# How To Calculate the Percentage Difference or Increase – Problems , Solutions and Applications in Real Life

### How To Calculate the Percentage Difference or Increase – Problems , Solutions and Applications in Real Life

The percentage difference also known as the percentage increase or decrease is a common calculation in maths, statistics, physics, business, commerce, economics and many other fields. It is used to indicate the change in the values of a particular measurement over a certain period of time. The delta symbol Δx is used to represent a change in quantity or values.

### Applications of Percentage Differences in Real Life

The following are some of the measurements whereby the percentage increase or decrease is applied or calculated:

Physics:

Change in speed (acceleration)

Change in mass

Change in pressure

Change in energy

Change in temperature

Change in volume

Change in density

Change in force

Change in momentum

Change in monthly sales volumes

Change in monthly revenue

Change in monthly profit

Change in monthly costs

Economics and Commerce:

Change in annual GDP

Change in monthly, quarterly and annual export values

Change in monthly, quarterly and annual import values

Change in median income

Change in median rent

Change in the cost of living index

Change in the consumer price index

Population Studies:

Change in national population

Change in city population

Change in male population

Change in female population

Change in demographics

Construction:

Change in the construction cost index

Change in the material price index

Change in median house prices

Change in number of monthly, quarterly and annual building permits

Change in construction output volumes

Change in construction tender price index

Change in construction input price index

Change in construction output price index

As you can see, the percentage difference is calculated and used in many fields whereby statistics and measurements are recorded.

### How To Calculate the Percentage Increase

In order to calculate a percentage increase, you must have two values measured over a time range, for example, let’s say you are measuring daily temperature. Recordings were taken in the morning, afternoon and night showing 25C°, 28C° and 23C° respectively. Calculate the percentage change in temperature between the morning and afternoon readings?

Solution:

Set up a logical equation. Let x1 be the first reading and x2 be the second reading. Let x3 be the difference between the readings. The delta symbol Δ  is used to represent a change in quantity. Therefore Δx = x3. The change in values in our example will be represented by Δx.

The percentage change formula is Δx =  (x2 – x1/x1) * 100

percentage change formula

Therefore,

>> x2 – x1 = x3

>> Δx = x3

>> x3 = (x2 – x1/x1) * 100

Making substitutions:   x1 = 25C°, x2 = 28C°

>> x3 = (28 – 25)/25)*100

>> x3 = (3/25)*100

>> x3 = 12%

Answer: The percentage change is Δx = 12%

**Note that in this example, the percentage change is positive, which is a percentage increase.

#### Problem 2:

Calculate the percentage change in temperature between the afternoon and night readings?

Solution:

Using the percentage formula Δx = (x2 – x1/x1) * 100

Making substitutions:

>> x1 = 28C°, and x2 = 23C°

>> Δx = (23 – 28/28) * 100

>> Δx = (-5/28) * 100

>> Δx = -17.86%

Answer: The percentage change is Δx = -17.86%

**Note that in this example, the percentage change is negative, which is a percentage decrease.

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### Problem 1

A car is travelling at 40km/hr, it then accelerates to 70km/hr. What is the percentage increase in speed?

Solution:

Set up a logical equation. Let x1 be the first reading and x2 be the second reading. Let x3 be the difference between the readings. The delta symbol Δ  is used to represent a change in quantity. Therefore Δx = x3. The change in values in our example will be represented by Δx.

The percentage change formula is Δx =  (x2 – x1/x1) * 100

Therefore,

>> x2 – x1 = x3

>> Δx = x3

>> x3 = (x2 – x1/x1) * 100

Making substitutions:   x1 = 40km/hr, x2 = 70km/hr

>> x3 = (70 – 40)/40)*100

>> x3 = (30/40)*100

>> x3 = 75

Answer: The percentage change in speed is Δx = 75%

**Note that in this example, the percentage change is positive, which is a percentage increase.

### Problem 2

A car is travelling at 120km/hr, it then decelerates to 85km/hr. By what percent did the speed decrease?

Solution:

Set up a logical equation. Let x1 be the first reading and x2 be the second reading. Let x3 be the difference between the readings. The delta symbol Δ  is used to represent a change in quantity. Therefore Δx = x3. The change in values in our example will be represented by Δx.

