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Percentage

percentage concept

Concept  | Practice set 1  | Practice set 2

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Percentage Quizzes

Quiz 1  |  Quiz 2  |  Quiz 3  |  Quiz 4  |  Quiz 5  |

 

In this article, we provide you a short and effective summary for Percentages. We cover a list of 15 formulas and short-cuts that you can use for Percentage questions. The following is a list of important formulas for Percentage:

1. Percent implies “for every hundred”.
% is read as percentage and x % is read as x per cent.

2. To calculate p % of y
(p/100) x y = (p x y)/100

p % of y = y % of p

3. To find what percentage of x is y: y/x × 100

4. To calculate percentage change in value
Percentage change = {change/(initial value)} x 100

5. Percentage point change = Difference of two percentage figures

6. Increase N by S % = N( 1+ S/100 )

7. Decrease N by S % = N (1 – S/100)

8. If the value of an item goes up/down by x%, the percentage reduction/increment to be now made to bring it back to the original point is 100x/ (100 + x) %.

9. If A is x% more /less than B, then B is 100x/(100 + x) % less/more than A.

10. If the price of an item goes up/down by x %, then the quantity consumed should be reduced by 100x/ (100 + x)% so that the total expenditure remains the same.

11. Successive Percentage Change
If there are successive percentage increases of a % and b%, the effective percentage increase is:
{(a + b + (ab/100)}%

12. Percentage – Ratio Equivalence:
formulas-and-tricks-11
formulas-and-tricks-21
N is Numerator
D is the Denominator

13. Product Stability Ratio:
A × B = P
If A is increased by a certain percentage, then B is required to be decreased by a certain percentage to keep the product P stable.

Expressing the percentage figures in ratios:

formulas-and-tricks-3

 

Basic Concepts of Percentages

In this lesson, we cover the absolute basics of Percentages. The purpose of this lesson is to help you answer one simple question: What are Percentages?

Basic Definition:
Percent implies “for every hundred” and the sign % is read as percentage and x % is read as x per cent. In other words, a fraction with denominator 100 is called a per cent. For example, 20 % means 20/100 (i.e. 20 parts from 100). This can also be written as 0.2.

Basic Formula:
In order to calculate p % of q, use the formula:
(p/100) x q = (pxq)/100

Also remember: p % of q = q % of p

Examples:
1. 100% of 60 is 60 x (100/100) = 60
2. 50% of 60 is 50/100 × 60 = 30
3. 5% of 60 is 5/100 × 60 = 3

Example: 60 % of a number is 360. What is 99 % of the same number?
Solution: Let the number be n.
Given (60/100) ×n = 360 => n = 600
99 % of 600 = (99/100) × 600 = 594

Example: 50 % of a number is 360. What is 99 % of the same number?
Solution: Let the number be y.
Given (50/100) x q = 360
=> q = 720
99% of 720 = (99/100) x 720 = 712.80

Expressing One Quantity as a Per Cent with respect to the other:

To express a quantity as a per cent with respect to other quantity, the following formula is used:

percentage-basic-concept

Example: What percent is 60 of 240?
Solution: First write the given numbers in the fraction form:
60/240 = ¼
Multiply the numerator and denominator with 25 to make the denominator equal to 100
(1×25)/(4×25) = 25/100
25 percent or 25 per 100 is called as 25%

Sample Question for the Basics of Percentage:
Example:A number exceeds 20% of itself by 40. The number is:
(a) 50
(b) 60
(c) 80
(d) 48

Solution: Let the number be p.
20% of itself means => p x (20/100)
Now, according to the question,
p – 20% of p = 40
=> {p – (20 x p)/100} = 40
=> {p-(p/5)} = 40
⇒ 5p – p = 200
∴ p = 50

Alternate Method:
Obviously, it is clear that difference is 80% i.e. 4/5 of number which is equal to 40
4/5p = 40
p = 40 x 5/4= 50.

