Percentage increase and decrease

Prices go up. Salaries go up. Some­times things get cheaper. We usu­ally talk about these changes in per­cent­ages: "Petrol is 10% dearer." In this les­son you will learn to find a new value after a per­cent­age change, to find the per­cent­age change itself, and some short results that save a lot of time. At the end you will see how a pop­u­la­tion grows year after year.

Increase by a per­cent­age#

The old price of a book is ₹80. The price goes up by 10%. What is the new price?

$$\displaystyle \text{increase} = 10\% \text{ of } 80 = \frac{10}{100} \times 80 = 8$$

$$\text{new price} = 80 + 8 = 88$$

The new price is ₹88.

Now turn it round. The old price was ₹80 and the new price is ₹88. What is the per­cent­age increase? The increase is ₹8. Com­pare it with the old price:

$$\displaystyle \left(\frac{8}{80} \times 100\right)\% = 10\%$$

A for­mula for increase#

Let the old value be $P$ and let it go up by $R\%$.

$$\displaystyle \text{increase} = \frac{R}{100} \times P$$

$$\displaystyle \text{new value} = P + \frac{R}{100}P = P\left(1 + \frac{R}{100}\right) = \left(\frac{100 + R}{100}\right)P$$

So to increase by 10%, mul­ti­ply by $\displaystyle \frac{110}{100}$. Check: $\displaystyle \frac{110}{100} \times 80 = 88$.

Decrease by a per­cent­age#

Let the old value be $P$ and let it go down by $R\%$.

$$\displaystyle \text{decrease} = \frac{R}{100} \times P$$

$$\displaystyle \text{new value} = P - \frac{R}{100}P = P\left(1 - \frac{R}{100}\right) = \left(\frac{100 - R}{100}\right)P$$

So to decrease by 25%, mul­ti­ply by $\displaystyle \frac{75}{100}$.

The main rules#

1. If $P$ is increased by $R\%$, the new value is $\displaystyle \left(\frac{100 + R}{100}\right)P$.
2. If $P$ is decreased by $R\%$, the new value is $\displaystyle \left(\frac{100 - R}{100}\right)P$.
3. To find the per­cent­age change, always com­pare with the orig­i­nal value:

$$\displaystyle \text{percentage increase} = \left(\frac{\text{increase}}{\text{original value}} \times 100\right)\%$$

$$\displaystyle \text{percentage decrease} = \left(\frac{\text{decrease}}{\text{original value}} \times 100\right)\%$$

4. To com­pare a greater and a smaller quan­tity:

$$\displaystyle \text{greater is more than smaller by } \left(\frac{\text{greater} - \text{smaller}}{\text{smaller}} \times 100\right)\%$$

$$\displaystyle \text{smaller is less than greater by } \left(\frac{\text{greater} - \text{smaller}}{\text{greater}} \times 100\right)\%$$

The dif­fer­ence is the same in both. Only the num­ber at the bot­tom changes: you divide by the quan­tity you are com­par­ing with.

Exam­ple 1#

$p$ is 6 times as large as $q$. By what per cent is $q$ less than $p$?

Let $q = x$. Then $p = 6x$. The dif­fer­ence is $5x$. We com­pare with $p$, the greater one.

$$\displaystyle \left(\frac{5x}{6x} \times 100\right)\% = \left(\frac{5}{6} \times 100\right)\% = \frac{250}{3}\% = 83\frac{1}{3}\%$$

Exam­ple 2#

A's salary is 20% less than B's salary. By what per cent is B's salary more than A's?

Take B's salary as ₹100. Then A's salary is ₹80. The dif­fer­ence is ₹20, and now we com­pare with A's salary, ₹80.

$$\displaystyle \left(\frac{20}{80} \times 100\right)\% = 25\%$$

So B earns 25% more than A. It is not 20%, because the com­par­i­son is made with a dif­fer­ent salary.

Two short­cut results#

Exam­ple 2 shows a pat­tern. Here it is as a rule.

Result 1. If A is $R\%$ more than B, then B is less than A by

$$\displaystyle \left(\frac{R}{100 + R} \times 100\right)\%$$

Result 2. If A is $R\%$ less than B, then B is more than A by

$$\displaystyle \left(\frac{R}{100 - R} \times 100\right)\%$$

Why Result 1 works. Let B be $x$. Then A is $\displaystyle \frac{100 + R}{100}x$. The dif­fer­ence is $\displaystyle \frac{R}{100}x$, and we com­pare it with A:

$$\displaystyle \left(\frac{\frac{R}{100}x}{\frac{100 + R}{100}x} \times 100\right)\% = \left(\frac{R}{100 + R} \times 100\right)\%$$

Check Exam­ple 2 with Result 2: $R = 20$ gives $\displaystyle \frac{20}{80} \times 100 = 25$, so 25%.

