Displaying items by tag: std 8
Wednesday, 15 September 2021 07:56
Geometry - Class 8 Mathematics Revision Notes
Geometry
Worked Exercise
- Find the value of x in the following.
Working
X+45+50=1800 (Angles on a straight lines are supplementary i.e. add up to 180º )
X+95=180º
X=85º
The value of x =85º - Find the sum of angle “a” and angle “b” in the figure below.
Working
Lines AB and C D are transversals are Therefore 90+b = 1800
Co-interior angles - supplementally
Therefore b=180-90
B = 90º
Angle a = 120º - (Corresponding angles)
Therefore a = 120º
Sum of a and b
=120 + 90
= 210º - Find the size of angle marked A B D in the figure below.
X+4x+x+30=180º (angles on a straight line are supplementary)
= 6x+30=180
6x=180-30
6x = 150
X = 25
Angle A B D =x + 4X
But x = 25
Therefore 25 + (4 x 25)
= 25 + 100
= 125º - Draw an equilateral triangle A B C where Line AB = 6cm.
Draw a circle touching the 3 vertices of the triangle. What is the radius of the circle?
Working
Steps:- Draw line A B = 6cm
- With A as the Centre with the same radius 6cm, mark off an arc above line A B.
- With B as the Centre with the same radius 6cm, mark off an arc above line A B to meet the arc in (II) above. Call the point of intersection point C
- Join C to A and C to B
- Bisect line A B and B C and let the bisectors meet at point X.
- With X as the Centre, draw a circle passing through points A, B and C.
- Measure the radius of the circle.
- Construct a triangle P Q R in which Q P = 6cm. Q R = 4cm and P R =8cm. Draw a circle that touches the 3 sides of the triangle, measure the radius of the circle.
Working- Draw line Q P 6cm
- With Centre Q, make an arc 4cm above line Q P.
- With Centre P, make an arc 8cm above line Q P and let the arc meet the one in (II) above. Label the point of intersection as R.
- Join R to P and R to Q.
- Bisect any two angles and let the bisectors meet at point Y.
- With Y as the Centre, draw a circle that touches the 3 sides of the triangle.
Construction
R = 3.5cm
- A rectangle measures 6cm by 2½ cm. What is the length of the diagonal?
Working
AC2 = AB2 + BC2 [ Pythagoras Theorem]
AC2 = 62 + 2 ½2
AC2 = 36 + 6.25
AC2 = 42.25
AC = √42.25
= 6.5 or 6 ½
NB: The Pythagoras theorem states
H2 =B2 +h2
h2 = H2 – b2
b2 = H2 –h2 - In the figure below, A B C is a straight line and B C D E is a quadrilateral. Angle CBD = 620 and lines EB = BD = DC. Line EB is parallel to DC.
What is the size of angle BDE?
Working
Consider triangle BCD (isosceles triangle)
Therefore base angles are equal
CBD = 62º
BCD = 62º
Therefore, BDC = 180 – 124 = 56º
Angle CDB = angle EBD [Alternate triangle]
Therefore EBD = 56º
Angle BDE =180 - 56/2
= 62º
Therefore, BDE = 62º - Find the size of the largest angle from the following triangle.
Working
4X – 10 + x – 20 + 3x + 10 = 180 [Angle sum of a triangle]
8x – 20 = 180
8x = 200
X = 25
4x – 10 = (100 – 10)º
= 90º largest angle.
Published in
Mathematics Class 8 Notes
Wednesday, 15 September 2021 07:28
Time, Speed and Temperature - Class 8 Mathematics Revision Notes
Time, Speed and Temperature
Worked Exercise
- An airplane took 4½ hours to fly from Cairo to Zambia. If it landed in Nairobi at Nairobi at 0215 h on Saturday, when did it take off from Cairo?
- Friday 2145 h
- Saturday 2245h
- Friday 2245h
- Saturday 2145 h
Working
The time the aeroplane took from midnight to 0215h of Saturday = 2h 15min
The difference (4h 30min – 2h 15min) is the time the aero plane took on Friday night.