The percentage change formula is Δx =  (x2 – x1/x1) * 100

Therefore,

>> x2 – x1 = x3

>> Δx = x3

>> x3 = (x2 – x1/x1) * 100

Making substitutions:   x1 = 120km/hr, x2 = 85km/hr

>> x3 = (85 – 120)/120)*100

>> x3 = (-35/120)*100

>> x3 = -29.17%

Answer: The percentage change in speed is Δx = -29.17%

**Note that in this example, the percentage change is negative, which is a percentage decrease.

### Problem 3

Wood is a material which absorbs and loses moisture depending on the temperature and relative humidity. A piece of wood weighs 750g, the following morning its swollen weight was 800g. What is the percentage increase in its weight and how much moisture did it gain?

Solution:

Set up a logical equation. Let x1 be the first reading and x2 be the second reading. Let x3 be the difference between the readings. The delta symbol Δ  is used to represent a change in quantity. Therefore Δx = x3. The change in values in our example will be represented by Δx.

The percentage change formula is Δx =  (x2 – x1/x1) * 100

Therefore,

>> x2 – x1 = x3

>> Δx = x3

>> x3 = (x2 – x1/x1) * 100

Making substitutions:   x1 = 750g, x2 = 800g

>> x3 = (800 – 750)/750)*100

>> x3 = (100/750)*100

>> x3 = 13

Answer: The percentage change in weight is Δx = 13%

**Note that in this example, the percentage change is positive, which is a percentage increase.

The wood gained (800 – 750) = 50g of moisture.

### Problem 4

Wood is a material which absorbs and loses moisture depending on the temperature and relative humidity. A piece of wood weighs 1000g, the following morning its shrunken weight was 940g. By what percent did its weight decrease and how much moisture did it lose?

Solution:

Set up a logical equation. Let x1 be the first reading and x2 be the second reading. Let x3 be the difference between the readings. The delta symbol Δ  is used to represent a change in quantity. Therefore Δx = x3. The change in values in our example will be represented by Δx.

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The percentage change formula is Δx =  (x2 – x1/x1) * 100

Therefore,

>> x2 – x1 = x3

>> Δx = x3

>> x3 = (x2 – x1/x1) * 100

Making substitutions:   x1 = 1000g, x2 = 940g

>> x3 = (940 – 1000)/1000)*100

>> x3 = (-60/1000)*100

>> x3 = -6%

Answer: The percentage change in weight is Δx = -6%

**Note that in this example, the percentage change is negative, which is a percentage decrease.

The wood lost (1000 – 940) = 60g of moisture.

### Problem 5

A jewellery store sold 60 necklaces in November and 105 necklaces in December. Calculate the percentage increase in sales volume?

Solution:

Set up a logical equation. Let x1 be the first reading and x2 be the second reading. Let x3 be the difference between the readings. The delta symbol Δ  is used to represent a change in quantity. Therefore Δx = x3. The change in values in our example will be represented by Δx.

The percentage change formula is Δx =  (x2 – x1/x1) * 100

Therefore,

>> x2 – x1 = x3

>> Δx = x3

>> x3 = (x2 – x1/x1) * 100

Making substitutions:   x1 = 60 necklaces, x2 = 105 necklaces

>> x3 = (105 – 60)/60)*100

>> x3 = (45/60)*100

>> x3 = 75%

Answer: The percentage change in sales volume is Δx = 75%

**Note that in this example, the percentage change is positive, which is a percentage increase.

### Problem 6

A jewellery store sold 105 necklaces in December and 33 necklaces in January of the New Year. Calculate the percentage decline in sales volume?

Solution:

Set up a logical equation. Let x1 be the first reading and x2 be the second reading. Let x3 be the difference between the readings. The delta symbol Δ  is used to represent a change in quantity. Therefore Δx = x3. The change in values in our example will be represented by Δx.

The percentage change formula is Δx =  (x2 – x1/x1) * 100

Therefore,

>> x2 – x1 = x3

>> Δx = x3

>> x3 = (x2 – x1/x1) * 100

Making substitutions:   x1 = 105 necklaces, x2 = 33 necklaces

>> x3 = (33 – 105)/105)*100

>> x3 = (-72/105)*100

>> x3 = -68.57%

Answer: The percentage change in sales volume is Δx = -68.57%

**Note that in this example, the percentage change is negative, which is a percentage decrease.

### Problem 7

A computer shop sold 112 USB bracelets in May and 140 USB bracelets in June. If each bracelet was priced at \$6.13 in May and a 10% promotional discount was offered in June, calculate the percentage change in the sales revenue?

