Home > Quantitative > Arithmetic > Percentages

Percentages

Key Concept

Solved Examples: 75

Practice Questions: 10

Solved Examples

Question 1: Arrange the following in ascending order: 0.4, 2/5, 38%, 45%

Convert all the numbers into percentages.
0.4 = 40%
2/5 = 40%
38% = 38%
45% = 45%
Therefore,
38% < 40% = 40% < 45%
Hence,
38%, 0.4, 2/5, 45%

Percentage
= 0.875 × 100
= 87.5%

Question 2: Which of the following is the greatest? 7/10, 68%, 0.69, 11/16

Convert each quantity into a percentage.
7/10 = 70%
68% = 68%
0.69 = 69%
11/16
= (11/16) × 100
= 68.75%
Comparing the values,
70% > 69% > 68.75% > 68%
Therefore, the greatest quantity is 7/10.

Question 3: Find 12.5% of 736.

12.5%
= 1/8
Therefore,
12.5% of 736
= (1/8) × 736
= 92

Shortcut method:
0.125 x 736 = 92

Question 4: A student scored 432 marks out of 540. What percentage of the total marks did the student score?

Percentage scored
= (432/540) × 100
= (4/5) × 100
= 80%

Question 5: The population of a town is 48,000. If 37.5% of the population is below the age of 18 years, how many people are below the age of 18 years?

37.5%
= 3/8
Therefore,
Number of people below 18 years
= (3/8) × 48,000
= 3 × 6,000
= 18,000

Shortcut method:
0.375 x 48000 = 18000

Question 6: If 25% of a number is 45, find the number.

Let the number be x.
25% of x = 45
(25/100) × x = 45
x = 45 × (100/25)
x = 45 × 4
x = 180

Shortcut method:
45/0.25 = 180

Question 7: A student secured 216 marks, which is 72% of the maximum marks. Find the maximum marks.

Maximum marks
= 216 × (100/72)
= 216 × (25/18)
= 12 × 25
= 300

Shortcut method:
216/0.72 = 21600/72 = 300

Question 8: The monthly salary of an employee after receiving 80% of the salary is ¤48,000. Find the full monthly salary.

et the full monthly salary be ¤S.
80% of S = 48,000
(80 ÷ 100) × S = 48,000
S = 48,000 × (100 ÷ 80)
S = 48,000 × (5 ÷ 4)
S = ¤60,000
∴ the full monthly salary is ¤60,000.

Shortcut method:
48000/0.80 = 60,000

Question 9: A shirt costs ¤800. Its price is increased by 15%. Find the new price.

Increase
= 15% of ¤800
= (15/100) × 800
= ¤120
New price
= ¤800 + ¤120
= ¤920

Shortcut method:
800 x 1.15 = 920

Question 10: The marked price of a bicycle is ¤16,000. During a sale, its price is reduced by 18%. Find the sale price.

Reduction
= (18/100) × 16,000
= ¤2,880
Sale price
= 16,000 − 2,880
= ¤13,120

Shortcut method:
16000 x 0.82 = 13,120

Question 11: A number is increased by 20% and then increased again by 10%. If the original number is 500, find the final value.

First increase
= 20% of 500
= 100
New value
= 500 + 100
= 600
Second increase
= 10% of 600
= 60
Final value
= 600 + 60
= 660

Shortcut method:
500 x 1.20 x 1.10 = 660

Question 12: The price of a laptop is increased by 25% and then reduced by 20%. If its original price is ¤40,000, find the final price.

Price after a 25% increase
= 40,000 × (125/100)
= ¤50,000
Price after a 20% decrease
= 50,000 × (80/100)
= ¤40,000

Shortcut method:
40,000 x 1.25 x 0.80 = 40,000

Question 13: The population of a town increases by 10% in the first year and by 20% in the second year. If the initial population is 50,000, find the population after two years.

Population after the first year
= 50,000 × (110/100)
= 55,000
Population after the second year
= 55,000 × (120/100)
= 66,000

Shortcut method:
50,000 x 1.10 x 1.20 = 66,000

Question 14: A student’s marks increase from 360 to 432 over one year. Find the percentage increase.

Increase in marks
= 432 − 360
= 72
Percentage increase
= (72/360) × 100
= (1/5) × 100
= 20%

Shortcut method:
Increase = 432 – 360 = 72
Increase % = (72/360) x 100 = 20%

Question 15: A shop sold 480 notebooks on Monday and 600 notebooks on Tuesday. By what percentage were the sales on Tuesday greater than those on Monday?

Increase in sales
= 600 − 480
= 120
Percentage increase
= (120/480) × 100
= (1/4) × 100
= 25%

Question 16: The price of Product A is ¤1,800 and the price of Product B is ¤1,500. By what percentage is Product B cheaper than Product A?

Difference in price
= 1,800 − 1,500
= ¤300
Percentage by which Product B is cheaper
= (300/1,800) × 100
= (1/6) × 100
= 16⅔%

Shortcut method:
(300 / 1,800) x 100 = 16.67%

Question 17: The price of an article is increased by 25% and then reduced by 20%. If the original price was ¤3,200, what is the final price?

