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Successive Percentage Change

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  • Last Updated : 12 May, 2022
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In mathematics, a percentage is a number or ratio expressed as a fraction of 100 that can be represented either in decimal form or fractional form. The percentage has no unit of measurement and is a dimensionless number since it is the ratio of two identical quantities. The word percentage originated from the Latin word “per centum”, which means “by a hundred” and is represented by the notation “%”. Percentages can be used to compare quantities. A percentage is computed by dividing the actual value by the total value, and the resultant is multiplied by 100. 

Percentage (%) = (Actual Value/total value) × 100%

Percentage Change

Percentage change means the change in the value of a quantity over a period of time. The percentage change of a quantity is the ratio of the change in value to its initial value multiplied by 100. The percentage change is computed by using the formula given below:

Percentage change = [(change in value)/original value] × 100%

Where Change in value = Final value – original value

Percentage Increase

A percentage increase refers to the increment in the value of a quantity over a period of time, i.e., the present value of the quantity is greater than its original value. To calculate the percentage increase, first, we have to find the difference between the original value and the increased value. Then the ratio between the obtained difference and the original value is multiplied by 100. The percentage increase is computed by using the formula below.

Percentage Increase = [(Final value – Original value)/Original value] × 100%

Percentage Decrease

A percentage decrease refers to the decrement in the value of a quantity over a period of time, i.e., the present value of the quantity is less than its original value. To calculate the percentage decrease, we have to first find the difference between the original value and the decreased value. Then the ratio between the obtained difference and the original value is multiplied by 100. The percentage decrease is computed by using the formula below.

Percentage decrease = [(Original value – Final value)/Original value] × 100%

Successive Percentage Change

When two or more percentage changes are applied to a quantity consecutively, the percentage change is called a “successive percentage change.” Here, the final change is not the simple addition of two or more percentages. In a successive percentage change, a quantity is changed by some percentage, and the obtained new quantity is changed by another percentage, i.e., both the percentages are not applied to the same actual value.

Successive Increment Percentage Change

When two or more increased percentage changes are applied to a quantity consecutively, the percentage change is called a “successive increment percentage change.” For example, if the population of the town increased by a% and then by b%, we now have to apply the first percentage, i.e., a%, to the initial value. Then we have to apply the second percentage, i.e., b%, to the resultant value obtained from the first percentage change.

Let the population of the town be z.

Now, the first percentage increase = z + (z × a/100) = z (1 + a/100) = X

Second percentage increase = X + (X × b/100) = {z (1 + a/100) + [z × (1 + a/100) × (b/100)]}

= z (1 + a/100) (1 + b/100) = Y

Therefore, the net percentage change in the population of the town after two successive increments = {(Y – z)/z} × 100

If the value of an object x is successively increased by a%, b%, and then by c% then the final value is x (1 + a/100) (1 + b/100) (1 + c/100).

Successive Decrement Percentage Change

When two or more decreased percentage changes are applied to a quantity consecutively, the percentage change is called a “successive decrement percentage change.” For example, if the price of a product is decreased by a% and then by b%, we now have to apply the first percentage, i.e., a%, to the initial value of the product. Then we have to apply the second percentage, i.e., b%, to the resultant value obtained from the first percentage change.

Let the price of the product be z.

Now, the first percentage decrease = z – (z × a/100) = z (1 – a/100) = X

Second percentage decrease = X – (X × b/100) = {z (1 – a/100) – [z × (1 – a/100) × (b/100)]}

= z (1- a/100) (1 – b/100) = Y

Therefore, the net percentage change in the price of the product after two successive decrements = {(Y – z)/z} × 100

If the value of an object x is successively decreased by a%, b%, and then by c% the final value is x (1 – a/100) (1 – b/100) (1 – c/100).

Other Successive Percentage Change

Both percentage increment and percentage decrement can also be applied to the initial value of an object successively, and the percentage changes can also be used multiple times.

If the value of an object x is successively changed by a%, b%, and then by c%, the final value is x (1 ± a/100) (1 ± b/100) (1 ± c/100), where the positive sign indicates an increment while the negative sign indicates a decrement.

Sample Problems

Problem 1: A number is first decreased by 15%, and then it is further decreased by 20%. The original number has been altogether decreased by?

