Concept: Percentage

Example:. Express the following in terms of percentage:
(a) 0.4            (b) 1.0            (c) 5/3            (d) 7x/y            (e) 1.23

Solution:

(a) multiply the decimal fraction by 100.

∴ 0.4 = (0.4 × 100)% = 40%

(b) 1.0 = (1.0 × 100)% = 100%

(c) 5/3 = (5/3 × 100)% = 166 2/3%

(d) 7x/y = (7x/y × 100)% = 700x/y%

(e) 1.23 = (1.23 × 100)% = 123%

Example: Express the following in terms of fraction:
(a) 22$\frac{1}{2}$%            (b) 35%            (c) a²/b%            (d) 0.3%            (e) 2/7%

Solution:
(a) Divide the given percentage by 100 to convert it into a fraction

∴ 22$\frac{1}{2}$% = (22$\frac{1}{2}$)/100 = 45/(2 × 100) = 9/40

(b) 35% = 35/100 = 7/20

(c) a²/b% = a²2/(b × 100) = a²/(100b)

(d) 0.3% = 0.3/100 = 3/1000

(e) 2/7% = 2/(7 × 100) = 1/350

Example: Find
(a) 9% of 27        (b) 0.02% of 6500        (c) 7/2% of 80        (d) 125% of 64        (e) 10% of 5% of 320

Solution:
(a) Multiply the number by P/100, if p% of the number is to be calculated.

∴ 9% of 27 = 9/100 × 27 = 243/100

(b) 0.02% of 650 = 0.02/100 × 6500 = 13/10

(c) 7/2% of 80 = 7/(2 × 100) × 80 = 14/5

(d) 125% of 64 = 125/100 × 64 = 80

(e) 10% of 5% of 320 = 10/100 × 5/100 × 320 = 1.6 = 8/5

Example: Find the following :
(a) 36 is what % of 144                (b) 7/8 is what % of 3/4    
(c) What % of 80 is 16                (d) 0.625 is equal to what % of 1$\frac{7}{28}$
(e) 36 × 14 is what % of 1400         (f) R is what % of N

Solution:
(a) Here we have to compare 36 with 144.

Value of percentage (%) = (Number to be compared)/(Number to be compared against) × 100

⇒ % = 36/144 × 100 = 25%

(b) Value of % =(7⁄8)/(3⁄4)×100 = 116$\frac{2}{3}$%

(c) Value of % = Result/(Original number) × 100 = 16/80 × 100 = 20%

(d) Value of % = 0.625/(1$\frac{7}{28}$) × 100 = 0.625/35 × 28 × 100 = 50%

(e) value of % = (36 × 14)/1400 × 100 = 36%

(f) value of % = R/N × 100 = 100R/N%

Example: 25% of a number is 20, what is 40% of that number? Also find the number

Solution:
25% → 20

⇒ 40%→ 20/25 × 40 = 32

Original number (N) = $\frac{20}{25}$ × 100 = 80

Example: A number A exceeds B by 25%. By what percent is B less than A?

Solution:
If B = 100, then A = 125

⇒ B = $\frac{100}{125}$ × 100 = 80% of A

∴ B is (100 - 80) = 20% less than A

Example: Increase 120 by 50%?

Solution:
We add 50% of 120 with 120.

∴ 120 + 120 × 50% = 120 + 60 = 180

Example: Decrease 120 by 50%?

Solution:
We subtract 50% of 120 from 120.

∴ 120 - 120 × 50% = 120 - 60 = 120.

Example: The daily wage is increased by 15%, and a person now gets Rs. 23 per day. What was his daily wage before the increase?

Solution:
Multiplication factor = $\left(1+\frac{\mathrm{15}}{100}\right)$ = $\left(1+\frac{3}{20}\right)$ = $\frac{23}{20}$

∴ Original daily wage =$\frac{23}{\mathrm{23/20}}$ = 20

Example: From a man’s salary, 10% is deducted on tax, 20% of the rest is spent on the education, and 25% of the rest is spent on food. After all these expenditures, he is left with Rs. 2,700 Find his salary.

Solution:
Let the salary be S.

10% of salary is deducted on tax.
∴ Remaining salary = $S\left(1-\frac{10}{100}\right)$

Now, 20% of this remaining salary is spent on education.
∴ Remaining salary = $S\left(1-\frac{10}{100}\right)\left(1-\frac{20}{100}\right)$

Now, 25% of this remaining salary is spent on food.
∴ Remaining salary = $S\left(1-\frac{10}{100}\right)\left(1-\frac{20}{100}\right)\left(1-\frac{25}{100}\right)$ which is equal to 2700

⇒  $S\left(1-\frac{10}{100}\right)\left(1-\frac{20}{100}\right)\left(1-\frac{25}{100}\right)$ = 2700

⇒ $S\times \frac{9}{10}\times \frac{4}{5}\times \frac{3}{4}$ = 2700

⇒ S = $\frac{2700\times 50}{27}$ = Rs. 5000

Example : A number is tripled. What is the percentage change.