Price after a 25% increase
= 3,200 × (125/100)
= ¤4,000
Price after a 20% decrease
= 4,000 × (80/100)
= ¤3,200

Shortcut method:
3,200 x 1.25 x 0.80 = 3,200

Question 18: The population of a town increased by 20% in one year and decreased by 10% in the following year. If the original population was 75,000, find the population after two years.

Population after the first year
= 75,000 × (120/100)
= 90,000
Population after the second year
= 90,000 × (90/100)
= 81,000

Shortcut:
75,000 x 1.20 x 0.90 = 81,000

Question 19: The value of a machine depreciates by 10% every year. If its present value is ¤2,40,000, what will be its value after two years?

Value after the first year
= 2,40,000 × (90/100)
= ¤2,16,000
Value after the second year
= 2,16,000 × (90/100)
= ¤1,94,400

Shortcut method:
2,40,000 x 0.90 x 0.90 = 1,94,400

Question 20: 30% of a number is 12 less than 45% of the same number. Find the number.

Difference between the two percentages
= 45% − 30%
= 15%
Therefore, 15% of the number = 12
Number
= 12 × (100/15)
= 80

Question 21: The difference between two numbers A and B (A > B) is 180. If 20% of A is equal to 30% of B, find the two numbers.

20% of A = 30% of B
20A = 30B
2A = 3B
A : B = 3 : 2
Difference
= 1 part
= 180
Therefore,
B = 2 × 180
= 360
A = 3 × 180
= 540
Answer: A = 540, B = 360

Question 22: At a marathon event, 80% of the registered participants reported for the race. Of those, 5% were disqualified during verification. The winner completed the race ahead of 1,824 participants, which represented 80% of the valid participants. Find the total number of registered participants.

80% of valid participants
= 1,824
Valid participants
= 1,824 × (100/80)
= 2,280
These are 95% of those who reported.
Participants who reported
= 2,280 × (100/95)
= 2,400
These are 80% of the registered participants.
Registered participants
= 2,400 × (100/80)
= 3,000

Question 23: A smartphone battery drains at two different rates. Running in High Performance mode for one hour consumes 45% of the total battery capacity, which is 600 mAh more than the battery consumed by running in Eco mode for one hour, where 30% of the battery capacity is consumed. Find the total battery capacity of the smartphone.

Difference in battery consumption
= 45% − 30%
= 15%
15% of the battery capacity
= 600 mAh
Total battery capacity
= 600 × (100/15)
= 4,000 mAh

Question 24: The price difference between a premium leather jacket and a denim jacket is ¤3,500. If 15% of the price of the leather jacket is equal to 40% of the price of the denim jacket, find the price of each jacket.

15% of Leather
= 40% of Denim
15L = 40D
3L = 8D
L : D = 8 : 3
Difference
= 5 parts
= ¤3,500
1 part
= ¤700
Leather jacket
= 8 × 700
= ¤5,600
Denim jacket
= 3 × 700
= ¤2,100
Answer: Leather Jacket = ¤5,600, Denim Jacket = ¤2,100

Question 25: In a local election, Candidate X secured 54% of the total votes, while Candidate Y secured 38% of the total votes. There were no invalid votes. If Candidate X won by a margin of 4,800 votes, find the total number of votes cast.

Difference in percentage
= 54% − 38%
= 16%
16% of total votes
= 4,800
Total votes
= 4,800 × (100/16)
= 30,000

Question 26: An investor divides a certain amount between two investment schemes, P and Q. The investment in Scheme P exceeds the investment in Scheme Q by $12,000. If 25% of the investment in Scheme P is equal to 55% of the investment in Scheme Q, find the amount invested in each scheme.

25% of P
= 55% of Q
25P = 55Q
5P = 11Q
P : Q = 11 : 5
Difference
= 6 parts
= $12,000
1 part
= $2,000
Investment in P
= 11 × 2,000
= $22,000
Investment in Q
= 5 × 2,000
= $10,000
Answer: Scheme P = $22,000, Scheme Q = $10,000

Question 27: Tank A has a capacity that is 240 litres greater than Tank B. When Tank A is 12% full, it contains exactly the same quantity of water as Tank B when it is 20% full. Find the capacity of each tank.

12% of Tank A
= 20% of Tank B
12A = 20B
3A = 5B
A : B = 5 : 3
Difference
= 2 parts
= 240 litres
1 part
= 120 litres
Capacity of Tank A
= 5 × 120
= 600 litres
Capacity of Tank B
= 3 × 120
= 360 litres
Answer: Tank A = 600 litres, Tank B = 360 litres

Question 28: The price of sugar rises by 25%. By what percentage must a household reduce its consumption of sugar so that its total expenditure remains unchanged?

Say original price of sugar = 100
Say sugar consumption = A units
Original expenditure = 100A
New rate of sugar per unit = 100 + (25/100)x100 = 125
New sugar consumption = B unit
New expenditure = 125B
Since expenditure remains same
125B = 100A
or, B = (100/125)A
So decrease = A – (100/125)A = A/5
Percentage decrease = [(A/5)/A]x100 = 20%

Question 29: A student needs 40% marks to pass an examination. If they secure 175 marks and fail the exam by exactly 25 marks, find the maximum possible marks for the examination.

