Percentage - Practice questions with conceptual Methods and Tricks

PERCENTAGE

Example 1: What percent decrease in salaries would exactly cancel out the 20% increase?

Solution:  Method 1:  Basic - Let salary = Rs 100
Increment = 20%
   ⇒ New salary = 100 + 20% of 100 = Rs 120
We need to decrease Rs 20 to get original salary.
   ⇒ Percentage decrease = (20/120) x 100 = (50/3) %

 Method 2:  Formula: Percentage decrease = (x/(100 + x)) x 100%
   ⇒ (20/120) x 100  = (50/3)%

Example 2: If the monthly salary of an employee is increased by 8/3%, he gets 72 rupees more. His monthly salary is?

Solution: (8/3)% increment = Rs 72
   ⇒ 1% increment will be = Rs 72 x (3/8)
   ⇒ 100% will be = 72 x 100 x 3/8 = Rs 2700
Salary is Rs 2700

Example 3: A sample of 50 litres of glycerine is found to be adulterated to the extent of 20%. How much pure glycerine should be added to it so as to bring down the percentage of impurity to 5%?

Solution:  Method 1:  Basic: Glycerine in the mixture is 80%.
Glycerine = 80% of 50 = 40 litres

Let glycerine to be added is x litres.
Now glycerine in mixture is 95%.
   ⇒ (40 + x)  =  95% of (50 + x)
   ⇒ x = 150 litres

 Method 2:  Method of Ratios: In this method we work on fixed quantity.
Glycerine : Water = 80% : 20%
After adding glycerine, new ratio = Glycerine : Water = 95% : 5%
Here fixed quantity is water.

Unfixed quantity to be added = Quantity of mixture x (Diff. of percentage of fixed quantity)/Final percentage of fixed quantity
   ⇒ Glycerine to be added = 50 x (20 – 5)/5 = 150 litres

Example 4: 1 litre of water is added to 5 litres of alcohol-water solution containing 40% alcohol strength. The new strength of alcohol in the new solution will be?

Solution:  Method 1:  Basic - Alcohol is 40% in the solution.
Alcohol in original solution = 40 x 5/100 = 2 Litres
Water in original solution = 5 - 2 = 3 Litres
On adding 1 litre water, it becomes 4 litres in solution.
   ⇒ New strength of alcohol = (Quantity of alcohol/Total Quantity of solution) x 100
   ⇒ (2/6) x 100 = 33.3%

 Method 2:  By ratios - Alcohol : Water = 40% : 60%
After adding 1 litre of water, let new ratio = Alcohol : Water = x% : (100 – x)%
Here Fixed quantity is Alcohol.

Unfixed Quantity to be Added: Quantity of mixture x (Difference of percentage of fixed quantity/Final percentage of fixed quantity)

Unfixed quantity added = 1 litre water
   ⇒ 1 = 5 x (40 - x)/x
   ⇒ x = 33.3%

Example 5: If 4 litres of water is evaporated on boiling from 12 litres of salt solution containing 7 percentage salt, the percentage of salt in the remaining solution is?

Solution:  Method 1:  Basic: Percentage of salt = 7%
   ⇒ Quantity of salt in 12 litres of solution = 7 x 12/100 = 0.84 units
   ⇒ Quantity of water = 12 – 0.84 = 11.16 units
After evaporation, solution left = 8 litres
   ⇒ Percentage of salt in the remaining solution = (0.84/8) x 100 = 10.5%

 Method 2:  As salt has not been evaporated from the solution, therefore:
   ⇒ Quantity of salt in original 12 litres of solution = Quantity of salt in remaining 8 litres of solution
   ⇒ 7% of 12 = x% of 8
   ⇒ x = 10.5%

Example 6: One type of liquid contains 20% water and the second type of liquid contains 35% of water. A glass is filled with 10 parts of first liquid and 4 parts of second liquid. The water in the new mixture in the glass is?

Solution:  Method 1:  Basic: Quantity of water in 10 parts of first liquid = 20% of 10 = 2 parts
Quantity of water in 4 parts of second liquid = 35% of 4 = 7/5 parts

Water in new mixture = (Quantity of water in first + Quantity of water in second)/Total quantity of liquid x 100
   ⇒ Water = (2 + 7/5)/(10 + 4) x 100 = 24.28%

 Method 2:  By Alligation:

   ⇒ (35 – x)/(x – 20) = 10/4
   ⇒ x = 24.28%

Example 7: If the price of a commodity is decreased by 20% and its consumption is increased by 20%. What will be the increase or decrease in the expenditure on the commodity?

Solution:  Method 1:  Basic: Let price of commodity = Rs 100
Consumption = 100 units
   ⇒ Initial expenditure = 100 x 100 = Rs 10000
New price decreased by 20% = Rs 80
New consumption increased by 20% = 120 units
   ⇒ Final expenditure = 80 x 120 = Rs 9600
Decrease in expenditure = Rs 10000 – Rs 9600 = Rs 400
   ⇒ Percentage decrease = (400/10000) x 100 = 4%

 Method 2:  Formula: We can use method of successive percentage change:
Net percent change = x% + y% + xy/100
   ⇒ 20% + (-20%) + 20 x (-20)/100 = -4%
Here negative sign means decrease in expenditure.

