SIMPLE PROBLEMS ON PERCENTAGE
SECOND TERM E NOTES
SUBJECT: MATHEMATICS
CLASS: BASIC FIVE
WEEK TOPIC
 RATIO AND PERCENTAGE
 SIMPLE PROBLEMS ON PERCENTAGE
 MONEY: profit and loss
 MONEY: SIMPLE INTEREST
 MONEY: DISCOUNT AND COMMISION
 PERIMETER OF PLANE SHAPES
 AREA OF RIGHT ANGLED TRIANGLE
 VOLUME
 PROBLEM ON WEIGHT
 AVERAGE SPEED
 WEEK ONE
RATIO AND PERCENTAGE
Meaning of ratio
The relation between two quantities (both of the same units) obtained by dividing one quantity with another is called ratio. Ratio can be denoted by using the symbol ( : ).
EXAMPLE 1
What is the ratio between the weight two bags of sugar of 4kg and 6kg respectively?
Solution
Ratio of weights of bags of sugar = 4kg/ 6kg = 2/3 = 2:3
EXAMPLE 2
A pole of length 165cm is divided into two parts such that lengths are in the ration 7:8. Find the length of each part of the pole?
Total ratio = 7 + 8 = 15
First part = ^{7/}_{15 }, second part = 8/15
Length of first part = 7/15 of 165cm
= 7/15 × 165
= 7 x 11 = 77cm
1 x 1
Length of second part = 165cm – 77cm = 88cm
DIRECT PROPORTION
Examples
 A marker costs $18. Calculate the cost of 5 markers.
 A student went to the market to purchase textbooks. He purchased two textbooks for $24
 What is the price of one notebook?
 What would be the price of 5 such note
Solution
 If one marker = $18
5 markers = $18 x 5 = $90
 2 textbooks = $24
1 textbook = m
2 × m = 1 x $24
2m = $24
M = $24 ÷ 2 = $12. Therefore 1 textbook = $24
(b) Since 1 textbook = $12
5 textbooks = $12 × 5 = $60
Equal and equivalent ratios
1. The two ratios below are equal or equivalent.
1/2 = 4/ 8
3/5 = 6/10
2. The two ratio below are not equal or equivalent
4/3≠6/10 3/5≠7/15
EVALUATION
 What is ratio?
 Find the ratio of each of the following in its lowest terms:
 24cm: 72cm
 425km: 750km
 75min: 150min
 85kg: 102kg
 A field is 50m in length and 60m in width. Find the ratio between its width and length.
 A scooter can travel 225km with 5 litres of petrol. How many litres of petrol is needed to travel 675km?
WEEK TWO
SIMPLE PROBLEMS ON PERCENTAGES
Percentages are fractions with 100 as denominator
EXAMPLE 1
Change the following fractions to percentages
 ^{2/}_{5 } = ^{2}/_{5 }of 100 = ^{2/}_{5 }× 100 = 200 = 40%
5
EXAMPLE 2
Change these percentage to fractions
 75% = 75 = 15 = 3
100 20 4
EXAMPLE 3
Change 7 ½% to fractions in their lowest terms
Solution
7 ½ % = 7 ½ out of 100
= 15/2 × 1/100
= (15 × 1) ÷ 200
= 15 ÷ 200 = 3/40
EXERCISE
Change the following fraction to percentage
 ^{2}_{/5}
 ^{2/}_{4}
 ^{3}/_{30}
Change to fractions in their lowest terms
 66 ½ %
 12 ¼ %
 16 ^{2}/_{3} %
WEEK THREE
MONEY: PROFIT AND LOSS
MEANING OF PROFIT
When the selling price of an article is higher or greater than the cost price, we have a profit or gain.
Profit = selling price – cost price
MEANING OF LOSS
When the selling price is less than cost price we have a loss.
Loss = cost price – selling price
EXAMPLE 1
If a clock is bought for $1,145 and sold for $1,170, what is the gain or losss
Solution
Cost price of clock = $1,145
Selling price = $1, 170
Since the selling price is more than the cost price, we have a profit
Profit = $1,170 – $1,145 = $25
EXAMPLE 2
By selling a tin of oil for $1,320, a man made a profit of $150. How much did he pay for it?
Solution
Selling price = $1,320
Profit = $150
Cost price = $1,320 – $150 = $1,170.
EXERCISE
 I bought an exercise book for $2500 and sold it for $3,200. How much profit did I make
 An article is sold for $4,000 and the loss is $1,050. Fin the cost price of the article
 A wrist watch is bought for $3,887 and sold for a profit of $722. Find the selling price
 A trader bought a dozen candles for $1,020 and sold them all at $960. What is the profit or loss of each candle?
WEEK FOUR
SIMPLE INTERREST
Meaning of interest: interest means the extra money that you [ay back when ypu borrow money that you receive when you invest money.
EXAMPLE 1
find the simple interst on $1,000 for 5 years at 3% per annum.
SOLUTION
Principal = $1,000 ( the amount borrowed
Time = 5 years ( the period for which the money is borrowed before it is paid back in full)
Rate = 3% ( extra money paid)
Simple interest = P x R x T
100
= 1,000 x 3 x 5
100
= 10 x 3 x 5 = $150
EXAMPLE 2
Seun deposit $14,500 in a savings Bank account at UBA which pays an interest rate of 15% per annum for 2 ½ years. What is the simple interest?