Secured marks = 175
Failed by = 25 marks
Pass marks = 175 + 25 = 200
Since 40% is the pass marks
so, 40% of the total marks is equal to 200
1% of the total marks is equal to 200/40
Therefore
100% of the total marks is (200/40)x100 = 500

Question 30: In an examination, Student A secures 32% marks and fails by 18 marks. In the same examination, Student B secures 41% marks, which is 9 marks more than the minimum passing score. Find the passing percentage for the examination.

The percentage gap between 2 students = 41% – 32% = 9%
The actual marks gap between the two = 9 – (-18) = 27
Now 9% of the total marks is equal to 27
So, 1% of the total marks is equal to (27/9) = 3
100% of the total marks is equal to (27/9) x 100 = 300
32% of the total marks is equal to (27/9) x 32 = 96
Therefore pass mark = 96 + 18 = 114
Pass percentage = (96/300) x 100 = 38%

Alternately
Since 3 marks is equivalent to 1%
So 18 marks is equal to (1/3) x 100 = 6%
Therefore pass percentage
= 32% + 18 marks
= 32% + 6%
38%

Question 31: Two friends, X and Y, took an aptitude test. Friend X scored 28% marks and failed by 20 marks. Friend Y scored 35% marks and still failed by 6 marks. Find the minimum passing marks required for the aptitude test.

The percentage gap between 2 students = 35% – 28% = 7%
The actual marks gap between the two = 20 – 6 = 14
Now 7% of the total marks is equal to 14
So, 1% of the total marks is equal to (14/7)
So 35% of the total marks is equal to (14/7) x 35 = 70
So pass marks = 70 + 6 = 76

Question 32: The population of a city increases by 10% in the first year and decreases by 10% in the second year. If the current population at the end of the second year is 99,000, find the initial population of the city two years ago.

Original population = 100
After 1 year, population = 100 x 10% of 100 = 110
After 2 years, population
= 110 – 10% of 110
= 110 – 11
= 99%
So 99% of original population is equal to 99,000
Therefore original population
= (99,000/99) x 100
= 100,000

Question 33: A manufacturing plant upgrades its equipment machinery in two stages. The production output increases by 25% after the first upgrade and expands by another 20% after the second upgrade. If the final optimized output is 4,500 units per week, find the baseline output before any upgrades were made.

Let the baseline output be A units per week.
After the first upgrade, the output increases by 25%
Output = A × 1.25
After the second upgrade, it increases by another 20%
Final output = A × 1.25 × 1.20
Given A × 1.25 × 1.20 = 4,500
Therefore, A × 1.5 = 4,500
A = 4,500 ÷ 1.5
A = 3,000
∴ the baseline production output was 3,000 units per week.

Question 34: An employee receives a performance-based salary hike of 15% in the first year, but due to an economic downturn, their salary is reduced by 10% in the following year. If their final monthly salary after these two adjustments is ¤46,575, find their initial monthly salary before the changes.

Let the initial monthly salary be ¤A.
After a 15% increase, Salary = A × 1.15
After a 10% decrease, Final salary = A × 1.15 × 0.90
Given:
A × 1.15 × 0.90 = 46,575
A × 1.035 = 46,575
∴ A = 46,575 ÷ 1.035
A = 45,000
∴ the initial monthly salary was ¤45,000.

Question 35: The length of a rectangle is increased by 20% and its width is decreased by 15%. Find the net percentage change in the area of the rectangle.

Let the original length be L and the original width be W.
Original area = L × W
After a 20% increase, the new length becomes L × 1.20
After a 15% decrease, the new width becomes W × 0.85
Therefore,
New area = L × 1.20 × W × 0.85
New area = L × W × 1.02
Thus, the new area is 102% of the original area.
Net percentage change 102% − 100% = 2%
∴ the area increases by 2%.

Question 36: The base of a triangle is increased by 30% and its height is reduced by 20%. Find the percentage change in the total area of the triangle.

Let the original base be b and the original height be h.
Original area = ½ × b × h
After a 30% increase, the new base = b × 1.30
After a 20% decrease, the new height = h × 0.80
Therefore,
New area = ½ × b × 1.30 × h × 0.80
New area = ½ × b × h × 1.04
Thus, the new area is 104% of the original area.
Net percentage change 104% − 100% = 4%
∴ the area increases by 4%.

Question 37: The radius of a circle is increased by 10%. Find the net percentage increase in the area of the circle.

Let the original radius be r.
Original area = πr²
After a 10% increase, the new radius = r × 1.10
Therefore, the new area is
New area = π(1.10r)²
= πr² × (1.10)²
= πr² × 1.21
Thus, the new area is 121% of the original area.
Net percentage increase 121% − 100% = 21%
∴ the area increases by 21%.

Question 38: The length of a rectangular storage box is increased by 20%, its width is increased by 10%, but its vertical height is reduced by 15%. Find the net percentage change in the total volume of the storage box.

Let the original length, width and height be l, w and h respectively.
Original volume = l × w × h
After the changes
New length = 1.20l
New width = 1.10w
New height = 0.85h
∴ New volume = 1.20l × 1.10w × 0.85h
= l × w × h × 1.20 × 1.10 × 0.85
= l × w × h × 1.122
Thus, the new volume is 112.2% of the original volume.
Net percentage change 112.2% − 100% = 12.2%
∴ the volume increases by 12.2%.

