Percentage is an important part of Quantitative aptitude. Whether it is DI, Profit & Loss, SI-CI, or Allegation, etc. all these chapters with the help of percentage can be solved easily. You can go through the basics of percentage and previous year asked questions.
A percentage is a number or ratio expressed as a fraction of 100.It is a proportion per hundred.
1.When we say 35 percent in mathematical notation we write 35%.
2.When we want to express this in mathematical form, 35% means 35 per 100 or (35/100).
Important: 50% of 20 can be written 20% of 50 as well.
You can also represent % into decimal, 50% = 0.5
Conversion of fraction into %.
to convert fraction into %, we multiply it by 100.
¼ = (¼)× 100 % = 25 %.
1/3 = (1/3) ×100 % = 33(1/3) %
1/14 = (1/14) ×100 % = (100/14)%=(50/7)%= 7 (1/7) %
Note: Never forget to express % notation in the percentage.
We suggest you that you must learn both tables given below.
Fraction |
Percentage |
Fraction |
Percentage |
Fraction |
Percentage |
1 | 100% | 1/7 | 14(2/7) % | 1/13 | 7 (9/13) % |
1/2 | 50% | 1/8 | 12(1/2) % | 1/14 | 7 (1/7) % |
1/3 | 33(1/3) % | 1/9 | 11(1/9) % | 1/15 | 6 (2/3) % |
1/4 | 25% | 1/10 | 10 % | 1/16 | 6 (1/4) % |
1/5 | 20% | 1/11 | 9 (1/11) % | ||
1/6 | 16(2/3) % | 1/12 | 8 (1/3) % |
Conversion of % into fraction.
To convert % into fraction, we divide it by 100. So, we can express in this way:
100% = (100/100) = 1 1% = (1/100) 2% = (2/100) = (1/50)
50% = 50/100 = ½
20% = 20/100 = 1/5
10% = 10/100 = 1/10
16(2/3)% = (50/3)% =50/(3×100) = 50/300 = 1/6
Percentage |
Fraction |
Percentage |
Fraction |
Percentage |
Fraction |
10% | 1/10 | 16 (2/3)% | 1/6 | 15% | 3/20 |
20% | 1/5 | 66 (2/3) % | 2/3 | 7(1/2)% | 3/40 |
40% | 2/5 | 6(1/4)% | 1/16 | 22(1/2)% | 9/40 |
60% | 3/5 | 18(3/4) % | 3/16 | 69(3/13) % | 9/13 |
80% | 4/5 |
In following examples we will try to avoid calculation using above table.
(i) 99% of 840
we can say 10% = 84,So 1% = 8.4
99% of 840 = 840-8.4=831.6
(ii)25% of 320 = (1/4)× 320
=80
(iii) 76% of 400?
76%=50%+25%+1%
= 200+100+4
= 304
(iv) 102% of 720?
1%= 7.2 so 2%= 14.4
102% = 100%+2%= 720+14.4 = 734.4
(v)18% of 300?
18% = 20%-2%= (1/5)×300-6
= 60-6 = 54
or 1% = 3 so 18%= 18×3=54
(vi) 12% of 540?
1%=5.4
12% = 10%+2+
= 54+10.8
= 64.8
Example1: Out of his total income, Mr. Sharma spends 20% on house rent and 70% of the rest on house hold expenses. If he saves Rs 1,800 what is his total income (in rupees)?
Solution: Let Income of Mr. Sharma is 100
then he spends 20% on house, so remaining amount is 80.
now he spends 70% of 80 on house hold expenses, so remaining amount left with him is 30% of 80
30% of 80 = 1800
24 = 1800
1 = 1800/24
1 = 75
100= 7500
hence total income is 7500 Rs.
Or, Let total income is P
(100%-20%)×(100%-70%)× P = 1800
80%× 30%× P=1800
((80×30)/(100*100)) × P = 1800
P = 7500
Example2: An army lost 10% its men in war, 10% of the remaining due to diseases died and 10% of the rest were disabled. Thus, the strength was reduced to 729000 active men. Find the original strength.