Tips & Tricks for Percentages:
Basic Tip-1:  If the new value of something is n times the previous given value, then the percentage increase is (n-1)  100%.

Derivation:
Let us consider two values p and q.
Let q be and original value and p be the new value.
According to conditions p= nq
We need to calculate the percentage increase.
You can either use direct formula= {(new value – old value)/old value} x 10
This value becomes= {(p – q)/q} x 100
{(nq – q)/q} x 100
=> (n-1) x 100%

Example: If X= 5.35 Y, then find the percentage increase when the value of something is from Y to X.
Solution:
Use the formula: (n-1)100%
Percentage increase from
Y to X = (5.35 -1)  100= 435%

Basic Tip-2:
When a quantity N is increased by K %, then the:
New quantity = N (1+ K/100 )
Examples:
Increase 150 by 20%= 150 {1+(20/100)} = 150 1.2= 180
Increase 300 by 30%= 300 {1+(30/100)}=  300 1.3= 390
Increase 250 by 27% = 250 {1+(27/100)} = 250 1.27 =317.5

Example: What is the new value when 265 is increased by 15%?
Solution: New quantity = N (1+ K/100)
= 265{1+(15/100)}
New quantity = 1.15 265= 304.75

Basic Tip 3:
When a quantity N is decreased by K %, then the:
New quantity =N (1 – K/100)

Examples:
Decrease 120 by 20%= 120 {1-(20/100)} = 120  0.8= 96
Decrease 150 by 40%=150 {1-(40/100)} = 150  0.6= 90
Decrease 340 by 27%= 340 {1-(27/100)}= 340  0.73= 248.2

Example: If the production in 2015 is 400 units and the decrease from 2014 to 2015 is 13%, find the production in 2014?
Solution:
Remember the formula:
New quantity =N (1 – K/100)
Let the production in 2014 be x.
It has been decreased by 13% , which then becomes 400 in 2015
[X{1-(13/100)}]= 400
Production in 2014= 400 / 0.87= 459.77 units

The purpose of this article is to provide you with a methodology to compare two percentages. Various applications and formulas based on this concept are explained here.

In this article, it is not about the results alone. The approach adopted while comparing percentages is important too. The derivations given in this article will help you understand this topic better.

Result 1: If A is P% more/less than B, then B is {100P/(100+P)}% less/more than A.

Let us say A is P% more than B
Therefore, we can say:
A = {(100 + P)/100}B
Now, to calculate by what percentage is B is less than A, we need to make the following calculation:
{(A-B)/A} x 100
percentages-comparing-two-percentages
Now, let us take up the reverse case.
Let say A is P% less than B
Therefore, we can say:
A = (100 – P/100)B
Now, to calculate by what percentage is B is more than A, we need to make the following calculation:
percentages-comparing-two-percentages-pic-21
Combining the two above, we arrive at our main result.

Result 2:
If the value of an item goes up/down by P%, the percentage reduction/increment that needs to be now made to bring it back to the original point is {100P/(100+P)}%

Result 3: If the price of an item goes up/down by P %, then the quantity consumed should be reduced/increased by{100P/(100+P)}%   so that the total expenditure remains the same.

Derivation for the result:
Expenditure on any quantity = price per piece × total consumed quantity
{For example, if a pen is of Rs. 5 and we have bought 10 such pens, the total expenditure is = 5 × 10 = Rs. 50}
Let P be the original price per time.
Let Q be the quantity consumed.
Original Expenditure= P × Q  ….  (1)
Let say price is increased by R%.
This means that the quantity has to be decreased in order to maintain expenditure constant.
Let’s assume the consumption is decreased by y%.
New Expenditure , E = P{(100+R)/100}Q{(100-y)/100} ………2
Since the original expenditure and the new expenditure are the same, we arrive at the following equation:
percentages-comparing-two-percentages-pic-3
Remember, each of these results effectively uses the basic concept of percentage and is a derivative of the same.

