# Formula - Percentage

**PERCENT**

The word “percent” is derived from the latin words “per centum”,which means “per hundred”.

A percentage is a fraction with denominator hundred.It is denoted by the symbol %.

Numerator of the fraction is called the **rate per cent.**

**VALUE OF PERCENTAGE**

Value of percentage always depends on the quantity to which it refers.

**Consider the statement :
**

“65% of the students in this class are boys”. From the context, it is understood that boys form 65% of the total number of students in the class. To know the value of 65% , the value of the total number of student should be known.

If the total number of students is 200, then, the number of boys \(= \frac{200 \times 65}{100} = 130;\) It can also be written as(200) × (0.65) = 130.

Note that the expressions 6%, 63%, 72%, 155% etc. do not have any value intrinsic to themselves. Their values depend on the quantities to which they refer.

**To express the fraction equivalent to % :** Express the fraction with the denominator 100, then the numerator is the answer.

**Example 1**

**1. Express the fraction 11/12 into the per cent.**

**Sol.** \(\frac{11}{12} = \frac{\frac{11}{12} \times 100}{100} = \frac{91\frac{2}{3}}{100} = 91\frac{2}{3}\) %

**To express % equivalent to fraction : **

a% = a/100

**Example 2**

**2. Express 45⅚ into fraction.**

**Sol.** \(45\frac{5}{6} = \frac{45\frac{5}{6}}{100} = \frac{275}{6 \times 100} = \frac{11}{24}\).

2% = 2/100

4% = 1/25

5% = 1/20

(6¼)% = 1/16

10% = 1/10

(11⅓)% = 17/150

(12½)% = ⅛

16% = 4/25

(16⅔)% = ⅙

20% = ⅕

25% = ¼

40% = ⅖

50% = ½

(66⅔)% = ⅔

60% = ⅗

75% = ¾

80% = ⅘

96% = 24/25

100% = 1

115% = 23/20

(133⅓)% = 4/3

**Example 3**

**3. Rent of the house is increased from 7000 to 7700. Express the increase in price as a percentage of the original rent.**

**Sol.** Increase value = 7700 – 7000 = 700

Increase % = \(\frac{Increase \; Value}{Original \; Value} \times 100 = \frac{700}{7000} \times 100\)

= 10

∴ Percentage rise = 10 %.

**Example 4**

**4. The cost of a bike last year was `19000. Its cost this year is `17000. Find the percent decrease in its cost.**

%decrease = \(\frac{19000 ? 17000}{19000} \times 100\)

= \(\frac{2000}{19000} \times 100 = 10.5\)%

∴ Percent decrease = 10.5 %.

If A is x% of C and Bis y % of C, then A is x/y × 100 % of B.

**Example 5**

**5. A positive number is divided by 5 instead of being multiplied by 5. By what percent is the result of the required correct value?**

**Sol.** Let the number be 1, then the correct answer = 5

The incorrect answer that was obtained = ⅕

∴ The required% = \(\frac{1}{5 \times 5} \times 100 = 4\)%

x % of a quantity is taken by the first, y % of the remaining is taken by the second and z % of the remaining is taken by third person. Now, if A is left in the fund, then the initial amount

= \(\frac{A \times 100 \times 100 \times 100}{\left ( 100 ? x \right )\left ( 100 ? y \right )\left ( 100 ? z \right )}\) in the begining.

x % of a quantity is added. Again, y % of the increased quantity is added. Again z % of the increased quantity is added. Now it becomes A, then the initial amount

\(= \frac{A \times 100 \times 100 \times 100}{\left ( 100 + x \right )\left ( 100 + y \right )\left ( 100 + z \right )}\)**Example 6**

**6. 3.5 % income is taken as tax and 12.5 % of the remaining is saved. This leaves 4,053 to spend. What is the income ?**

**Sol.** By direct method,

If the price of a commodity increases by r %, then reduction in consumption, so as not to increase the expenditure is

\(\left ( \frac{r}{100 + r} \times 100 \right )\)%.If the price of a commodity decreases by r %, then the increase in consumption so as not to decrease the expenditure is \(\left ( \frac{r}{100 ? r} \times 100 \right )\)%.

**Example 7**

**7. If the price of coal be raised by 20%, then find by how much a householder must reduce his consumption of this commodity so as not to increase his expenditure ?**

**Sol.** Reduction in consumption = \(\left ( \frac{20}{100 + 20} \times 100 \right )\)%

= \(\left ( \frac{20}{120} \times 100 \right )\)% = 16.67%

Population Formula If the original population of a town is P,and the annual increase is r %, then the population after n years is \(P\left ( 1 + \frac{r}{100} \right )^{n}\) and population before n years = \(\frac{P}{\left ( 1 + \frac{r}{100} \right )^{n}}\) If the annual decrease be r %, then the population after n years is \(P\left ( 1 ? \frac{r}{100} \right )^{n}\) and

population before n years = \(\frac{P}{\left ( 1 ? \frac{r}{100} \right )^{n}}\)

**Example 8**

**8. The population of a certain town increased at a certain rate percent per annum. Now it is 456976. Four years ago, it was 390625. What will it be 2 years hence ?**

**Sol.** Suppose the population increases at r% per annum. Then,

∴ \(\left ( 1 + \frac{r}{100} \right )^{2} = \sqrt{\frac{456976}{390625}} = \frac{676}{625}\)

Population 2 years hence \(= 456976\left ( 1 + \frac{r}{100} \right )^{2}\) \(= 456976 \times \frac{676}{625} = 494265\) approximately.