Solution:
When a number is tripled, it is effectively multiplied with 3.

∴ % change = (3 - 1) × 100 = 200%

Example : A number is increased by 50%. By what percentage should it be decreased to get back the original number.

Solution:
For 50% increase, multiplication factor = $\left(1+\frac{50}{100}\right)$ = $\frac{3}{2}$

∴ New number = Initial number × $\frac{3}{2}$

To get the original number, this new number must be multiplied with $\frac{2}{3}$

∴ % change = (2/3 - 1) × 100 = -33.33%

∴ The new number should be decreased by 33.33% to get back the original number.

Example : When a number is successively changed by a% and b%, what is the overall percentage change.

Solution:
When a number is successively changed by a% and b%, the overall multiplication factor = $\left(1+\frac{a}{100}\right)\left(1+\frac{b}{100}\right)$ = f

∴ percentage change = (f - 1) × 100% = $\left(1+\frac{a}{100}+\frac{b}{100}+\frac{a\times b}{{100}^2}-1\right)$ × 100% = a + b + $\frac{a\times b}{100}$

Example : When a number is successively changed by 20% and 25%, what is the overall percentage change.

Solution:
Method 1: (Multiplication Factor)
Overall multiplication factor = $\left(1+\frac{20}{100}\right)\left(1+\frac{25}{100}\right)$ = $\frac{6}{5}\times \frac{5}{4}$ = $\frac{3}{2}$

∴ % change = (3/2 - 1) × 100 = 50%

Method 2: (Formula)
Overall % change = 20 + 25 + $\frac{\mathrm{20}\times \mathrm{25}}{100}$ = 50%

Example : When a number is successively changed by -20% and 25%, what is the overall percentage change.

Solution:
Method 1: (Multiplication Factor)
Overall multiplication factor = $\left(1-\frac{20}{100}\right)\left(1+\frac{25}{100}\right)$ = $\frac{4}{5}\times \frac{5}{4}$ = 1

∴ % change = (1 - 1) × 100 = 0% i.e. no change

Method 2: (Formula)
Overall % change = -20 + 25 + $\frac{\mathrm{-20}\times \mathrm{25}}{100}$ = 0%

Example : When a number is successively changed by -20% and -25%, what is the overall percentage change.

Solution:
Method 1: (Multiplication Factor)
Overall multiplication factor = $\left(1-\frac{20}{100}\right)\left(1-\frac{25}{100}\right)$ = $\frac{4}{5}\times \frac{3}{4}$ = $\frac{3}{5}$

∴ % change = (3/5 - 1) × 100 = - 40%

Method 2: (Formula)
Overall % change = - 20 - 25 + $\frac{\mathrm{-20}\times \mathrm{-25}}{100}$ = - 40%

Example : When sugar price increase by 25%, by what percentage should a family reduced its consumption so that monthly expenditure does not change.

Solution:
Method 1:
Let the initial price of sugar = Rs. 100/kg and quantity consumed = 10kg / month
∴ Total Expenditure = 100 × 10 = Rs. 1,000

Now, price increases by 25%, hence, new price = 125
New quantity consumed = 1,000/125 = 8 kg

∴ quantity reduces from 10 kg to 8 kg i.e. reduction of 20%

Method 2: (Multiplication factor)
Expenditure = Price × Quantity
⇒ multiplication factor for expenditure = multiplication factor for price × multiplication factor for quantity

Overall multiplication factor for expenditure = 1 (no change)
Multiplication factor for price = $\left(1+\frac{25}{100}\right)$ = $\frac{5}{4}$
Let multiplication factor for quantity = f

∴ 1 = 5/4 × f
⇒ f = 4/5
∴ % change in quantity = (4/5 - 1) × 100 = -20%

Example : When sugar price increase by 50%, by what percentage should a family reduced its consumption so that monthly expenditure goes up by only 20%.