Time on Friday night
h min
4 30
- 2 15
2 15
= 2h 15min before midnight
Time of takeoff from Cairo
h min
24 00
- 2 15
21 45 on Friday
The correct answer is A (Friday 2145 h)
- A train let Mombasa on Monday at 2125 h and took sixteen and half hours to reach
Kisauni. When did the train reach Kisumu?- Tuesday 1.55 a.m
- Tuesday 1.55 p.m
- Wednesday 1.55 p.m
- Monday 1:55 a.m
Working
Monday: from 2125h to midnight = 2400h - 2125h
= 2h 35min
Tuesday: Number of hours traveled from midnight
= 16h 30min - 2h 35 min
= 13h 55min
The train arrived at Kisumu on Tuesday at 1355h
This is the same as 1.55p.m
The correct answer is B (Tuesday 1.55pm)
- A meeting started at quarter to noon. If the meeting lasted for 2 h 35min, what time in 24-h clock system did the meeting end?
- 1320h
- 1420h
- 1310h
- 1410h
Working
The meeting started at 11.45
Add the meeting time
h min
11 45
+ 2 35
14 20
The meeting ended at 1420h
The correct answer is B (1420 h)
- A wall clock gains 3 seconds every one hour. The clock was set correct at 1pm on Tuesday. What time was it showing at 1pm on Friday on the following week?
Working
The number of days from Tuesday 1 pm to Friday 1pm the following week = 10days.
Number of hours = (24 x 10) = 240 hrs.
The clock gains 3 seconds after every hour in ten days.
240 x 3 = 720 seconds
Min = 720/60 = 12 min
Hence it will show 1 p.m. + 12 min = 1.12 pm
In 24 h clock system
= 1312h
The correct answer is B (1312h) - A cyclist traveled from Nairobi to Nyeri for 4h 30min at a speed of 80km/h. He drove back to Nairobi taking 4 hours. What is his speed, in km/h?
- 90
- 72
- 80
- 100
Working
Distance = speed x time
= 80 x 4½
= 360 km
From Nyeri - Nairobi distance = 360km
Time taken = 4hrs
Therefore speed = Distance/Time
= 90km/h
The correct answer is A (90km/hr)
- A motorist crosses a bridge at a speed of 25m/s. What is his speed in km/hr?
- 80
- 90
- 60
- 30
Working
When working out this kind of question we use a relationship,
If 10 m/s = 36 km/h
25m/s = ?
= ( x 36) km/h
= 90 km/h
The correct answer is B (90km/h)
- The distance between Mombasa and Mtito Andei is 290km. A bus left Mombasa at 1035h and traveled to Mtito Andei at a speed of 50km/h. At what time did it arrive at Mtito Andei?
- 1623h
- 1523h
- 1423h
- 1723h
Working
Time = Distance/Speed
= 290/50
= 5 4/5hours or 5h 48min
Arrival time = Departure time = Time taken + Time taken
h min
10 35
+5 48
16 23
The arrival time 1623 h
The correct answer is A (1623h)
- Kamau drove from town M to town N a distance of 150 km. He started at 9.30 am and arrived at town N at 11.00 am. He stayed in town for one hour and 50 minutes. He drove back reaching town M at 2.30pm. Calculate Kamau’s average speed for the whole journey.
- 90km/h
- 100km/h
- 60km/h
- 150 km/h
Working
Total distance from M to N and back
= 150 x 2
= 300 km
Total time taken
From 9.30 - 11.00 = 1 h 30 min
Time spent in town
= 1 h 50 min
Time taken from N to M
= 1430h – 1250h
= 1h 40min
Total time = 5 hours
Average speed = Total distance/Total time taken
=(60km/h)
The correct answer is C (60km/h)
- The temperature of an object was 20º C below the freezing point. It was warmed until there was a rise of 40º in temperature. What is the reading in the thermometer?