**Example 9**

**9. The population of a city increases at the rate of 4% per annum. There is an additional annual increase of 1% in the population due to the influx of job seekers. Therefore, the % increase in the population after 2 years will be :**

(a) 10

(b) 10.25

(c) 10.55

(d) 10.75

**Sol.(b)** The net annual increase = 5%.

Let the initial population be 100.

Then, population after 2 years = 100 × 1.05 × 1.05

= 110.25

Therefore, % increase in population

= (110.25 – 100) = 10.25%

**First Increase and then decrease**

If the value is first increased by x% and then decreased by y % then there is \(\left ( x - y - \frac{xy}{100} \right )\)% increase or decrease, according to the +ve or –ve sign respectively.

**Example 10**

**10. A number is increased by 10%. and then it is decreased by 10%. Find the net increase or decrease percent.**

**Sol.** % change = \(10 ? 10 ? \frac{10 \times 10}{10} = 1\)%

i.e 1% decrease.

**Average percentage rate of change over a period.**

The percentage error = \(= \frac{\left ( The \; Error \right )}{True \; Value} \times 100\)%

**Successive increase or decrease**

If the value is increased successively by x % and y % then the final increase is given by

\(\left ( x + y + \frac{xy}{100} \right )\)%If the value is decreased successively by x % and y % then the final decrease is given by

\(\left ( - x - y - \frac{xy}{100} \right )\)%**Example 11**

**11. The price of a car is decreased by 10 % and 20% in two successive years. What percent of price of a car is decreased after two years ?**

**Sol.** Put x = –10 and y = –20, then

∴ The price of the car decreases by 28%.

**Student and Marks**

The percentage of passing marks in an examination is x%. If a candidate who scores y marks fails by z marks, then the maximum marks M = \(\frac{100(y + z)}{x}\)

A candidate scoring x % in an examination fails by ‘a’ marks, while another candidate who scores y% marks gets ‘b’ marks more than the minimum required passing marks. Then the maximum marks \(M = \frac{100(a + b)}{y - x}\).

**Example 12**

**12. Vishal requires 40% to pass. If he gets 185 marks, falls short by 15 marks, what was the maximum he could have got ?**

**Sol.** If Vishal has 15 marks more, he could have scored 40% marks.

Now, 15 marks more than 185 is 185 + 15 = 200

Let the maximum marks be x, then

40% of x = 200

=> \(\frac{40}{100} \times x = 200\)

=> \(x = \frac{200 \times 100}{40} = 500\)

Thus, maximum marks = 500

**Quicker method :**

Maximum marks \(= \frac{100 \times (185 + 15)}{40} = \frac{100 \times 200}{40} = 500\).

**Example 13**

**13. A candidate scores 15% and fails by 30 marks, while another candidate who scores 40% marks, gets 20 marks more than the minimum required marks to pass the examination. Find the maximum marks of the examination.
**

**Sol.** Quicker method :

Maximum marks \(= \frac{100 \times (30 + 20)}{40 ? 15} = 200\)

**2–dimensional figure and area**

If the sides of a triangle, square, rhombus or radius of a circle are increased by a%, its area is increased by

\(= \frac{a(a + 200)}{100} = 200\)%

**Example 14**

**14. If the radius of a circle is increased by 10 %, what is the percentage increase in its area ?**

**Sol.** Radius is increased by 10%.So, Area is increased by \(= \frac{10(10 + 200)}{100} = 21\)%

**Example 15**

**15. If the length and width of a rectangular garden were each increased by 20%, then what would be the percent increase in the area of the garden ?**

(a) 20%

(b) 24%

(c) 36%

(d) 44%

**Sol.(d)** Let the original length and width of the garden be x and y units, respectively.

Then, the original area = x × y = xy square units

New area = 1.2 x × 1.2y = 1.44xy square units % increase in area \(= \frac{\left ( 1.44xy ? xy \right )}{xy} \times 100 = 44\)%

**NOTE:-**

If A’s income is r% more than that of B, then B’s income is less than that of A by

\(\left ( \frac{r}{100 + r} \times 100 \right )\)%If A’s income is r %less than that of B, then B’s income is more than that of A by

\(\left ( \frac{r}{100 ? r} \times 100 \right )\)%If the both sides of rectangle are changed by x% and y% respectively, then % effect on area \(= x + y + \frac{xy}{100}\) (+/– according to increase or decrease)

**Example 16**

**16. If A’s salary is 50% more than B’s,then by what percent B’s salary is less than A’s salary ?**

**Sol.** By direct method,B’s salary is less than A’s salary by

**Example 17**

**17. Ravi’s weight is 25% that of Meena’s and 40% that of Tara’s. What percentage of Tara’s weight is Meena’s weight.**

**Sol.** Let Meena’s weight be x kg and Tara’s weight be y kg.

Then Ravi’s weight = 25% of Meena’s weight

\(= \frac{25}{100} \times x\) ……..(i)Also, Ravi’s weight = 40% of Tara’s weight

\(= \frac{40}{100} \times y\) ……..(ii)From (i) and (ii), we get

\(= \frac{25}{100} \times x = \frac{40}{100} \times y\)=> 25x = 40y

=> 5x = 8y => x = &frac85;y

Meena’s weight as the percentage of Tara’s weight

\(= \frac{x}{y} \times 100 = \frac{\frac{8}{5}y}{y} \times 100 = \frac{8}{5} \times 100 = 160\)Hence, Meena’s weight is 160% of Tara’s weight.