Area of 1 smaller slab = 50cm  × 50cm = 2500cm2
No of smaller slabs = 720,000 = 288 slabs
2500
2. Total cost = Sh(600×4) + Sh(288×50)
= Sh(2400 + 14400)
= Sh. 16,800

24.
1.
1.
 x 0 1.5 3 4.5 y −4 −3.5 −3 −2.4
2.
 x 0 1 2 3 y −4 −2.5 −1 0.5
2.
3.
x = 0
y = −4

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