Question 39: A person’s monthly income is ¤60,000. Their expenditure is 75% of their income. If their income increases by 20% while their expenditure increases by 10%, find the percentage increase in their monthly savings.

Original income = ¤60,000
Original expenditure
= 75% of ¤60,000
= ¤45,000
Original savings = ¤60,000 − ¤45,000
= ¤15,000
New income = ¤60,000 × 1.20
= ¤72,000
New expenditure = ¤45,000 × 1.10
= ¤49,500
New savings = ¤72,000 − ¤49,500
= ¤22,500
Increase in savings = ¤22,500 − ¤15,000
= ¤7,500
Percentage increase in savings
= (7,500 ÷ 15,000) × 100
= 50%
∴ the monthly savings increase by 50%.

Question 40: A person’s income increases by 25%, while their expenditure increases by 20%. Their original income was ¤40,000, and their original savings were ¤8,000. Find their new monthly savings.

Original expenditure = ¤40,000 − ¤8,000
= ¤32,000
New income = ¤40,000 × 1.25
= ¤50,000
New expenditure = ¤32,000 × 1.20
= ¤38,400
New savings = ¤50,000 − ¤38,400
= ¤11,600
∴ the new monthly savings are ¤11,600.

Question 41: A person’s monthly income is ¤50,000, and their monthly savings are ¤8,000. If their income increases by 20% and their expenditure increases by 15%, find the percentage increase in their savings.

Original expenditure = ¤50,000 − ¤8,000
= ¤42,000
New income = ¤50,000 × 1.20
= ¤60,000
New expenditure = ¤42,000 × 1.15
= ¤48,300
New savings = ¤60,000 − ¤48,300
= ¤11,700
Increase in savings = ¤11,700 − ¤8,000
= ¤3,700
Percentage increase = (3,700 ÷ 8,000) × 100
= 46.25%
∴ the savings increase by 46.25%.

Question 42: A person’s income is 25% more than their expenditure. If their monthly savings are ¤9,000, find their monthly income and expenditure.

Let the monthly expenditure be ¤E.
Income is 25% more than expenditure
Income = 1.25E
Savings = Income − Expenditure
= 1.25E − E
= 0.25E
Given 0.25E = 9,000
E = 9,000 ÷ 0.25
E = ¤36,000
Therefore, Income = ¤36,000 × 1.25
= ¤45,000
∴ the monthly expenditure is ¤36,000 and the monthly income is ¤45,000.

Question 43: A person’s income increases by 20%, while their expenditure increases by 10%. As a result, their savings increase by 50%. If their original monthly savings were ¤12,000, find their original monthly income.

Original savings = ¤12,000
New savings = ¤12,000 × 1.50
= ¤18,000
Therefore, the increase in savings is
= ¤18,000 − ¤12,000
= ¤6,000
Let the original monthly income be ¤M.
Then the original monthly expenditure is
= ¤(M − 12,000)
After the changes new income
= ¤M × 1.20
New expenditure = ¤(M − 12,000) × 1.10
Therefore,
1.20M − 1.10(M − 12,000) = 18,000
1.20M − 1.10M + 13,200 = 18,000
0.10M = 4,800
M = 48,000
∴ the original monthly income was ¤48,000.

Question 44: In a school, 60% of the students are girls. Of the girls, 25% participate in a science competition. If 90 girls participate in the competition, find the total number of students in the school.

Let the total number of students be N.
Girls = 60% of N
Participants = 25% of 60% of N
Therefore,
(25 ÷ 100) × (60 ÷ 100) × N = 90
(15 ÷ 100) × N = 90
N = 90 × (100 ÷ 15)
N = 600
∴ the school has 600 students.

Question 45: In a survey, 80% of the people contacted responded to the survey. Of those who responded, 75% supported a particular proposal. If 1,800 people supported the proposal, find the total number of people contacted.

Let the total number of people contacted be N.
People who responded = 80% of N
People who supported the proposal
= 75% of 80% of N
Therefore,
(75 ÷ 100) × (80 ÷ 100) × N = 1,800
0.60N = 1,800
N = 1,800 ÷ 0.60
N = 3,000
∴ 3,000 people were contacted.

Question 46: The number of members of a club increases by 20% in the first year and by 25% in the second year. After these two increases, the club has 9,000 members. Find the original number of members.

Let the original number of members be N.
After the first increase
Members = N × 1.20
After the second increase final members = N × 1.20 × 1.25
Given:
N × 1.20 × 1.25 = 9,000
N × 1.50 = 9,000
N = 9,000 ÷ 1.50
N = 6,000
∴ the original number of members was 6,000.

Question 47: The price of an electronic device is increased by 20% and then reduced by 25%. The final price is ¤1,800 less than the original price. Find the original price.

Let the original price be ¤P.
After a 20% increase
Price = P × 1.20
After a 25% decrease
Final price = P × 1.20 × 0.75
Final price = P × 0.90
Thus, the final price is 90% of the original price.
Therefore, the decrease is
100% − 90% = 10%
Given that the decrease is ¤1,800
10% of P = ¤1,800
P = 1,800 × (100 ÷ 10)
P = ¤18,000
∴ the original price was ¤18,000.