Example: If A’s income is 60% less than that of B’s, then B’s income is what percent more than that of A?
percentages-comparing-two-percentages-example-pic-1

In this article, we deal with the concept of successive percentage change. This is a problem type in Percentages and using the formula in this article, you can easily solve questions based on this concept in matter of seconds.

What is Successive Percentage Change?

The concept of successive percentage change deals with two or more percentage changes applied to quantity consecutively. In this case, the final change is not the simple addition of the two percentage changes (as the base changes after the first change).

Formula for Percentage Change:

Suppose a number N undergoes a percentage change of x % and then y%, the net change is:
New number = N × (1 + x/100) × (1 + y/100)
Now, (1 + x/100) × (1 + y/100) = 1 + x/100 + y/100 + xy/10000
If we say that x + y + xy/100 = z, then (1 + x/100) × (1 + y/100) = 1 + z/100
Here, z is the effective percentage change when a number is changed successively by two percentage changes.

Various cases for Percentage Change:
Both percentage changes are positive:
x and y are positive and net increase = (x+y+xy/100) %.

One percentage change is positive and the other is negative:
x is positive and y is negative, then net percentage change = (x-y-xy/100)%

Both percentage changes are negative:
x and y both are negative and imply a clear decrease= (-x-y+xy/100)%
Percentage Change involving three changes:
If value of an object/number P is successively changed by x%, y% and then z%, then final value.
percentage-successive-percentage-change

Example 1: The capacity of a ground was 100000 at the end of 2012. In 2013, it increased by 10% and in 2014, it decreased by 18.18%. What was the ground’s capacity at the end of 2014?

Solution:
When One percentage change is positive and the other is negative:
x is positive and y is negative, then net percentage change = (x-y-xy/100)%
Final Percentage Change over the original value = 10-18.18 – (10 × 18.18/100)= -9.998
(the difference above is cause by using exact values).
So the capacity of the ground is decreased by 9.998%
Hence, net capacity = 9000.2

Example 2: A’s salary is increased by 10% and then decreased by 10%. The change in salary is
Solution:
Percentage change formula when x is positive and y is negative = {x – y – (xy/100)}%
Here, x = 10, y = 10
= {10 – 10 – (10 x 10)/100} = -1%
As negative sign shows a decrease, hence the final salary is decreased by 1%.

Example 3: A number is first increased by 10% and then it is further increased by 20%. The original
number is increased altogether by:
Answers:
Percentage change formula when both x and y are positive ={x – y – (xy/100)}%
Here, x = 10 and y = 20
Hence net percentage change == {10 + 20 – (10 x 20)/100} = 32%

In this lesson, we cover the absolute basics of Percentages. The purpose of this lesson is to help you answer one simple question: What are Percentages?

Basic Definition:
Percent implies “for every hundred” and the sign % is read as percentage and x % is read as x per cent. In other words, a fraction with denominator 100 is called a per cent. For example, 20 % means 20/100 (i.e. 20 parts from 100). This can also be written as 0.2.

Basic Formula:
In order to calculate p % of q, use the formula:
(p/100) x q = (pxq)/100

Also remember: p % of q = q % of p

Examples:
1. 100% of 60 is 60 x (100/100) = 60
2. 50% of 60 is 50/100 × 60 = 30
3. 5% of 60 is 5/100 × 60 = 3

Example: 60 % of a number is 360. What is 99 % of the same number?
Solution: Let the number be n.
Given (60/100) ×n = 360 => n = 600
99 % of 600 = (99/100) × 600 = 594

Example: 50 % of a number is 360. What is 99 % of the same number?
Solution: Let the number be y.
Given (50/100) x q = 360
=> q = 720
99% of 720 = (99/100) x 720 = 712.80

Expressing One Quantity as a Per Cent with respect to the other:

