# Successive Percent Change

Suppose a number x undergoes a percentage change of a% and then b%, then:

Net percentage change ≠ (a + b)%

Can you guess why?

The net percentage change is not the simple addition of the two percentage changes as the base changes after the first change.

New number after a% change = x × (1 + $\frac{a}{100}$)
This is the new base, i.e. the base changes.

So, New number after b% change = [x × (1 + $\frac{a}{100}$)] × (1 + $\frac{b}{100}$)

So, the new number after successive a% and b% changes = x × ($\frac{a}{100}$) × ($\frac{b}{100}$) = x × [1 + $\frac{𝑎 + 𝑏 + \frac{ab}{100}}{100}$]

So, net percentage change = 𝑎 + 𝑏 + $\frac{ab}{100}$

Net Change is an increase or decrease according to the positive or negative sign of the final result.

If there are more than 2 successive percentage changes then we can apply net percent change formula successively in pair of two.

In successive percentage increase/decrease, the order of the percentages does not matter. You can apply anyone first; the answer will be the same.

As, x × (1 + $\frac{a}{100}$) × (1 + $\frac{b}{100}$) = x × (1 + $\frac{b}{100}$) × (1 + $\frac{a}{100}$)

##### Q. If Mragank’s salary increases by 20% in the first year and by 10% the next year, then what will be the net percentage increase in his salary after 2 years?

Explanations :

Let Mragank’s salary be Rs x

Mragank’s salary after first year = x × (120/100) = 1.2x

Mragank’s salary after second year = 1.2x × (110/100) = 1.2x × 1.1 = 1.32x

So, Net percentage increase = [(1.32x - x)/x] × 100 = (0.32x/x) × 100 = 0.32 × 100 = 32%

Explanation 2: Using the concept of Multiplying Factors

First multiplying factor = 1.2
Second multiplying factor = 1.1
Net Multiplying Factor = 1.2 × 1.1 = 1.32
So, Net percentage increase = 32%

Explanation 3: Formula Approach

Net percentage change ≠ (a + b)%

Net percentage increase = (𝑎 + 𝑏 + $\frac{ab}{100}$)% = (20 + 10 + $\frac{20 × 10}{100}$)% = (30 + 2)% = 32%

Explanation 4: Percentage Method

Let the initial salary of Mragank be Rs 100

Salary after 20% increase = 100 + 20 = Rs 120
Salary after further 10% increase = 120 + 12 = Rs 132
So, Net percentage increase = 32%

Explanation 5: Using Fractions

20% ≡ 1/5
New salary after first year = 5 + 1 = 6
Ratio of salary after first year /original salary = 6/5

10% ≡ 1/10
New salary after second year = 10 + 1 = 11
Ratio of salary after second year / salary after first year = 11/10

Ratio of salary after second year / original salary = (11/10) × (6/5) = 33/25

So, Net percentage increase = [(33 - 25)/25] × 100 = (8/25) × 100 = 32%

##### Q. A person had invested a certain amount in a share. The share rose by 100% one day and fell by 50% the next day. What is the percentage profit/loss made by the person?

(a) 50% profit
(b) 0% profit
(c) 50% loss
(d) Cannot be determined

Explanation:

If any number is increased by 100%, it doubles. Similarly, on decreasing by 50%, it halves.

Hence, the final sum he had will be the same as the original. Hence, no profit, no loss.

Tips:

Use net percent change formula method if the percentage data given is in integers but kind of ugly, e.g. 17% increase, 19% increase (in such type of data it will be hard to use either the percentage method or fraction method)

Use percent method if the percentage data given is in integers that are beautiful, e.g. 20% increase, 15% increase (in such type of data you can also use either the formula method or fraction method, but percentage method will be the fastest)

Use fraction method if the percentage data given is in decimals (that are percentage equivalents of fractions), e.g. 16.67% increase, 12.5% increase (in such type of data it will be hard to use either the percentage method or formula method)

##### Q. If the price of a book is decreased by 22.22% and then increased by 50%, then what must be the net percent change in its price?

Explanation: Using Fractions

22.22% ≡ 2/9
New price = 9 - 2 = 7
Ratio of new price/original price = 7/9

50% ≡ 1/2
New price = 2 + 1 = 3
Ratio of new price/original price = 3/2

Ratio of final price / original price = (7/9) × (3/2) = 7/6
So, Net percentage increase = [(7 - 6)/6] × 100 = (1/6) × 100 = 16.67%

## Special case

We sometimes encounter a special case when we use net percent change formula.

• If a number is first increased by a% and then it is decreased by a%, then net effect is always a decrease which is equal to:
a – a – $\frac{a^2}{100}$ = - $\frac{a^2}{100}$%

• The net percentage change will be the same even if a number is first decreased by a% and then it is increased by a%.
-a + a – $\frac{a^2}{100}$ = - $\frac{a^2}{100}$%

So, in both the cases the answer is the same, i.e. - $\frac{a^2}{100}$%

##### Q. If price of a jute bag is increased by 20% and then decreased by 20%, then what will be the net percentage change in its price?