Solution:

Step 1

Calculate the monthly sales revenue:

May >> 112 x \$6.13 = \$686.56

June >> 140 x \$6.13 = \$858.20

Calculate June Discount >> 10% of \$858.20 = (10/100)*\$858.20 = \$85.82

Therefore, the monthly sales in June were >> \$858.20 – \$85.82 = \$772.38

Step 2

Set up a logical equation. Let x1 be the first reading and x2 be the second reading. Let x3 be the difference between the readings. The delta symbol Δ  is used to represent a change in quantity. Therefore Δx = x3. The change in values in our example will be represented by Δx.

The percentage change formula is Δx =  (x2 – x1/x1) * 100

Therefore,

>> x2 – x1 = x3

>> Δx = x3

>> x3 = (x2 – x1/x1) * 100

Making substitutions:   x1 = \$686.56, x2 = \$772.38

>> x3 = (772.38 – 686.56)/686.56)*100

>> x3 = (85.82/686.56)*100

>> x3 = 12.5%

Answer: The percentage change in sales revenue is Δx = 12.5%

**Note that in this example, the percentage change is positive, which is a percentage increase.

### Problem 8

An ISP Telecom company increased the cost of its prepaid unlimited monthly data plan from \$65 to \$75 a month. By what percentage was the price increased?

Solution:

Set up a logical equation. Let x1 be the first reading and x2 be the second reading. Let x3 be the difference between the readings. The delta symbol Δ  is used to represent a change in quantity. Therefore Δx = x3. The change in values in our example will be represented by Δx.

The percentage change formula is Δx =  (x2 – x1/x1) * 100

Therefore,

>> x2 – x1 = x3

>> Δx = x3

>> x3 = (x2 – x1/x1) * 100

Making substitutions:   x1 = \$65, x2 = \$75

>> x3 = (75 – 65)/65)*100

>> x3 = (10/65)*100

>> x3 = 15.38%

Answer: The percentage change in the price of monthly data is Δx = 15.38%

**Note that in this example, the percentage change is positive, which is a percentage increase.

### Problem 9

The population of New York City declined from 8,469,153 in 2016 to 8,336,817 people in 2018. Calculate the percentage decline in population over this period?

Solution:

Set up a logical equation. Let x1 be the first reading and x2 be the second reading. Let x3 be the difference between the readings. The delta symbol Δ  is used to represent a change in quantity. Therefore Δx = x3. The change in values in our example will be represented by Δx.

The percentage change formula is Δx =  (x2 – x1/x1) * 100

Therefore,

>> x2 – x1 = x3

>> Δx = x3

>> x3 = (x2 – x1/x1) * 100

Making substitutions:   x1 = 8,469,153 people, x2 = 8,336,817 people

>> x3 = (8,336,817 – 8,469,153)/8,469,153)*100

>> x3 = (-132,336/8,469,153)*100

>> x3 = (-0.0156)*100

>> x3 = -1.56%

Answer: The percentage change in the population was Δx = -1.56%

**Note that in this example, the percentage change is negative, which is a percentage decrease.

### Problem 10

The Construction Tender Price Index (TPI) in the city of Dublin was 160.5 in the second quarter of 2020 and 152.7 in the second quarter of 2019. Calculate the percentage increase in the tender price index over this period?

Solution:

Set up a logical equation. Let x1 be the first reading and x2 be the second reading. Let x3 be the difference between the readings. The delta symbol Δ  is used to represent a change in quantity. Therefore Δx = x3. The change in values in our example will be represented by Δx.

The percentage change formula is Δx =  (x2 – x1/x1) * 100

Therefore,

>> x2 – x1 = x3

>> Δx = x3

>> x3 = (x2 – x1/x1) * 100

Making substitutions:   x1 = 152.7, x2 = 160.5

>> x3 = (160.5 – 152.7)/152.7)*100

>> x3 = (7.8/152.7)*100

>> x3 = 5.1%

Answer: The percentage change in the tender price index was Δx = 5.1%

**Note that in this example, the percentage change is positive, which is a percentage increase.