Question 48: The actual length of a metal rod is 80 cm, but it is measured as 76 cm. Find the percentage error in the measurement.

Error in measurement = 80 − 76 = 4 cm
Percentage error = (4 ÷ 80) × 100 = 5%
∴ the percentage error is 5%.

Question 49: The actual weight of a package is 2.5 kg, but it is recorded as 2.65 kg. Find the percentage error in the recorded weight.

Error = 2.65 − 2.5 = 0.15 kg
Percentage error
= (0.15 ÷ 2.5) × 100 = 6%
∴ the percentage error is 6%.

Question 50: The radius of a circular plate is measured as 21 cm instead of its actual value of 20 cm. By what percentage is the calculated area greater than the actual area?

Actual area = π × 20² = 400π cm²
Calculated area = π × 21² = 441π cm²
Increase in calculated area
= 441π − 400π
= 41π cm²
Percentage increase
= (41π ÷ 400π) × 100
= 10.25%
∴ the calculated area is 10.25% greater than the actual area.

Question 51: The length and width of a rectangular sheet are each measured 5% greater than their actual values. Find the percentage error in the calculated area of the sheet.

Let the actual length and width be L and W.
Actual area = L × W
Since each measurement is 5% greater
Measured length = 1.05L
Measured width = 1.05W
Therefore, calculated area = 1.05L × 1.05W = 1.1025LW
Thus, the calculated area is 110.25% of the actual area.
Percentage error = 110.25% − 100% = 10.25%
∴ the calculated area is 10.25% greater than the actual area.

Question 52: The side of a square is measured as 4% greater than its actual value. Find the percentage error in the calculated perimeter of the square.

Let the actual side of the square be s.
Actual perimeter = 4s
Measured side = 1.04s
Calculated perimeter
= 4 × 1.04s
= 4.16s
Therefore, the calculated perimeter is
(4.16s ÷ 4s) × 100 = 104%
Thus, the percentage error is 104% − 100% = 4%
∴ the calculated perimeter is 4% greater than the actual perimeter.

Question 53: The radius of a spherical object is measured 10% less than its actual value. By what percentage is the calculated volume less than the actual volume?

Let the actual radius be r.
Actual volume = (4/3)πr³
Measured radius = 0.90r
Calculated volume
= (4/3)π(0.90r)³
= (4/3)πr³ × 0.729
Thus, the calculated volume is 72.9% of the actual volume.
Percentage decrease = 100% − 72.9% 27.1%
∴ the calculated volume is 27.1% less than the actual volume.

Question 54: The length of a rectangular field is measured 8% greater than its actual length, while its width is measured 5% less than its actual width. Find the percentage error in the calculated area of the field.

Let the actual length and width be L and W.
Actual area = L × W
Measured length = 1.08L
Measured width = 0.95W
Therefore, calculated area
= 1.08L × 0.95W
= 1.026LW
Thus, the calculated area is 102.6% of the actual area.
Percentage error = 102.6% − 100% = 2.6%
∴ the calculated area is 2.6% greater than the actual area.

Question 55: A quantity is increased by 20% and then decreased by 25%. By what percentage must the resulting quantity be increased to return to the original value?

Let the original quantity be N.
After a 20% increase = 1.20N
After a 25% decrease = 1.20N × 0.75 = 0.90N
Thus, the resulting quantity is 90% of the original quantity.
To return to the original quantity, the increase required is
= (100 − 90)% = 10% of N
But the increase is applied to 90% of N.
Therefore, Required percentage increase
= (10 ÷ 90) × 100
= 11⅑%
∴ the resulting quantity must be increased by 11⅑%.

Question 56: A student scored 20% more marks than another student. The first student’s score was 30% more than the second student’s score, when measured relative to the second student’s score. If the difference between their scores is 24 marks, find their scores.

First student’s score = 130% of second student’s score
Let the second student’s score be M.
Then the first student’s score is = 1.30M
Difference = 1.30M − M = 0.30M
Given
0.30M = 24
M = 24 ÷ 0.30
M = 80
First student’s score = 1.30 × 80 = 104
∴ the two scores are 104 marks and 80 marks.

Question 57: The price of an article is increased by 10%. The quantity that can be purchased for a fixed amount of money therefore decreases by what percentage?

Let the original price per unit be ¤P.
After a 10% increase new price = ¤1.10P
For the same amount of money, the quantity purchased is inversely proportional to the price.
∴ New quantity ÷ Original quantity = 1 ÷ 1.10
= 10 ÷ 11
Thus, the new quantity is (10 ÷ 11) × 100 = 90.909…% of the original quantity.
Percentage decrease:
= 100% − 90.909…%
= 9.09…%
= 9¹⁄₁₁%
∴ the quantity purchased decreases by 9¹⁄₁₁%.

Question 58: A number is increased by p% and then decreased by p%. The final value is 9% less than the original value. Find p.

Let the original number be N.
After a p% increase
New value = N × (1 + p ÷ 100)
After a p% decrease
Final value = N × (1 + p ÷ 100) × (1 − p ÷ 100)
Using (1 + a)(1 − a) = 1 − a²
we get final value = N × [1 − (p ÷ 100)²]
The final value is 9% less than the original:
Final value = 91% of N
Therefore, 1 − (p ÷ 100)² = 0.91
or, (p ÷ 100)² = 0.09
or, p ÷ 100 = 0.3
or, p = 30
∴ p = 30%.