Solution: Let army has 100 men.
10% loss in war, so remained are 90 men
then,10% of 90 died due to diseases, remained 90-9 = 81
then again, 10% of 81 again disabled
So, remained men = 90% of 81
90% of 81 = 729000
(90×81)/100 =729000
1= 10000
100 = 1000000
hence total men are 1000000.
Example3: In a village three people contested for the post of village Sarpanch. Due to their own interest, all the voters voted and no one vote was invalid. The losing candidate got 30% votes. What could be the minimum absolute margin of votes by which the winning candidate led by the nearest rival, if each candidate got an integral per cent of votes?
Solution: As given, no vote was invalid i.e. 100% votes were polled and all candidate got votes in integer value. There were 3 candidates, one losing candidate got 30%, so remaining two candidates got 70% vote of the total.
Candidate 1 + candidate 2 = 70%
Important point which is given in question is minimum absolute margin and integral value.
Case 1: Suppose candidate 1 got 40%, then candidate 2 had got 30%. But this is not mininmum absolute margin.
Case 2: Both got 35% votes, If both got equal votes then there will be no winning candidate.
Case 3: One candidate must have got 34% and another one have got 36%.
Hence absolute margin is 2%.
Example4: The difference between 4/5 of a number and 45% of the number is 56. What is 65% of the number?
Solution: Let number is P.
we can say 4/5 = 80%
so, (80%-45%) of P = 56
35% of P = 56
P = (56/35%)
65% of P = 56/35 ×65 = 104
Example5: Deeksha’s science test consist of 85 questions from three sections- i.e. A, B and C. 10 questions from section A, 30 questions from section B and 45 question from section C. Although, she answered 70% of section A, 50% of section B and 60% of section C correctly. She did not pass the test because she got less than 60% of the total marks. How many more questions she would have to answer correctly to earn 60% of the marks which is passing grade?
Solution: If she has done 60% of total questions she would have passed,.
So, no. of question to be done to pass= 60% of 85 = (3/5)×85 = 51
But she done 70% of A = 70% of 10 = 7
50% of B = 50% of 30 = 15
60% of C = (3/5) of 45 = 27
So , total questions she attempted = (7+15+27) = 49
If she has attempted (51-49) = 2 more questions she would have passed.
Example6: In an election between 2 candidates, 75% of the voters cast their votes, out of which 2% votes were declared invalid. A candidate got 18522 votes which were 75% of the valid votes. What was the total number of voters enrolled in the election?
Solution: Let total number of voters enrolled are P.
Number of votes casted = 75% of P = (75/100) P = 0.75 P
Important: Those votes which were declared invalid are 2% of casted voted not 2% of total votes.
So, valid votes are = (100%-2%) of 0.75P = 98% of 0.75P
Given Candidates got 75% of valid votes = 18522
(75%) × 98% × 0.75 P = 18522
(3/4) * (98/10) * (3/4) P = 18522
P = 42 × 800
P = 33600 votes.
Example7: An ore contains 20% of an alloy that has 85% iron. Other than this, in the remaining 80% of the ore, there is no iron. What is the quantity of ore (in kg) needed to obtain 60 kg of pure iron?
Solution: Let quantity of ore is P kg
P × 20% × 85% = 60kg
P × (1/5) × (17/20) = 60
P = (60×5× 20)/17
P = 6000/17 Kg
Example8: 5% of one number (X) is 25% more than another number (Y). If the difference between the numbers is 96 then find the value of X?
Solution : Given: 5% of X = Y + 25% of Y
0.05 X = 1.25 Y
X = 25 Y
X-Y=96
25Y-Y =96
24Y=96
Y = 4 so, X =100
Important Short tricks and Concepts of Percentage for SSC Exams
In our previous article,we discussed the basics of Percentages with you. Today we will share some important concepts and shortcut tricks that will help you solve questions based on percentages easily and within less time.
Concept 1
1. A+B+AB/100 When A and B both are positive change
2. A-B-AB/100 When A is positive change and B is negative change
3. -A+B-AB/100 When A is negative change and B is positive change.
