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 right-angled 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 = 78cm2
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 = πr2h
= 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
= 462cm3
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