To express a quantity as a per cent with respect to other quantity, the following formula is used:

percentage-basic-concept

Example: What percent is 60 of 240?
Solution: First write the given numbers in the fraction form:
60/240 = ¼
Multiply the numerator and denominator with 25 to make the denominator equal to 100
(1×25)/(4×25) = 25/100
25 percent or 25 per 100 is called as 25%

Sample Question for the Basics of Percentage:
Example:A number exceeds 20% of itself by 40. The number is:
(a) 50
(b) 60
(c) 80
(d) 48

Solution: Let the number be p.
20% of itself means => p x (20/100)
Now, according to the question,
p – 20% of p = 40
=> {p – (20 x p)/100} = 40
=> {p-(p/5)} = 40
⇒ 5p – p = 200
∴ p = 50

Alternate Method:
Obviously, it is clear that difference is 80% i.e. 4/5 of number which is equal to 40
4/5p = 40
p = 40 x 5/4= 50.

Tips & Tricks for Percentages:
Basic Tip-1:  If the new value of something is n times the previous given value, then the percentage increase is (n-1)  100%.

Derivation:
Let us consider two values p and q.
Let q be and original value and p be the new value.
According to conditions p= nq
We need to calculate the percentage increase.
You can either use direct formula= {(new value – old value)/old value} x 10
This value becomes= {(p – q)/q} x 100
{(nq – q)/q} x 100
=> (n-1) x 100%

Example: If X= 5.35 Y, then find the percentage increase when the value of something is from Y to X.
Solution:
Use the formula: (n-1)100%
Percentage increase from
Y to X = (5.35 -1)  100= 435%

Basic Tip-2:
When a quantity N is increased by K %, then the:
New quantity = N (1+ K/100 )
Examples:
Increase 150 by 20%= 150 {1+(20/100)} = 150 1.2= 180
Increase 300 by 30%= 300 {1+(30/100)}=  300 1.3= 390
Increase 250 by 27% = 250 {1+(27/100)} = 250 1.27 =317.5

Example: What is the new value when 265 is increased by 15%?
Solution: New quantity = N (1+ K/100)
= 265{1+(15/100)}
New quantity = 1.15 265= 304.75

Basic Tip 3:
When a quantity N is decreased by K %, then the:
New quantity =N (1 – K/100)

Examples:
Decrease 120 by 20%= 120 {1-(20/100)} = 120  0.8= 96
Decrease 150 by 40%=150 {1-(40/100)} = 150  0.6= 90
Decrease 340 by 27%= 340 {1-(27/100)}= 340  0.73= 248.2

Example: If the production in 2015 is 400 units and the decrease from 2014 to 2015 is 13%, find the production in 2014?
Solution:
Remember the formula:
New quantity =N (1 – K/100)
Let the production in 2014 be x.
It has been decreased by 13% , which then becomes 400 in 2015
[X{1-(13/100)}]= 400
Production in 2014= 400 / 0.87= 459.77 units

What is the Product Constancy Ratio?
In this concept, we essentially refer to the practice wherein two or more quantities make up a third quantity. With the variation in the numbers of one quantity, the other quantities need to undergo change in order to maintain the same product.

Let’s take an example.
Let there be two quantities A and B that multiply to form a quantity P. We can say:
A × B = P

Now if A is increased by a certain percentage, then B is required to be decreased by a certain percentage to keep the product P stable.
The following table illustrates the varying values of A and B that will maintain the same product P.
product-constancy-ratio-article-pic-1

Application of Product Constancy: Expenditure Questions
If the price of a commodity increases or decreases by a%, then, the percentage decrease or increase in consumption, so as not to increase or decrease the expenditure is equal to:
(a/100+a) x 100%

Example: Length of a rectangle is increased by 33.33%. By what percentage should the breadth be decreased so that area remains constant?
Solution: Using the table above:
Since length is increased by 33.33%, the breadth will decrease by 25% to keep area constant.
Let’s make these calculations also.
Let original length= L
Original Breadth= B
Increase length= 4/3 L
Since the area remains same, we can say
L x B = Increased Breadth x 4/3L
Therefore,
Increased breadth = ¾ Original Breadth = 25% reduction in breadth.

