Percentage imp formulas and shortcut
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.
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
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 = 8408.4=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)×3006
= 606 = 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 909 = 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 (5149) = 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.
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 909 = 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 (5149) = 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
XY=96
25YY =96
24Y=96
Y = 4 so, X =100
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
XY=96
25YY =96
24Y=96
Y = 4 so, X =100
Percentage(%)
Percentage is per‑cent which means parts per hundred(1/100).
If we have to convert percentage into fraction then it is divide by 100.
Example 1:‑ If we write 45% then its equal to 45/100 or in fraction 9/20 or in decimal 0.45
If we have to convert fraction into percentage we have to multiple with 100.
Example 2:‑ if we write 3/5 in fraction it is equal to 60% =3/5x100=60.
Convert Percentage into Decimal:
 20% = 20/100 = 0.5
Convert Decimal Into Percentage:
 0.25 = (0.25 × 100) % = 25%
 1.50 = (1.50 × 100) % = 150%
Types of Formulas and Short Tricks
Type 1: Percentage Increase/Decrease:If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is: [R/ (100 + R)] x 100%
If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is: [R/ (100  R)] x 100%
If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is: [R/ (100  R)] x 100%
Type 2: Results on Population:Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:1.Population after n years = P(1 + R/100)n2.Population n years ago =P/(1 + R/100)n
Type 3: Results on Depreciation:
Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:
1.Value of the machine after n years = P(1  R/100)n
2.Value of the machine n years ago = P/[(1  R/100)]n
3.If A is R% more than B, then B is less than A by= [R/ (100 + R)] x 100%
4.If A is R% less than B, then B is more than A by= [R/ (100  R)] x 100%
Note: For two successive changes of x% and y%, net change = {x + y +xy/100}%
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