- 60 Cº
- 40Cº
- 20Cº
- 20Cº
Working
Below freezing point means; - 20
Rose by 40º
Therefore - 20º + 40 = 20 C
The correct answer is C (20º C)
Published in
Mathematics Class 8 Notes
Tagged under
Wednesday, 15 September 2021 07:02
Money - Class 8 Mathematics Revision Notes
Money
Worked Exercise
- Mutiso paid sh.330 for an item after the shopkeeper gave him a 12% discount. What was the marked price of the radio?
- sh300
- sh369.60
- sh375
- sh350
Working
Marked price = 100%
Discount = 12%
S.P = 100% - 12%
= 88%
If 88 % = 330
100% = ?
100 x 300/88 = Sh375
The correct answer is C (375)
- Olang’ borrowed sh.54000 from a bank which charged interest at the rate of 18% p.a. He repaid the whole loan after 8 months .How much did he pay back?
- sh6480
- sh60, 480
- sh14580
- sh77760
Working
I = PRT/100
= 54000 x 18 x 8/100 x 12
= sh6480
Amount = P + I
= (54,000 + 6,480) shillings
= Ksh 60, 480
The correct answer is B
- The cash price of a microwave is sh. 18000. The hire purchase price of the microwave is 20% more than the cash price. Bernice bought it on hire purchase terms by paying 40% of the hire purchase price as the deposit and the balance equal monthly installments of sh1620. How many installments did she pay?
- 12
- 10
- 9
- 8
Working
Let the cash price be 100%
Hire purchase = 100% + 20%
= 120% of the cash price
= 120/100 x 1800
= sh.21, 600
Deposit = 40% of HPP
= 40/100 x 21,600
= sh.8, 640
HPP = D + MI
I = HPP - D/MI
= 21600 – 8640/1620 - = 8 Months
The correct answer is D (8)
- Salim deposited sh25000 in a bank which paid compound interest at the rate of 10% per annum. If he withdraws all his money after years, how much interest did his money gain?
- sh5250
- sh2500
- sh1375
- sh387
Working
Interest for year 1
I = PRT/100
= 25000 x 10 x 1/100
= Sh2500
Amount = 25000 + 2500
= 27,500
Interest for 2nd year
I = PRT/100
= 27,500 x 10 x ½/100
= Sh13775
Total interest (2,500 + 1,375)
= Sh3875
The correct answer is D (Sh 3875)
- Kamaru bought bananas in groups of 20 at sh20 per group. He grouped them into smaller groups of 5 bananas each and sold them at sh10 per group. What percentage profit did he make?
- 40%
- 50%
- 60 %
- 70%
Working
For every 20 bananas = sh 25
One group produces 4 smaller groups of 5 bananas each
S. P = 4x 10
= sh40
B.P price = sh25
Profit = 40 – 25
= sh15
% profit = P/BP x 100
= 60%
The correct answer is C (60).
- A shopkeeper bought 3 trays of eggs at sh 150 per tray. On the way to the shop, he realized 20% of the eggs were broken. He sold the rest at sh 72 per dozen. How much loss did he make?
- sh450
- sh432
- sh18
- sh28
Working
B.P for 3 trays = 3 x 150
= sh450
Number of eggs = 3 x 30
= 90 eggs
20% eggs broke = 20/100 x 90
= 18 eggs broken
Therefore remained = (90 - 18) eggs
= 72 eggs
1 dozen = 12 eggs
? = 72 eggs
= 6 dozens
1 dozen = sh.72
6 dozens = ?
Loss = B.P – S.P
= 450 - 432
sh18
The correct answer is C (sh18)
- A Salesperson earns a basic salary of sh7500 per month. He is also paid a 5% commission on all sales above sh30, 000. In a certain month his total earnings were sh.14250. What was his total sales for that month?
- sh135000
- sh285000
- sh165000
- sh315000
Working
Commission = sh14250 – sh7500
= sh6750
5% = sh6750
100% = ?
