Percentage Quantitative Aptitude PDF
The word “percent” is derived from the latin words “per centum”, which means “per hundred”.
- A percentage is a fraction with denominator hundred, It is denoted by the symbol %.
- Numerator of the fraction is called the rate per cent.
VALUE OF PERCENTAGE:
Value of percentage always depends on the quantity to which it refers: Consider the statement, “65% of the students in this class are boys”. From the context, it is understood that boys from 65% of the total number of students in the class. To know the value of 65%, the value of the total number of student should be known. If the total number of student is 200, then,
The number of boys =130;
It can also be written as (200) × (0.65) =130.
Note that the expressions 6%, 63%, 72%, 155% etc. Do not have any value intrinsic to themselves. Their values depend on the quantities to which they refer.
To express the fraction equivalent to %:
Express the fraction with the denominator 100, then the numerator is the answer.
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Example 1:
Express the fraction 11/12 into the per cent.
Solution:
11/12=(11/12×100)/100=(91 2/3)/100=912/3%
To express % equivalent to fraction:
a% =a/100
Example 2:
Express 45 5/6% into fraction.
Solution:
45 5/6% = (45 5/6)/100=275/(6×100)=11/24.
Example 3:
Rent of the house is increased from ` 7000 to `7700. Express the increase in price as a percentage of the original rent.
Solution:
Increase value = Rs 7700 – Rs 7000 = Rs 700
Increase % = (Increas value)/(Original value)×100= 700/7000×100=10
∴ Percentage rise = 10%
Example 4:
The cost of a bike last year was Rs19000. Its cost this year is Rs 17000. Find the per cent decrease in its cost.
Decrease % = (Decreas value)/(Original value) × 100
% decrease = (19000-17000)/19000×100
=2000/19000×100= 10.5%.
∴ Percentage decrease = 10.5%.
If A is x % if C and B is y % of C, then A is x/y × 100% of B.
Example 5:
A positive number is divided by 5 instead of being multiplied by 5. By what per cent is the result of the required correct value?
Solution:
Let the number be 1, then the correct answer = 5
The incorrect answer that was obtained =.
The required % = = 4%
If two numbers are respectively x% and y% more than a third number, then the first number is % of the second and the second is % of the first.
If two numbers are respectively x% and y% less than a third number, then the first number if % of the second and the second is % of the first.
x% of a quantity is taken by the first, y% of the remaining is taken by the second and z% of the remaining is taken by third person. Now, if A is left in the fund, then the initial amount
=(A×100×100×100)/((100-x)(100-y)(100-z)) in the beginning.
x% of a quantity is added. Again, y% of the increased quantity is added. Again z% of the increased quantity is added. Now it becomes A, then the initial amount
=(A×100×100×100)/((100+x)(100+y)(100+z))
Example 6:
3.5% income is taken as tax and 12.5% of the remaining is saved. This leaves Rs. 4,053 to spend. What is the income?
Solution:
By direct method,
Income = (4053×100×100)/((100-3.5)(100-12.5)) = Rs 4800.
If the price of a commodity increases by r%, then reduction in consumption, so as not to increase the expenditure is (r/(100+r)×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)×100)%.
Example 7:
If the price of coal be raised by 20%, then find by how much a householder must reduce his consumption of this commodity so as not to increase his expenditure?
Solution:
Reduction in consumption = (20/(100+20)×100)%
= (20/(100+20)×100)% = 16.67%
POPULATION FORMULA
If the original population of a town is P, and the annual increase is r%, then the population after n years is P(1+r/100)^n and population before n years = P/(1+r/100)^n
If the annual decrease be r%, then the population after n years is P(1-r/100)^n and population before n years = P/(1+r/100)^n
Example 8:
The population of a certain town increased at a certain rate per cent annum. Now it is 456976. Four years ago, it was 390625. What will it be 2 years hence?
Solution:
Suppose the population increases at r% per annum. Then, 390625 (1+r/100)^4 = 456976
(1+r/100)^2 = √(456976/390625)= 676/625
Population 2 years hence = 456976 (1+r/100)^2
= 456976 × 676/625 = 494265 approximately.
Example 9:
The population of a city increase at the rate of 4% per annum. There is an additional annual increase of 1% in the population due to the influx of job seekers. Find percentage increase in the population after 2 years.
Solution:
The net annual increase = 5%
Let the initial population be 100.
Then, population after 2 years = 100×1.05×1.05 = 110.25
Therefore, % increase in population = (110.25-100) = 10.25%
If a number A is increased successively by x% followed by y% and then z%, then the final value of A will be A(1+x/100)(1+y/100)(1+z/100)
In case a given value decreases by an percentage then we will use negative sign before that.
First Increase and then decrease:
If the value is first increased by x% and then decreased by y% then there is (x-y-xy/100)% increase or decrease, according to the +ve or –ve sign respectively.
If the value is first increased by x% and then decreased by x% then there is only decrease which is equal to (x^2/100).
Example 10:
A number is increased by 10% and then it is decreased by 10%. Find the net increase or decrease per cent.
Solution:
% change = (10×10)/100=1%
i.e. 1% decrease.