Example: When speed of a car is increased by 25%, time taken reduces by 40 minutes in covering a certain distance.What is the actual time taken to cover the same distance by actual speed?
Solution: We have Speed × Time = Distance
Since speed has been increased by 25%, time will reduce by 20%.
Now, 20% of actual time = 40 min
Actual time = 200 min

Example: Price of a commodity has increased by 60%. By what percent must a consumer reduce the consumption of the commodity so as not to increase the expenditure ?
(a) 37%
(b) 37.5%
(c) 40.5%
(d) 60%
product-constancy-ratio-article-pic-2

Common Fractions with Decimal and Percent Equivalents

The purpose of this article is very simple: to provide you an exhaustive conversion table that you can use as a short-cut for calculating percentages.

Remember, it is extremely vital to save every second possible in the examination and what could be quicker in Mathematics than using some clever calculation techniques. One such technique is the conversion tables for Percentages.

In order to convert a fraction into percentage, multiply the fraction by 100 and put the “%” sign.
Example: ½ can be written as ½ x 100 = 50%.

To convert a percentage into a fraction, employ the reverse technique: divide the number by 100 and remove the “%” sign.
Example: 25% can be written as 25/100 = 1/4

To convert a percentage into a ratio, first convert the percentage into a fraction, and then the fraction into a ratio.
Example: Convert 25% into a ratio.
25% = 25/100 = ¼ = 1 : 4

Conversion of Fraction into percentage table:
The following table will help you solve percentage questions quicker:

percentages-fractions-and-ratios-1
percentages-fractions-and-ratios-2

In this article, we simply deal with the different types of problem based on Percentages. This lesson does not have any problems of its own but simply is based on the concepts we have studied so far.

Problem Type-1:

If a reduction of p % in the price of an article enables a person to buy y kg more for Rs R,then we can arrive at the following results:
Reduced price=> Rp/100y per kg
Original prices per kg= Rp/(100-p)y per kg

Example: A reduction of 21% in the price of an item enables a person to buy 3 kg more for 100. Thereduced price of item per kg is:
(a) Rs. 5.50
(b) Rs. 7.50
(c) Rs. 10.50
(d) Rs. 7.00

Solution:(d)
Reduced price will be:
Rp/100y per kg
In our case R= Rs. 100 , x=21% , y=3kg
{(100 x 21)/(100 x 3)} = Rs. 7
Alternate method:
Expenditure = price per quantity × consumption

E = P × Q =100……. (1)
Now, as per the question,
percentages-types-of-problems-questions-1-article-1-pic-1

Problem Type-2: Problems based on mixtures

Example: A vessel has 60 L of solution of acid and water having 80% acid. How much water is to be added to make it solution in which acid forms 60%?
(a) 48 L
(b) 20 L
(c) 36 L
(d) None of these

Solution: (b)
Given, percentage of acid = 80%
Then, percentage of water = 20%
In 60L of solution, water = (60 x 20)/ 100 = 12L
Let p liter of water is to be added.
According to the question,
=>{(12 + p)/(60 + p)} x 100 = 40  (∵   100 – 60 = 40% water)
=>1200 + 100p = 2400 + 40p
⇒ 60p = 1200
p= 20L

Problem Type-3: Problems based on Ratios and Fractions
Example: If the numerator of a fraction is increased by 20% and the denominator is decreased by 5%, the value of the new fraction become 5/2. The original fraction is:
(a) 24/19
(b) 3/18
(c) 95/48
(d) 48/95

Solution: (c)
Let original fraction be p/y
According to the question,
{(120/100)p/(95/100)y} = 5/2
120p/95y = 5/2
=> p/y = (5/2)x (95/120) = 95/48.