= 100/5 x 6750
= Sh. 135,000
Total sales = (135,000 + 30,000)
= sh165000
The correct answer is C (sh 165,000)
- Shiku bought the following items from a shop
6kg of sugar @ sh45
½ of tea for sh90
3 kg of rice @ sh30
2kg of fat @ sh70
If she used one thousand shillings to pay for the items, what balance did she receive- sh410
- sh455
- sh590
- sh765
Working
Shiku’s Bill
Item Sh ct 6kg sugar @ sh45 270 00 ½ kg tea for sh90 90 00 3kg rice @ sh30 90
00 2kg fat @ sh70 140 00 Total 590 00
Balance = sh1000 – sh590
The correct answer is = sh410 (A)
- Maranga paid sh4, 400 for a bicycle after he was given a 12% discount. James bought the same item from a different shop and was given a 15%. How much more than James did Maranga pay for the bicycle?
- sh250
- sh300
- sh750
- sh150
Working
Maranga B.P = 100% - 12%
= 88%
4400/88 x 100 = sh5000
James B.P = 100% - 15%
= 85%
85 x 100/4400 = sh4,250
How much more? = (5000-4250) shillings
= sh750
The correct answer is C (750)
- The table below shows postal charges for sending letters;
Mass of letter Sh ct Up to 20g 25 00 Over 20g up to 50g 30 00 Over 50g up to 100g 35 00 Over 100g up to 250g 50 00 Over 250g up to 500g 85 00 Over 500g up to 1kg 135 00 Over 1kg up to 2kg 190 00
Namu posted two letters each weighing 95g and another one weighing 450g. How much did he pay at the post office?- sh120
- sh135
- sh155
- sh240
Working
Two letters
95g → Sh35.00
95g → Sh35 .00
Another 450g → Sh85.00
The correct answer is C (sh155)
Published in
Mathematics Class 8 Notes
Wednesday, 15 September 2021 06:44
Volume, Capacity and Mass - Class 8 Mathematics Revision Notes
Volume, Capacity and Mass
Worked Exercises
- A Jerry can contains 5 litres of juice. This juice is used to fill 3 containers each of radius 7 cm and height of 10cm. How many milliliters of juice are left in the jerry can?
- 38
- 480
- 400
- 420
Working
Volume of container: = Πr2 h
=22/7 x 7 x 7 x 10
= 1540 cm3
Volume of 3 such containers
= (1540x3) cm3
= 4620 cm3
Volume of juice in jerry can = (5 x 1000)
= 5000cm3
Volume of juice left = (5000-4620) cm3
= 380 cm3
= 380 ml
The correct answer is A (380ml)
- The diagram below represents a solid whose dimensions are shown.
What is the volume in cm3?- .30000
- 300000
- 3000
- 3000000
Working
Volume = Area of the Cross-section x length
Volume of the top = (20 x 10 x 150)
= 30,000cm3
Volume of the bottom = 60 x 30 x150
= 270,000cm3
Whole solid = top + bottom
= 30,000 + 270,000
= 300,000cm3
The correct answer is B (300 000)
- In the month of October, a farmer delivered 48750kg of maize to a miller. In November the amount of maize delivered was 1850kg more than that of October. The amount delivered in December was 2450kg less than that of November. What was the total mass, in tonnes, was delivered by the farmer in the 3 months?
- 145.65
- 147.5
- 152.4
- 150.55
Working
October = 48750 kg
November = (48750+1850) kg
= 50,600 kg
December = 50,600-2,450) kg
= 48,150 kg
Total mass = 48750+50600 +48150
= (147500/1000) tonnes
= 147.5 tonnes.
The correct answer is B (147.5)
- A rectangular tank measures 1.2m by 80cm by 50cm. water is poured into the tank to a height of 15cm. How many more liters of water are needed to fill the tank?
- 144
- 14.4
- 33.6
- 336
Working
Capacity of the tank = 120 x 80 x 50
= 480,000cm3
Convert to litres = 480,000/1000
= 480litres
Volume of the water poured = 120 x 80 x 50
= 144000cm3
Convert to litres = 144000/1000
= 144 litres
Volume of water needed = 480 – 144 = 366litres.