Solution:
Method 1:
Let the initial price of sugar = Rs. 100/kg and quantity consumed = 10kg / month
∴ Total Expenditure = 100 × 10 = Rs. 1,000

Now, price increases by 50%, hence, new price = 150
Expenditure increases by 20%, hence, new expenditure = 1200
New quantity consumed = 1,200/150 = 8 kg

∴ quantity reduces from 10 kg to 8 kg i.e. reduction of 20%

Method 2: (Multiplication factor)
Expenditure = Price × Quantity
⇒ multiplication factor for expenditure = multiplication factor for price × multiplication factor for quantity

Overall multiplication factor for expenditure = $\left(1+\frac{20}{100}\right)$ = $\frac{6}{5}$
Multiplication factor for price = $\left(1+\frac{50}{100}\right)$ = $\frac{3}{2}$
Let multiplication factor for quantity = f

∴ 6/5 = 3/2 × f
⇒ f = 4/5
∴ % change in quantity = (4/5 - 1) × 100 = -20%

Example: Groundnut oil is being sold at Rs. 27 per kg. During last month its cost was Rs. 24 per kg. Find by how much % a family should reduce its consumption, so as to keep the expenditure the same.

Solution:
Let the initial consumption of groundnut oil be = x and new expenditure = x₁

Expense (Initial) = 24x and Expense (Final) = 27x₁

∴ 24x = 27 × x₁    [since expense has to remain the same]

x₁ = 8x/9

In the above fraction we can see that the new consumption is 8/9^th fraction of the old consumption.

Thus, the reduction is 1/9^th of the original consumption.

Hence, percentage reduction in consumption is 1/9 × 100%.

Example: If the radius of the circle is increased by 20%, its area will increase by what %?

Solution:
We know that Area of circle = 𝜋 × r × r

overall multiplication factor for area = $\left(1+\frac{20}{100}\right)\times \left(1+\frac{20}{100}\right)$ = ${\left(1+\frac{20}{100}\right)}^2$ = ${\left(\frac{6}{5}\right)}^2$ = $\frac{36}{25}$

∴ % change in area (36/25 - 1) × 100 = 44%

Hence, increase in area = 44%

Example: If the sides of a square are increased by 30%. Find the resultant increase in area.

Solution:
Let A ∝ a² where ‘a’ = initial side

New area A₁ ∝ (1.3a)² 

⇒ A₁ ∝1.69a²

Thus, it can be seen that new area is 69% more than the old area.

Example: A reduction of 25% in the price of Cadbury chocolates, enables a person to buy 5 kg more for Rs. 120. Find the original price of Cadbury chocolates per kg.

Solution:
Method 1 :
In this question the fixed expense = Rs. 120

If P is the initial price, reduced price = 0.75P

Now, new quantity (at the reduced price) – initial quantity = 5

120/0.75P - 120/P = 5

⇒ P = Rs. 8

Alternate Method:
Since the price reduces by 25%, the initial quantity can be bought for 120 × 0.75 = Rs. 90 at the reduced price.

∴ Remaining amount = 120 - 90 = Rs. 30

Now, 5 kg extra can be bought for this Rs. 30

∴ Reduced price = 30/5 = Rs. 6 = 0.75P

∴ Initial price = 6/0.75 = Rs. 8

Example: When N is reduced by 4, it becomes 80% of itself. What is the value of N?

Solution:
The problem implies that

20% of N = 4    (since 100 – 80 = 20)

N = 4/20 × 100

∴ N = 20

Example: If 10% of an electricity bill is deducted, Rs. 45 is still to be paid. How much was the bill?

Solution:
If 10% of the bill is reduced, still 90% has to be paid

Here the decreased bill = Rs. 45

⇒ 90% of Total Bill = 45

⇒ N = (45 × 100)/90

∴ N = 50

Hence, the bill was Rs. 50

Example: If the price of 1 kg of cornflakes is increased by 25%, the increase is Rs. 10. Find the new price of cornflakes per kg.

Solution:
Due to 25% increase, the price changes by Rs. 10

⇒ 25% of price = 10

⇒ price = (10 × 100)/25 = Rs. 50

∴ New price of cornflakes per kg is Rs. 50

Example: In an election between two candidates, a candidate who gets 62% of the total votes polled is elected by a majority of 144 votes, Find the number of votes polled and votes secured by the candidate. [All votes polled are valid votes]

Solution:
Let, the total number of votes be 100x

Winning candidate gets 62% = 62x votes

Losing candidate gets (100 - 62 =) 38% = 38x votes

Now, difference of votes is 144

∴ 62x - 38x = 144

⇒ 24x = 144

⇒ x = 6

Thus, total number of votes = 100x = 600 votes and winner gets = 62x = 372 votes

Example: The rate for admission to an exhibition was Rs. 5 and was later reduced by 20%. As a result, the sale proceeds increased by 44%. What was the % increase in attendance?

