# SIMPLE PROBLEMS ON PERCENTAGE

SECOND TERM E NOTES

SUBJECT: MATHEMATICS

CLASS: BASIC FIVE

WEEK TOPIC

1. RATIO AND PERCENTAGE
2. SIMPLE PROBLEMS ON PERCENTAGE
3. MONEY: profit and loss
4. MONEY: SIMPLE INTEREST
5. MONEY: DISCOUNT AND COMMISION
6. PERIMETER OF PLANE SHAPES
7. AREA OF RIGHT ANGLED TRIANGLE
8. VOLUME
9. PROBLEM ON WEIGHT
10. AVERAGE SPEED
11. 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

1. A marker costs \$18. Calculate the cost of 5 markers.
2. A student went to the market to purchase textbooks. He purchased two textbooks for \$24
3. What is the price of one notebook?
4. What would be the price of 5 such note

Solution

1. If one marker = \$18

5 markers = \$18 x 5 = \$90

1. 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

1. What is ratio?
2. Find the ratio of each of the following in its lowest terms:
3. 24cm: 72cm
4. 425km: 750km
5. 75min: 150min
6. 85kg: 102kg
7. A field is 50m in length and 60m in width. Find the ratio between its width and length.
8. 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

1. 2/5 = 2/5 of 100 = 2/5 × 100 = 200 = 40%

5

EXAMPLE 2

Change these percentage to fractions

1. 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

1. 2/5
2. 2/4
3. 3/30

Change to fractions in their lowest terms

1. 66 ½ %
2. 12 ¼ %
3. 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

1. I bought an exercise book for \$2500 and sold it for \$3,200. How much profit did I make
2. An article is sold for \$4,000 and the loss is \$1,050. Fin the cost price of the article
3. A wrist watch is bought for \$3,887 and sold for a profit of \$722. Find the selling price
4. 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

1. Square L

L L perimeter = L + L + L + L = 4L

L

1. Rectangle L

L L perimeter = L + L + B + B

= 2L + 2B = 2 ( L+B)

L

1. 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

1. Length = 9cm, breadth = 5cm
2. Length= 12cm, breadth = 7cm
3. Length = 14cm, breadth = 1ocm
4. Length = 16cm, breadth =11cm
5. Length = 11.5cm, breadth = 4.5cm

EXERCISE 2

Calculate the perimeter of the following circles. Take π = 22/7

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

1. Base = 14cm, height = 16cm
2. Base = 10cm, height = 8cm
3. Base = 25cm, height = 22cm
4. Base = 30cm, height = 15cm
5. Base = 24cm, height = 18cm
6. Base = 14cm, height = 16cm
7. Base = 34cm, height = 15cm
8. Base = 52cm, height = 48cm

WEEK EIGHT

VOLUME OF CYLINDER

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:

1. Radius 3.5cm, height = 10cm
4. Diameter = 1o½cm, height 16cm
5. 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:

1. 2000g
2. 5000g
3. 3500g
4. 1205g
5. 750g

Change to grams

1. 6.5 kg
2. 0.5kg
3. 4.298kg
4. 8.765kg
5. 13.020kg
6. 0.07kg