The correct answer is D (366)
- The diagram below represents a solid triangular prism.
What is the volume in cm3?- 2400
- 2000
- 5200
- 576
Working
Apply Pythagorean relation in triangle ABC
BC =√262 -102
=√576
= 24cm
Volume = Area of the Cross section x length
= ½ x 24 x 10x 20
= 2400cm3
The correct answer is A (2400cm3)
- A cylindrical tank has a radius of 2m and a height of 1.5m. The tank was filled with water to a depth of 0.5M. What is the volume of water in the tank, in litres? (П = 3.14)
- 6280
- 628
- 9240
- 18840
Working
Volume = П r 2h
= 3.14 x 2 x 2 x 0.5
= 6.28 m3
In litres = (6.28 x1000) litres
= 6280 litres
The correct answer A (6280)
- When processed, 7kg of coffee beans produce 1kg of processed coffee. Processed coffee is then packed in 50kg bags. A farmer delivered 5.6 tonnes of coffee berries in one month. How many bags were obtained?
- 12
- 16
- 40
- 20
Working
Mass of coffee berries = 5.6tonnes
= 5.6x1000
= 5600kg
Mass obtained = 5600/7
= 800kg
Number of bags = 800 ÷ 50
= 16 bags
The correct answer is B (16)
- A rectangular container whose base measures 40cm by 60cm has 30 liters of water when full. Find the height of the container in cm.
- 0125
- 1.25
- 12.5
- 125
Working
V = base area x height
Height = volume/base area
Volume = 30 litres
= 30x1000
= 30,000cm3
Height = 30,000/2400
= 12.5cm
The correct answer is C (12.5)
- A shopkeeper had 43 litres sand 5 litres and 5 dl of paraffin. He packed all the paraffin in 7.5 dl-containers. How many containers did he fill?
- 58
- 5.8
- 6
- 60
Working
Convert decilitres into litres
1 dl =1/10 litres
5 dl =5/10 litres
7.5 dl =7.5/10 litres = 0.75 litres
Hence 43 litres 5dl = 43.5 litres
No of containers = 43.5/0.75 = 58 containers
The correct answer is 58 (A)
- The figure below shows a cylindrical solid of diameter 28cm and length 20 cm. A square hole of side 1.5 cm has been removed. What is the volume of the material in the solid, in 3cm3?
- 12320
- 4500
- 8400
- 7820
Working
Volume of solid = volume of a cylinder - volume of the square hole
= ( x 14 x 14x 20) - (15 x 15 x 20)
= 12320 - 4500
= 7,820 cm3
The correct answer is D (7,820cm3)
Published in
Mathematics Class 8 Notes
Tuesday, 14 September 2021 13:10
Numbers - Class 8 Mathematics Revision Notes
In this section you will need the following hints to solve the exercises:
- Place value of whole numbers
- Total value of whole numbers
- Multiplication of whole numbers/tables
- BODMAS
- LCM and GCD
Worked Exercise
- What is four million seventy thousand and five hundred and thirty three?
- 4,070,353
- 4,070,533
- 4,007,533
- 4,700,533
Working
Using the place value table, the question can be solved as follows:
MillionsHundred
ThousandsTen
thousands
Thousands
Hundreds
tens
Ones4 0 7 0 5 3 3
- What is the square root of 7 9/16
- 7 ¾
- 2 ¾
- 1 3/8
- 21/16
Working
Step 1: Change the mixed fraction to improper Find the square root of both numerator and denominator.
Step 2: Find the square root of both numerator and denominator
= √121
√16
=11/4
Step 3: Change the improper fraction to mixed fraction
= 2¾
The correct answer is B
- What is 25% as a fraction?
- 1/5
- ¾
- ½
- ¼
Working
Step1: Express the percentage with 100 as a denominator.