Solution:
Sale proceeds = price × attendance

Multiplication factor for sales = $\left(1+\frac{44}{100}\right) = \frac{36}{25}$

Multiplication factor for price = $\left(1-\frac{20}{100}\right) = \frac{4}{5}$

Let the multiplication factor for attendance = f

⇒ 36/25 = 4/5 × f

⇒ f = 36/25 × 5/4 = 9/5

∴ % change in attendace = (9/5 - 1) × 100 = 80%

∴ increase in attendance = 80%

Alternate Method:
Let the initial attendace = 100 people

∴ Initial sale proceeds = 5 × 100 = Rs. 500

Now, new price = 5 × 0.8 = Rs. 4

and, new sales proceeds = 500 × 1.44 = Rs. 720

∴, new attendance = 720/4 = 180

∴, Increase in attendance = (180 - 100)/100 × 100 = 80%

* The information of price is redundant in this question.

Example: A piece of cotton cloth, 20m long, shrinks by 0.5% after washing. Find the length of the cloth after washing.

Solution:
Let the length of the cloth after washing be x metro or Now it left 100 – 0.5 = 99.5%

∴ x = 20 × 99.5%

⇒ x = 19.9 meter

Hence, original length = 19.9 meter

Example: By mistake, a line was measured as 11.25 cm, which was 2.5% more than the actual length. Find the actual length of the line?

Solution:
let actual length of line be x cm

∴ x(100 + 2.5)% = 11.25

⇒ x = (11.25 × 100)/102.5 = 10.9

Hence, actual length = 10.9 cm

Example: When the price of a pressure cooker was increased by 15%, the sale of pressure cookers (quantity) decreased by 15%. What was the net effect on the sales?

Solution:
Let the original price be Rs. 100 also the number of item 100.

∴ Revenues = 100 × 100 = 10000

After change price = Rs. 115 and item = 85

Now, Revenue = 115 × 85 = 9775

∴ decrease % =(10000 - 9775)/10000 × 100

Hence, decreased by = 2.25%

Alternate Method :
1.15 × 0.85 = 0.9775

Thus, decrease = (0.9775 - 1) × 100 = 2.25%

Example:. A medical student has to secure 40% marks to pass. He gets 80 marks and fails by 60 marks. Find the maximum marks?

Solution:
Let the maximum number of marks be x

∴ x × 40% = 80 + 60 (he needs to pass)

⇒ x = 350

Hence, maximum marks = 350

Example: When 75% of a number is added to 75, the result is the number again. The number is

Solution:
let the number be x

∴ 75% of x + 75 = x

⇒ 75 = x – 75% of x

⇒ x = 7500/25 = 300

Hence, the original number is 300

Example: What is the total number of candidates in an examination, if 31% fail, and the number of those who pass exceeds the number of those who fail by 247?

Solution:
Let the total number of candidates be x

Pass % = 100 – 31 = 69

Now, 69% of x – 31% of x = 247

⇒ 38% of x = 247 ⇒ x = (247 × 100)/38 = 650

Hence, the total number = 650

Example: Rita loses 12$\frac{1}{2}$% of her money and after spending 70% of the remainder, has Rs. 210 left. How much money did she have at first?

Solution:
Let Rita have Rs. x at first

30% of (87$\frac{1}{2}$% of x) = 210

or 3/10 × 7/8 × x = 210

⇒ x = Rs. 800. Hence, Rita had Rs. 800 at first.

Example 5% people of a village died by bombardment, 15% of the remainder left the village on account of fear. If now the population is reduced to 3553. How much was it in the beginning?

Solution:
Let the population at beginning be x

∴ people left after bombardment = 95% of x

and number after leaving 15% people = 85% of (95% of x) = 3553 (given)

⇒ 85/100 × 95x/100 = 3553 ⇒ x = 4400

Example: In a factory, there are 40% technicians and 60% non-technicians. If the 60% of the technicians and 40% of non-technicians are permanent employees, then find the percentage of worker who are temporary?

Solution:
Let the number of employees be 100

∴ Number of technicians = 40

∴ Number of non-technicians = 60

Now, Number of temporary employee

= 40% of 40 + 60% of 60 = 16 + 36 = 52

Hence, 52% of the workers is temporary

Example: A salesman receives a salary of Rs. 1000 per month and commission of 8% on all sales in excess of Rs. 2000. What should be the amount of his sales in a particular month if he were to earn Rs. 5000 in that month

Solution:
Let amount of his sales be x

∴ 1000 + 8% (x – 2000) = 5000

⇒ x = Rs. 52000

Example: During one year, the population of a town increased by 10% and during the next year, it diminished by 10%. If at the end of the second year the population was 89100, what was it at the beginning?

Solution:
Let the population at the beginning = x

net decrease % (10)²/10% = 1%

∴ Population at the end = x × 99% = 89100

x = (89100 × 100)/99

x = 90,000