Example 8: In an examination, a student who gets 20% of the maximum marks fails by 5 marks. Another student who scores 30% of the maximum marks gets 20 marks more than the pass marks. The necessary percentage required for passing is?

Solution:  Method 1:  Basic: Let maximum marks be x.
   ⇒ 20% of x + 5 = Passing marks = 30% of x -20
   ⇒ (30% - 20%) of x = 25
   ⇒ x = 250 marks
Passing marks = 20% of 250 + 5 = 55
   ⇒ Passing percentage = (55/250) x 100 = 22%

 Method 2:  Formula - Maximum marks = 100 x (Difference of their scores)/(Difference of their percentage marks)
   ⇒ Maximum marks = 100 x (20 + 5)/(30 – 20) = 250
   ⇒ Passing marks = 20% of 250 + 5 = 55
   ⇒ Passing percentage = (55/250) x 100 = 22%

Example 9: 72% of the students of a certain class took Biology and 44% took Mathematics. If each student took at least one subject from Biology or Mathematics and 40 took both, then the total number of students in the class is?

Solution:  Method 1:  72% of the students took biology = n(B) = 72%
44% of the student took mathematics = n(M) = 44%
Each student took at least one subject = n(B∪M) = 100%
Students who took both the subjects = n(B∩M)

Formula of set theory: n(B∪M) = n(B) + n(M) - n(B∩M)
   ⇒ 100 = 72 + 44 - n(B∩M)
   ⇒ n(B∩M) = 16%

Students who took both the subjects = 40
   ⇒ 16% = 40
   ⇒ 100% = (40/16) x 100 = 250 students

 Method 2:  Total number of student = 100%

• 72% took biology, it means (100 – 72)% did not took biology, they took only mathematics = 28%

• 44% took mathematics, it means (100 – 44)% did not took maths, they took only biology = 56%

   ⇒ Total percent of students who took both biology and mathematics = 100% – (28 + 56)% = 16%
   ⇒ 16% = 40
   ⇒ 100% = (40/16) x 100 = 250 students

Example 10: The length of a rectangle is increased by 10% and breadth decreased by 10%. Then area of the new rectangle will increase or decrease?

Solution:  Method 1:  Everything is 100% in itself whether length, Breadth or any other.
                                Old                             New
Length                   100%  → +10%  →    110%
Breadth                 100%  → -10%  →     90%
Area = L x B =      10000                          9900
⇒ Percentage decrease in Area = [(10000-9900)/10000] x 100 = 1%

 Method 2:  Direct formula: We can use successive percentage change formula:
   ⇒ Net % change = 10% - 10% - 10 x 10/100% = -1%
Here negative sign means area decreased by 1%.

Example 11: In an election between two candidates, 75% of the voters cast votes, out of which 2% votes were declared invalid. A candidate got 9361 votes which were 75% of the valid votes. The total number of voters enrolled in that election was?

Solution: Let the total number of voters enrolled be x.
   ⇒ Number of votes polled = 75% of x = 3x/4
   ⇒ Number of valid votes = 3x/4 – 2% of 3x/4 = 147x/200
Now 75% of 147x/200 = 9261
   ⇒ x = 16800

Example 12: In a town, the population was 8000, In one year, male population increased by 10% and female population increased by 8% but the total population increased by 9%. The number of males in the town was?

Solution:  Method 1:  Basic: Let the population of males be x

Female population will be (8000 - x)
   ⇒ 110% of x + 108% of (8000 - x) = 109% of 8000
   ⇒ x = 4000
Number of males in the town was 4000.

 Method 2:  By Alligation:

Men : Women = 1:1
Number of Men = ½ x 8000 = 4000

Example 13: A reduction of 10% in the price of sugar enables a housewife to buy 6.2 Kg more for Rs 1116. The reduced price per kg is?

Solution:   Method 1:  Basic: Let the original price of sugar = Rs x/kg
Quantity of sugar purchased at original price = 1116/x kg
Reduced price = 90% of x = 9x/10
Quantity of sugar purchased at reduced price = 1116/(9x/10) kg
Difference of quantity is given as = 6.2 Kg
   ⇒ 1116/(9x/10) - 1116/x = 6.2
   ⇒ x = Rs 20
   ⇒ Reduced price = 9 x 20/10 = Rs 18

 Method 2:  By Formula -
Reduced price = XZ/100y
x = percentage change
z = Price
y = Quantity
   ⇒ Reduced price = [(10 x 1116)/(100 x 6.2)] = Rs 18

 Method 3:  Conceptual - 6.2 Kg more sugar is available after 10% reduction in overall price.
   ⇒ 6.2 kg sugar = Rs 10% of 1116
   ⇒ 1 kg sugar price = [(10 x 1116)/(100 x 6.2)] = Rs 18

Example 14: Fresh fruit contains 68% water and dry fruit contains 20% water. How much dry fruit can be obtained from 100 kgs of fresh fruits?

Solution: Fresh fruits lose water and become dry fruit, It means pulp remains same in both fresh and dry fruit. Let x kg dry fruits can be obtained.
   ⇒ Pulp in fresh fruit = (100 – 68)% = 32%
   ⇒ Pulp in dry fruit = (100 – 20) = 80%
Pulp of fresh fruit = Pulp of dry fruit
   ⇒ 32% of 100 = 80% of x
   ⇒ x = 40 kgs