Question 59: In a class, 40% of the students are boys. If 25% of the boys and 20% of the girls are absent on a particular day, what percentage of the entire class is present?

Let the total number of students be 100.
Number of boys = 40
Number of girls = 100 − 40 = 60
Boys absent = 25% of 40 = 10
Girls absent = 20% of 60 = 12
Total absent = 10 + 12 = 22
Therefore, students present = 100 − 22 = 78
∴ 78% of the class is present.

Question 60: A number is 40% less than a second number. By what percentage is the second number greater than the first number?

Let the second number be 100.
The first number is 40% less = 100 − 40 = 60
Difference between the numbers = 100 − 60 = 40
The percentage by which the second number is greater than the first is calculated using the first number as the base.
Therefore, percentage increase
= (40 ÷ 60) × 100
= 66⅔%
∴ the second number is 66⅔% greater than the first number.

Question 61: A container is 60% full. After adding 24 litres of liquid, it becomes 75% full. Find the capacity of the container.

Increase in the percentage filled = 75% − 60% = 15%
Therefore, 15% of the container’s capacity = 24 litres
Let the capacity be ¤C — since this is a volume, we should not use the currency symbol. Let the capacity be C litres.
(15 ÷ 100) × C = 24
C = 24 × (100 ÷ 15)
C = 160 litres
∴ the capacity of the container is 160 litres.

Question 62: In an election, candidate A receives 20% more votes than candidate B, while candidate B receives 25% fewer votes than candidate C. If candidate C receives 12,000 votes, find the difference between the numbers of votes received by candidates A and C.

Candidate B receives 25% fewer votes than C.
Therefore, B = 75% of C
B = 0.75 × 12,000
B = 9,000
Candidate A receives 20% more votes than B.
Therefore, A = 120% of B
A = 1.20 × 9,000
A = 10,800
Difference between C and A
= 12,000 − 10,800
= 1,200
∴ the difference is 1,200 votes.

Question 63: A 400-gram sugar solution contains exactly 30% sugar. How many grams of pure sugar must be added to this mixture to turn it into a 40% sugar solution?

Sugar initially present = 30% of 400
= (30 ÷ 100) × 400
= 120 grams
Let the amount of pure sugar added be S grams.
After adding S grams of pure sugar
Total sugar = 120 + S grams
Total solution = 400 + S grams
The final solution contains 40% sugar.
Therefore, (120 + S) ÷ (400 + S) = 40 ÷ 100
100(120 + S) = 40(400 + S)
12,000 + 100S = 16,000 + 40S
60S = 4,000
S = 66⅔
∴ 66⅔ grams of pure sugar must be added.

Question 64: A laboratory container holds 12 litres of a salt solution with a concentration of 25%. If 2 litres of pure water are added to dilute the mixture, what is the new percentage concentration of salt in the solution?

Initial amount of salt = 25% of 12 litres
= (25 ÷ 100) × 12
= 3 litres
Since only pure water is added, the amount of salt remains 3 litres.
New total volume = 12 + 2 = 14 litres
New percentage concentration = (3 ÷ 14) × 100
= 21.428…%
= 21⁵⁄₇%
∴ the new concentration of salt is 21⁵⁄₇%.

Question 65: A 60-litre solution of milk and water contains 10% water. How many litres of pure water must be added to this mixture to make it contain 20% water in the final solution?

Initial amount of water = 10% of 60 litres
= (10 ÷ 100) × 60
= 6 litres
Let the amount of pure water added be W litres.
After adding W litres of water, amount of water
= 6 + W litres
Total solution = 60 + W litres
The final solution contains 20% water.
Therefore, (6 + W) ÷ (60 + W) = 20 ÷ 100
or, 100(6 + W) = 20(60 + W)
or, 600 + 100W = 1,200 + 20W
or, 80W = 600
or, W = 7.5
∴ 7.5 litres of pure water must be added.

Question 66: In a mathematics competition, 70% of the participants solved problem A correctly, 75% solved problem B correctly, and 80% solved problem C correctly. What is the absolute minimum percentage of participants who must have solved all three problems correctly?

The percentages who did not solve each problem are
Problem A: 100% − 70% = 30%
Problem B: 100% − 75% = 25%
Problem C: 100% − 80% = 20%
Total percentage of failures
30% + 25% + 20% = 75%
Therefore, at most 75% of the participants can have failed at least one of the three problems.
Hence, at least 100% − 75% = 25% of the participants must have solved all three problems correctly.
∴ the absolute minimum percentage is 25%.

Question 67: If the numerical length of a rectangle is increased by 40%, by what exact percentage must its width be decreased so that the overall perimeter of the rectangle remains completely unchanged, given that the original length was exactly twice the original width?

Let the original width be W.
Therefore, the original length is L = 2W
Original perimeter:
= 2(L + W)
= 2(2W + W)
= 6W
After a 40% increase, the new length is = 1.40 × 2W = 2.80W
Let the percentage decrease in width be p%.
New width = W × (1 − p ÷ 100)
Since the perimeter remains unchanged
2(2.80W + W(1 − p ÷ 100)) = 6W
Divide both sides by 2W
2.80 + 1 − p ÷ 100 = 3
3.80 − p ÷ 100 = 3
p ÷ 100 = 0.80
p = 80
∴ the width must be decreased by 80%.