Solution
Principal = $14,500
Rate= 15%
Time= 2 ½ or 2.5 years
SI = P x R x T
100
= 14500 x 15 x 2.5 = 145 x 15 x 2.5 = $5, 437.50
100
EXERCISE
Copy and complete the table
S/N  Principlal  Rate  Time  Simple interest 
$23,00  5%  4 years  

$19,000  6%  7 years  

$28,000  3%  2 years  
$30,000  4%  $6,000 
WEEK FIVE
PROFIT AND LOSS PERCENT
Profit or losss percent is expressed as a percentage of the cost price
EXAMPLE 1
An article bought for $3,ooo was sold for $3,300. Find the profit percent
Solution
Cost price = $3,000
Selling price = $3,300
Profit = $3300 – $3,000 = $300
Profit % = Profit x 100
Cost price 1
= 300 x 100
3000 1
= 300 x 1 = 300 = 10%
30 x 1 30
EXAMPLE 2
A book bought for $25,00 was sold for $22,000. What is theloss percent?
Solution
Cost price =$25,000
Selling price= $22,000
Loss = $25,000 – $22,000 = $3,000
Loss % =loss x 100
Cost price 1
=3000 x 100 = 12%
25000 1
EXERCISE
Cost price  Selling price  Profit %  Loss%  
1.  $1,300  $1,365  
$33,555  $32,446  
$56,000  $59,540  
$100,680  $99,960 
WEEK SIX
PERIMETER OF PLANE SHAPES
 Square L
L L perimeter = L + L + L + L = 4L
L
 Rectangle L
L L perimeter = L + L + B + B
= 2L + 2B = 2 ( L+B)
L
 Circle
Perimeter = 2 x π x r
= 2πr
Where π = 22/7 or 3.142
EXAMPLE
Calculate the perimeter 9cm
9cm 9cm
9cm
Perimeter = 4 x length
= 4 x 9cm
= 36cm
EXAMPLE 2
Calculate the Perimeter of the rectangle
15cm
9cm
Perimeter = 2 ( L+B)
= 2 (15cm + 9cm)
= 2(24cm) = 2x 24 = 28cm
EXAMPLE 3
Calculate the perimeter of the circle of radius 14cm. Take π = 22/7
Perimeter = 2 π r
= 2 x π x r
= 2 x 22 x 14
7
= 2 x 22 x 2 = 88cm
EXERCISE 1
CALCULATE THE PEREIMETER OF THE RECTANGLES WITJHH THE FOLLOWING DIMENSIONS
 Length = 9cm, breadth = 5cm
 Length= 12cm, breadth = 7cm
 Length = 14cm, breadth = 1ocm
 Length = 16cm, breadth =11cm
 Length = 11.5cm, breadth = 4.5cm
EXERCISE 2
Calculate the perimeter of the following circles. Take π = 22/7
 Radius = 7cm
 Radius = 14cm
 Radius = 3.5
 Radius = 21cm
 Radius =28cm
 Radius = 12cm
 Radius = 35cm
 Radius = 4.9cm
WEEK SEVEN
AREA OF RIGHT ANGLED TRIANGLE
A rightangled triangle has 3 – sides
A
Height
B C
Example 1
Calculate the area of a triangle of height 12cm and base 13cm
Solution
12cm
13cm
Area of triangle = ½ x base x height
= ½ x 12cm x 13cm
1 1
= 1 x 6cm x 13cm = 78cm^{2}
EXERCISE
Calculate the area of the triangles with the followinh dimensions
 Base = 14cm, height = 16cm
 Base = 10cm, height = 8cm
 Base = 25cm, height = 22cm
 Base = 30cm, height = 15cm
 Base = 24cm, height = 18cm
 Base = 14cm, height = 16cm
 Base = 34cm, height = 15cm
 Base = 52cm, height = 48cm
WEEK EIGHT
VOLUME OF CYLINDER
Radius
Height
Volume of cylinder = πr^{2}h
= π x r x r x h
EXAMPLE
Find the volume of a cylinder of radius 3 ½ cm and height 12cm
Solution
3 ½
12cm
Volume = π x r x r x h
=22 x 7 x 7 x 12cm
7 x 2 x 2 x 1
= 22 x 7 x 3
1 x 1 x 1
= 462cm^{3}
Exercise
(take π= 22/7) Find the volume of cylinders whose radii and heights are given as:
 Radius 3.5cm, height = 10cm
 Radius 10.5, height 10m
 Radius 4.2cm, height15cm
 Diameter = 1o½cm, height 16cm
 Diameter = 14cm, height 15cm
WEEK NINE & TEN
WEIGHT
The units used for measuring weight are kjilogram (kg) and gram (g).
1000 gram = 1kg
1000kg = 1 ton
EXAMPLE1
Change to kilogram
2000g to kilogram
1000g = 1kg
2000g = m
Cross multiply
1000 × m = 2000 x 1
1000m = 2000
m = 2000 = 2kg
1000
EXAMPLE 2
Change 5kg to grams
Solution
To change from kg to grams, multiply by 1000
5kg = 5 x 1000
= 5000 grams
EXERCISE
Change the following to kilograms:
 2000g
 5000g
 3500g
 1205g
 750g
Change to grams
 6.5 kg
 0.5kg
 4.298kg
 8.765kg
 13.020kg
 0.07kg