Explanations :

Explanation 1: Formula Approach

Net percentage increase = - $\frac{a^2}{100}$% = - $\frac{20^2}{100}$% = - $\frac{400}{100}$% = -4%

Explanation 2: Percentage Method

Let the initial price of jute bag be Rs 100
Price after 20% increase = 100 + 20 = Rs 120
Price after further 20% decrease = 120 - 24 = Rs 96
So, Net percentage decrease = -4%

Explanation 3: Fraction Method

20% increase ≡ 1/5
New price = 5 + 1 = 6
Ratio of new price/original price = 6/5

20% decrease ≡ 1/5
New price = 5 - 1 = 4
Ratio of new price/original price = 4/5

Ratio of final price / original price = (6/5) × (4/5) = 24/25
So, Net percentage decrease = [(25 - 24)/25] × 100 = (1/25) × 100 = 4%

## z = x × y

Let there be two quantities x and y that multiply to form a quantity z. We can say:
z = x × y

Say x is changed by a% and y is changed by b%, then:
z = x (1 + $\frac{a}{100}$) × y (1 + $\frac{b}{100}$) = x y × [1 + $\frac{𝑎 + 𝑏 + \frac{ab}{100}}{100}$ ] = z × [1 + $\frac{𝑎 + 𝑏 + \frac{ab}{100}}{100}$]

So changing x and y by a% and b% respectively means that z is changed by:
𝑎 + 𝑏 + $\frac{ab}{100}$.

So, the question basically reduces to:
If z is successively changed by a% and then b%, then what is the net percentage change?

That is, changing x by a% and y by b% ≡ changing z successively by a% and b%

Let’s see a couple of examples:

##### Q. If in a rectangle the length increases by 40% and breath decreases by 20%, then what must be the net change in area?

Explanations :

Explanation 1: Formula Method

Net percentage change = (𝑎 + 𝑏 + $\frac{ab}{100}$)% = (40 - 20 - $\frac{40 × 20}{100}$)% = (20 − 8)% = 12%

Explanation 2: Using the concept of Multiplying Factors

First multiplying factor = 1.40
Second multiplying factor = 0.80
Net Multiplying Factor = 1.4 × 0.8 = 1.12

So, Net percentage increase = 12%

Explanation 3: Percentage Method

Let the initial area of the rectangle be 100
Area after 40% increase in length = 100 + 40 = 140
Area after further 20% decrease in breath = 140 - 28 = 112

So, Net percentage increase = 12%

Explanation 4: Fraction Method

40% increase ≡ 2/5
New length = 5 + 2 = 7
Ratio of new length/original length = 7/5

20% decrease ≡ 1/5
New breath = 5 - 1 = 4
Ratio of new breath /original breath = 4/5

Ratio of final area / original area = (7/5) × (4/5) = 28/25

So, Net percentage increase = [(28 - 25)/25] × 100 = (3/25) × 100 = 12%

##### Q. The length and breadth of a cuboid is increased by 10% and 20% respectively, while its height is reduced by 30%. What must be the total percent increase/decrease in the volume of the cuboid?

Explanations :

Explanation 1: Formula Method

Volume of cuboid = l × b × h

As here 3 variables are involved, we will have to use the formula twice.

Net percentage increase of l & b = (𝑎 + 𝑏 + $\frac{ab}{100}$)% = (10 + 20 + $\frac{10 × 20}{100}$)% = (30+2)% = 32%

Net percentage increase of (l & b) & h = (𝑎 + 𝑏 + $\frac{ab}{100}$)% = (32 - 30 - $\frac{32 × 30}{100}$)% = (2 − 9.6)% = -7.6%

So, percent decrease in the volume of the cuboid = 7.6%

Explanation 2: Percentage Method

Let initial volume be 100.
After 10% increase, volume = 100 + 10% of 100 = 100 + 10 = 110
After further 20% increase, volume = 110 + 20% of 110 = 110 + 22 = 132
After further 30% decrease, volume = 132 - 30% of 132 = 132 – (3 × 13.2) = 132 – 39.6 = 92.4

So, percent decrease in the volume of the cuboid = 100 – 92.4 = 7.6%

Explanation 3: Fraction Method

10% increase ≡ 1/10
20% increase ≡ 1/5
30% decrease ≡ 3/10

Ratio of final volume / original volume = (11/10) × (6/5) × (7/10) = 462/500

So, Net percentage decrease = [(500 - 462)/500] × 100 = (38/500) × 100 = 7.6%

## Successive Percent Change for three or more years

### Constant Growth Rate

If the population of a town is P and it is increased at the rate of r% per annum (i.e. growth rate is constant over the years), then

(a) Population after n years, $P_n = P (1 + \frac{r}{100})^n$

(b) Population n years ago, P = $\frac{P_n}{(1 + \frac{r}{100})^n}$

### Variable Growth Rate

If the population of a town is P and increases at the rate of $r_1%$ in first year, $r_2%$ in second year and $r_3%$ in third year (i.e. growth rate varies over the years), then

(a) Population after 3 years, $P_3 = P (1 + \frac{r_1}{100}) (1 + \frac{r_2}{100}) (1 + \frac{r_3}{100})$

(b) Population 3 years ago, P = $\frac{P_3}{(1 + \frac{r_1}{100}) (1 + \frac{r_2}{100}) (1 + \frac{r_3}{100})}$

If present value of a machine is P and it’s depreciation rates are $r_1%$, $r_2%$ and $r_3%$ yearly (i.e. depreciation rate varies over the years), then

(a) Value of machine after 3 years, $P_3 = P (1 - \frac{r_1}{100}) (1 - \frac{r_2}{100}) (1 - \frac{r_3}{100})$

(b) Value of machine 3 year ago, P = $\frac{P_3}{(1 - \frac{r_1}{100}) (1 - \frac{r_2}{100}) (1 - \frac{r_3}{100})}$

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