Question 68: In a sequence of three numbers, the first number is 20% less than the second number, and the third number is 50% more than the sum of the first two numbers. By what percentage is the third number greater than the second number?

Let the second number be N.
The first number is 20% less than the second
So, 1st number = 80% of N = 0.80N
The sum of the 1st two numbers= 0.80N + N
= 1.80N
The third number is 50% more than this sum
So 3rd number = 1.50 × 1.80N
= 2.70N
Therefore, the third number is 2.70N, while the second number is N.
Increase from the 2nd number to the 3rd = 2.70N − N
= 1.70N
Percentage increase = (1.70N ÷ N) × 100
= 170%
∴ the third number is 170% greater than the second number.

Question 69: An auditorium has a certain number of rows. Due to renovation, the number of rows is reduced by 20%, but the number of seats per row is increased by 20%. If the total seating capacity drops by exactly 16 seats, find the original total seating capacity of the auditorium.

Let the original total seating capacity be S.
A 20% reduction in the number of rows means the number of rows becomes
80% of the original
A 20% increase in the number of seats per row means the seats per row become
120% of the original
Therefore, the new seating capacity is S × 0.80 × 1.20
= 0.96S
Thus, the new capacity is 96% of the original capacity.
Therefore, the decrease in capacity is 100% − 96% = 4%
Given that this decrease is 16 seats
4% of S = 16
(4 ÷ 100) × S = 16
S = 16 × (100 ÷ 4)
S = 400
∴ the original total seating capacity was 400 seats.

Question 70: In a test, 80% of the students passed Mathematics, 70% passed Science, and 60% passed both subjects. What percentage of the students passed exactly one of the two subjects?

Students who passed Mathematics only = 80% − 60% = 20%
Students who passed Science only = 70% − 60% = 10%
Therefore, students who passed exactly one subject
= 20% + 10%
= 30%
∴ 30% of the students passed exactly one subject.

What are Percentages?

A percentage is a way of expressing a number as a fraction of 100. The word percent means “per hundred”, where “per” means “for every” and “cent” means “hundred”.
For example, 25% means 25 out of every 100, which can also be written as:
25% = 25/100 = 1/4 = 0.25

Percentages provide a convenient way to compare quantities and express proportions, irrespective of their actual values. They are widely used in mathematics as well as in everyday life to represent marks, discounts, profits, losses, interest rates, taxes, population growth, election results, and statistical data.
For example, if a student scores 72 marks out of 90, the percentage is calculated as:
(72/90) × 100 = 80%
This means the student has scored 80 marks out of every 100.

Percentages can also be converted into fractions and decimals, and vice versa. Understanding these conversions is essential because many aptitude problems require switching between these forms to simplify calculations.

A sound understanding of percentages forms the foundation for several important arithmetic topics, including Profit & Loss, Discount, Simple Interest, Compound Interest, Ratio & Proportion, Data Interpretation, and Successive Percentage Change. Mastering percentages, therefore, makes many advanced quantitative aptitude problems much easier to solve.

Meaning of a Percentage

The symbol % means “per hundred” or “out of every 100.”
Examples:
1% = 1/100
25% = 25/100 = 1/4
100% = 1
200% = 2

Percentage, Fraction and Decimal

A percentage can always be written as a fraction or a decimal.
Example:
40% = 40/100 = 2/5 = 0.4
12.5% = 12.5/100 = 1/8 = 0.125
Being able to convert quickly between these forms greatly simplifies calculations.

Percentages Greater than 100%

A percentage can be greater than 100.
Examples:
100% means the whole quantity.
150% means one and a half times the quantity.
250% means two and a half times the quantity.
Such percentages frequently appear in profit, growth and comparison problems.

Finding a Percentage of a Number

To find x% of a number,
Convert the percentage into a fraction or decimal.
Multiply it by the given number.
Example:
15% of 240
= (15/100) × 240
= 36

Percentage Increase and Decrease

If a quantity increases by p%,
New Value
= Original Value × (100 + p)/100
If a quantity decreases by p%,
New Value
= Original Value × (100 − p)/100
These relationships are used extensively in later chapters such as Profit & Loss, Discount and Compound Interest.

Competitive Exam Shortcuts

Use Multiplication Factors for Percentage Increase and Decrease

Percentage Change	MF for Increase	MF for Decrease
5%	× 1.05	× 0.95
10%	× 1.10	× 0.90
15%	× 1.15	× 0.85
20%	× 1.20	× 0.80
25%	× 1.25	× 0.75
30%	× 1.30	× 0.70
40%	× 1.40	× 0.60
50%	× 1.50	× 0.50

Memorise Common Percentage–Fraction Equivalents

Percentage	10%	12.5%	20%	25%	33⅓%	37.5%	50%	62.5%	66⅔%	75%	80%	87.5%
Fraction	1/10	1/8	1/5	1/4	1/3	3/8	1/2	5/8	2/3	3/4	4/5	7/8

Numerator Swapping Trick

For any two numbers,
x% of y = y% of x
Choose the easier calculation.
Example
20% of 50
= 50% of 20
= 10

The 10% and 1% Rules

10% of a number is obtained by moving the decimal point one place to the left.
1% of a number is obtained by moving the decimal point two places to the left.
From these,
5% = Half of 10%
15% = 10% + 5%
30% = 3 × 10%
Example
15% of 640
= 10% + 5%
= 64 + 32
= 96

Successive Percentage Change

If a quantity changes by a% and then by b%, the net percentage change is
Net Change = a + b + (ab/100)
Use a negative sign for decreases.
Example
Increase by 20%, then decrease by 10%
= 20 − 10 − (20 × 10)/100
= 8%

Reverse Percentage

If the final value after an increase is known, divide by the multiplication factor.

