# Study Notes for CAT 2021: Basics of Percentage

Updated : March 22nd, 2021
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The 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.

The 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 proportion per hundred.
1. When we say 35 per cent 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 a 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.

### 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 a 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

### 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 household 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 the house, so the remaining amount is 80.
now he spends 70% of 80 on household expenses, so the 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 the 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 an integer value. There were 3 candidates, one losing candidate got 30%, so the remaining two candidates got 70% vote of the total.
Candidate 1 + candidate 2 = 70%

An important point which is given in the question is the minimum absolute margin and integral value.
Case 1: Suppose candidate 1 got 40%, then-candidate 2 had got 30%. But this is not the minimum 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 has got 36%.
Hence the 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 consists of 85 questions from three sections- i.e. A, B and C. 10 questions from section A, 30 questions from section B and 45 questions 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 the 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 the 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 cast 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

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 the 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

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