Average percentage rate of change over a period.
=((New Value-Old Value))/(Old Value)×100/n% where n = period.
The percentage error = (The Error)/(True Value)×100%
SUCCESSIVE INCREASE OR DECREASE
In the value is increased successively by x% and y% then the final increase is given by (x+y+xy/100)%
In the value is decreased successively by x% and y% then the final decrease is given by (-x-y-xy/100)%
Example 11:
The price of a car is decreased by 10% and 20% in two successive years. What per cent of price of a car is decreased after two years?
Solution:
Put x = -10 and y = -20, then
-10-20+
The price of the car decreases by 28%.
STUDENT AND MARKS
The percentage of passing marks in an examination is x%. If a candidate who scores y marks fails by z marks, then the maximum marks M = 100(y+z)/x
A candidate scoring x% in an examination fails by ‘a’ marks, while another candidate who scores y% marks gets ‘b’ marks more then the minimum required passing marks. Then the maximum marks M = 100(a+b)/x
In an examination x% and y% students respectively fail in two different subjects while z% students fail in both subjects then the % age of student who pass in both the subjects will be {100-(x + y – z)}%
Example 12:
Vishal requires 40% to pass. If he gets 185 marks, falls short by 15 marks, what was the maximum he could have got?
Solution:
If Vishal has 15 marks more, he could have scored 40% marks.
Now, 15 marks more then 185 is 185+15 = 200
Let the maximum marks be x, then 40% of x = 200
⇒ × x = 200 ⇒ x =500
Thus, maximum marks = 500
Alternate method:
Maximum marks = (100(185+15))/40=(100×200)/40 = 500
Example 13:
A candidate scores 15% and fails by 30 marks, while another candidate who scores 40% marks, gets 20 marks more then the minimum required marks to pass the pass the examination. Find the maximum marks of the examination.
Solution:
By short cut method:
Maximum marks = (100(30+20))/(40-15) =200
2-DIMENSIONAL FIGURE AND AREA
If the sides of a triangle, square, rectangle, rhombus or radius of a circle are increased by a%, its area is increased by(a(a+200))/100 %
If the sides of a triangle, square, rectangle, rhombus or radius of a circle are decreased by a %
Then its area is decreased by (a(200-a))/100%.
Example 14:
If the radius of a circle is increased by 10%, what is the percentage increase in its area?
Solution:
Let R be the radius of circle.
Area of Circle, A =πR^2
Now, radius is increased by 10%
New radius, R’ = R + 10% of R = 1.1 R
New Area, A’ = π(1.1R)^2= 1.21 πR^2%
increase in area =(1.21πR^2-πR^2)/(πR^2 )×100=21%
Shortcut Method:
Radius is increases by 10%.
So, Area is increased by (10(10+200))/100 = 21%
If the both sides of rectangle are changed by x% and y% respectively, then % effect on area = x + y+xy/100 (+/- according to increase or decrease.
Example 15:
If the length and width of a rectangular garden were each increased by 20%, then what would be the per cent increase in the area of the garden?
Solution:
By direct formula
% increase in area =(20 (20+200))/100=44%
If A’s income is r% more than that of B, then B’s income is less than that of A by (r/(100+r)×100)%
If A’s income is r% less than that of B, then B’s income is more than that of A by (r/(100-r)×100)%
Example 16:
If A’s salary is 50% more than B’s, then by what percent B’s salary is less than A’s salary?
Soluti
Let B’s salary be Rs x
Then, A’s salary = x + 50% of x = 1.5x
B’s salary is less than A’s salary by ((1.5x-x)/1.5x×100)% = 100/3 = 33.33%
Shortcut method,
B’s salary is less than A’s salary by (50/(100+50)×100)%
=50/150×100% = 33.33%
Example 17:
Ravi’s weight is 25% that of Meena’s and 40% that of Tara’s. What percentage of Tara’s weight is Meena’s weight.
Solution:
Let Meena’s weight be x kg and Tara’s weight be y kg. Then Ravi’s weight = 25% of Meena’s weight
= 25/100×x …..(i)
Also, Ravi’s weight = 40% of Tara’s weight
= 40/100×y …..(ii)
From (i) and (ii), we get
25/100×x=40/100×y
⇒ 25x = 40y
⇒ 5x = 8y ⇒ x = 8/5 y
Meena’s weight as the percentage of Tara’s weight
= x/y×100= ( 8/5 y)/y×100
= 8/5×100=160
Hence, Meena’s weight is 160% of Tara’s weight.
Example 18:
The monthly salaries of A and B together amount to `50,000. A spends 80% of his salary and B spends 70% of his salary. If now their saving are the same, then find the salaries of A and B.
Solution:
Let A’s salary by x, then B’s salary (50,000-x)
A spends 80% of his salary and saves 20%
B spends 70% of his salary and saves 30%
Given that
20% of x = 30% of (50,000-x)
20/100×x=30/100×(50,000-x)
50x/100=(30×50,000)/100
⇒ x = (30×50,000×100)/(100×50)=30,000
A’s salary Rs 30,000
B’s salary = Rs 50,000 – Rs 30,000 = Rs20,000
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