Price and how much we buy#

If the price of rice goes up, a fam­ily can buy a lit­tle less and still spend the same money. How much less?

Result 3. If the price of a thing goes up by $R\%$, then to keep the spend­ing the same, the amount used must go down by

$$\displaystyle \left(\frac{R}{100 + R} \times 100\right)\%$$

Result 4. If the price goes down by $R\%$, then the amount used can go up, for the same spend­ing, by

$$\displaystyle \left(\frac{R}{100 - R} \times 100\right)\%$$

Why Result 3 works. Let the old price be $p$ and the old amount used be $x$, so the spend­ing is $px$. The new price is $\displaystyle \frac{100 + R}{100}p$. For the same spend­ing, the new amount is

$$\displaystyle \frac{px}{\frac{100 + R}{100}p} = \frac{100x}{100 + R}$$

The amount went down by $\displaystyle x - \frac{100x}{100 + R} = \frac{Rx}{100 + R}$. As a per­cent­age of $x$, that is $\displaystyle \frac{R}{100 + R} \times 100$.

Exam­ple 3#

The price of petrol goes up by 25%. By how much must a man cut his use of petrol so that he spends no more than before?

$$\displaystyle \left(\frac{25}{100 + 25} \times 100\right)\% = \left(\frac{25}{125} \times 100\right)\% = 20\%$$

He must use 20% less petrol.

Pop­u­la­tion growth#

Result 5. The pop­u­la­tion of a town is $P$ now, and it grows at $R\%$ every year. Then

1. the pop­u­la­tion after $n$ years is $\displaystyle P\left(1 + \frac{R}{100}\right)^n$,
2. the pop­u­la­tion $n$ years ago was $\displaystyle \frac{P}{\left(1 + \frac{R}{100}\right)^n}$.

Why it works. Each year the pop­u­la­tion is mul­ti­plied by $\displaystyle \left(1 + \frac{R}{100}\right)$.

After 1 year: $\displaystyle P + \frac{R}{100}P = \left(1 + \frac{R}{100}\right)P$.

After 2 years: $\displaystyle \left(1 + \frac{R}{100}\right)\left(1 + \frac{R}{100}\right)P = P\left(1 + \frac{R}{100}\right)^2$.

After $n$ years it has been mul­ti­plied $n$ times. For part 2, let the pop­u­la­tion $n$ years ago be $x$. Then $\displaystyle x\left(1 + \frac{R}{100}\right)^n = P$, so $\displaystyle x = \frac{P}{\left(1 + \frac{R}{100}\right)^n}$.

Exam­ple 4#

A town has 20,000 peo­ple. The pop­u­la­tion grows by 10% each year. What will it be after 2 years?

$$\displaystyle 20000\left(1 + \frac{10}{100}\right)^2 = 20000 \times \frac{11}{10} \times \frac{11}{10} = 24200$$

Exam­ple 5#

A town has 24,200 peo­ple now. It has grown by 10% each year. What was its pop­u­la­tion 2 years ago?

$$\displaystyle \frac{24200}{\left(\frac{11}{10}\right)^2} = 24200 \times \frac{10}{11} \times \frac{10}{11} = 20000$$

Prac­tice#

1. Increase 250 by 20%.
2. Decrease 640 by 15%.
3. A price goes up from ₹400 to ₹460. Find the per­cent­age increase.
4. A salary goes down from ₹25,000 to ₹22,000. Find the per­cent­age decrease.
5. A is 25% more than B. By what per cent is B less than A?
6. A's income is 10% less than B's. By what per cent is B's income more than A's?
7. The price of sugar goes up by 20%. By what per cent must a fam­ily cut its use of sugar so that it spends no more?
8. The price of oil goes down by 20%. By what per cent can a fam­ily use more oil for the same money?
9. A vil­lage has 50,000 peo­ple and grows by 4% a year. What will its pop­u­la­tion be after 2 years?
10. A town has 13,310 peo­ple now. It has grown by 10% a year. What was its pop­u­la­tion 3 years ago?

Answers#