Problem Type-4: Problems based on Income, salary, expenditure

Example:The monthly income of a person was Rs 13500 and his monthly expenditure was Rs 9000. Nextyear his income increased by 14% and his expenditure increased by 7%. The per cent increase inhis savings was:
(a) 7%
(b) 21%
(c) 28%
(d) 35%

Solution: (c)
Given, monthly income = 13500 and expenditure = 9000
Then, original savings= Rs. (13500-9000) = Rs 4500
New income = 114% of Rs. 13500 = Rs 15390
New expenditure= 107% of Rs 9000 = Rs 9630
New saving = Rs. (15390 – 9630) = Rs 5760
NS = new savings
OS = Original savings
Percentage increase in savings = {(NS – OS)/OS} X 100
{(5760 – 4500)/4500} X 100 = (1260/4500)X 100 = 28%

In this article, we simply deal with the different types of problem based on Percentages. This lesson does not have any problems of its own but simply is based on the concepts we have studied so far. This article is an extension of the previous article. You can read it here.

Problem Type-5: Problems based on students and marks

Shortcut-1 for the problem type:
A candidate scores x% is an examination fails by ‘a’ marks, while another candidate who scores y% marks and get b marks more than the maximum required passing marks, then the maximum marks for the examination is given by
percentages-exercises-question-2

Shortcut-2 for the problem type:
In an examination a% of total number of candidates failed in a subject X and b% of total number of candidates failed in subject Y and c% failed in both subjects, then percentage of candidates, who passed in both the subjects, is [100 – (a + b – c)]%.

Example: In a quarterly examination a student secured 30% marks and failed by 12 marks. In the same examination another student secured 40% marks and got 28 marks more than minimum marks to pass. The maximum marks in the examination is:
(a) 300
(b) 500
(c) 700
(d) 400

percentages-exercises-question-example-pic-1

Example: In an examination, 52% of the candidates failed in English and 42% failed in Mathematics. If17% failed in both the subjects, then the percentage of candidates who passed in both the subjects, was:
(a) 23%
(b) 21%
(c) 25%
(d) 22%

Solution: (a)
In an examination a% of total number of candidates failed in a subject X and b% of total number of
candidates failed in subject Y and c% failed in both subjects.
Then, percentage of candidates who passed in both the subjects: [100–(a + b– c)%].
Why so?
We need to subtract failed candidate from total to calculate pass candidates, but in this process, we subtract people who were fail in both of subjects twice, so we added c again.
Now equation can be read as [ 100—a-b+c]
Here, a = 52, b = 42 and c = 17, then Percentage of candidates who passed in both the subjects
= [100 –(52 +42–17)]
= 100–77=23%

Problem Type-2: Problems based on percentage in excess or in short

Example:
lf A has 4/5 of the number of books that shelf B has. If 25% of the books A are transferred to B and then 25 % of the books from B are transferred to A then the percentage of the total number of books that on shelf A is:

(a) 25%
(b) 50%
(c) 75%
(d) 100%

Solution: (b)

Let the number of books in shelf B be 100.
∴ Number of books in shelf A = {(100 x 4)/5} = 80
On transferring 25% i.e.,of books of shelf A to shelf B, the books on
shelf B = {100 + (80 x 25)/100}
B = 100 + 20 = 120
Books left in shelf A = 80-20 = 60
Again, on transferring 1/4th books of shelf B to shelf A, the books on
shelf A = {60 + (120/4)} = 90
Total no of books in A and B = 120 +60 = 180
Required percentage of books on shelf A = (90/180) x 100 = 50%

Problem Type-3: Problems based on percentages and number system

Example: A number, on subtracting 15 from it, reduces to its 80%. What is 40% of the number?
(a) 75
(b) 60
(c) 30
(d) 90

Solution: (c)
Let the number be p.
According to the question,
p – 15 = p(80/100)
p -15 = 4p/5
⇒ 5p – 75 = 4p ⇒ p = 75
Now, 40% of 75 = (75 x 40)/ 100 = 30