Solution:

Let the number be x,

Now the number is decreased by 15%

= x – (x × 15/100) = 85x/100

Now the number is further decreased by 20%

= 85x/100 – (85x/100 × 20/100)

= 68x/100

The final value of number = 68x/100

Now calculate the percentage change

Percentage change = {(Final value – Initial value)/Initial value} × 100%

 = {(68x/100 – x)/x} × 100%

= (-32x/100) × 100%

= – 32%

Here, the negative sign indicates decrement. Hence, the net decrease is 32%

Problem 2: The population of a town increased by 5% in 2020. It will again be increased by 10% in 2021. Find the net change in the population increment of the town.

Solution:

Let the population of the town be x,

Given that the population of a town is increased by 5% and it is increased by 10%.

We know that if the value of an object x is changed by a% and then by b% then the final value = x (1 ± a/100) (1 ± b/100),

Where the positive sign indicates an increment while the negative sign indicates the decrement.

Hence, the final population of the town = x (1 + 5/100) × (1 + 10/100) = (x) × (105/100) × (110/100) = 1.155x

Now, the percentage change = {(Final value – Initial value)/Initial value} × 100

= {(1.155x – x)/x} 100

= 15.5%

Hence, the net increment in the population of the town is 15.5%

Problem 3: What must be the total percentage change in the volume of a cuboid if its length and breadth are decreased by 20% and 30%, respectively, while its height is increased by 40%?

Solution: 

Let the volume of the cuboid be x,

Volume of the cuboid = l × b × h

Given that the length and breadth of a cuboid are decreased by 20% and 30% respectively, and height is increased by 40%.

We know that if the value of an object x is changed by a%, b%, and then c% successively, 

Then the final value = x (1 ± a/100) (1 ± b/100) (1 ± c/100), where the positive sign indicates an increment while the negative sign indicates the decrement.

Hence, the final volume of the cuboid = x (1 – 20/100) × (1 – 30/100) × (1 + 40/100)

= (x) × (80/100) × (70/100) × (140/100) = 0.784x

Now, the percentage change = {(Final value – Initial value)/Initial value} × 100

= {(0.784x – x)/x} × 100 = – 2.16% (negative sign indicates the decrement)

Hence, the volume of cuboid is decreased by 2.16%

Problem 4: Ram invested a certain amount in shares. The shares rose by 25% one day and fell by 15% the next day. What is the percentage profit or loss made by Ram?

Solution:

Let the amount invested by Ram in shares be x,

Given that the shares rose by 25% on the first day and then fell by 15% the next day.

We know that if the value of an object x is changed by a% and then by b% then the final value w= x (1 ± a/100) (1 ± b/100),

Where the positive sign indicates an increment while the negative sign indicates the decrement.

Hence, the final value of shares = x(1 + 25/100) (1 – 15/100) = (x) × (125/100) × (85/100) = 1.0625x

Now, the percentage change = {(Final value – Initial value)/Initial value} × 100

= {(1.0625x – x)/x} × 100

= 6.25%

Hence, Ram had a profit of 6.25%

Problem 5: What must be the net percentage change in the area of a rectangle if its length is increased by 10% while its breadth is decreased by 7%?

Solution: 

Let the area of the rectangle be x,

We know that, the area of rectangle = l × b

Given that, the length is increased by 10% and the breadth is decreased by 7%.

We know that if the value of an object x is changed by a% and then by b% then the final value w= x (1 ± a/100) (1 ± b/100),

Where the positive sign indicates an increment while the negative sign indicates the decrement.

Hence, final value of area = x (1 + 10/100) (1 – 7/100) = x × (11/10) × (93/100) = 1.023x

Now, the percentage change = {(Final value – Initial value)/Initial value} × 100

= {(1.023x – x)/x) × 100 = 2.3%

Hence, the area of the rectangle is increased by 2.3%

Problem 6: The shopkeeper gave 20% and 30% off on bags for an anniversary sale. What is the final discount given by the shopkeeper?

Solution:

Let the price of a bag be x,

Given that, the shopkeeper gave 20% and 30% off on bags.

We know that if the value of an object x is changed by a% and then by b% then the final value,

w = x (1 ± a/100) (1 ± b/100),

Where the positive sign indicates an increment while the negative sign indicates the decrement.

Hence, the final price = x(1 – 20/100)(1 – 30/100) = x × (80/100) × (70/100) = 0.56x

Now, the percentage change = {(Final value – Initial value)/Initial value} × 100

= {(0.56x – x)/x} × 100 = – 44% (negative sign indicates the decrement)

Hence, the final discount gave by the shopkeeper is 44%.


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