=25/100
Step 2: Simplify
=¼ correct answer is D
- What is the value of of 1/3 of(½ + 1/9) ÷1/6
- 11/324
- 1/99
- 12/9
- 4/11
Working
Step1: Using the order of operation, BODMAS, solve the brackets first.
1/2 + 1/9 = 11/18
Step 2: Open brackets and calculate ‘of ‘
=1/3 of (11/18) ÷ 1/6
=1/3 x (11/18) ÷ 1/6
=11/54 ÷ 1/6
Step3: Calculate the division part
=11/54 ÷ 1/6
=11/54 x 6/1(multiply by the reciprocal of 1/6)
=11/9
Step 4: Change the improper fraction to mixed fraction.
= 1 2/9
The correct answer is C.
- The price of radio is Sh1800. The price was reduced by 15% during an auction. How much is the price after the reduction?
- Sh270
- Sh2070
- sh1530
- sh1785
Working
Marked price = Sh1800
Percentage decrease = 15%
New price
85% of Sh1800 (100% - 15%)
= 85 x 1800/100
= Sh1, 530
The correct answer is Sh 1530 (C)
- In a certain year a tea factory produced 2500 tonnes of tea leaves. The following year the tonnes increased to 4000. What is the percentage increase?
- 160%
- 62½ %
- 60%
- 37½ %
Working
First year = 2500 tonnes
Second year = 4000 tonnes
Increase = 1500 tonnes (4000-2500)
% Increase = Increase x 100/Original
= 60%
The correct answer is C (60%)
- What is the next number in the sequence below.
6, 10, 19, 35, ….. - 60
- 84
- 71
- 51
Working
The next difference is 5² = 25
The next number is 35 + 25 = 60
The correct answer is A (60)
Published in
Mathematics Class 8 Notes
Tuesday, 14 September 2021 12:15
Properties of Geometric Shapes - Class 8 Mathematics Revision Notes
General Geometric Shapes
Square
- All sides are equal
- Opposite sides are parallel
- Each interior angle is a right angle (90º)
- The interior angles total up to 360º
- Diagonals bisect each other at right angles.
- Diagonals measure the same length and bisect interior angles.
Rectangle
- Each interior angle is 90º and they all add up to 360º
- Diagonals are equal
- Diagonals bisect each other but NOT at right angles
Parallelogram
- Opposite sides are equal and parallel
- Opposite angles are equal
- Diagonals bisect each other
- Diagonals are not equal
- Adjacent angles are supplementary (add up to 180º)
Rhombus
- All sides are equal
- Opposite sides are parallel
- Opposite angles are equal
- Diagonals bisect each other at 90º
- Diagonals bisect the interior angles
Trapezium
- The sum of the interior angles is 360º
- Has a pair of parallel lines which are not of the same length
- Has a perpendicular height joining the two parallel lines
Right-angled Triangle (Pythagorean relationship)
- H2 = b2 + h2
- b2 = H2 – h2
- H2 = H2 - b2
Examples of relationships
Base Height Hypotenuse 3 4 5 6 8 10 5 12 13 7 24 25 8 15 17 9 40 41
Properties of Triangles and Parallel Lines
Triangle
Exterior angles & interior angles
- Angles x, y, and z are exterior angles while a, b, and c are interior angles.
- Exterior angles add up to 360º while interior angles add up to 180º.
- Angles x, a; b, z; and c, y; are adjacent to each other and they add up to 180º (supplementary angles)
Parallel Lines and Transversal
- Angles at a point e.g. a + b+ c + d = 360º
- Vertically opposite e.g. a/d, b/c, f/g, e/h. They are equal
- Corresponding angles e.g. b/f, a/e, c/g, d/h. They are equal
- Alternate angles e.g. c/f, d/e are always equal.
- Co-interior angles e.g. c/e, d/f, are always equal.
- Co-interior/allied angles e.g. c/e, d/f are formed by parallel lines. They are supplementary.
Speed, Distance and Time
The formulae related to speed, distance and time can be derived from the following triangle.
Published in
Mathematics Class 8 Notes
Tagged under