4. -A-B+AB/100 When A and B both are negative change.
Important: There is no need to remember above formulas, you have to just remember
±A ± B ±AB/100 and put the sign of change, if negative, then (-) and positive then, (+) but keep in mind that sign of AB is product of signs of A and B.
Example1: The price of a book is reduced by 10% and sale of the book is increased by 15%. Find the net effect on revenue.
Example2 : If length and breadth of a rectangle is increased by 5% and 8% respectively. Find the % change in area of the rectangle.
Concept 2:
New solution × new % = old solution × old %
This formula is applicable for the commodity which is constant in the solution or mixture, its quantity doesn’t change after mixing in solution.
Example3: A mixture of sand and water contains 20% sand by weight. Of it 12 kg of water is evaporated and the mixture now contains 30% sand.
Solution: In this sand is constant in the mixture. So we will apply this formula on sand not on water.
(a)Find the original mixture.
Let the original mixture is P kg, So new mixture = (P-12) kg
old% = 20 and new % = 30
new solution × new % = old solution × old %
(P-12)× 30% = P × 20%
3P – 36 = 2P
P = 36 Kg.
(b) Find the quantity of sand and water in the original mixture.
Quantity of sand in original mixture = 20% of 36 = 7.2 Kg
Quantity of water in original mixture = 80% 0f 36 = 28.8 Kg
OR = Quantity of mixture – quantity of sand = 36 -7.2 = 28.8 Kg
Example 4: 30 litres of a mixture of alcohol and water contains 20% alcohol. How much litre of water must be added to make the alcohol 15% in the new mixture?
Solution: only water is added in the mixture so, there is no change in alcohol. We will apply above formula on alcohol. Let water added is P litres.
Old mixture = 30 litres, old % of alcohol = 20%
New mixture = 30+P litres, new % of alcohol = 15
using, new solution × new % = old solution × old %
(30+P) × 15% = 30 × 20%
P = 10 litres , hence 10 litres of water is added.
Concept 3
Example 5: If the price of milk increased by 25%, by how much percent must Rahul decrease his consumption, so as his expenditure remains same.
Solution: Let price of milk is 20 Rs/litre and Rahul consumes 1 litre milk.
Expenditure of Rahul = price × consumption
Now price of milk is increased by 25%, so new price is (125/100)× 20 = 25 Rs.
but his expenditure remains same
So, new consumption × new price = old price × old consumption
new consumption × 25 = 20 × old consumption
new consumption =(20/25) × old consumption
new consumption% = (20/25)× old consumption × 100
new consumption% = 80% of old consumption
decrease in consumption = 20 %
Using above trick: Given, price % is increased so sign will be (+) and consumption % will decrease.
Decrease in consumption =(25/125) × 100 = 20%
Example 6: If the price of milk decrease by 25%, by how much percent must Rahul increase his consumption, so as his expenditure remains same.
Solution: Let price of milk is 20 Rs/litre and Rahul consumes 1 litre milk.
Expenditure of Rahul = price × consumption
Now price of milk is decreased by 25%, so new price is × 20 = 15 Rs.
but his expenditure remains same
So, new consumption × new price = old price × old consumption
new consumption × 15 = 20 × old consumption
new consumption = (20/15)× old consumption
new consumption% = (20/15)× old consumption × 100
new consumption% = 133(1/3)% of old consumption
increase in consumption = 33(1/3) %
Using above trick: given price % is decreased so sign will be (-) and consumption % will increase.
Increase in consumption = (25/75) x 100 = 33(1/3)%
Concept 4:
Example 7: The population of a town is 6000. It increases 10% during 1^{st}year, increases 25% during 2^{nd} year and then again decreases by 10% during 3^{rd} year. What is the population after 3 year.
Example 8: The population of a village increases by 10% during the first year, decreased by 12% during 2^{nd} year and again decreased by 15% during 3^{rd} year. If the population at the end of 3^{rd} year is 2057.