Final Situation	Divide By
After 20% increase	1.20
After 25% increase	1.25
After 10% decrease	0.90
After 20% decrease	0.80

Example
A price becomes ₹1,500 after a 25% increase.
Original price
= 1,500 ÷ 1.25
= ₹1,200

Percentage Comparison

When comparing two quantities,
Percentage by which A is greater than B
= ((A − B)/B) × 100
Percentage by which A is less than B
= ((B − A)/B) × 100
Always divide by the quantity used as the reference.

Expenditure Remains Constant

If the price of an item changes and the total expenditure is to remain the same:
If the price increases by R%
Required reduction in consumption
= R/(100 + R) × 100%
If the price decreases by R%
Required increase in consumption
= R/(100 − R) × 100%

Think Before Calculating

Instead of directly finding awkward percentages, rewrite them using easier values.
Examples:
19% = 20% − 1%
47% = 50% − 3%
98% = 100% − 2%
This approach often reduces calculations significantly.

Summary of Percentages

Concept	Formula / Rule	Key Point
Meaning of Percentage	x% = x/100	A percentage represents a value out of every 100.
Percentage of a Number	(Percentage/100) × Number	Convert the percentage into a fraction or decimal before multiplying.
Increase by p%	New Value = Original Value × (100 + p)/100	Used in population growth, salary increments and appreciation.
Decrease by p%	New Value = Original Value × (100 − p)/100	Used in depreciation, discounts and reductions.
More than 100%	Values greater than 100% are valid.	For example, 150% means one and a half times the original quantity.
Fraction to Percentage	Fraction × 100%	Multiply the fraction by 100 to obtain the equivalent percentage.
Decimal to Percentage	Decimal × 100%	Move the decimal point two places to the right.
Percentage to Decimal	Percentage ÷ 100	Move the decimal point two places to the left.

Common Mistakes

Confusing a percentage with a decimal: Remember that 25% = 0.25, not 25. Always divide the percentage by 100 before using it in calculations.
Using the wrong base while comparing percentages: When finding the percentage increase or decrease, always divide the difference by the original quantity, not the new quantity.
Adding successive percentage changes directly: An increase of 20% followed by 10% is not the same as a 30% increase. Each percentage is calculated on the updated value.
Assuming an increase followed by the same percentage decrease results in the original value: A 20% increase followed by a 20% decrease does not restore the original value because the second percentage is calculated on the increased amount.
Converting recurring percentages incorrectly: Values such as 33⅓% and 66⅔% should be recognised as 1/3 and 2/3 respectively for quicker calculations.
Using lengthy decimal calculations instead of simple fractions: Percentages such as 12.5%, 37.5%, 62.5%, and 87.5% are more easily handled as 1/8, 3/8, 5/8, and 7/8.
Forgetting to simplify fractions before multiplying: Simplifying the numerator and denominator first makes calculations faster and reduces the chance of arithmetic errors.
Finding a percentage of the wrong quantity: In word problems, carefully identify the quantity on which the percentage is to be calculated before performing any calculations.
Ignoring the wording of comparison questions: “By what percentage is A greater than B?” and “By what percentage is B less than A?” generally produce different answers because the base quantity is different.
Rounding values too early: Keep calculations exact until the final step. Premature rounding may lead to incorrect answers, especially in multi-step percentage problems.

Practice Questions

Question 1: The price of an article is increased by 18%. If its original price was ₹2,750, find the new price.

Question 2: A machine depreciates by 15% every year. If its present value is ₹96,000, find its value after one year.

Question 3: A number is first increased by 25% and then decreased by 20%. If the original number was 960, find the final value.

Question 4: The population of a city increased by 12% in the first year and by 25% in the second year. If the original population was 1,20,000, find the population after two years.

Question 5: The price of Product A is ₹2,400, while the price of Product B is ₹2,100. By what percentage is Product A more expensive than Product B?

Question 6: The salary of an employee was increased by 20% and later reduced by 20%. If the original salary was ₹75,000, find the final salary.

Question 7: The value of a machine first depreciates by 20% and then appreciates by 25%. If its original value was ₹3,60,000, find the final value and the overall percentage change from the original value.

Question 8: A trader marks an article 30% above its cost price. During a sale, a discount of 20% is offered on the marked price. If the cost price of the article is ₹2,500, find the selling price.

Question 9: A school’s student strength increased by 20% in one year. In the following year, 10% of the students left the school. If the school initially had 2,500 students, find the final number of students.

Question 10: The price of a commodity is increased by 20%. By what percentage must the new price be reduced so that it becomes equal to the original price?